This document was uploaded by user and they confirmed that they have the permission to share it. If you are author or own the copyright of this book, please report to us by using this DMCA report form. Report DMCA

Download & View **Mathematical Methods Cas - Maths Quest 11** as PDF for free.

**Words:**264,862**Pages:**600

Loading documents preview...

MATHS QUEST 11 Mathematical Methods CAS ROBYN WILLIAMS | NICOLAOS KARANIKOLAS | KYLIE BOUCHER GAYLE ROBERTS | JENNIFER NOLAN | GEOFF PHILLIPS CONTRIBUTING AUTHORS ROBERT CAHN | DOUGLAS SCOTT | HOWARD LISTON SUPPORT MATERIAL JOHN DOWSEY | DENNIS FITZGERALD | EMILY HUI | CAROLINE MEWS VINOD NARAYAN | PETER SWAIN | DAVID TYNAN | IAN YOUNGER WAYNE YOUNGS | SIMONE RICHARDSON | DINA ANTONIOU | NORRENE HILL

3RD EDIT ION

VCE M AT HEM AT I CS U N I T S 1 & 2

Third edition published 2013 by John Wiley & Sons Australia, Ltd 42 McDougall Street, Milton, Qld 4064 First edition published 2009 Second edition published 2010 Typeset in 10/12pt Times LT Std © John Wiley & Sons Australia, Ltd 2009, 2010, 2013 The moral rights of the authors have been asserted. National Library of Australia Cataloguing-in-Publication data Title: Edition: ISBN: Notes: Target audience: Subjects: Other authors/ contributors: Dewey number:

Maths quest 11 mathematical methods CAS/Robyn Williams . . . [et al.]. 3rd ed. 978 1 118 31058 8 (paperback) 978 1 118 31067 0 (flexisaver) 978 1 118 31060 1 (ebook) Includes index. For secondary school age. Mathematics — Textbooks. Williams, Robyn (Robyn Ellen) 1967– 510

Reproduction and communication for educational purposes The Australian Copyright Act 1968 (the Act) allows a maximum of one chapter or 10% of the pages of this work, whichever is the greater, to be reproduced and/or communicated by any educational institution for its educational purposes provided that the educational institution (or the body that administers it) has given a remuneration notice to Copyright Agency Limited (CAL). Reproduction and communication for other purposes Except as permitted under the Act (for example, a fair dealing for the purposes of study, research, criticism or review), no part of this book may be reproduced, stored in a retrieval system, communicated or transmitted in any form or by any means without prior written permission. All inquiries should be made to the publisher. Illustrated by Aptara and Wiley Composition Services Typeset in India by Aptara Printed in Singapore by Craft Print International Ltd 10 9 8 7 6 5 4 3 2 1

Contents 2K Using technology to solve quadratic equations

Introduction vi About eBookPLUS viii Acknowledgements ix

Exercise 2K 82 2L Simultaneous quadratic and linear equations Exercise 2L 87

Chapter 1

Linear functions

1

1A Solving linear equations and inequations 1B 1C 1D 1E 1F 1G

1H ■ ■ ■ ■

1 Exercise 1A 3 Rearrangement and substitution 4 Exercise 1B 6 Gradient of a straight line 8 Exercise 1C 10 Sketching linear functions 12 Exercise 1D 14 Simultaneous equations 15 Exercise 1E 17 Finding the equation of a straight line 18 Exercise 1F 20 Distance between two points and midpoint of a segment 21 Exercise 1G 22 Linear modelling 24 Exercise 1H 25

Summary 27 Chapter review 28 ICT activities 34 Answers 35

Chapter 2

Quadratic functions 2A Polynomials 2B 2C 2D 2E 2F

2G 2H 2I

2J

39

39 Exercise 2A 40 Expanding quadratic expressions 41 Exercise 2B 43 Factorising quadratic expressions 44 Exercise 2C 46 Factorising by completing the square 47 Exercise 2D 50 Solving quadratic equations — Null Factor Law 50 Exercise 2E 53 Solving quadratic equations — completing the square 55 Exercise 2F 57 The quadratic formula 58 Exercise 2G 61 The discriminant 62 Exercise 2H 66 Graphs of quadratic functions as power functions (turning point form) 66 Exercise 2I 68 Graphs of quadratic functions (intercepts method) 70 Exercise 2J 78

■ ■ ■ ■

80 83

Summary 90 Chapter review 92 ICT activities 95 Answers 96

Chapter 3

Cubic and quartic functions

105

Polynomials of degree 3 and 4 105 105 Exercise 3A 106 3B Long division of polynomials 107 Exercise 3B 109 3C Polynomial values 109 Exercise 3C 110 3D The remainder and factor theorems 111 Exercise 3D 113 3E Factorising polynomials 114 Exercise 3E 116 3F Sum and difference of two cubes 117 Exercise 3F 118 3G Solving polynomial equations 118 Exercise 3G 120 3H Cubic graphs — intercepts method 121 Exercise 3H 124 3I Quartic graphs — intercepts method 126 Exercise 3I 127 3J Graphs of cubic functions in power function form 128 Exercise 3J 130 3K Domain, range, maximums and minimums 132 Exercise 3K 134 3L Modelling using technology 136 Exercise 3L 137 3m Finite differences 138 Exercise 3M 142 3A Expanding

■ ■ ■ ■

Summary 144 Chapter review 146 ICT activities 150 Answers 151

ExAm prACtICE 1

Based on Chapters 1–3

159

Chapter 4

Relations, functions and transformations 4A Set notation

161 Exercise 4A 162 4B Relations and graphs Exercise 4B 165

163

161

4C Domain and range 167

Exercise 4C 170 4D Types of relations (including functions) 172 Exercise 4D 174 4E Power functions (hyperbola, truncus and square root function) 175 Exercise 4E 182 4F Function notation 183 Exercise 4F 186 4G Special types of function (including hybrid functions) 187 Exercise 4G 189 4H Inverse relations and functions 192 Exercise 4H 193 4I Circles 193 Exercise 4I 195 4J Functions and modelling 197 Exercise 4J 197 ■■ ■■ ■■ ■■

Summary 200 Chapter review 202 ICT activities 209 Answers 210

Chapter 5

Exponential and logarithmic functions 217 Introduction 217 5A Index laws 217 Exercise 5A 220 5B Negative and rational powers 221 Exercise 5B 224 5C Indicial equations 224 Exercise 5C 226 5D Graphs of exponential functions 227 Exercise 5D 230 5E Logarithms 231 Exercise 5E 233 5F Solving logarithmic equations 234 Exercise 5F 236 5G Logarithmic graphs 236 Exercise 5G 237 5H Applications of exponential and logarithmic functions 238 Exercise 5H 239 ■■ ■■ ■■ ■■

Summary 243 Chapter review 246 ICT activities 250 Answers 251

Chapter 6

Circular functions 255 6A Trigonometric ratio revision 255

Exercise 6A 256 6B The unit circle 258

Exercise 6B 261 6C Radians 262

Exercise 6C 264 6D Symmetry 265

Exercise 6D 268 iv Contents

6E Identities 269

Exercise 6E 272 6F Sine and cosine graphs 273

Exercise 6F 277 6G Tangent graphs 280

Exercise 6G 282 6H Solving trigonometric

equations 283 Exercise 6H 287 6I Applications 288 Exercise 6I 290 ■■ ■■ ■■ ■■

Summary 293 Chapter review 296 ICT activities 301 Answers 302

Exam practice 2

Based on Chapters 1– 6 307 Chapter 7

Matrices 309 7A Addition and subtraction of

matrices 309 Exercise 7A 315 7B Multiplying matrices 316 Exercise 7B 318 7C Solving matrix equations 320 Exercise 7C 324 7D Matrices and transformations 325 Exercise 7D 328 ■■ ■■ ■■ ■■

Summary 330 Chapter review 331 ICT activities 335 Answers 336

Exam practice 3

Based on Chapters 1– 7 339 Chapter 8

Rates of change 341 8A Identifying rates 341

Exercise 8A 342 8B Constant rates 343

Exercise 8B 345 8C Variable rates 347

Exercise 8C 348 8D Average rates of change 349

Exercise 8D 351 8E Instantaneous rates 353

Exercise 8E 355 8F Motion graphs (kinematics) 357

Exercise 8F 359 8G Relating the gradient function to the original

function 364 Exercise 8G 364 8H Relating velocity–time graphs to position–time graphs 365 Exercise 8H 366

8I Rates of change of polynomials 369

11B Calculating probabilities 478

Exercise 11B 481

Exercise 8I 373 ■■ ■■ ■■ ■■

Summary 375 Chapter review 377 ICT activities 383 Answers 384

Chapter 9

Differentiation 389 9A Introduction to limits 389

Exercise 9A 392 9B Limits of discontinuous, rational and hybrid functions 393 Exercise 9B 395 9C Differentiation using first principles 397 Exercise 9C 399 9D Finding derivatives by rule 400 Exercise 9D 405 9E Rates of change 407 Exercise 9E 410 9F Sketching graphs containing stationary points 412 Exercise 9F 417 9G Solving maximum and minimum problems 418 Exercise 9G 421 ■■ ■■ ■■ ■■

Summary 423 Chapter review 425 ICT activities 431 Answers 432

Chapter 10

Antidifferentiation (integration) 437 10A Antidifferentiation 437

Exercise 10A 439 10B Deriving the original function from the gradient

function 440 Exercise 10B 442 10C Approximating areas enclosed by functions 444 Exercise 10C 448 10D The fundamental theorem of integral calculus 452 Exercise 10D 456 10E Applications of antidifferentiation 457 Exercise 10E 459 ■■ ■■ ■■ ■■

Summary 463 Chapter review 465 ICT activities 468 Answers 469

Exam practice 4

Based on Chapters 1– 10 473 Chapter 11

Introductory probability 475 11A Introduction to experimental probability 475

Exercise 11A 477

11C Tree diagrams and lattice diagrams 483

Exercise 11C 485 11D The Addition Law of Probabilities 487

Exercise 11D 490 11E Karnaugh maps and probability tables 492

Exercise 11E 495 11F Conditional probability 497

Exercise 11F 500 11G Transition matrices and Markov chains 501

Exercise 11G 506 11H Independent events 507

Exercise 11H 511 11I Simulation 513

Exercise 11I 515 ■■ ■■ ■■ ■■

Summary 516 Chapter review 518 ICT activities 523 Answers 524

Chapter 12

Combinatorics 529 12A The addition and multiplication principles 529

Exercise 12A 531 12B Permutations 533

Exercise 12B 534 12C Factorials 536

Exercise 12C 537

12D Permutations using nPr 538

Exercise 12D 540

12E Permutations involving restrictions 541

Exercise 12E 544 12F Arrangements in a circle 545

Exercise 12F 546

12G Combinations using nCr 547

Exercise 12G 549

12H Applications to probability 551

Exercise 12H 554 ■■ ■■ ■■ ■■

Summary 557 Chapter review 558 ICT activities 561 Answers 562

Exam practice 5

Based on Chapters 1–12 565 ■■ Answers 567

Cumulative exam practice 569 Functions and graphs 569 Algebra 570 Rates of change and calculus 573 Probability 575 ■■ Answers 585

Index 587

Contents v

Introduction Maths Quest 11 Mathematical Methods CAS Third edition is specifically designed for the VCE Mathematical Methods CAS course and based on the award-winning Maths Quest series. This resource contains: • a student textbook with accompanying eBookPLUS • a TI-Nspire CAS calculator companion • a Casio ClassPad CAS calculator companion • a solutions manual • flexisaver versions of all print products • teacher support material available on eGuidePLUS.

Student textbook Full colour is used throughout to produce clearer graphs and headings, to provide bright, stimulating photos and to make navigation through the text easier. Cumulative exam practice questions assist students with exam preparation. Clear, concise theory sections contain worked examples and highlighted important text and remember boxes. Icons appear for the eBookPLUS to indicate that interactivities and eLessons are available online to help with the teaching and learning of particular concepts. Worked examples in a Think/Write format provide clear explanation of key steps and suggest presentation of solutions. Many worked examples have eBookPLUS icons to indicate that a Tutorial is available to elucidate the concepts being explained. Worked examples also have calculator icons that indicate support in the Calculator Companion books, which contain comprehensive step-by-step CAS calculator instructions, fully integrated into the examples, for the TI-Nspire CAS and Casio ClassPad calculators. Exercises contain many carefully graded skills and application problems, including multiple choice questions. Cross-references to relevant worked examples appear with the first ‘matching’ question throughout the exercises. A selection of questions are tagged as technology-free to indicate to students that they should avoid using their calculators or other technologies to assist them in finding a solution. Exam practice sections contain exam style questions, including time and mark allocations for each question. Fully worked solutions are available on the eBookPLUS for students. Each chapter concludes with a summary and chapter review exercise containing examination-style questions (multiple choice, short answer and extended response), which help consolidate students’ learning of new concepts. Also included are questions from past VCE exams along with relevant Exam tips.

Student website — eBookpLUS The accompanying eBookPLUS contains the entire student textbook in HTML plus additional exercises. Students may use the eBookPLUS on laptops, tablets and school or home computers, and cut and paste material for revision, assignments or the creation of notes for exams. Career profiles and History of mathematics place mathematical concepts in context. Investigations, often suggesting the use of technology, provide further discovery learning opportunities. WorkSHEET icons link to editable Word documents, and may be completed on screen or printed and completed by hand. SkillSHEET icons link to printable pages designed to help students revise required concepts, and contain additional examples and problems. Interactivity icons link to dynamic animations, which help students to understand difficult concepts. vi

Introduction

eLesson icons link to videos or animations designed to elucidate concepts in ways other than what the teacher can achieve in the classroom. Tutorial icons link to one-way engagement activities, which explain the worked examples in detail for students to view at home or in the classroom. Test yourself tests are also available and answers are provided for students to receive instant feedback.

Teacher website — eGuidePLUS The accompanying eGuidePLUS contains everything in the eBookPLUS and more. Two tests per chapter, fully worked solutions to WorkSHEETs, the work program and other curriculum advice in editable Word format are provided. Maths Quest is a rich collection of teaching and learning resources within one package. The eGuidePLUS provides NEW material from which teachers may set school-assessed coursework (SAC), such as application tasks, analysis tasks and semester exams, as well as additional investigations, worksheets and technology files.

Introduction vii

About eBookPLUS Next generation teaching and learning This book features eBookPLUS: an electronic version of the entire textbook and supporting multimedia resources. It is available for you online at the JacarandaPLUS website (www.jacplus.com.au).

Using the JacarandaPLUS website To access your eBookPLUS resources, simply log on to www.jacplus.com.au using your existing JacarandaPLUS login and enter the registration code. If you are new to JacarandaPLUS, follow the three easy steps below. Step 1. Create a user account The first time you use the JacarandaPLUS system, you will need to create a user account. Go to the JacarandaPLUS home page (www.jacplus.com.au), click on the button to create a new account and follow the instructions on screen. You can then use your nominated email address and password to log in to the JacarandaPLUS system. Step 2. Enter your registration code Once you have logged in, enter your unique registration code for this book, which is printed on the inside front cover of your textbook. The title of your textbook will appear in your bookshelf. Click on the link to open your eBookPLUS. Step 3. View or download eBookPLUS resources Your eBookPLUS and supporting resources are provided in a chapter-by-chapter format. Simply select the desired chapter from the drop-down list. Your eBookPLUS contains the entire textbook’s content in easy-to-use HTML. The student resources panel contains supporting multimedia resources for each chapter. Once you have created your account, you can use the same email address and password in the future to register any JacarandaPLUS titles you own.

viii About eBookPLUS

Using eBookPLUS references eBookPLUS logos are used throughout the printed books to inform you that a multimedia resource is available for the content you are studying. Searchlight IDs (e.g. INT-0001) give you instant access to multimedia resources. Once you are logged in, simply enter the searchlight ID for that resource and it will open immediately.

Minimum requirements • A modern internet browser such as Internet Explorer 7+, Mozilla Firefox 3+, Google Chrome 8+, Safari 3+ or Opera 9+ • Adobe Flash Player 10+ • Javascript must be enabled (most browsers are enabled by default).

Troubleshooting • Go to the JacarandaPLUS help page at www.jacplus.com.au/jsp/help.jsp. • Contact John Wiley & Sons Australia, Ltd. Email: [email protected] Phone: 1800 JAC PLUS (1800 522 7587)

Acknowledgements The authors and publisher would like to thank the following copyright holders, organisations and individuals for their assistance and for permission to reproduce copyright material in this book.

Images • iStockphoto: pp. 23 (upper)/© Danny Warren, 25/© MichaelSvoboda/, 26 (right)/© Aneese, 33/© Felix Alim, 41/© technotr, 135 (upper)/© Steve Maehl, 135 (lower)/© Scott Cramer, 136/© Keiichi Hiki, 161 (lower left)/© Adnan Ahmad, 163/© TIM MCCAIG, 240 (upper)/© Hal Bergman, 257/© Mac99, 429/© Grafissimo, 452/© Don Bayley, 460/© JamesReillyWilson/, 461/© Doug Sims, 515/uatp2, 522/ © Ivan Bajic, 544 (lower)/© Clerkenwell_Images • © Corbis Corporation: p. 240 (lower); p. 535/Jack Hollingsworth • © Digital Stock/Corbis Corporation: pp. 8 (left, right), 83, 198 • © Image Source: p. 241 • © ImageState: p. 421; p. 497/© John Foxx Images • John Wiley & Sons Australia: pp. 316, 541 (four images)/Photo by Renee Bryon; pp. 486, 544 (upper)/Malcolm Cross; p. 521 (lower)/Taken by Jo Patterson • © Newspix: p. 309/Robert McKell; p. 389/News Ltd/3rd Party Managed Reproduction & Supply Rights • © Photodisc, Inc.: pp. 40, 82, 161 (lower middle, lower right), 165, 196, 239, 242, 291, 292, 300, 360, 363, 425, 451, 482 (upper), 559 • photolibrary.com: p. 23 (lower); p. 217/Sydney/ SPL/CNRI; p. 467/Simon Fraser • Photolibrary Royalty Free: p. 11/© Huntstock/Photolibrary Inc. • Shutterstock: pp. 26 (left)/© Hannu Liivaar, 62/© Rudy Balasko, 88/© Marie Lumiere, 161 (upper left)/© C, 208/© Brendan Howard, 289/© Jeff Davies, 299 (lower)/© G Tipene, 299 (upper)/© Patricia Hofmeester, 341/© Colin Stitt, 355/© Kalim, 369/© Germanskydiver, 407 (left)/© 1971yes, 407 (right)/© Pete Niesen, 408/© Jamie Roach, 409/© James Thew, 411/© Gelpi, 478/© Andrejs Pidjass, 482 (lower)/© Kitch Bain, 487/© Roger Browning, 501/© Rich Carey, 512/© Claudio Gennari, 521 (upper)/© ronfromyork, 530/© arkasha1905, 540/© Lincoln Rogers, 545/© arkasha1905, 556/ © Lario Tus • © Stockbyte: p. 161 (upper right) • © Viewfinder Australia Photo Library: p. 199 • Visy Recycling: p. 197 Every effort has been made to trace the ownership of copyright material. Information that will enable the publisher to rectify any error or omission in subsequent editions will be welcome. In such cases, please contact the Permissions Section of John Wiley & Sons Australia, Ltd.

Acknowledgements

ix

Chapter 1

Linear functions diGital doC doc-9695 10 Quick Questions

Chapter ContentS 1a 1B 1C 1d 1e 1F 1G 1h

Solving linear equations and inequations Rearrangement and substitution Gradient of a straight line Sketching linear functions Simultaneous equations Finding the equation of a straight line Distance between two points and midpoint of a segment Linear modelling

Solving linear equations and inequations 1a

A linear equation is one that involves a variable raised to the power of 1 only. Recall that x1 is the same as x, so a linear equation involving x would contain only x’s, and not x , x2, x3, x −1, x −2 and so on. For 8 example, y = 7x − 3, ax + by = c and 3 x + 1 = 9 are all linear equations. 1 1 Note that y = is not a linear equation, as the is really x −1. x x To isolate a particular variable — known as ‘making x (or whatever the variable is) the subject’ — we focus on the variable by ‘undoing’ other terms and operations. Remember to do the same operation to both sides of an equation, in the reverse order to that originally used to make up the equation. Though the focus of this chapter is linear equations, some other types of equations will be included for skills practice. When there is only one variable involved in an equation, we may attempt to solve and find a numerical value by rearranging to make the variable the subject. 1 2

Worked example 1

Solve the following linear equations. 3x − 2 4x +5=1 a 7x − 4 = 17 b c 2 − 1 = 6 5 4 think

a 1 Write the equation. 2

Add 4 to both sides.

3

Divide both sides by 7.

b 1 Write the equation. 2

Subtract 5 from both sides.

Write

a 7 x − 4 = 17

7 x = 21 21 7 x=3 x=

b

3x − 2 +5=1 4 3x − 2 − = 4 4 Chapter 1 • Linear functions

1

3

Multiply both sides by 4.

4

Add 2 to both sides.

5

Divide both sides by 3.

3 x − 2 = − 16 3 x = − 14 x=

−14

3

4 x − 1 = 6 5 4x −1 = 3 5 4x =4 5

c 1 Write the equation.

c 2

2

Divide both sides by 2.

3

Add 1 to both sides.

4

Multiply both sides by 5.

5

Divide both sides by 4.

4 x = 20 20 4 x=5 x=

When an equation has variables on both sides, at some stage they must be gathered together on the same side of the equation. Worked example 2

Solve: a 4x − 3 = 3(6 − x)

b

9 x + 3 13 x + 7 = 2 3

c

3 − x 9( x + 7) = + 1. 4 10

think

a 1 Write the equation.

a 4x − 3 = 3(6 − x)

2

Expand the right-hand side (RHS).

4x − 3 = 18 − 3x

3

Collect x’s on one side, for example, the side that results in a positive x term, in this case, the left-hand side (LHS). (That is, add 3x to both sides.)

7x − 3 = 18

4

Add 3 to both sides.

5

Divide both sides by 7.

b 1 Write the equation.

7x = 21 21 7 x=3 x=

b

9 x + 3 13 x + 7 = 2 3 LCD = 6

2

Find the lowest common denominator for both terms.

3

Write all terms with the common denominator, adjusting numerators accordingly (so that numerator and denominator have been multiplied by the same amount).

3(9 x + 3) 2(13 x + 7) = 6 6

4

Now that all terms have the same denominator, the numerators must be equal. (Multiply each side by 6.)

3(9 x + 3) = 2(13 x + 7)

5

Expand all brackets.

6

Collect x’s on the LHS and numbers on the RHS.

7

Simplify and solve.

c 1 Write the equation. 2

2

Write

Find the lowest common denominator for all three terms.

Maths Quest 11 Mathematical Methods CAS

27 x + 9 = 26 x + 14 27 x − 26 x = 14 − 9 x=5 c

3 − x 9( x + 7) = +1 4 10 LCD = 20

3

Write all terms with the common denominator, adjusting numerators accordingly (so that numerator and denominator have been multiplied by the same amount).

5(3 − x ) 2 × 9( x + 7) 20 = + 20 20 20

4

Now that all terms have the same denominator, the numerators must be equal. (Multiply each term by 20.)

5(3 − x) = 18(x + 7) + 20

5

Expand all brackets.

15 − 5x = 18x + 126 + 20

6

Collect x’s on the RHS and numbers on the LHS.

15 − 126 − 20 = 18 x + 5 x − 131 =

7

−131

Simplify and solve.

23

23 x

=x −131

x=

23

There is only one solution to a linear equation. For example, the equation 3x + 2 = 8 has the solution x = 2. The solution to a linear equation can be represented as a single point on a number line. If the equals sign is replaced with an inequality sign, the solution is a portion of a number line. These expressions are referred to as linear inequations (sometimes called inequalities). Worked example 3

Solve the following linear inequations. a 6x − 7 ≥ 3x + 5 b −3m + 5 < −7 think

Write

a 6x − 7 ≥ 3x + 5

a 1 Write the inequations. 2

Combine the variable terms by subtracting 3x from both sides.

3

Add 7 to both sides.

4

Divide both sides by 3.

3x − 7 ≥ 5 3x ≥ 12 x≥4 b −3m + 5 < −7

b 1 Write the inequations.

−3m

2

Subtract 5 from both sides.

3

Divide both sides by −3. Reverse the inequality sign, as you are dividing by a negative number.

exercise 1a

m>4

Solving linear equations and inequations

1 We1 Solve the following linear equations. a 3x − 19 = −13 b 4x + 25 = −7

12 − 3 x =5 3 3 x − 1 = 10 h 2 5 2 We2 Solve the following linear equations. a 2x − 9 = 3(2x − 11) b 7x − 1 = 17(3x − 13) x+2 x−5 x + 11 2( x + 14) = = e f 6 3 3 9 6 x + 7 5x + 1 2 x + 29 x + 44 = +1 = +2 i j 5 4 3 8 3x − 1 =5 4 7x + 4 g − 8 = −9 3 d

< −12

diGital doC doc-9696 equation solvers

c 9x + 19 = −2

4x − 6 −7=3 3 x i 7 8− = 77 4

e

f

c x + 11 = 2(x + 12)

4 x + 66 13 − 3 x = 3 4 7 x − 9 21 − x = − 18 k 9 3 g

d 3x − 7 = 3(35 − 2x)

x + 10 2(7 − 3 x ) = 9 5 17 − x 49 + 2 x = +5 l 2 5

h

Chapter 1 • Linear functions

3

Solve the following linear inequations. 11 − 3 x b ≥6 a 15 − 6x ≤ 2 7

3 We3

d 16 − 4x < 7(1 − x)

1B

e 3x − 7 ≥ 2(35 − 2x)

19 − x + 3 > −1 − 4 4 − x 2( x + 1) ≤ f 3 6 c

rearrangement and substitution

When there is more than one variable involved in an equation, we may rearrange the equation to make a particular variable the subject using the same rules of equation solving described in the previous section. Worked example 4

Rearrange each of the following to make the variable in parentheses the subject. 1 m a 6x + 8y − 48 = 0 (y) b s = ut + at 2 k(u) c T = 2π k(k) 2 k think

a 1 Write the equation.

a 6x + 8y − 48 = 0

6x + 8y = 48

2

Add 48 to both sides.

3

Subtract 6x from both sides.

4

Divide both sides by 8.

y=

48 − 6 x 8

5

Cancel if possible. Here, divide the numerator and denominator by 2.

y=

24 − 3 x 4

6

Other ways of representing the answer are shown opposite.

b 1 Write the equation.

8y = 48 − 6x

24 3 − x 4 4 3 = 6− x 4 −3 = x+6 4

or y =

s = ut + 12 at 2

b

2

Subtract 12 at 2 from both sides.

s − 12 at 2 = ut

3

Multiply both sides by 2.

2s − at 2 = 2ut

4

Divide both sides by 2t.

5

Write the equation with the desired variable on the left.

c 1 Write the equation.

4

Write

2s − at 2 =u 2t 2s − at 2 u= 2t c

T = 2π T = 2π

m k

2

Divide both sides by 2π.

3

Square both sides.

T2 m = 4π 2 k

4

Form the reciprocal of both sides to make k the numerator.

4π 2 k = T2 m

Maths Quest 11 Mathematical Methods CAS

m k

5

Multiply both sides by m.

6

Write the equation with the desired variable on the left.

4π 2m =k T2 k=

4π 2m T2

Once a variable is isolated, we may substitute values of other variables to calculate various values of the isolated variable. The following worked example illustrates some practical applications. Worked Example 5

a The formula for converting temperature in degrees Fahrenheit (F) (which is the system used in

5( F − 32) the USA) to degrees Celsius (C ) is C = . 9 i Make F the subject. ii What is the temperature in degrees Fahrenheit when the temperature measured in degrees Celsius is 21° C? b The area (A) of a circle is given by A = π r 2, where r is the radius. Calculate the value of r correct to 2 decimal places when A = 20 cm2. Think

a i 1 Write the equation.

Write

C=

a i

5( F − 32) 9

2

Multiply both sides by 9.

9C = 5( F − 32)

3

Divide both sides by 5.

9C = F − 32 5

4

Add 32 to both sides.

5

Write the equation with F first. Sometimes it may be appropriate to use a common denominator.

9C + 32 = F 5 9C F= + 32 5 9C + 160 or F = 5

ii 1 Replace C with 21. Note that 9C means 9 × C. 2

9 × 21 + 32 5 189 + 32 F= 5 F = 37.8 + 32 F=

ii

Evaluate F.

F = 69.8 so 21°C = 69.8 °F b

1

Write the equation. Solve for r (make r the subject) as follows.

2

Divide both sides by π.

3

Take the square root of both sides, and write r first. As r is the radius, we take the positive root only.

4

Substitute A = 20 into the new formula.

b

A = πr2 A = r2 π r=

If A = 20, r=

5

Evaluate r.

A π

20 π

r = 2.52 cm, correct to 2 decimal places

Chapter 1 • Linear functions 5

Worked example 6

Note: In this example, m represents metres, m/s represents metres per second (velocity), and m/s2 represents metres per second per second (acceleration). The final velocity, v m/s, of an object that begins with velocity u m/s and accelerates at a m/s2 over a distance of s m is given by the equation v2 = u2 + 2 as. a Find the value of v when u = 16, a = 2 and s = 36. b Rearrange the given equation to make s the subject. c Find the distance travelled by an object that begins with a velocity of 10 m/s and has a final velocity of 4 m/s while accelerating at −1 m/s2. think

Write

a 1 Write the given equation and then make v the subject

a v 2 = u 2 + 2as

by finding the square root of both sides.

v = ± u 2 + 2as

2

Substitute u = 15, a = 2 and s = 36.

= ± 162 + 2 × 2 × 36

3

Simplify and evaluate v.

= ± 256 + 144 = ± 400 = ± 20 The final velocity is 20 m/s.

b 1 Begin with the given equation.

v 2 = u 2 + 2as

b

2

Subtract u2 from both sides.

v 2 − u 2 = 2as

3

Divide both sides by 2a.

v 2 − u2 =s 2a

4

Reverse so that s is given on the left.

c 1 Match the variables with the given information.

s=

v 2 − u2 2a

c u = 10

v=4 a = −1

3

Write the formula that has s as the subject (see part b above). Substitute the values given in step 1.

4

Simplify and evaluate.

5

Explain the answer in words.

2

exercise 1B

s= =

v 2 − u2 2a 4 2 − 10 2 2 × −1

16 − 100 − 2 − 84 = − 2 = 42 =

The object travels 42 m in its initial direction.

rearrangement and substitution

Each of the following is a real equation used in business, mathematics, physics or another area of science. Make the variable shown in parentheses the subject in each case. a A=L+P (P) b A = lw (l) d (t) c v= d C = 2πr (r) t

1 We4

6

Maths Quest 11 Mathematical Methods CAS

(β)

f

kQq r2

(r)

g Fd =

(v)

h v = γ rT

(γ)

i

(w)

j

e E = αθ + βθ 2

1 2 1 2 mv − mu 2 2 S = 2w(l + h) + 2lh

F=

S = 2πr2 + 2πrH

(H)

2 We5 Calculate the value of the subject (the first mentioned variable), given the values of the other

variables. k a I= 2 d b E = K + mgh c D = (n −

3

k = 60, d = 15 K = 250, m = 2, g = 10, h = 5

1 )λ 2

n = 3, λ = 2.8

d E = hf0 − W

h = 6.62, f0 = 5000, W = 20 000

e v = ω r 2 − y2

ω = 2, r = 1.6, y = 1

Make the variable in parentheses the subject and find its value using the given information. a A = l2

(l)

A = 60

4 πr3 3

(r)

V = 1000

c v = u + at

(a)

v = 25, u = 0, t = 6

(l)

T = 4, g = 9.8

(c)

K = 6.9, α = 0.05

b V=

d T = 2π e K=

l g

cα 2 1− α

4 We6 The perimeter, P, of a rectangle of length l and width w may be found using the equation

P = 2(l + w). a Find the perimeter of a rectangle of length 16 cm and width 5 cm. b Rearrange the equation to make w the subject. c Find the width of a rectangle that has perimeter 560 mm and length 240 mm.

5 The area of a trapezium (Figure A) is given by A =

a + b , where a and b are the lengths h 2

of the parallel sides and h is the height. a Find the area of the trapezium shown in Figure B. b Using Figure A, find an equation for determining side a in terms of the area A and side b. c Find a in Figure C. 9m a h

Area A

b Figure A

16 m

50 cm a

62 cm Area = 2000 cm2

21 m Figure B

Figure C

r 2 , 100 where A is the amount ($) in the account after two years, D is the initial deposit ($) and r is the interest rate (%). a Find the amount in such an account after two years if the initial deposit was $1000 and the interest rate was 6%. b Make r the subject of the equation. c Find the rate required for an initial deposit of $1000 to grow to $2000 after 2 years.

6 The size of a 2-year investment account with a particular bank is given by A = D 1 +

Chapter 1 • Linear functions

7

7 The object and image positions for a lens of focal length f are related by

1 1 1 + = , where u is the distance of the object from the lens u v f and v is the distance of the image from the lens. 4 2 3 a Make f the subject of the equation. b Make u the subject of the equation. c How far from the lens is the image when an object is 30 cm in front of a lens of focal length 25 cm? b 8 The length of a side of a right-angled triangle can be found using Pythagoras’ theorem: c2 = a2 + b2, where c is the length of the longest side, and a and b are the lengths of the two shorter sides. Find the value of b in the triangle above. 1 9 The volume of a cone is given by the rule V = 3 π r2h, where r is the radius of the widest r part of the cone and h is the vertical height of the cone. Given that the volume of a cone is 100 cm3 and its radius at the widest point is 12 cm, find the height of the h cone, expressing your answer in terms of π. the formula

diGital doC doc-9697 Career profile rick morris – vigneron

Gradient of a straight line

1C

cone

The gradient of a line describes its slope or steepness. y

y

y Negative gradient

Positive gradient

x

x

y (x2, y2) Rise (x1, y1) Run x

y

Zero gradient

Infinite gradient

x

x

The gradient may be calculated using the formula: y −y rise m = run or m = 2 1 x 2 − x1 These terms are illustrated at left. Here are two examples of where gradient can affect our everyday lives. Can you think of others?

A cliff face with a steeper gradient provides a greater challenge for climbers.

Scientists calculate the required gradient of solar panels so that the maximum amount of energy is absorbed.

8

Maths Quest 11 Mathematical Methods CAS

Worked example 7

Calculate the gradient of this linear graph using the intercepts shown. think 1

2

y

Write

Identify the rise and run. Rise = 14, run = 2. Calculate m = rise . run

14

14 2 =7

m=

−2

x

Worked example 8

Calculate the gradient of the line passing through the points (3, −6) and (−1, 8). think

Write

y2 − y1 . x 2 − x1

m=

1

Use the formula m =

2

Match up the terms in the formula with the values given. (x1, y1) (x2, y2) (3, −6) (−1, 8)

3

Substitute the given values.

4

Simplify.

5

Cancel if possible.

y2 − y1 x 2 − x1

8 − −6 − 1− 3 14 14 = = −− 44 − 7 = −7 = 2 2

m=

If the angle a line makes with the positive direction of the x-axis is known, the gradient may be found using trigonometry applied to the triangle shown below. y rise

θ run

x

tan (θ ) = opposite = rise = m adjacent run Worked example 9

y

a Calculate the gradient (accurate to 3 decimal places) of a line making

an angle of 40° to the positive x-axis.

b Calculate the gradient of the line shown at right. Express your answer

60°

to 2 decimal places.

think

a Since the angle the line makes with

the positive x-axis is given, the formula m = tan (θ) can be used.

Write

x

y

a m = tan (θ)

= tan (40°) = 0.839, correct to 3 decimal places

40°

x

Chapter 1 • Linear functions

9

b

1

2

b θ = 180° − 60°

The angle given is not the one between the graph and the positive direction of the x-axis. Calculate the required angle θ.

60° θ

Use m = tan (θ) to calculate m to 2 decimal places.

exercise 1C 1 We7

y

= 120°

m = tan (θ) = tan (120°) = −1.73

Gradient of a straight line

Calculate the gradient of each of the following linear graphs using the intercepts shown.

a

b

y

y

6

2

x

−3

6

x

Without drawing a graph, calculate the gradient of the line passing through: a (2, 4) and (10, 20) b (4, 4) and (6, 14) c (10, 4) and (3, 32) d (5, 31) and (− 7, 25). 3 We9a Calculate the gradient (accurate to 3 decimal places) of a line making the angle given with the positive x-axis. a 50° b 72° c 10° d − 30° e 150° f 0° g 45° h 89°

2 We8 diGital doC doc-9698 Gradient of a straight line

4 We9b

Calculate the gradient of each line below. Give answers to 2 decimal places. y

a

c

y

b

43°

x

y

69°

x

y

d

x 28°

15°

x

5 Which of these lines has: a a non-zero positive gradient? b a negative gradient? c a zero gradient? d an undefined gradient?

y

Maths Quest 11 Mathematical Methods CAS

B

5 4 3 2 1

A

10

x

−5−4−3−2−1 0 1 2 3 4 5 −1 −2 −3 −4 −5

D

C x

6 mC a Which of the following lines has a

b Which of the following lines has a

gradient of −2?

B

A

y 5 4 3 2 1

gradient of 3?

C

A

D

5 4 3 2 1

E

−5−4−3−2−1 0 1 2 3 4 5 −1 −2 −3 −4 −5

D

x

y

B C

−5−4−3−2−1 0 1 2 3 4 5 −1 −2 −3 −4 −5 E

x

7 Burghar plots the coordinates of a proposed driveway on a plan that is shown below. What is the

gradient of the proposed driveway?

Garage

way Drive 2 m

17 m 8 An assembly line is pictured below. What is the gradient of the sloping section? (Give your answer as a

fraction.) BOFFO Made in Austra lia

BOFFO Made in Austra lia

BOFFO Made in Austra lia

0.85 m 15 m

9 Determine the value of a in each case so the gradient joining the points is equal to the value given. a (3, 0) and (5, a), gradient 2 b (2, 1) and (8, a), gradient 5 c (0, 4) and (a, −11), gradient 3 d (a, 5) and (5, 1), gradient −2 10 For safety considerations, wheelchair

ramps are constructed under regulated specifications. One regulation requires that the maximum gradient of a ramp exceeding 1200 mm in length is 1 . 14 a Does a ramp 25 cm high with a horizontal length of 210 cm meet the requirements? 1 b Does a ramp with gradient meet 18 the specifications? c A 16 cm high ramp needs to be built. Find the horizontal length of the ramp required to meet the specifications.

diGital doC doc-9699 SkillSHEET 1.1 Using a gradient to find the value of a parameter

Chapter 1 • Linear functions

11

1d

Sketching linear functions

The general form for linear equations is y = mx + c, where m is the gradient of the line and c is the y-intercept. y ie

ad

Gr

y = mx + c Gradient

y

m nt

y-intercept x

y-intercept

x

x-intercept These lines have identical gradients (equal m values).

diGital doC doc-9700 WorkSHEET 1.1

To sketch a graph from a linear equation expressed in general form, follow these steps. Step 1 Plot the y-intercept on a set of axes. Step 2 Find and plot a second point on the line. Do this by substituting any value of x into the equation and finding the corresponding y-value. Step 3 Join the two points. Alternatively, you can use a CAS calculator or other graphing technology. To find the equation of a line given the gradient and y-intercept, simply substitute the values of m and c into y = mx + c.

Sketching linear graphs using intercepts

y

To draw a graph, only two points are needed. A line may then be drawn through the two points, and will include all other points that follow the given rule. The two points can be chosen at random; however, it is often easier to sketch a graph using the points where the graph crosses the axes. These points are called x- and y-intercepts. The x-intercept occurs when y = 0, and the y-intercept occurs when x = 0.

x

Worked example 10

Sketch the graph of y = −x + 6, showing x- and y-intercepts. think 1

2

Write/draW

Find the y-intercept (when x = 0). Substitute x = 0 into the equation.

If x = 0: y = −1 × 0 + 6 y=6

Find the x-intercept (when y = 0). Substitute y = 0 into the equation.

If y = 0: 0 = −x + 6

3

Mark the intercepts on a set of axes.

4

Join the intercepts with a straight line.

x=6

(6, 0)

y (0, 6)

(6, 0)

12

Maths Quest 11 Mathematical Methods CAS

(0, 6)

x

Worked example 11

Sketch the graph of 3x − 2y = 12. think

Write/draW

1

Find the y-intercept (when x = 0). Substitute x = 0 into the equation.

If x = 0: 3 × 0 − 2y = 12 −2y = 12 12 y= − 2 y = −6

2

Find the x-intercept (when y = 0). Substitute y = 0 into the equation.

If y = 0: 3x − 2 × 0 = 12 3x = 12 x=4

3

Mark the intercepts on a set of axes.

4

Join the intercepts with a straight line.

y

x

4

−6

The graphs of some equations do not have two intercepts, as they pass through the origin (0, 0). Such equations are of the form y = k x or a x + by = 0. To sketch graphs of such equations, we use (0, 0) and any other point, for example the point where x = 1. (We could choose any other non-zero value.) Worked example 12

Sketch the graph for the equation 4 x − 3y = 0. think 1

Try substituting x = 0 to find the y-intercept.

2

Note that the graph passes through (0, 0). There is no point substituting y = 0, as we know we’ll get x = 0.

3

Substitute another x-value. In this example we use x = 1.

4

Plot the points (0, 0) and (1, 43 ) on a set of axes. Note that 4 is 1 1 , which is a little less 3 3 than 1 1 . 2

Write/draW

4 × 0 − 3y = 0 −3y = 0 y=0

If x = 0:

tUtorial eles-1404 Worked example 12

(0, 0)

If x = 1:

4 × 1 − 3y = 0 4 − 3y = 0 4 = 3y 4 y= 3 4 (1, ) 3

y 2 (1, 4–3 )

1 (0, 0) 1

x

Chapter 1 • Linear functions

13

Sketching linear functions

exercise 1d

1 Use a CAS calculator or other method to sketch graphs of the following equations on the same set of axes. a y=x b y = 2x c y = 3x d y = −x e y = −2x 2 In question 1, what is the effect on the graph of the number in front of the x (the ‘x-coefficient’ or

‘gradient’)? 3 Use a CAS calculator or other method to sketch graphs of the following equations on the same set of axes. a y=x+1 b y=x+2 c y=x+3 d y=x−4 diGital doC doc-9701 linear graphs

4 Use a CAS calculator or other method to sketch graphs of the following equations on the same set of

axes. a y = 2x + 1

c y = −3x + 6

b y = 2x − 7

d y = 3x − 5

5 In questions 3 and 4, what is the effect on the graph of the number at the end of the equation (the

‘y-intercept’)? 6

Write the equation of a line having the following properties (where m = gradient and c = y-intercept). a m = 2, c = 7 b m = −3, c = 1 c m = 5, c = −2 −1 −3 2 1 f y-intercept 12, gradient −2 d m= ,c= e m = 4, c = 2 3 3 Rearrange the following equations and state the gradient and y-intercept for each.

7

a 2y = 8x + 10 e 21x + 3y = −27 i 15 − 6y + x = 0

b 3y = 12x − 24 f −10x + 5y = 25 j 2y + 7 + 5x = 0

−y

c g

= 3x − 1 − 2x = 66

d 16 − 4y = 8x h 8x + 3y − 2 = 0

−11y

8 State the equation for each of the following graphs. a

y − 1– 2

−1

y

b

6 5 4 3 2 1

3 2 1 −2

0 −1 −2

1

x

2

y

c

e

−1

0 −1 −2 −3 −4 −5

1

5 4 3 2 1

x

2

−1

0 −1 −2 −3 −4 −5

f

y

14

0 −1 −2 −3 −4 −5

1

2

3

Maths Quest 11 Mathematical Methods CAS

4

1

2

3

4

5

1

2

3x

y

9 8 7 6 5 4 3 2 1

5 4 3 2 1 −1

x

2

y

d

3 2 1 −2

0 −1 −2 −3 −4 −5

x −3

−2

−1 −10 −2 −3 −4 −5 −6 −7 −8 −9

6

x

9 We10 Sketch graphs of the following linear equations, showing x- and y-intercepts. a y = 6x + 18 b y = 3x − 21 c y = −2x − 3 d y = 10 − 5x e y = −9x + 30 f y = 2(x − 8) 10 We11 Sketch graphs for each of the following equations. a 2x + 3y = 6 b 4x + 5y = 20 c 6x − 3y = −18 d 7x − 5y = 35 11 Sketch the graph for each equation. a 6x + 7y + 42 = 0 c −3x + 4y − 16 = 0 12 We12 Sketch the graph for each equation. a x+y=0 c 2x + y = 0

b 5x − 2y + 20 = 0 d y − 3x + 6 = 0 b x−y=0

13 mC Which of the following is in the form ax + by = c? a 2x − 3y − 1 = 0 C 2x + 3y = x e y=

2 3

B 2x + 3y + 1 = 0 d 2x + 3y = 1

x−1

The x- and y-intercepts for the equation 2y = −3x + 12 are (respectively): a 2 and 3 B −3 and 12 − C 4 and 6 d −4 and −6 e 4 and 6

14 mC

15 mC Which of the following has a y-intercept of −3? a y = −3x − 3 C x + 3y = 9

B y = −3x + 3 d x − 3y + 9 = 0

e 3x + y + 9 = 0

1e

Simultaneous equations

Simultaneous equations are groups of equations containing two or more variables. In this section, we look at pairs of linear equations involving the variables x and y. Each equation, as we have learned in previous sections, may be represented by a linear graph that is true for many x- and y-values. If the graphs intersect (when wouldn’t they?), the values of x and y at the intersection are those that make both equations true.

y Both graphs have the same x- and x y-values here.

Graphical solution Finding the point of intersection of two straight lines can be done graphically; however, the accuracy of the graph determines the accuracy of the solution. Consequently, using a calculator to solve the equations graphically is more reliable than reading the solution from a hand-drawn graph.

Worked example 13

Use a CAS calculator to solve the following simultaneous equations graphically. y = −3x + 5 4x − 7y + 8 = 0 think 1

Using a CAS calculator, make y the subject of the second equation. Complete the entry line as: solve (4x − 7y + 8 = 0, y)

Write

solve (4x − 7y + 8 = 0, y) 4( x + 2) Result: y = 7

Chapter 1 • Linear functions

15

2

On a Graphs page, complete the function entry lines as: f 1(x) = −3x + 5 4 × ( x + 2) f 2( x ) = 7 Use the calculator to find the intersection point.

f 1(x) = −3x + 5 4( x + 2) f 2(x) = 7

3

Write the answer.

Solving y = −3x + 5 and 4x − 7y + 8 = 0 simultaneously produces the solution (1.08, 1.76).

algebraic solution It is possible to solve simultaneous equations without graphs, that is, algebraically. The methods of substitution and elimination taught in earlier years may be used. Worked example 14

Use the substitution method to solve the following simultaneous equations. y = 2x + 5 −x + 3y = 25 think 1

Write down and label the equations.

2

Substitute [1] into [2] and label the resulting equation [3].

3

Solve [3] for x and label the solution as [4].

4

Write

Use the solution to solve for y and label the solution as [5].

−x −x

+ 3(2x + 5) = 25

−x

+ 6x + 15 = 25 5x + 15 = 25 5x = 10 x=2

Substitute [4] into [1]. y = 2(2) + 5 y=4+5 y=9

5

State the complete answer.

Solution: (2, 9)

6

Optional check: substitute [4] and [5] into [2] to check that these values for x and y make [2] true.

Check in [2]. LHS = −x + 3y = −2 + 3(9) = 25 = RHS ✓

y = 2x + 5 + 3y = 25

[5]

Worked example 15

Use the elimination method to solve these simultaneous equations. 2 x + 9y = −5 5 x − 2y − 12 = 0 think

16

1

Write down and label the equations.

2

Rearrange [2] so it is in a similar form to [1]. Call this [3]. Write down [1] again.

Maths Quest 11 Mathematical Methods CAS

Write

2x + 9y = −5 5x − 2y − 12 = 0

[1] [2]

5x − 2y = 12 2x + 9y = −5

[3] [1]

[1] [2] [3]

[4]

3

Obtain 10x in both [1] and [3].

2 × [3] 10x − 4y = 24 5 × [1] 10x + 45y = −25

[4] [5]

4

Eliminate x as shown.

[5] − [4]:

5

Solve for y.

49y = −49 − 49 y= 49 y = −1

[6]

6

Substitute [6] into [1] to find x.

Substitute [6] into [1]. 2x + 9 (−1) = −5 2x − 9 = −5 2x = −5 + 9 2x = 4 x=2

[7]

−1)

7

State the solution.

Solution: (2,

8

Again, [6] and [7] may be checked in [2] if desired.

Check in [2]. LHS = 5x − 2y − 12 = 5(2) − 2(−1) − 12 = 10 + 2 − 12 =0 = RHS ✓

Worked example 16

Two shoppers buy the following at a fruit shop, paying the amounts given. What was the cost of each apple and each banana? Shopper 1: 4 apples and 3 bananas for $2.59 Shopper 2: 6 apples and 5 bananas for $4.11 think

Write

1

Decide on variable names for the unknown quantities.

Let a = cost of an apple (in cents). Let b = cost of a banana (in cents).

2

Write equations involving these variables. Work in terms of cents.

4a + 3b = 259 6a + 5b = 411

3

Choose a variable to eliminate, in this case b.

5 × [1]: 20a + 15b = 1295 3 × [2]: 18a + 15b = 1233

[3] [4]

4

Find [3] − [4] and solve for a.

[3] − [4]: 2a = 62 a = 31

[5]

Substitute [5] into [1]. 4 × 31 + 3b = 259 124 + 3b = 259 3b = 135 b = 45

[6]

5

6

Solve for b.

State the answer using [5] and [6] as a guide.

exercise 1e

[1] [2]

The cost of an apple is 31 cents, and the cost of a banana is 45 cents.

Simultaneous equations

1 We13 Use a CAS calculator to solve the following simultaneous equations. a y = −2x, y = −4x − 6 b y = 3x − 5, y = 20 c y = 3x + 5, y = 7x − 4 e y = 10x + 1, 2x + y = −6

+ y = −4, y = 6x + 5 f 9x + y = 17, x + y = 14 d

−3x

diGital doCS doc-9702 Simultaneous linear equations — graphical method

Chapter 1 • Linear functions

17

2 We14 Use the substitution method to solve the following simultaneously. a y = 3x + 1, y = 2x + 2 b y = 5x + 5, y = −x − 19 c y = x + 2, 3x − 4y = −1

− 3y = 2, y = −6x + 7 f y = 10 − x, 2x + 7y = 5 We15 Use the elimination method to solve these simultaneous equations. a 9x + 10y = 153, 3x − y = 12 b 7x − 11y = −13, x + y = 11 − c 6x − 2y = 10, 2x + 5y = 8 d −3x + y = 8, 4x + 2y = 21 e 7y − x = 11, x + y = 10 f x − 11y = −15, y + 6x = 9 We16 At the conclusion of a tour of Wonky Willy’s confectionery factory, Nutrina buys 10 choc balls and 8 fizz wizzers for $4.30, and her friend purchases 6 choc balls and 9 fizz wizzers for $4.05. Determine the cost of each type of lolly. The sum of two whole numbers, x and y, is 41. The difference between them is 3. Write two equations involving x and y and solve them to find the numbers. A farmer counts emus and sheep in a paddock, and notes there are 57 animals and 196 feet. Assuming no animal amputees, how many of each animal are there? A sports store supplies 24 basketballs and 16 cricket balls to one school for $275.60, and delivers 12 basketballs and 32 cricket balls to another school for $211. If delivery is free, how much did the supplier charge for each type of ball? e

3

4

5 6 7

d y = −2x + 3, −5x + 2y = 1

−4x

8 A businessperson hires a stretch limousine for 2 days and a sedan for 3 days while on an interstate trip.

If the total car hire cost of the trip was $675, and the limousine cost triple the price of the sedan, find the cost per day of the limousine. 9 mC A manufacturing plant produces square and circular metal panels in fixed sizes. If the mass of a square panel is 13 kg and that of a circular panel is 22 kg, how many of each panel are there in a truck loaded with 65 panels of total mass 1205 kg? The equations to solve are: a 13s + 22c = 1205, s + c = 65 B 22s + 13c = 1205, s + c = 65 C 13s + 22c = 65, s + c = 1205 d 22s + 13c = 65, s + c = 1205 e 13s + 22c = 1205, s + c = 35 10 mC Which of the following is a solution of 11x + 2y = −121 and 10x + 12y = −222? a (11, 2) C (10, 12) e (6, 10) B (−121, −222) d (−9, −11)

1F

Finding the equation of a straight line

Consider a general linear graph containing the particular points (x1, y1), (x2, y2) and the general point (x, y) (which could be any point). Using the first two of these points in the formula for gradient, we have y2 − y1 [1] m= x −x 2 1 Using the first point and the general point in the same formula yields y − y1 [2] m= x − x1 Putting [2] = [1] gives

y − y1 y2 − y1 , which may be rearranged to = x − x1 x 2 − x1 y −y y − y1 = 2 1 ( x − x1 ) [3] x 2 − x1

y

(x2, y2) (x, y) (x1, y1) x

y2 − y1 , equation [3] may be written x 2 − x1 y − y1 = m(x − x1) [4] This last formula may be used to find the equation of a straight line when two points are given or when the gradient and only one point are given. When two points are given, the gradient m may be found y2 − y1 first using m = x − x and substituted into the formula y − y1 = m(x − x1) along with one of the points. 2 1 Since m =

18

Maths Quest 11 Mathematical Methods CAS

Worked Example 17

Find the equation of the line having gradient 34 that passes through (7, 11). Express your answer in the forms i ax + by + c = 0 and ii y = mx + c. Think

Write

1

As one point and the gradient are known, use the formula y − y1 = m(x − x1).

y − y1 = m(x − x1)

2

List the given information.

m = 4 (x1, y1) (7, 11)

3

Substitute for all variables except x and y.

4

Simplify, expressing in the form ax + by + c = 0.

i 4y − 44 = 3(x − 7) 4y − 44 = 3x − 21 3x − 4y + 23 = 0

5

Express your answer in the form y = mx + c.

ii 3x + 23 = 4y

3

y − 11 =

3 4

(x − 7)

3

y= 4x+

23 4

Worked Example 18

Find the equation of the straight line containing the points (2, −5) and (−3, 1). Express your answer in the forms i ax + by + c = 0 and ii y = mx + c. Think

Write

1

Write down the points so they match the variables in the formula.

(x1, y1) (x2, y2) (2, −5) (−3, 1)

2

As two points are known, first use the y2 − y1 formula m = to find m. x 2 − x1

m=

y2 − y1 x 2 − x1

1− −5 − 3− 2 6 = − 5 =

= 3

Write the formula y − y1 = m(x − x1).

4

Substitute the calculated gradient m = 5 and the first point (x1, y1) = (2, −5). Leave x and y as they are.

5

Simplify and express in the two forms required.

−6 5

y − y1 = m(x − x1) −6

y − −5 =

−6 5

( x − 2)

6

y + 5 = −5 ( x − 2) −5y − 25 = 6(x − 2) −5y − 25 = 6x − 12 −5y = 6x + 13 i So 6x + 5y + 13 = 0, or ii

y=

−6 5

x−

13 5

Perpendicular lines Perpendicular lines are lines that meet at right angles. The gradients of two perpendicular lines, when multiplied together, equal −1. Stated mathematically: m1 × m2 = −1 Chapter 1 • Linear functions 19

−1 Another way to write this relationship is m1 = m . This type of relationship is known as a negative 2

reciprocal. For example, the negative reciprocal of gradients are negative reciprocals.

exercise 1F

2 3

is

−3 2

. So, two lines are perpendicular if their

Finding the equation of a straight line

1 Copy and complete the table below.

Gradient of equation 1 (m1)

Equation 1 f1(x)

diGital doC doc-9703 SkillSHEET 1.2 reciprocals and negative reciprocals

a

y = 2x + 1

y=

b

y = 3x − 4

y=

c

y = 4x + 6

d

y = 5x − 3

e

y=

−1 2 −1 3

x+2

y = −4x − 9

2

y=

7

−5 2

m1 × m2

x

1

−9

Gradient of equation 2 (m2)

Equation 2 f2(x)

x+4

7

y = 9x + 1

x

2 Sketch f1(x) and f2(x) for part a above using a calculator or using graph paper with identical scales on

each axis to produce accurate graphs. What do you notice about the graphs?

3 Repeat question 2 for graphs b to e. Do you notice anything special about each pair of graphs? 4 Find the gradient of a line perpendicular to another line that has a gradient of: a 4

b

−9

c

1 7

d

−8

9

e

7 2

f

1.

5 Find the gradient of a line that is perpendicular to the line with equation: a y = −5x + 2 d y=

−7 6

x−2

2 3

b y=x−1

c y=

x+1

e 2x + y = 5

f 3x − 4y = 7.

Find the equation of a straight line having the gradient given and passing through the point listed. Express your answer in the forms i ax + by + c = 0 and ii y = mx + c. a (1, 2) gradient 3 b (4, 1) gradient 5 c (3, −2) gradient −1 d (−3, 2) gradient 1

6 We17 diGital doC doc-9704 equation of a straight line

2

Find the equation of the line containing each pair of points. Express your answers in the forms i ax + by + c = 0 and ii y = mx + c. a (5, 2) (3, 1) b (1, 1) (5, 5) c (6, 3) (8, 2) d (2, −2) (0, 1) − Find the equation of the line passing through (3, 3) that makes an angle of 45° with the positive x-axis. Find the equation of the line containing (7, −2) that makes an angle of 71.565° with the positive x-axis. Find the equation of the line (in y = mx + c form) that: a is perpendicular to the line with equation y = 3x + 1, passing through (−3, 6) 2 b is parallel to the line with equation y = x − 9, passing through (4, −7) 5 c is parallel to the line with equation 3x + 6y = 8, passing through (2, 2) d is perpendicular to the line with equation −6x + 7y − 2 = 0, passing through (4, 0) e has gradient 2, passing through the intersection of the lines with equations y = 3x − 5 and y = −2x + 5 −3 f has gradient 4 , passing through the intersection of the lines with equations x + 4y = −14 and −5x + 2y = 4. Find the equation of the line that passes through the point of intersection of the lines whose equations are 7x − 3y − 19 = 0 and 3x + 2y + 5 = 0, given that the required line is parallel to the line with equation −5x − 2y = 3. Find the equation of a line containing the intersection of the lines with equations y = −3x + 4 and 5x − 3y + 40 = 0 that: 6 2 a has a gradient of 7 b is perpendicular to the line with gradient 3

7 We18

8 9 10

11

12

c passes through the point (−1, 9)

20

Maths Quest 11 Mathematical Methods CAS

d is parallel with the line joining (−8, 5) and (0, 4).

13 A line passes through the points (−8, −5), (4, −3) and (a, 12). Find the value of a. 14 The points (2, 7) and (6, 9) lie on the same straight line. Does the point (4, 8) also lie on this line? 15 The height of a particular young pine tree is found to increase in a linear manner each month in the first

year after planting. Find an equation connecting height with time in months after planting, using the information supplied in the diagram below.

diGital doC doc-9705 WorkSHEET 1.2

52 cm 34 cm

After 5 months

After 2 months

distance between two points and y (x , y ) midpoint of a segment y 1G

2

The distance, d, between any two points on the Cartesian plane may be found using Pythagoras’ theorem applied to a right-angled triangle as shown at right. Using Pythagoras’ theorem: c2 = a2 + b2 or c = a 2 + b 2 and replacing c with d, a with (x2 − x1) and b with (y2 − y1),

2

2

d

y1

(x1, y1)

(y2 − y1)

interaCtiVitY int-0260 distance between two points and midpoint of a segment

(x2 − x1)

we have d = ( x 2 − x1 )2 + ( y2 − y1 )2 . x1

x2 x

Worked example 19

Find the distance between the points (−3, 7) and (5, −2) correct to 3 decimal places. think

Write

1

Match up (−3, 7) and (5, −2) with (x1, y1) and (x2, y2).

(x1, y1) (−3, 7)

2

Substitute into the formula for d and simplify.

d = ( x 2 − x1 )2 + ( y2 − y1 )2

tUtorial eles-1405 Worked example 19

(x2, y2) (5, −2)

= (5 − − 3)2 + ( − 2 − 7)2 = (8)2 + ( − 9)2 = 64 + 81 = 145 = 12.042, correct to 3 decimal places

Chapter 1 • Linear functions

21

midpoint of a segment The middle or midpoint M (xm, ym) of a segment joining two general points A (x1, y1) and B (x2, y2) is shown on the Cartesian plane below. y

B (x2, y2) y2 − ym M (xm, ym) A (x1, y1)

ym − y1 xm − x1

x2 − xm

C

E D x

Consider the triangles ACM and MEB. ∠MAC = ∠BME (since AM and BM have the same slope) ∠CMA = ∠EMB (since ∠ACM and ∠MEB are both 90°) AM = BM (given, as M is the midpoint) Therefore ΔACM ≡ ΔMEB. Since the horizontal and vertical sides of the two triangles must be equal, we have: xm − x1 = x2 − xm and ym − y1 = y2 − ym Simplifying these, 2xm = x2 + x1 2ym = y2 + y1 xm =

x 2 − x1 2

ym =

y2 + y1 2

x +x y +y So the point M has coordinates 1 2 , 1 2 . 2 2 Worked example 20

Find the midpoint of the segment joining (5, 9) and (−3, 11). think 1

Match (5, 9) and (−3, 11) with (x1, y1) and (x2, y2).

2

Substitute values into the formula for M and simplify.

Write

(x1, y1) (5, 9) M=

(x2, y2) (−3, 11)

x1 + x 2 y1 + y2 , 2 2

5 + − 3 9 + 11 = , 2 2 2 20 = , 2 2 = (1,10)

distance between two points and midpoint of a segment exercise 1G

diGital doC doc-9706 distance between two points

1 We19 Find the distance between each of the following pairs of points. a (4, 5) and (1, 1) b (7, 14) and (15, 8) c (2, 4) and (2, 3)

d (12, 8) and (10, 8)

2 Calculate the distance between each of the pairs of points below, accurate to 3 decimal places. a (−14, 10) and (−8, 14) b (6, −7) and (13, 6) c (−11, 1) and (2, 2) 3 Find the distance between each of the following pairs of points in terms of the given variable(s). a (a, 1), (2, 3) b (5, b), (0, 6) c (c, 2), (4, c) d (d, 2d), (1, 5) 22

Maths Quest 11 Mathematical Methods CAS

4 Two hikers are about to hike from A to B (shown on the map below). How far is it from A to B ‘as the

crow flies’, that is, in a straight line?

N 50 m 100 m 200 m

100 m 200 m

B (E7, N4)

300 m Lake Phillios

E

W

A (W12, S5)

S

Grid spacing : 1 km

5 Using the coordinates shown

y (in metres)

on the aerial photo of the golf course, calculate (to the nearest metre): a the horizontal distance travelled by the golf ball for the shot down the fairway b the horizontal distance that needs to be covered in the next shot to reach the point labelled A in the bunker.

(225, 96)

A (320, 148) x (in metres)

(80, –64)

6 We20 Find the midpoint of the segment joining each of the following pairs of points. a (1, 3) and (3, 5)

b (6, 4) and (4, −2)

c (2, 3) and (12, 1)

7 Find the midpoint of the segment joining each of the following pairs of points. a (7, −2) and (−4, 13) b (0, 22) and (−6, −29) c (−15, 8) and (−4, 11)

d (6, 3) and (10, 15) d

(−3, 40) and (0, −27)

8 Find the coordinates of the midpoint of each of the following pairs of points, in terms of a variable or

variables where appropriate. a (2a, a) and (6a, 5a)

diGital doC doc-9707 midpoint of a segment

c (3f, 5) and (g, −1)

b (5, 3c) and (11, 3c)

9 Find the value of a in each question below so that the point M is the midpoint of the segment joining

points A and B. a A (−2, a), B (−6, 5), M (−4, 5)

3

b A (a, 0), B (7, 3), M (8, 2 ) Chapter 1 • Linear functions

23

y

10 A fun-run course is drawn (not to scale) at right.

If drink stations D1, D2 and D3 are to be placed at the middle of each straight section, give the map coordinates of each drink station. 11 Find the equation of a line that has a gradient of 5 and passes through the midpoint of the segment joining (−1, − 7) and (3, 3). 12 Find the equation of a line parallel to the line with equation 9x − 3y = 5 that passes through the midpoint of the segment connecting (0, −4) and (−2, 10).

1H

(1.5, 3.5)

(−4.5, 5)

D1

Official tent

D2

x

(1.5, −2) START/ FINISH

Coordinates are in kilometres.

(3, −7)

D3

(13, −8)

Linear modelling

Many real-life applications, such as fees charged for services, cost of manufacturing or running a business, patterns in nature, sporting records and so on, follow linear relationships. These relationships may take the form of a linear equation; for example, F = 50 + 30t may be used by a tradesperson to calculate her fee (in dollars) for t hours of work. Here, F is the fee in dollars, and t the time in hours. The 50 represents an initial fee for simply turning up, while the 30t is the amount charged for the time spent on the job. For example, if t = 2 hours, 30t = 60, so the total charge for the work would be $(50 + 60) = $110. Equations like F = 50 + 30t are sometimes referred to as ‘linear models’, a common form of which is: Total cost = Fixed cost + Cost per unit × Number of units. This is, of course, equivalent to y = mx + c. y = c + mx. Worked Example 21

A generator company charges a $200 delivery fee, and a rental fee of $1500 per day. a Find an expression relating total charge to the number of days for which the generator is hired. b Sketch a graph of the relationship. c What would be the charge for 4 weeks of rental? Think

a 1 Define convenient variables. 2

The fixed cost is $200, and the cost per unit is $1500. (c = 200, m = 1500)

b 1 Find the vertical intercept (when n = 0). 2

The total cost rises $1500 each day, so the graph must show this.

Write/draw

a Let T = total charge (in dollars) and

n = number of days the generator is hired. T = 200 + 1500n

b If n = 0, T = 200 T ($) 3200 1700 200 1

c 1 After 4 weeks, n = 28. Substitute this in the equation from part a.

2

Write the answer in words.

24 Maths Quest 11 Mathematical Methods CAS

2

n (Days)

c If n = 28

T = 200 + 1500 × 28 = 200 + 42 000 = 42 200 After 4 weeks, the total cost is $42 200.

Worked example 22

‘Rent-a-Chef’ provides food cooked and served by a qualified chef at parties. The company charges $120 as a booking fee, and an additional $30 per hour. Another company, ‘Greased Lightning’, provides fast food served by two students at a cost of $65 per hour, with no booking fee. Under what conditions would it be cheapest to hire ‘Greased Lightning’? think

tUtorial eles-1406 Worked example 22

Write

1

Define convenient variables.

Let C = cost (total) in dollars and t = time in hours.

2

Write an equation for the cost of hiring both organisations.

Rent-a-Chef C = 120 + 30t Greased Lightning C = 65t

3

Use simultaneous equations to find when the cost is the same with each group.

Put [1] = [2] 120 + 30t = 65t 120 = 35t 120 t= 35 = 3.4 hours

4

At 3.4 hours, the cost is the same. Since Greased Lightning has the higher per hour cost, after 3.4 hours, they would be more expensive.

It is cheaper to hire Greased Lightning for food preparation and service of less than 3.4 hours (3 hours and 26 minutes) duration.

[1] [2]

Notes 1. 0.4 hours = 0.4 × 60 minutes = 24 minutes. 2. An alternative approach would be to use a CAS calculator and find the point at which the two graphs crossed.

exercise 1h

linear modelling

The cost of hiring a floodlit tennis court consists of a booking fee and an hourly rate. a Use the photo to write an equation for the total hire cost in terms of the hourly rate. b Sketch a graph of the relationship. c What would be the charge for 3 hours?

1 We21

diGital doC doc-9708 Simultaneous linear equations

2 A singing telegram service charges a $60

appearance fee, and $8 per minute sung. a Write an equation for the total cost of a singing telegram in terms of the number of minutes sung. b Sketch a graph of the relationship. c What would be the charge for a 5-minute singing telegram? 3 Colleen delivers junk mail and is paid $32 to traverse a particular route, and a further 10 cents per

leaflet delivered. a Write an equation for the total payment she receives. b Sketch a graph of the relationship expressed in a. c What would be Colleen’s pay if she delivers 1650 leaflets along the route? 4 A pay-TV salesperson receives $300 per week plus $20 for every household he signs up to have pay-

TV connected. How much does the salesperson receive for a week in which he signs up 33 households? 5 We22 A computer firm, SuperComputers Inc., offers a back-up plan covering the ongoing service and

troubleshooting of its systems after sale. The cost of signing up for the service plan is $215, and there is an Chapter 1 • Linear functions

25

hourly rate of $65 for the serviceperson’s time. Purchasers not signing up for the plan are charged a flat rate of $150 per hour for service. Would it be advisable to sign up for the service plan if you expected to need 3 hours of service assistance during the life of a computer purchased from SuperComputers Inc? 6 Two amusement parks show the following information for school-age tourists in a promotional brochure. After how many rides does an excursion to Fun World become the cheaper option for the same number of rides?

A qua Wor ld $8.00 entry $2.50 per ride

$12 entry $1.50 per ride

7 A telephone company, Opus, offers calls to Biddelonia for a connection fee of $14, and thereafter

$1 per minute. Its rival, Elstra, offers calls for $2 per minute (no connection fee) to the same country. a Compare the cost of a 10-minute call to Biddelonia using each company. b At what point would it be cheaper to use Opus? 8 It costs you $6 to get into a taxi (the ‘flagfall’), and $1.50 per kilometre if you use PinkCabs, while NoTop taxis charge $8 flagfall, and $1.20 per kilometre. a How much would it cost with each company to travel 15 kilometres in one of its cabs? b When would it cost the same to use both companies? 9 Medirank, a health insurance company, charges $860 per year (for a single person) and requires customers to pay the first $100 of any hospital visit. HAB, on the other hand, charges an annual fee of $560 and requires its members to pay the first $150 of any hospital visit. Determine the number of hospital visits in a year for which the cost of health services is the same whichever company insures you. 10 Nifty is a car hire firm that charges insurance of $135 and $50 per day car hire. A competitor, Savus,

simply charges $65 per day and offers ‘free’ insurance. You are planning a holiday, and would prefer to use Savus. Under what conditions (days hired) could you justify this choice? 26 Maths Quest 11 Mathematical Methods CAS

Summary Solving linear equations and inequations

• Do the same to both sides and remember inverse operations + and −, × and ÷, √ and 2. • Aim to get a single variable by itself. • Solve inequations the same way as equations, keeping the original inequality sign at each step, unless multiplying or dividing by a negative number.

rearrangement and substitution

• ‘Make x the subject’ means manipulate into the form ‘x = …’. • ‘Substitute’ means to replace a variable with a value.

Gradient of a straight line

• m=

Sketching linear functions

• • • • •

Simultaneous equations

• Simultaneous equations can be solved with a calculator. • For equations of the form y = ax + b, y = cx + d, consider using substitution. • For equations of the form ax + by = c, dx + ey = f, consider using elimination.

Finding the equation of a straight line

• Formulas for finding the equation of a straight line: y −y y − y1 = 2 1 ( x − x1 ) y = (mx + c) y − y1 = m(x − x1) x 2 − x1 • For perpendicular lines, m1 × m2 = −1.

distance between two points and midpoint of a segment

• d = ( x 2 − x1 )2 + ( y2 − y1 )2

linear modelling

y2 − y1 m = tan (θ ) x 2 − x1 where θ is the angle the line makes with the positive direction of the x-axis.

The general equation for a straight line of gradient m and y-intercept c is y = mx + c. Lines with the same gradient (m) are parallel. To find the y-intercept, let x = 0 and find y. To find the x-intercept, let y = 0 and find x. If y = 0 when x = 0, substitute another x-value (for example x = 1) to find another point on the line. • Join two points and/or intercepts with a straight line.

x1 + x 2 , y1 + y2 • M= 2 2 • Total cost (y) = fixed cost (c) + cost per unit (m) × number of units (x) • y = c + mx

Chapter 1 • Linear functions

27

Chapter review S h ort anS Wer

3(5 x − 4) 6(4 x + 3) . = 7 5 7x + 8 = 4 x − 9. 2 Find the value of x where 3 10 1 Solve the equation

− 3( x

1− x . 4 2 4 The following formula may be used to study planetary motion. 3 Solve the inequality

+ 4)

≤

GmM m 4π 2 R = R2 T2 Make T the subject of the equation.

c

6

5 Using Pythagoras’ theorem, find the length of the hypotenuse of the

triangle shown. Express your answer: a in surd form b to 3 decimal places.

9

6 Calculate the gradient of each of the following lines. y a 6

c

1 grid square = 1 unit

6 5 4 3 2 1

−5 −4 −3 −2 −1 0 −1 −2 −3

x

−8

y

b

y

1 2 3 4 5 6x

d

y x (−12, −3)

x

(−4, −10) 7 Find the gradient of the line joining (−7, 15) and (2, −6). 8 Find the gradient of the line shown. y

77° x 9 State the gradient of the line below. y

5

28

Maths Quest 11 Mathematical Methods CAS

x

10 State the gradient and y-intercept (in that order) for each of the following. a y = 3x − 7

b 5x + 3y = 30

11 Find the equation for a linear graph having gradient 12 Sketch graphs of the following, showing intercepts. a y = −3x + 24

b −x + 8y = 40

c 2x − 4y − 8 = 0

and y-intercept −3.

2 5

c 9x − 7y − 63 = 0

d y + 6x = 0

−2x

13 Solve y = 3x + 10 and y = − 15 graphically. a Sketch the solution on a set of axes.

b State the solution (point of intersection).

14 Solve y = −3x, y = 6x − 15 using substitution. 15 Use the method of elimination to solve 4x − 7y = 21, −2x + y = 6. 16 A piggybank contains 67 coins. If there are only one- and two-dollar coins in the piggybank,

and there are 25 more one-dollar coins than two-dollar coins, how many of each type are there? 17 Find the gradient of a line perpendicular to 3x − 9y = 7. 18 Find the equation of the line containing (−4, 8) and (3, 1). −6

19 Find the equation of the line having gradient

7

that passes through (1, 4). 1 4

20 Find the equation of the line perpendicular to y = x − 5 that passes through (−8, 6). 21 The distance between (2, −7) and (a, −2) is

41 units. Find the value of a if it is positive.

22 Show that the triangle with vertices (3, 7), (3, 3) and (6, 3) is a right-angled triangle. 23 The midpoint of the line joining (k, 2h) and (9k, 6h + 2) is (20, −11). Find k and h. 24 The washing machine repair company ‘Washed out’ charges $75 to come to your house, as well as an

hourly charge of $65, calculated to the nearest half hour. a Write an equation that may be used to calculate the cost of any service call if the time taken by the repairer is known. b Sketch a graph of the relationship between repair cost and time taken to do a repair. 1 c How much would it cost to have a repair done that takes 3 2 hours? 7 x − 23 = 99 would be to: 3 a add 23 to both sides B divide both sides by 3 C divide both sides by 7 d multiply both sides by 3 e multiply both sides by 7

m U ltip l e C ho iC e

1 The first step in solving

2 x = −5 is a solution to the equation: a 3x + 7 = −8

B 2x − 7 = −5

d 2(x + 3) = 10

e

3 The solution to a x ≤ 10

− 5x

9

x + 14 ≥ 2( x + 2) is: 3 2 B x≥5

x + 25 =5 6

C

= 45

2

C x≤5

d x≤

−6

−2 5

e x≤ 5

4 When c2 = a2 + b2 is rearranged to make a the subject, the equation becomes: a c=

a2 + b2

B a2 = b2 + c2

d a=

c2 − b2

e a=b+c

C a2 =

c2 − b2

1

5 Which values, when substituted into K = 2 mv 2, give a value for K of 4? a m = 2, v = 4

B m = 4, v = 2

d m = 8, v = 1

e m = 1, v = 16

1

C m = 8, v = 2

6 Using the equation P = m1v1 + m2v2, if P = 10, m1 = 2, m2 = 6 and v1 = 4, v2 would equal: a

1 3

B

1 2

C 1

d 2

e 3

Chapter 1 • Linear functions

29

y

7 The line shown has a gradient of: a B C d e

−6

6

−3 −2

2 6 3

x

3

8 The gradient of the line shown at right is 3. The value of a must be: a −2 B −1 C 5 d 7 e 11

y (6, a)

9 The gradient of the line joining (−1, 0) and (4, −10) is: a

−4

B

d 4

−2

x

C 2

e 5

(2, −5)

10 Which of the graphs below has a gradient of y

a

6 ? 7

C y

y

B

7

6 6 7 x

y

d

x

−7

x

−6

y

e

x

−7

6

−6

−7

11 The gradient of the line with equation y = a −1

B

6 7

x − 1 is: 7

6 7

C 6

12 The y-intercept of the line with equation y = 12 x + a

x

2 3

B

3 2

2 3

d 6

e 7

d 3

e 12

is:

C 2

13 The gradient and y-intercept (in that order) of the line with equation 2x − 3y = 7 are: a 2 and −3 d

2 3

and

B 2 and 7

−7

e

3

−3

and

C 7 2

−2 3

and 7

14 Which of the following could be the graph of y = 2x + c? a

y

B

y

C

y

c x −c

30

Maths Quest 11 Mathematical Methods CAS

c x

x

y

d

−c

e y

c

x

x y

15 The equation of the line shown at right is: a 2x − 5y = 1 C 15x + 6y = −30 e

x y − =1 2 5

B 2x − y = 4 d 10y − x = −2

2

x

−5

16 To solve the equations 2x + y = 5 and 3x − 6y = 12 graphically on a calculator, you would enter the

equations in the function entry line as: f 1(x) = 2x + y and f 2(x) = 3x − 6y f 1(x) = 5 and f 2(x) = 12 f 1(x) = 5 − 2x and f 2(x) = 12 + 6y f 1(x) = 2x + 5 and f 2(x) = 3x + 12 x f 1(x) = −2x + 5 and f 2(x) = 2 − 2

a B C d e

17 Which of the following would be the most effective way to solve the following equations

simultaneously? y = 2x − 13 [1] y = 7x + 2 [2] a Multiply [1] by 2 and [2] by 13 and add the newly formed equations. B Multiply [1] by 7 and put it equal to [2]. C Multiply [2] by 2 and put it equal to [1]. d Multiply [1] by 2 and [2] by 7 and subtract the newly formed equations. e Put [1] equal to [2]. 18 The gradient of a line perpendicular to a line with a gradient of 7 is: a −1

B −7

d 7

e

7

7 1

19 The gradient of a line perpendicular to y = a

−1 7 5

d 16

20 The equation of the line containing (1,

C

+5 is: 16

1

C

−7

−2)

1

− 7x

B 7 e

−7

16 7

16

and (2, −3) could be expressed as:

a y−2=x−1

B y+2=1−x

d x − 2y = −3

e 3x − 5y = 1

21 A line with equation y − 7 = 5(x − 1) has: a gradient 5 and contains the point (7, 1) C gradient 5 and contains the point (1, 7) e gradient −1 and contains the point (5, 7)

C y+3=x−1

B gradient −7 and contains the point (−1, −7) d gradient −5 and contains the point (1, −7)

22 The distance between (4, 3) and (−2, 1) is equal to: a (−2 − 4)2 + (1 − 3)2

B

(4 − 3)2 + ( − 2 − 1)2

(4 + 2)2 + (3 − 1)2

e

(− 2 − 4)2 − (1 − 3)2

d

C

(4 2 + 32 ) − ( − 22 − 12 )

Chapter 1 • Linear functions

31

23 The midpoint of the segment joining (11, −3) and (−5, 17) is: 1 1 2 2 1 1 ( − 2 2 ,8 2 )

a ( − 5 , −1 )

B (3, 7)

d

e (4, 6)

C (6, 14)

24 Bote lives 5 kilometres from the nearest post office. At noon one day he begins cycling (from home) at

20 kilometres per hour in a constant direction away from the post office. At t hours after Bote begins cycling, the distance, D km, that he is from the post office is given by: a D = 5t B D = 20t C D = 5t + 20 d D = 20t + 5

e D = 20t − 5

25 The linear function f : D → R, f ( x ) = 6 − 2 x has range [−4, 12]. The domain D is: a [−3, 5] d e x t ended r e SponS e

[−14,

B [−5, 3]

18]

e

[−18,

C R

14]

1 The graph at right is a rough sketch of three points on a section of

y

sheet metal that are to be drilled by a programmed robotic drilling arm. Any deviation from a straight path, no matter how slight, means the arm must be programmed for more than one direction. 16 The coordinates marked are correct. Will the robotic arm be able to 12 move in one direction only to drill all three holes? 2 Points A, B and C have the coordinates (1, 6), (0, 0) and (−2, 2).

Find the coordinates for a point D so that the four points form a parallelogram. (−4, −2),

−1)

6

C B A

8

14

20

x

−7).

3 Consider the points (6, 2), (4, and (0, a Find the coordinates of the midpoints of each side of the quadrilateral formed by the points. b Show that the shape formed by the midpoints is a parallelogram. c Repeat parts a and b for a different set of starting points. d What can you conjecture based on your answers? 4 The cost of a parachuting course consists of a charge of $250, which covers equipment hire and tuition,

and a further expense of $55 per jump. a Express the total cost, C, as a function of j, the number of jumps. b How many jumps could a person doing the course afford if she was prepared to spend up to $1000? 5 A physics student conducts an experiment to find out how much a spring stretches when various

weights are hung from it. Her results are shown in the table below.

a b c d 32

Length of spring (cm)

Force applied (N)

4

0

7

10

12

20

16.5

30

20.5

40

25

50

What is the natural or ‘unstretched’ length of the spring? Plot a graph of the student’s results. Draw a straight line through the points that best describes the data. Select two points on the line and use them to fit a linear equation to the line.

Maths Quest 11 Mathematical Methods CAS

A second student conducts the same experiment on a similar spring. His results are shown below. Length of spring (cm)

Force applied (N)

5

0

10

10

16

20

21

30

24

40

28

50

e On the same set of axes you used in part b, plot the results of

the second experiment and join the points with a line of best fit. f Write an equation that describes the relationship between the

force applied and the length of the second spring. The gradients of graphs such as the ones you have drawn give an indication of the stiffness of a spring. The greater the gradient, the harder it is to stretch the spring. The lower the gradient, the easier it is to stretch the spring. g Comment on the stiffness of the two springs investigated by the students. h Is it likely that these two springs will ever be the same length at a given force; that is, is it likely that the intersection of the two graphs could ever really happen? Explain your answer.

diGital doC doc-9709 Test Yourself Chapter 1

Chapter 1 • Linear functions

33

ICT activities Chapter opener diGital doC • 10 Quick Questions doc-9695: Warm up with ten quick questions on linear functions (page 1)

1a

Solving linear equations and inequations

diGital doC • doc-9696: Use trial and error to balance an equation and hence solve for the unknown variable (page 3)

1B

rearrangement and substitution

diGital doC • Career profile doc-9697: Learn how a vigneron uses substitution and other areas of mathematics in his work (page 8)

1C

Gradient of a straight line

diGital doCS • doc-9698: Calculate the gradient between two given points (page 10) • SkillSHEET 1.1 doc-9699: Practise using the gradient to find the value of a parameter (page 11)

1d

Sketching linear functions

diGital doCS • WorkSHEET 1.1 doc-9700: Solve linear equations, use substitution and calculate gradients of straight line graphs (page 12) • doc-9701: Investigate the effect of changing the gradient and y-intercept for a linear graph (page 14) tUtorial • We 12 eles-1404: Watch a tutorial on how to sketch a linear graph by hand (page 13)

1e

Simultaneous equations

diGital doC • doc-9702: Use the graphical method to solve simultaneous linear equations (page 17)

34

Maths Quest 11 Mathematical Methods CAS

1F

Finding the equation of a straight line

diGital doCS • SkillSHEET 1.2 doc-9703: Practise writing positive and negative reciprocals of rational numbers (page 20) • doc-9704: Investigate the equations of straight lines (page 20) • WorkSHEET 1.2 doc-9705: Sketch linear graphs, determine equations of linear graphs and application problems (page 21)

1G distance between two points and midpoint of a segment interaCtiVitY • Distance between two points and midpoint of a segment int-0260: Consolidate your understanding of how to calculate the distance between two points and the midpoint of a segment (page 21) tUtorial • We 19 eles-1405: Watch how to calculate the distance between two points on a Cartesian plane (page 21) diGital doCS • doc-9706: Investigate the distance between two points (page 22) • doc-9707: Investigate the midpoint of a segment (page 23)

1h

linear modelling

diGital doC • doc-9708: Investigate simultaneous linear equations (page 25) tUtorial • We 22 eles-1406: Watch how to apply linear algebra skills to determine the conditions for which it would be cheapest to hire a catering company (page 25)

Chapter review diGital doC • Test Yourself doc-9709: Take the end-of-chapter test to test your progress (page 33)

To access eBookPLUS activities, log on to www.jacplus.com.au

Answers CHAPTER 1 linear FUnCtionS

3 a d g 4 a c 5 a c 6 a 2 7

exercise 1a

Solving linear equations and inequations b −8

c

7 −1 6

e −1 h 10 b 5

d 112

e 12

f i c f

−9

h

j −4

k

76 59 −9

1 a 2 d g 2 a

9

g

13

b x≤

3 a x≥ 6

d x < −3 exercise 1B

substitution

− 31

c x>3

e x ≥ 11

f x≥2

b 5 a b 6 a

f r=

j H= b 350 e 2.498

kQq F

S − 2π r 2 2π r

c 224 cm

x

7 −21 y

c

x

− 3–2

y=x+2

y

3 a–d

y=x+1

y

d

10

e

y=x+3

y

30 1 x

4 y=x− 4

1

c 18 cm

−1

−5

01

y = 2x − 7 x

−16

10 a

2

b

g 11 , 1 5 , 6 2

−6

y 4

y-axis (hence the name ‘y-intercept’). 6 a y = 2x + 7 b y = −3x + 1 2 1 c y = 5x − 2 d y= 3x+3

−2

x

3

5 The number is where the graph cuts the

1

y

y = −3x + 6

−7

e y= 4x− 2 7 a 4, 5 c −3, 1 e −7, −9

x

8

y = 2x + 1 y = 3x − 5

i

y

y

c 40 mm

d

x

10 — 3

−4

6

1 2

x

2

f

−1

c

x y

y = 2x y=x

4 a–d

Gradient of a straight line b5

x−5

y

b

graph. Positive values make the graph slope up when moving (or tracing) to the right; negative values make the graph slope down when moving to the right.

c 7

25 9 h= cm π

−4

2 −5

d y = 6x + 5 f y = −5x

−3

y = 3x

−5

18

2 The higher the number, the steeper the

−3

3

x−2

9 a

− 17 300

y

y = −2x

b y=

−3

8 b=2

b

4 3

Sketching linear functions

3V , 6.204 br=3 4π T 2 d l = g , 3.972 2π

A, 7.746

c y=

x

A A− D − 1 = 100 b r = 100 D D c 41.4% fv uv 7 a f = b u= v− f u+v c 150 cm

2 a 2

1 a–e

8 a y = 4x + 2

e y = 2x − 1

y = −x

v2 hγ = rT

v−u , 4.167 t (1 − α ) K c= , 2622 α2 42 cm P P − 2l w = − l or w = 2 2 240 m2 2A 2 A − bh a= − b or a = h h $1123.60

1 a 2

b 31 d 3 b Yes

c 0.176 f0

A w C d r= 2π

E − αθ θ2

exercise 1C

4 − 5 No

exercise 1d

3.078 − 0.577 57.290 2.61 − 3.73 C B E

b l=

c a=

4 a

3

b e h b d b d b 8

rearrangement and

2 a 0.267 d 13 100

e

9 a c 10 a

3

2 Fd + mu 2 g v= m S − 2lh i w= 2(l + h)

3 a l=

17

l −7

d v

e β=

3

9 −12 −13 −5

i 3

1 a P=A−L c t=

−7

1.192 − 0.577 1 0.93 − 0.53 D A B

f y = −2x + 12

4, −8 −2, 4 2, 5 −8 2 h , 3 3 b d f j

−5 − 7 , 2 2

5

c

x

y 6

−3

x

Chapter 1 • Linear functions

35

d

y

b

y

y = 20

x

5

exercise 1F Finding the equation of a straight line

11 a

e x

−5 −5

4 a

y

10

5 a 5

( 94 , 47 4)

d

x

y = 3x + 5

x

y

d

y 4 — − 16 3

5 y = 6x + 5 0

x

y = −4 + 3x x

y

f −1

7

e

c

1 2

f

ii y = 2 x + 2

7 a i x − 2y − 1 = 0 b i x−y=0

ii y = 2 x − 2 ii y = x

c i x + 2y − 12 = 0

ii y = 2 x + 6

d i 3x + 2y − 2 = 0

ii y = 2 x + 1 9 y = 3x − 23

−1

3

x

(1, −1)

−5

7 , − 29 (− 12 6 )

(1, 1)

f

y = 17 − 9x

y

x 14

17

( 38 ,

109 ) 8 y = 14 − x

y

2 a (1, 4)

(1, −2)

13 D

14 E

c (−7, −5)

15 A

Simultaneous equations y

1 a

e

23 − 20 (14 , 7 )

3 a (7, 9)

(−3, 6) y = −4x − 6

x

17 9

x

c e

1 −1−1 1

x y = −2x

4 6 7 8 9

(1,

−2)

59 21 (8, 8)

9 9

f (13,

−3)

b (6, 5) d

1 19 ( 2, 2 )

f

84 99 ( 67 , 67 )

15 cents and 35 cents 5 22 and 19 16 emus, 41 sheep Basketballs $9.45, cricket balls $3.05 Limousine $225 (sedan $75) A 10 D

Maths Quest 11 Mathematical Methods CAS

1

2

43

−7

14

−3

9

−3 2

x +7

b y= 5x− 5

x+3

d y= 6x+ 3

−1 2

f y= 4x − 2

3 82

12 a y = 7 x + 7

b y=

−x

39

d y= 8 + 4 14 Yes

exercise 1G distance between two points and midpoint of a segment 1 a 5 b 10 c 1 d 2 2 a 7.211 b 14.765 c 13.038

3 a b (−4, −15) 5 17 d( , )

1

x+5

13 94 15 H = 22 + 6t

y

7

−3

c y = −x + 8

y = −6 − 2x

1

−1

11 y = 2 x − 2

5

3

d i x − 2y + 7 = 0

6

x

2 −4

ii y = 3x − 1 ii y = 5x − 19 ii y = −x + 1

−5

y = 10x + 1

−3

6 a i 3x − y − 1 = 0 b i 5x − y − 19 = 0 c i x+y−1=0

e y = 2x − 3 5

36

6 7

c y=

y

e

y

exercise 1e

−2

b −1

10 a y =

(−3, −13)

−6

c

c −7

8 y=x−6

x

2

b

e

1

−4

12 a

1

b 9

4

d 8

y = 7x − 4

d

−1 9

−6

c

−9 7 , , −1 7 9

y

b

d 5 , 2 , −1

2 They are perpendicular. 3 They are perpendicular.

x

5

c

2 −5

1

y −7

b 3, 3, −1

c 4 , −4, −1

y = 3x − 5

−7

−1

−1

1 a 2, 2 , −1

, 20 ) ( 25 3

20

c

a2 − 4a + 8

b

b 2 − 12b + 61

2(c 2 − 6c + 10)

d

5d 2 − 22d + 26

4 21.024 km 5 a 216 6 a (2, 4) c (7, 2) 1 2

1 2

7 a (1 , 5 ) 1

1

− c ( 9 2 , 9 2)

8 a (4a, 3a)

c 3 f + g , 2

b 108 b (5, 1) d (8, 9) − − 1 b ( 3, 3 2 ) 1

1

− d ( 1 2 , 6 2)

b (8, 3c)

2 9 a 5 b 9 10 D1 (−1.5, 4.25), D2 (−1.5, 1.5), D3 (8, −7.5) 11 y = 5x − 7 12 y = 3x + 6

linear modelling

9 Undefined

1 a C = 25 + 5t

−

10 a 3, 7

b Cost ($)

1 2

c

35 30 25

b

−5

3,

22 Teacher to check. 23 k = 4, h = −3 24 a C = 75 + 65t b C ($)

10

, −2 2

11 y = 5 x − 3 12 a

1 2 Time (h)

140

y

75

24

c $40 2 a C = 60 + 8m

1

76 68 60

1 5 9 13 17 21 25

x

8 y

b 1 2 Time (min)

5

c $100 3 a P = 32 + 0.1n

x

−40

b Payment ($) 34 33 32

c $197

$960 Yes ($410 compared to $450) After 4 rides a Opus $24, Elstra $20 b After 14 minutes 8 a PinkCabs $28.50, NoTop $26 2 b After 6.7 km (6 3 km) 9 6 visits 10 Savus would be cheaper for up to 9 days hire.

13 a

5 a 6 a

3 4 5

c 11 7

−7 3

b (−5, −5)

y f(x) = 3x + 10 5

Chapter reVieW

x

−5 (−5, −5) −5

f(x) = −2x − 15

4π 2 R3

or 2π R

R GM b 3.873 −7

b 11 d

−7

− 63 − 33

5 14 ( , − 5)

15 ( 10 , 5 )

3

16 17 18

21 two-dollar and 46 one-dollar coins −3 y = −x + 4 −6

C C B E C B

4 8 12 16 20 24

D D A E B D

a parallelogram.

(1, −6)

1 26 3 x ≥ −14

3 7 11 15 19 23

4 a C = 250 + 55j b 13 jumps 5 a 4 cm b, c, e

x

−2

A A B B A D

d The midpoints of any quadrilateral form

y

4 5 6 7

Short anSWer

2 6 10 14 18 22

be shown by calculating gradients or equations for lines joining different pairs of points. 2 (−1, 8) −9 1 3 a Midpoints: (1, 0), (5, 2 ), (2, −4), (−2, 2 )

x

7

d

D D B D E C A

1 No, the points are not collinear. This may

−9

15

t (h)

extended reSponSe

y

c

10 20 Number of leaflets

GM

2

mUltiple ChoiCe

b Cost ($)

4 T=

c $302.50

205

8

19 y = 7 x + 7

8 4.331

20 y = −4x − 26

34

21 6

Force applied to spring (N)

exercise 1h

60 50 40 30 20

Student 1 Student 2

10 0

5

10 15 20 25 Length of spring (cm)

30

d Answers will vary. One possible answer 30

95

is y = 13 x − 13 . f Answers will vary. One possible answer is y = 2x − 10. g The first spring is stiffer than the second, as the gradient of its graph is greater than that of the second spring. h The graphs intersect at the point (−8.75, −27.5). It is not possible for the springs to have a negative length, so this point is not achievable.

Chapter 1 • Linear functions

37

Chapter 2

Quadratic functions diGital doC doc-9710 10 Quick Questions

Chapter ContentS 2a 2B 2C 2d 2e 2F 2G 2h 2i 2J 2k 2l

Polynomials Expanding quadratic expressions Factorising quadratic expressions Factorising by completing the square Solving quadratic equations — Null Factor Law Solving quadratic equations — completing the square The quadratic formula The discriminant Graphs of quadratic functions as power functions (turning point form) Graphs of quadratic functions (intercepts method) Using technology to solve quadratic equations Simultaneous quadratic and linear equations

2a

polynomials

A polynomial in x, sometimes denoted by P(x), is an expression containing only non-negative whole number powers of x. The degree of the polynomial is given by the highest power of the variable x. Examples of some polynomials are: 10 3x + 1 x2 + 4x + 7 −5x3 + 0.5x2 2x4 − x3 + 2x2 + x + 5

is a polynomial of degree 0 as 10 can be written as 10x0 is a polynomial of degree 1, or a linear polynomial is a polynomial of degree 2, or a quadratic polynomial is a polynomial of degree 3, or a cubic polynomial is a polynomial of degree 4, or a quartic polynomial.

2 − Expressions such as 3x2 − 4x + , −5x4 + x3 − 2 x and x 2 + sin (x) + 1 are not polynomials. x The general form of a degree n polynomial is given by P(x) = an x n + an − 1xn − 1 + . . . + a2x2 + a1x + a0 where n is a positive whole number and an, an − 1, . . . a2, a1, a0 are called coefficients (or constants) provided the leading coefficient an ≠ 0. This chapter will deal with polynomials of degree 2, or quadratics. The general form of a quadratic polynomial is P(x) = a2x2 + a1x + a0. It is also commonly written as y = ax2 + bx + c when graphing applications are involved. In the polynomial equation P(x) = 13x2 − x + 1: x is the variable. 13 is the coefficient of x2 and 13x2 is the quadratic term. −1 is the coefficient of x and −x is the linear term. The constant term is 1. The degree of the polynomial is 2. The leading term is 13x2 as it is the term with the highest power of x. Chapter 2 • Quadratic functions

39

An example of where polynomials are useful is shown below.

The surface area, S, of a plant hothouse of length L and height x can be approximated by the quadratic polynomial S ( x ) = π x 2 + Lπ x − 4.

Value of a polynomial diGital doC doc-9711 History of mathematics Évariste Galois

A polynomial may be evaluated by substitution of a number for the variable. For example, for the polynomial P(x) = 13x2 − x + 1, the value of P(x) when x = −3 is given symbolically as P(−3) and is found by substitution: P(− 3) = 13(− 3)2 − (− 3) + 1 = 13 × 9 + 3 + 1 = 121 This process can be completed on your CAS calculator. A CAS calculator is particularly useful for performing multiple substitutions simultaneously. 1. Using a CAS calculator, define the polynomial 1. Define t(x) = 3x2 − 2x + 16. t(x) = 3x2 − 2x + 16. 2. To evaluate t(x) for x-values of −3, 2 and 5, 2. t({− 3, 2, 5}) complete the entry line as: The result is (49, 24, 26.5279). t({− 3,2, 5})

exercise 2a 1

polynomials

State the degree of each of the following polynomials. b 65 + 2t u4 1 d x6 − 3x5 + 2x4 + 6x + 1 e 2u − − 6 + u5 3 2 a x3 − 9x2 + 19x + 7

c 2x − 8 + 3x2

2 State the variable for each polynomial in question 1. 3 Which polynomial in question 1 is: a linear? b quadratic?

c cubic?

4 For each of the following expressions:

• if it is a polynomial, write P • if it is not a polynomial, write N and circle the term(s) that made it so. x2 5 +x a 7x + 6x2 + b c 3x 4 − 2 x 3 − 3 x − 4 9 x d k 40

−

2

+ k − 3k 3 + 7

Maths Quest 11 Mathematical Methods CAS

e 2x − 8x + 1

5 Consider the polynomial P( x) = −2x3 − 3x + 4x2 + 5. a What is the degree of the polynomial? b What is the variable? c What is the coefficient of the linear term? d What is the value of the constant term? e Describe the role of the ‘4’ using the terminology of polynomials. 6 For the polynomial T(m) = 5m − 2m2 − 1, evaluate the following. a T(2) b T(−2)

c T(x + 1)

7 A sports scientist determines the following equation for the action of a breaststroke swimmer during

one complete stroke: v(t) = 63.876t 6 − 247.65t 5 + 360.39t 4 − 219.41t 3 + 53.816t 2 + 0.4746t where t = time (sec) and v(t) = velocity (m/s). a b c d

What is the degree of the polynomial? What is the variable? Evaluate v(0.2) and interpret the meaning of your answer. How much faster is she swimming at 0.7 seconds than she is at 0.5 seconds?

2B

Expanding quadratic expressions

A quadratic expression is a polynomial of degree 2. It must contain a quadratic term; any others (a linear term and a constant term) are optional. Examples of quadratic expressions include: x2 − 3x − 1

−5x

− 10x2

3t 2 + 1

Quadratic expressions can be produced when two linear expressions are multiplied. Consider the expansion (4x + 9)(2x − 3). When expanding brackets, ‘multiply everything by everything else’ as shown on the diagram at right. That is, first term × everything in the second brackets, then second term × everything in the second brackets. The results of each part of the expansion are: 1 4x × 2x = 8x2 2 4x × −3 = −12x 3 9 × 2x = 18x Parts 2 and 3 may be combined to give 6x, so we have:

4

2

1

(4x + 9)(2x − 3) 3

4

9 × −3 = −27

(4x + 9)(2x − 3) = 8x2 + 6x − 27 This method can be easily extended to deal with brackets containing more than two terms. The above method can be used on all types of binomial expansions, though a couple of shortcuts for special cases are shown in worked examples 1, 2 and 3. Chapter 2 • Quadratic functions 41

Worked Example 1

Expand the following. a (3x + 5)(6x − 7) b (2x − 9)2 c (5x − 3)(5x + 3) d −2(x − 9)(4 − x) Think

a 1 Write the expression and mark the required

multiplications.

Write

a (3x + 5)(6x − 7)

2

First term × everything in the second brackets gives 18x2 − 21x.

3

Second term × everything in the second brackets gives 30x − 35.

= 18x2 − 21x + 30x − 35

4

Combine the middle (x) terms.

= 18x2 + 9x − 35

b 1 Write the expression. Recognise it as a ‘perfect square’. 2

Remember the shortcut: Square the first term, double the product of the two terms and square the last term.

3

Square the first term to get 4x2.

4

Double the product of the two terms to get 2 × (2x) × (−9) = −36x.

5

Square the last term to get 81.

c 1 Write the expression. Recognise that the only difference

between the two sets of brackets is the sign in the middle, so the answer is a ‘difference of two squares’. 2

The shortcut for the expansion of this type is (first term)2 − (second term)2.

d 1 Write the expression.

b (2x − 9)2

= 4x2 − 36x + 81 c (5x − 3)(5x + 3)

= (5x)2 − 32 = 25x2 − 9 d −2(x − 9)(4 − x)

2

Rewrite the question so x is the first term in both brackets. This is not essential as long as all combinations of terms are multiplied in the next step.

= −2(x − 9)(−x + 4)

3

Expand the brackets first.

= −2(−x2 + 13x − 36)

4

Multiply the brackets’ contents by −2.

= 2x2 − 26x + 72

Worked Example 2

Expand (x − 7)(x + 2) − (2x − 1)(x + 4) and simplify. Think

Write

1

Write the expression.

(x − 7)(x + 2) − (2x − 1)(x + 4)

2

Expand the first pair of brackets.

(x − 7)(x + 2) = x2 − 5x − 14

3

Expand the second pair of brackets.

(2x − 1)(x + 4) = 2x2 + 7x − 4

4

Subtract the two expanded groups in the order given. Use new brackets for clarity as shown.

(x − 7)(x + 2) − (2x − 1)(x + 4) = (x2 − 5x − 14) − (2x2 + 7x − 4)

5

Apply the negative sign to the contents of the second brackets.

= x2 − 5x − 14 − 2x2 − 7x + 4

6

Collect like terms and simplify.

= −x2 − 12x − 10

42 Maths Quest 11 Mathematical Methods CAS

Worked example 3

If P(x) = (2x − 3)2 − (2x + 1)(3x + 9): a evaluate P(1) b expand and simplify P(x) c use your result in b to confirm your answer in a. think

Write

a 1 Substitute x = 1 into the rule for P(x).

a P(1) = (2 × 1 − 3)2 − (2 × 1 + 1)(3 × 1 + 9)

2

Simplify each bracketed term.

P(1) = (−1)2 − (3)(12)

3

Simplify P(1).

P(1) = 1 − 36 = −35

b 1 Expand the first bracketed term.

b (2x − 3)2 = 4x2 − 12x + 9

2

Expand the second and third bracketed terms.

(2x + 1)(3x + 9) = 6x2 + 21x + 9

3

Subtract the second result from the first result to obtain P(x).

P(x) = 4x2 − 12x + 9 − (6x2 + 21x + 9)

4

Simplify.

P(x) = −2x2 − 33x

c 1 Evaluate P(1) using the formula from part b.

2

c P(1) = −2(1)2 − 33 × 1

= −2 − 33 = −35

The answers are the same.

Compare with the result from part a.

exercise 2B

expanding quadratic expressions

Simplify your answers to questions in this exercise as fully as possible. 1 We1a Expand the following. a (2x + 6)(x + 5) c (5x − 7)(5x + 2) e (x + 4)(7x − 9) g (6x − 13)(2 − 3x) i (x − 9)(9x − 1)

b d f h j

(3x + 1)(4x − 3) (8x − 3)(6x − 1) (x + 6)(x + 15) (5 − x)(6 − x) (4x + 21)(x − 3)

2 We1b Expand using the shortcut for perfect squares. a (2x + 3)2 b (3x − 5)2 2 c (6x + 1) d (7x − 6)2 2 e (x − 8) f (x + 13)2 − 2 g ( 2x + 9) h (2x + 9)2 2 i (4 − 3x) j (6 − x)2 3 We1c Expand (remember the shortcut for difference of squares). a (2x − 6)(2x + 6) b (3x + 5)(3x − 5) c (6x + 1)(6x − 1) d (2x − 9)(2x + 9) e (11x + 3)(11x − 3) f (x − 12)(x + 12) g (x + 6)(x − 6) h (7 − 2x)(7 + 2x) i (1 − x)(1 + x) j (5x + 1)(5x − 1) 4 We1d Expand. a 2(x + 6)(3x + 5) c −4(x + 2)(2x − 5) e 8(3x − 1)(4x − 1) g 4(4 − x)(7 − x) i a(2x + 9)(x − 6)

diGital doCS doc-9712 SkillSHEET 2.1 expanding perfect squares doc-9713 SkillSHEET 2.2 expanding differences of squares

b 3(x − 4)(2x + 7) d 3(4x − 9)(2x − 1) f −7(2x + 3)(5x − 10) h −5(7x − 4)(2 − x) j −b(8 − 2x)(x + 4) Chapter 2 • Quadratic functions

43

5

Expand the following. a (6x − 13)(6x + 13) c −3(7 + 2x)(x − 8) e (2x − 14)(2x + 14) g (x + 16)(x − 9) i −4(5x − 3)2

b d f h j

6 We2 Expand and simplify. a (x − 9)(x + 2) + (x + 4)(x − 4) c 2(x + 5)2 + 5(2x + 7)(x − 3) e 4(x − 3)(3x + 5) − 2(2x + 1)2 g ( x + 3)( x − 3) i 12 (2x + 1)(4x + 7)

(5x + 2)(12x − 5) (3x + 11)2 (6x − 5)2 −(x − 4)(x + 4) 5(2x + 7)(2x − 7)

b (3x + 7)(2x − 1) + (4x − 3)(3x − 4) d (x − 12)(x + 12) − 3(x + 1)(x + 5) f 6(2x − 5)(2x + 5) − (x − 6)(x + 6) h (3 x + 2)(4 x − 2) j (5x + 13)(3x − 5)

7 We3 Expand. Use a calculator to verify the answers. a (2a + 4)(2a + b) b (x + 2y)(3x − 5y) c (6 − 7c)(2 − 7c) d (u + 4v)2 e (6r − s)(2r + 5s) f (3u − 2t)(3u + 2t) g 2(h − 8k)(h + 8k) h −3(m − 6n)2

Factorising quadratic expressions

2C

Factorising is the reverse process to expanding. It involves writing an expression as a product of two or more factors. Four methods of factorising will be considered.

highest common factor (hCF) Always look for a common factor first. Take out the highest common factor from every term in the expression and place it in front of the expression. This makes the factorisation process simpler if further factorisation is required. For example, 4x2 + 8x − 12 can be factorised to 4(x2 + 2x − 3). We would then use inspection (see below) on the bracketed quadratic.

Factorisation by inspection A quadratic polynomial of the form ax2 + bx + c is called a quadratic trinomial. It should first be checked to see if it is a perfect square: a2x2 − 2abx + b2 = (ax − b)2 or a2x2 + 2abx + b2 = (ax + b)2 If not, factorisation is carried out by inspection or ‘trial and error’. This method involves finding factors of a × c that add up to b, i.e. the coefficient of the x term.

perfect squares a2x2 + 2abx + b2 = (ax + b)2 A test for a perfect square is as follows: 1. Arrange the expression in order of decreasing powers of x. 2. Does first term coefficient × last term × 2 = middle term coefficient? (Coefficients are numbers or variables in front of x2 and x terms.) 3. If yes, you have a perfect square.

For example: 4x 2 + 20x + 25 Double 2

10 Multiply (2 × 5)

difference of two squares (dotS) ax2 − b =

(

ax + b

For example, 44

49x2

)(

)

a x − b where a and b may simplify to a rational number.

− 9 = (7x + 3)(7x − 3).

Maths Quest 11 Mathematical Methods CAS

5

Worked example 4

Factorise the following. a −6x2y + 15xy b x2 + 7x + 12 2 d 27x − 75 e 9x2 − 30x + 25

c 6x2 + 28x − 48

think

Write

a −6x2y + 15xy

a 1 Write the expression. 2

Take out a common factor of −3xy. Make the common factor negative so the leading term inside the brackets will be positive.

= −3xy(2x − 5)

b x2 + 7x + 12

b 1 Write the expression. 2

Look for a common factor. (There isn’t one.)

3

Write (x )(x ) and look for factors of +12 that add to give +7. +4 and +3 achieve this.

= (x + 4)(x + 3) c 6x2 + 28x − 48

c 1 Write the expression. 2

Is there a common factor? Yes (2).

3

Attempt to factorise by inspection. Write 2(3x and try factors of −24. A few possibilities are: (3x + 12)(x − 2) = 3x2 + 6x − 24 NO (3x + 1)(x − 24) = 3x2 − 71x − 24 NO (3x + 4)(x − 6) = 3x2 − 14x − 24 NO (Nearly, just a wrong sign on the x term.) (3x − 4)(x + 6) = 3x2 + 14x − 24 YES

= 2(3x2 + 14x − 24) )(x

)

= 2(3x − 4)(x + 6) d 27x2 − 75

d 1 Write the expression. 2

Look for a common factor. There is one (3).

= 3(9x2 − 25)

3

Recognise the difference of squares in the brackets.

= 3(3x + 5)(3x − 5) e 9x2 − 30x + 25

e 1 Write the expression. 2

There are no common factors.

3

There seem to be a few square numbers in the expression, which looks suspiciously like a perfect square. The square root of the first term is 3x, and the square root of the last term is 5 or −5. Since we need a negative middle term, take −5. Double the product of these is 2 × 3 × −5 = −30, which is the middle term, so we have a perfect square.

= (3x − 5)2

Worked example 5

Factorise: a (x + 3)2 − 24

b (x − 6)2 + 5(x − 6) + 6.

think

Write

a 1 Write the expression, and recognise a difference of

squares. 2

( (

) second term )

a (x + 3)2 − 24

first term + second term ×

= ( x + 3) + 24 ( x + 3) − 24

first term −

= x +3+2 6 x +3−2 6

(

)(

)

Chapter 2 • Quadratic functions

45

b (x − 6)2 + 5(x − 6) + 6

b 1 Write the expression, and notice the linear expression

(x − 6), which is in the usual position of the variable.

2

Let X = (x − 6) and rewrite the expression. Note: This is often called a substitution method.

Let X = (x − 6) X2 + 5X + 6

3

Factorise the new version of the expression.

= [X + 2][X + 3]

4

Replace X with x − 6.

= [(x − 6) + 2][(x − 6) + 3]

5

Simplify.

= (x − 4)(x − 3)

Worked example 6

Factorise the following using an appropriate method. a −6x2y + 15xy b 50 − 2(x − 4)2 think

Write

a −6x2y + 15xy

a 1 Remove the common factor of 3xy. 2

= 3xy(−2x + 5)

= 3xy(5 − 2x) or −3xy(2x − 5)

Simplify.

b 50 − 2(x − 4)2 = 2(25 − (x − 4)2)

b 1 Remove the common factor of 2. 2

Factorise the expression within the brackets using the difference of two squares method.

= 2(52 − (x − 4)2) = 2(5 − (x − 4))(5 + (x − 4))

3

Simplify.

= 2(5 − x + 4)(5 + x − 4) = 2(9 − x)(1 + x)

exercise 2C

Factorising quadratic expressions

1 We4a Factorise the following. a 5x2y3 + 20xy2 b −8ax2 + 14ax 4 3 2 d 22r s + 11r s e 5x2 + 10x + 75 2 We4b, c Factorise the following. a x2 + 16x + 63 b x2 − 17x + 66 − 2 d 3x − 6x + 24 e −x2 − 4x + 96 2 g 6x − 37x + 45 h −40x2 + 46x + 14

c 2x2 + 22x + 48 f 6x2 − 11x − 35

3 We4d Factorise the following. 2 a 4x − 9 b 81x2 − 25 d 18x2 − 162 e (x + 1)2 − 4 g 98 − 2(x + 3)2 h −3(5x − 8)2 + 27

c 2x2 − 32 f (3x − 4)2 − 1

4 We4e Factorise the following. a x2 + 14x + 49 b 25x2 − 10x + 1

− 12x + 12 2 g 4x − 4 5 x + 5 d

3x2

−72x2

+ 24x − 2 2 h 4x − 12 2x + 18 e

5 We5a Factorise the following. a (x + 3)2 − 9 b (x − 3)2 − 16 2 d (3x − 2) − 81 e 2(x + 1)2 − 8 2 g 50(3x + 2) − 98 h 1 − (6 − x)2 46

c 7p2q − 21p + 7 f −3x2 − 18x + 6

Maths Quest 11 Mathematical Methods CAS

c 2x2 + 12x + 18 f

x2 + 2 3 x + 3

c 36 − (2x + 7)2 f −3(x − 4)2 + 48 i 64 − (2 − 3x)2

6 We5b Factorise the following. a (x + 7)2 + 9(x + 7) + 20 c (2x + 7)2 − 9(2x + 7) + 8

b 6(x − 3)2 + 13(x − 3) + 5 d 12(3x − 11)2 − 19(3x − 11) − 18

7 We6 Factorise the following, using an appropriate method. a 3x2 − 24x − 27 b 25x2 + 10x + 1 2 c (x − 13) − 2(x − 13) + 1 d x2 − 28x + 196 2 e 18 x − 6 x f 60x2 + 40x + 5 2 g 60x − 5x h 9 − 9x2y2

4(3x − 1)2 − (x + 2)2 j −12x2 + 70x + 98 8 mC a Which of the factors below could be multiplied by (x + 5) to get x2 + 14x + 45? a (x − 19) B (x + 5) C (x + 9) d (x + 14) e (x + 31) i

b The factorised form of 6x2 − 67x − 60 is: a 2(3x + 5)(x + 6) d (6x − 5)(x + 12)

B 2(3x − 5)(x − 6) e (6x + 5)(x − 12)

c A factor of 6x2 − 54 is: a (x − 3)

B (x − 6)

C (3x + 6)(2x + 10)

C (x − 9)

d (6x − 9)

e 6x

9 Quadratic polynomials are said to be identical if they have the same factorised forms and the same

expanded forms; the relevant coefficients must be equal. Suppose that P(x) = −5(x + 3)2 + 35(x + 3), Q(x) = 60 − 5x2 + kx and R(x) = −5(x + 3)(x + m). Use written algebra to find the values of k and m, if it is known that the three polynomials are identical. Use a mental arithmetic substitution technique to verify your answers.

2d

Factorising by completing the square

Factorising by inspection works best when whole numbers are involved — for example (2x + 3)(x − 7). Sometimes it is clear that it is impossible to find whole numbers to complete a factorisation. For example, with x2 + 6x − 1, there is no pair of factors of 6 that add to −1. In such cases, completing the square may be used to factorise a quadratic. The method of completing the square involves ‘manufacturing’ a perfect square so that an expression may be factorised as a difference of squares. Because this method often produces surds, factorising this way is sometimes referred to as ‘factorising over R’, where R is the set of real numbers that includes surds. Worked example 7

Use the method of completing the square to factorise the following over R. a x2 + 6x − 1 b x2 − 7x + 8 think

a 1 Write the expression.

Write

a x2 + 6x − 1

x 2 + 6 x + ( 62 )2 − ( 62 )2 − 1

2

Halve and square the x-coefficient (6), and then add and subtract it from the equation. Since the same value has been added and subtracted, the expression is equivalent to that in the question.

3

Evaluate the brackets.

= x2 + 6x + 9 − 9 − 1

4

Combine the first three terms as a perfect square, as x2 + 6x + 9 = (x + 3)2.

= (x + 3)2 − 10

5

Recognise a difference of two squares.

= ( x + 3) + 10 ( x + 3) − 10

6

Simplify.

= x + 3 + 10 x + 3 − 10

(

)(

)

Chapter 2 • Quadratic functions

47

b x2 − 7x + 8

b 1 Write the expression. 2

Halve and square the x-coefficient (−7), and then add and subtract it from the equation.

3

Evaluate the brackets.

4

Combine the first three terms as a perfect square, 7 2. since x 2 − 7 x + 49 4 = (x − 2 )

5

Combine the last two terms.

x 2 − 7x + (

) − ( −27 )2 + 8

−7 2 2

49 = x 2 − 7 x + 49 4 − 4 +8 8 = ( x − 72 ) − 49 4 +1 2

32 = ( x − 72 ) − 49 4 + 4 2

= ( x − 72 ) − 17 4 2

6

Recognise a difference of squares.

= ( x − 72 ) +

7

Simplify the surd part if possible.

= x − 72 +

8

Use a common denominator for the last two terms if desired.

2

(

17 2

(

17 4

( x − 7 )2 − 2

)( x −

7 2

)(

−

17 2

or x − 7 −2 17 x − 7 +2 17

17 4

)

)

Worked example 8

Consider the quadratic polynomials represented by P(x) = x2 + bx + 4, where b can be any real number. Answer the following questions. a Factorise P(x). b Find the linear factors of P(x), if they exist for b = 3, 4, 5 and 6. Discuss the differences in the formats of your answers. think

a 1 Write the expression.

Write

a x2 + bx + 4

b2 b2 = x 2 + bx + − + 4 2 2

2

Complete the square: halve and square the x-coefficient (b), and then add and subtract it in the expression.

3

Recognise the first three terms as a perfect square.

b 2 b2 = x + − + 4 2 4

4

Simplify.

b 2 b 2 − 16 = x + − 2 4

5

Recognise a difference of two squares.

b b 2 − 16 b b 2 − 16 =x+ + x+ − 2 4 2 4

6

Simplify the surds.

b b 2 − 16 b b 2 − 16 =x+ + x+ − 2 2 2 2

7

Put all the terms over a common denominator (2).

8

State the factors.

2 x + b + b 2 − 16 2 x + b − b 2 − 16 = 2 2 The factors of P(x) are x+

48

Maths Quest 11 Mathematical Methods CAS

± b 2 − 16 + b . 2

2 x + b ± b 2 − 16 or 2

b 1 Evaluate the factors for b = 3.

b x+

± 32 − 16 + 3 2

32 − 16 =

−7

2

Evaluate the surd.

3

Since the surd is the square root of a negative number, there are no possible linear factors.

There are no linear factors for b = 3.

4

Evaluate the factors for b = 4.

x+

5

Simplify.

6

State the factors.

When b = 4, x + 2 is a repeated factor, so P(x) is a perfect square.

7

Evaluate the factors for b = 5.

x+

8

Simplify.

± 4 2 − 16 + 4 2

± 0 +4 2 =x+2 = x+

± 52 − 16 + 5 2

= x+

± 25 − 16 + 5 2

± 9 +5 2 ±3 + 5 = x+ 2 8 2 = x + or x + 2 2 = x + 4 or x + 1 = x+

State the factors.

The factors are x + 4 and x + 1.

10

Evaluate the factors for b = 6.

x+

11

Simplify.

9

± 62 − 16 + 6 2

= x+

± 36 − 16 + 6 2

= x+

± 20 + 6 2

= x+

±2 5 + 6 2

= x ± 5+3 12

State the factors.

The factors are x + 5 + 3 and x − 5 + 3.

13

Note the differences in the number and format of the linear factors; when b = 3, there were no linear factors. Write your observations.

b = 3: There are no linear factors. b = 4: x + 2 is a repeated linear factor so P(x) is a perfect square. b = 5: The linear factors are x + 1 and x + 4. b = 6: The linear factors are x + 5 + 3 and x − 5 + 3; they are found by completing the square.

Chapter 2 • Quadratic functions 49

Exercise 2D

Factorising by completing the square

1 WE7a Use the method of completing the square to factorise the following over R. a x2 + 4x − 3 b x2 + 10x + 20 2 c x + 6x + 7 d x2 + 2x − 7 e x2 + 8x + 13 f x2 − 4x − 1 g x2 − 12x + 19 h x2 − 2x − 5 2 i x − 8x + 10 j x2 − 6x − 4 2 WE7b Use the method of completing the square to factorise the following over R. a x2 + 3x + 1 b x2 + 5x − 3 2 c x − 7x + 2 d x2 − x − 1 2 e x + 9x + 4 f x2 + 11x − 6 2 g x − 3x + 5 h x2 + 5x + 2 2 i x − 13x − 1 j x2 + x − 3 3 MC Which of the following could be added to and subtracted from x2 + 36x + 1 to assist factorising

using completing the square? A 6 b 18

c 72

d 144

4 MC Which of the following could be added to and subtracted from

using completing the square? A 4 b 17 4 5 MC The expression

x2

c 16

+ 6x − 2 factorises to:

( x − 3 + 2 )( x − 3 − 2 ) c ( x + 3 + 7 )( x + 3 − 7 ) e ( x + 3 + 6 )( x + 3 − 6 ) A

x2

e 324

− 17x − 8 to assist factorising

d 17

e 289

2

(x + 3 + d ( x − 3 + b

)( 10 )( x − 3 −

4

) 10 )

11 x + 3 − 11

6 MC The expression x2 + 4x + 12: A b c d e

can be factorised using whole numbers can be factorised using the method of completing the square cannot be factorised using the methods covered in this topic so far is a linear expression, and so cannot be factorised is already factorised

7 Challenge: Factorise the following. a 3x2 − 18x − 3 c −5x2 − 10x + 15

b 2x2 + 10x + 4 d −12x2 + 4x − 8

Solving quadratic equations — Null Factor Law

2E

As a quadratic equation is a degree 2 polynomial (highest x power of 2), it will have at most two solutions. To solve the quadratic equation x2 − 4x + 3 = 0, factorise by inspection. x2 − 4x + 3 = 0 (x − 1)(x − 3) = 0 or A × B = 0 where A = (x − 1) and B = (x − 3). For A × B to equal zero, either A or B or both must be zero. This is known as the ‘Null Factor Law’. That is, A = 0 or B=0 So (x − 1) = 0 or (x − 3) = 0 Solving these two equations gives: x − 1 = 0 or x−3=0 x = 1 or x=3 50 Maths Quest 11 Mathematical Methods CAS

Worked example 9

Solve the following. a (x + 5)(7x − 11) = 0 c 4x2 + 25 = 20x

b 12x2 − 11x − 15 = 0 d 6x2 = 54

think

a 1 Write the equation. 2

Note that the equation is already factorised.

3

Apply the Null Factor Law and solve two equations.

b 1 Write the equation. 2

Factorise by inspection. This may take several attempts.

3

Solve two equations.

c 1 Write the equation. 2

Rearrange (transpose) so all terms are on the side on which the x2 term is positive. Write terms in order of decreasing powers of x.

3

Factorise (in this case by recognising a perfect square).

4

Solve (one equation here, or two identical equations if you like).

d 1 Write the equation.

Write

a (x + 5)(7x − 11) = 0

x+5=0 x = −5

or or

7x − 11 = 0 7x = 11 x = 11 7

or or or

3x − 5 = 0 3x = 5 x = 53

or or

x−3=0 x=3

b 12x2 − 11x − 15 = 0

(4x + 3)(3x − 5) = 0 4x + 3 = 0 4x = −3 − x = 43 4x2 + 25 = 20x

c

4x2 − 20x + 25 = 0

(2x − 5)2 = 0 2x − 5 = 0 2x = 5 x = 52 6x2 = 54

d

2

Rearrange.

6x2 − 54 = 0

3

Take out the common factor, 6.

6(x2 − 9) = 0

4

Notice a difference of squares.

6(x + 3)(x − 3) = 0

5

Solve.

x+3=0 x = −3

Worked example 10

The height of a triangle is 5 cm more than its base length. If the area of the triangle is 18 cm2, find the base length and height. think 1

Define the length and height in terms of a variable. Here, x is used.

2

Draw a diagram.

Write

tUtorial eles-1407 Worked example 10

Let base length = x Then height = x + 5 x+5 x

Chapter 2 • Quadratic functions

51

3

Use the given information about area to form an equation involving the variable.

Area = 12 bh 18 = 12 (x)(x + 5) 36 = x(x + 5) 36 = x2 + 5x x2 + 5x − 36 = 0

4

Rearrange and form a quadratic equation.

5

Factorise.

(x + 9)(x − 4) = 0

6

Solve.

x = −9 or x = 4

7

Discard the negative answer, as length in this context must be positive.

x=4

8

Calculate the height (x + 5).

h=x+5 =9

9

Write the answer in words.

Base length = 4 cm, height = 9 cm.

(as x > 0)

Worked example 11

Consider the quadratic equations below. Equation 1: (2x + 1)(x − 3) = −4x Equation 2: (2x + 1)(x − 3) = −4x2 Use a calculator to determine which equation(s) has rational solutions. think

Write

Equation 1 (2x + 1)(x − 3) = −4x 2x2 − 5x − 3 = −4x

1

Expand the brackets.

2

Put all the terms on the left-hand side (add −4x to both sides).

3

Factorise.

(2x − 3)(x + 1) = 0

4

Apply the Null Factor Law.

2x − 3 = 0 3 x=2

5

Evaluate the solutions.

Both solutions are rational.

2x2 − x − 3 = 0

x+1=0 x = −1

or or

Equation 2 1

Expand the brackets.

2

Put all the terms on the left-hand side (add −4x2 to both sides).

3

Divide all the terms by 6.

4

Complete the square.

(2x + 1)(x − 3) = −4x2 2x2 − 5x − 3 = −4x2 6x2 − 5x − 3 = 0 x2 − 56 x − 12 = 0 x2 − 5 x + 6

25 144

25 − 144 − 12 = 0

+ 72 =0 ( x − ) − 25144 (x − ) − ( ) = 0 5 2 12

5 2 12

52

5

97 12

97 12

2

5

Apply the Null Factor Law.

x − 12 +

6

Evaluate the solutions.

Both solutions are irrational. Only equation 1 has rational solutions.

Maths Quest 11 Mathematical Methods CAS

= 0 or x − 12 −

97 12

5

=0

Fixed point iteration Fixed point (or simple) iteration is a way of solving equations numerically rather than algebraically. To use this method, the equation to be solved must be manipulated into the form x = g(x). For example, if the equation to be solved is f (x) = x2 − 7x + 1 = 0, a possible manipulation is x2 − 7x + 1 = 0 x2 + 1 = 7x x2 + 1 =x 7 In this case, we have x = g(x), where g( x ) = Fixed point iteration works as follows: 1. Make an initial guess, say x = 1.

x2 + 1 . 7

2. Substitute the guess into g(x).

g( x ) =

x 2 + 1 12 + 1 2 = = 7 = 0.2857 7 7

3. Substitute the improved guess, 0.2857.

g( x ) =

x 2 + 1 0.28572 + 1 = = 0.1545 7 7

4. Substitute the improved guess, 0.1545.

g( x ) =

x 2 + 1 0.15452 + 1 = = 0.1463 7 7

5. Substitute the improved guess, 0.1463.

g(x) =

x 2 + 1 0.14632 + 1 = = 0.1459 7 7

6. Repeat the process until successive values for g(x) are equal (within a tolerance of, say, 0.0001).

g( x ) =

x 2 + 1 0.14592 + 1 = = 0.14589 7 7

7. In this example, the equal successive values of g(x) are 0.1459 and 0.14589, within the tolerance of 0.0001.

x = 0.1459 is a solution to x2 − 7x + 1 = 0.

Solving quadratic equations — null Factor law exercise 2e

Solve the following.

1 We9a

a (x + 6)(2x + 3) = 0

b (x + 1)(6x − 1) = 0

c (x − 5)(x + 2) = 0

d (7x + 12)(9x + 2) = 0

e (16x + 8)(2x − 6) = 0

f

g (6 − 7x)(x + 6) = 0

h

−3x(x

2 We9b Factorise and solve the following. a x2 + 6x − 72 = 0 b 2x2 − 7x − 4 = 0 d 2x2 + 12x − 110 = 0

e 3x2 − 27 = 0

g 12 −

h 15x −

75x2

3 We9c, d

=0

d

=1

g 32x2 = 162 j

f

32x2 − 16x + 2 = 0

=0

b 6x2 + 7x = 49 e

−32x

=

6x2

+ 10

h 5x2 = 8x

+5 =x 13

6x 2

c 1 + 8x = −16x2 f i

27 + 12x2 = 36x 10 − x x2 = 2

The solutions to the equation (x − 3)(x + 7) = 0 are: x = and x = −7 x = −3 and x = 7 x = 3 and x = 7 x = 3 and x = −7 x = 4 and x = −21

4 mC a B C d e

3x2

c 4x2 + 13x + 10 = 0

diGital doC doc-9714 Quadratic expressions

Rearrange and solve the following.

a x2 + 45 = 14x

36x2

(7 + x)(23 − x) = 0

− 1) = 0

−3

Chapter 2 • Quadratic functions

53

5 mC a

−2 13

Which of the following is a solution of 2x2 − 11x = 13? B

− 13 2

2 C 13

d 1

e 13 2

The solutions to a quadratic equation are x = −9 and x = 25 . The equation could be: a (x − 9)(x − 25 ) = 0 B 5(x − 9)(x − 2) = 0 C 2(x + 9)(x − 5) = 0 d (x + 9)(5x − 2) = 0 e (x + 9)(2x − 5) = 0 x 7 We10 The width of a rectangle is 3 cm less than its 2 length. If the area of the rectangle is 40 cm , find the length and width. 8 We11 A rectangular piece of metal with length 6 cm and width 4 cm is to be enlarged by increasing both of these 4 cm dimensions by x cm as shown at right. a If the value of x is 2, describe the change in area that occurs. b Find the exact dimensions of the enlarged rectangle if its 6 cm area is (20 2 + 32) cm2. 6 mC

x

9 The temperature, T (°C), in an office follows the equation T = 85 t 2 − 5t + 30, where t is the time (in

10

11

12 diGital doC doc-9715 WorkSHEET 2.1

13

14

15

hours) the office’s climate control system has been operating. How long does it take the temperature to reach 20 °C? A number pattern follows the rule n2 + 3n + 2; that is, the number pattern is 6, 12, 20, . . . (found by substituting n = 1, n = 2, n = 3 into the rule). a Find the next value by substituting n = 4 into the rule. b What positive n value gives the number 210 when substituted into the rule? The diagram at right demonstrates the idea of ‘rectangular’ numbers. The formula N = w(w + 1) gives the value of a rectangular number based on a shape with a width of w dots. w=1 w=2 w=3 a Which rectangular number has a N=1 N=6 N = 12 width of 6? b What is the width of the rectangular number 272? The amount of bending, B mm, of a particular wooden beam under a load is given by B = 0.2m2 + 0.5m + 2.5, where m kg is the mass (or load) on the end of the beam. What mass will produce a bend of 8.8 mm? A window washer drops a squeegee from a scaffold 100 m off the ground. The relationship between the height of the squeegee (h), in metres, and the length of time it has been falling (t), in seconds, is given by h = 100 − 5t 2. a When does the squeegee pass a window 30 m off the ground? b Exactly how long does it take for the squeegee to hit the ground? The following equations are already in the form x = g(x). Find a solution using fixed point iteration with an initial guess of x = 1. x2 + 6 x2 − 2 a x= b x= 5 3 Find a solution to each of the following (correct to 3 decimal places) using fixed point iteration. First manipulate each equation into the form x = g(x) in a similar way to that shown on page 53. a x2 − 9x + 2 = 0 b x2 + 5x − 7 = 0 2 c x − 4x + 1 = 0 d 2x2 − 11x + 8 = 0

16 Show that an alternative manipulation of x2 − 7x + 1 = 0 into the form x = g(x) is x =

7 x − 1. Does this form converge using fixed point iteration? If so, state the solution found. 17 Find a manipulation of 4x2 + x − 3 = 0 that converges using fixed point iteration. State the solution. 18 Investigate when equations of the form x2 − bx + 1 are likely to converge to a solution using x2 + 1 fixed point iteration with a manipulation of the form x = . b

54

Maths Quest 11 Mathematical Methods CAS

Solving quadratic equations — completing the square 2F

Recall that when you cannot factorise quadratics by the method of sensibly guessing whole numbers, the method of completing the square may be used. Completing the square may also be used to solve quadratic equations that don’t appear to easily factorise. Worked example 12

Solve the following, giving answers in exact (surd) form. a x2 + 10x + 2 = 0 b x2 − 7x + 1 = 0 c x2 − 8 = 0 think

a 1 Write the equation. Notice that there are no

factors of 2 that add to 10, so this equation is a candidate for completing the square.

tUtorial eles-1408 Worked example 12

Write

a

x2 + 10x + 2 = 0

x 2 + 10 x + ( 102 )2 − ( 102 )2 + 2 = 0

2

Halve and square the x-coefficient and then add and subtract it.

3

Evaluate the terms in brackets.

4

Partially simplify (the last two values) as shown.

(x2 + 10x + 25) − 23 = 0

5

Form a perfect square with the first three terms.

(x + 5)2 − 23 = 0

6

Take the constant term (in this case the −23) to the other side of the equation, remembering to change the sign.

7

Take the square root of both sides.

8

Solve for x.

b 1 Write the equation.

x2 + 10x + 25 − 25 + 2 = 0

(x + 5)2 = 23

x + 5 = ± 23 x = −5 ± 23 b x2 − 7x + 1 = 0

2

Decide that completing the square is appropriate.

3

Halve and square the x-coefficient. Add and subtract this term.

4

Evaluate the terms in the brackets.

49 x 2 − 7 x + 49 4 − 4 +1 = 0

5

Partially simplify.

49 4 x 2 − 7 x + 49 4 − 4 +4 =0

x 2 − 7 x + ( 72 )2 − ( 72 )2 + 1 = 0

45 ( x 2 − 7 x + 49 4 )− 4 = 0

6

Form a perfect square.

7

Solve for x. Note that the surd part has been simplified where possible.

( x − 72 )2 − 45 4 =0 ( x − 72 )2 =

45 4

x − 72 = ± =±

45 4 9×5 4

= ± 3 25 x = 72 ± 3 2 5 =

7±3 5 2

Chapter 2 • Quadratic functions

55

c 1 The equation may be solved for x immediately,

without completing the square. 2

Rearrange the equation.

3

Take the square root of both sides and simplify.

c x2 − 8 = 0

x2 = 8 x=± 8 = ± 4×2 = ±2 2

Worked example 13

Use the method of completing the square to solve the following equations, or to explain why there are no solutions. a 2(2x − 3)2 − 2(2x − 3) − 2 = 0 b 3x2 + 6x + 4 = 0. think

a 1 Write the equation. Notice the repeated appearance

of the linear expression (2x − 3), suggesting a substitution technique. Let a = 2x − 3.

Write

a 2(2x − 3)2 − 2(2x − 3) − 2 = 0

Let a = 2x − 3 ∴ 2a2 − 2a − 2 = 0

2

Divide all terms (i.e. both sides) of the equation by 2, the coefficient of a2.

a2 − a − 1 = 0

3

Halve and square the coefficient of a and then add and subtract it.

a2 − a +

4

Solve for a by completing the square.

(a

2

( ) − ( ) −1 = 0 1 2

2

1 2

1

2

)

1

− a + 4 − 4 −1 = 0

(a − )

1 2 2

5

−4=0

(a − )

1 2 2

=

5 4

1

a− 2 = ± 1

a= 2± 5

Replace a with 2x − 3, and solve for x.

1

2x − 3 = 2 ±

1

5 2

1

5 4

x = 32 + 4 ±

6

Write the answer.

Maths Quest 11 Mathematical Methods CAS

5 4

The solution to 2(2x − 3)2 − 2(2x − 3) − 2 = 0 is x=

56

5 2

5 2

2x = 3 + 2 ±

x = 47 ±

5 4

7± 5 . 4

b 1 Write the equation. 2

Divide all terms of the equation by 3, the coefficient of x2.

3

Apply the process of completing the square. Notice here that the next step, finding the square root of both sides of the equation, is not possible as negative does not exist. Note: You will soon learn a way to predict when this will happen.

4

3x2 + 6x + 4 = 0

b

Write the answer.

x 2 + 2x + 4 = 0 3

x 2 + 2 x + ( 22 )2 − ( 22 )2 + 43 = 0 x 2 + 2 x + 1 − 1 + 43 = 0 ( x + 1)2 + 13 = 0

−1

( x + 1)2 =

3

There are no solutions to 3x2 + 6x + 4 = 0.

Solving quadratic equations — completing the square exercise 2F

diGital doCS doc-9716 SkillSHEET 2.3 Solving equations in the complete square form doc-9717 SkillSHEET 2.4 Simplifying surds

1 We12a Solve, giving the solution in exact (surd) form. a x2 + 8x + 1 = 0 b x2 + 12x + 3 = 0 2 c x + 4x − 2 = 0 d x2 − 6x + 4 = 0 e x2 − 10x + 18 = 0 f x2 − 6x + 6 = 0 2 We12b Solve, giving the solution in surd form. a x2 + 3x − 1 = 0 b x2 + 5x + 2 = 0 2 c x − 7x + 5 = 0 d x2 − 9x − 2 = 0 2 e x + 11x + 4 = 0 f x2 − x − 6 = 0

Rearrange and solve, giving the solution in exact form. b x2 + 2 = 6x d 4 − x2 = 7x f x2 − 3(5x − 2) = 0 x 2 + 3x − g 14x − x2 = −1 h = 2 4 4 We12c Solve. 2 a x − 14 = 0 b 6 − x2 = 0 c 3x2 = 36 d −2x2 + 18 = 0 3

a x2 = 4x + 1 c 9x − 2 = x2 e 2(3x + 5) = x2

When completing the square to solve x2 + 16x + 1 = 0, the perfect square part of the expression will be: a 8 B 16 C 64 d (x + 8) e (x + 8)2

5 mC

When completing the square to solve x2 − 20 x − 4 = 0, the value to be simultaneously added and subtracted is: a 4 B 10 C 5 d 100 e 20

6 mC

7 mC

An equation that is being solved using completing the square is at the stage shown below. (x − 9)2 − 15 = 0

The solution is: a (x − 9)2 = 15

B x = ± 24

d x = 9 ± 15

e

C x = − 9 ± 15

x = 3 ± 15

8 We13 For each of the following, find exact solutions (if they exist) or prove there are no solutions. a 5x2 − 2x + 3 = 0 b 3(2x − 5)2 + 6(2x − 5) − 4 = 0 c 4x2 − 2x = 6 d 5(4x + 3) = (4x + 3)2 + 9 Chapter 2 • Quadratic functions

57

2G

The quadratic formula

An alternative to the methods of factorising by inspection or completing the square is to use the quadratic formula. The derivation of the formula follows, and is based on the method of completing the square, but all you have to remember is the formula in the last step. ax2 + bx + c = 0 (where a ≠ 0) Consider the general quadratic equation. b c Divide every term by a. x2 + x + = 0 a a Use the completing-the-square method. 2 2 b b2 b2 c b x2 + x + 2 − 2 + = 0 Halve and square to get b = b a 4a 4a a a 2a 4a2 and add and subtract it to the equation. b b2 b2 c x2 + x + 2 = 2 − Rearrange. a 4a 4a a The left side is now a perfect square. Get 2 2− x + b = b 4 ac a common denominator on the right side. 2a 4a2 Take the square root of both sides. Remember b b 2 − 4 ac x+ =± to take both positive and negative square roots. 2a 4a2 x+

Simplify the RHS denominator.

b ± b 2 − 4 ac = 2a 2a

b from both sides. 2a

x=

Write using the one denominator (2a).

x=

Subtract

−b

2a −

±

b 2 − 4 ac 2a

b ± b 2 − 4 ac 2a

The following examples show how the quadratic formula may be used. Worked Example 14

Use the quadratic formula to solve the following without a calculator. a 3x2 − 8x − 9 = 0 b x2 + 5x + 6 = 0 Give answers in exact (surd) form and comment on the usefulness of this formula for each example. Think

a 1 Write the equation and match up a, b and c.

Write

c a a b

3x2 − 8x − 9 = 0

2

Write the quadratic formula.

x=

3

Substitute the matched values for a, b and c.

x=

4

Simplify.

x=

5

At this stage, it is tempting to cancel 2 out of the 8 and 6, but both terms of the numerator must possess this factor, and initially they don’t. However, in simplifying 172 into 4 × 43 , the factor of 2 emerges; then the cancelling is possible in step 6.

x=

6

Cancel down a common factor of 2 from the numerator and denominator.

x=

58 Maths Quest 11 Mathematical Methods CAS

−b ±

b 2 − 4 ac 2a

− ( − 8) ± +8 ±

(− 8)2 − (4 × 3 × − 9) 2×3

64 + 108 6

8 ± 172 6 8 ± 4 × 43 x= 6 8 ± 2 43 x= 6 4 ± 43 3

4 + 43 4 − 43 or x = 3 3

7

Write the two solutions separately.

x=

8

In the original equation, the coefficient of x2 was 3; also, there are surds in the answer. These facts suggest that neither completing the square nor standard factorisation would have been straightforward methods.

Using the quadratic formula was the most appropriate method for solving this equation.

b 1 Write the equation and match up a, b and c.

ba

b c 1x2 + 5x + 6 = 0

2

Write the quadratic formula.

x=

3

Substitute the matched values for a, b and c.

x=

4

Simplify.

x=

−b ±

b 2 − 4 ac 2a

−5 ±

52 − (4 × 1 × 6) 2 ×1

x= x= x= x =

−

5 ± 25 − 24 2

−5±

1

2 −5±1

2 −5+1

2 −

or x =

−5−1

2

−

4 6 or x = 2 2

x = − 2 or x = − 3 5

In the original equation, the coefficient of x2 was 1; also, there are no surds in the answer. This suggests that an easier factorisation method (i.e. inspection) would be more suitable. Note: This applies only if the question does not specify a particular method.

Using the quadratic formula is not the most suitable method for solving this equation. Factorisation of the left side in the format (x + a)(x + b) would be more efficient.

Worked example 15

Use the quadratic formula to solve 2x2 − 2kx + 4k = 0, and use your solution to solve 2x2 + 6x − 12 = 0. think

Write

1

Recognise a common factor of 2 (divide each term by 2).

2x2 − 2kx + 4k = 0 x2 − kx + 2k = 0

2

Identify a, b and c for the quadratic formula.

a = 1, b = −k, c = 2k

3

Substitute the values for a, b and c into the quadratic formula.

x=

k ± ( − k )2 − 4 × 1 × 2 k 2 ×1

4

Simplify.

x=

k ± k 2 − 8k 2 Chapter 2 • Quadratic functions

59

5

Compare 2x2 − 2kx + 4k = 0 to 2x2 + 6x − 12 = 0.

6

Substitute k = −3 into the solution found in step 4 and simplify.

−2k

= 7 and 4k = −12, so k = −3.

( − 3)2 − 8 × − 3 2 −3 ± 9 + 24 x= 2 −3 ± 33 x= 2 x=

−3 ±

Worked example 16

Consider the set of quadratic equations in the form 4x2 + bx + 9 = 0, where b is 11, 12 or 13. Use the quadratic formula to explain the solutions that occur with each b-value. think

Write

1

Write the equation and match up a, b and c.

a b c 4x2 + bx + 9 = 0

2

Substitute the values for a, b and c into the quadratic formula.

x=

3

Simplify.

x=

4

Substitute b = 11 and simplify.

x= x= x=

−b ±

b2 − 4 × 4 × 9 2×4

−b ±

b 2 − 144 8

− 11 ±

112 − 144 8

− 11 ±

121 − 144 8

− 11

− 23

± 8

5

Evaluate the result.

There is no solution, as we cannot evaluate

6

Substitute b = 12 and simplify.

x=

− 12 ±

122 − 144 8

− 12 ±

144 − 144 8 − 12 ± 0 x= 8 −3 x= 2 x=

7

Evaluate the result.

There is one solution, x =

8

Substitute b = 13 and simplify.

x= x= x= x=

60

Maths Quest 11 Mathematical Methods CAS

− 13 ±

132 − 144 8

− 13 ±

169 − 144 8

− 13

± 25 8

− 13 ± 5

8

−3 2

.

− 23

.

9 10

and x = −1.

There are two solutions, x =

The b2 values have had a clear impact on the solutions; recall that ± 0 = 0 and ± 25 = ±5.

The structure of the quadratic formula means the changing value of the expression under the square root sign, b2 − 144, is critical. b = 11: There was no solution (as negative cannot be resolved). b = 12: There was one solution (as ± 0 = 0). b = 13: There are 2 solutions (as ± 25 = ±5).

exercise 2G 1

−9

Evaluate the result.

4

the quadratic formula

Identify a, b and c (the coefficients of x2, x and the constant respectively) in each of the following quadratic equations. a x2 + 4x − 3 = 0 b x2 − 7x + 9 = 0 2 c 9 − 4x + x = 0 d 7 − 3x − 6x2 = 0 2 1 e 2 x − 7x + 5 = 0 f (x + 1)2 = 0 g (2x − 3)2 = 0 h 5 − 2(x2 + 2) = 0 − 6x 2 + 4 i 3x2 − 10x + 4 + 4x2 − 11x = 0 j =0 2 Use the quadratic formula to solve the following (even though the equations may be factorised by inspection). a x2 + 9x + 20 = 0 b x2 − 10x + 16 = 0 c 2x2 − 13x − 24 = 0 d −4x2 + 13x − 3 = 0

2 We14

3 We15 Find exact (surd) solutions to each of the following. a x2 + 5x + 3 = 0 b x2 + 8x + 5 = 0 2 c x + 3x + 1 = 0 d x2 + 10x + 12 = 0 e x2 − 6x + 2 = 0 f x2 − 7x + 6 = 0 g x2 − 4x − 2 = 0 h x2 − 9x − 8 = 0 − 2 i 2x + 3x + 1 = 0 j −4x2 + 12x − 1 = 0 − 2 k 2kx + 4x + 6k = 0 l 2kx2 − (k + 1)x + 4 = 0 4 a Use a CAS calculator and your answer for question 3k above to solve x2 + 4x − 3 = 0, and write the

value of k you used to do this.

b Use a CAS calculator and your answer for question 3l above to solve 62x2 − 32x + 4 = 0, and write

the value of k you used to do this. 5 Use a CAS calculator to find approximate solutions (3 decimal places) to each of the following. a x2 − 6x − 2 = 0 b x2 + 3x − 9 = 0 c −2x2 + 7x + 1 = 0 d −9x2 − 2x − 2 = 0 e −x2 − 8x + 1 = 0 f 3x2 + x + 9 = 0 6 Use a CAS calculator to find approximate solutions to the following. a x2 + 6x = 11 b 2x2 = 7 − 4x − 2 c 10x + 2 = 5x d x2 = 8x − 6 2 e 5 = 9x − 2x f x2 − 2 = 7x + 4 2 g 5x + 6x + 2 = 0 h −x2 + 4x = 8 7 The population of a colony of rare African ants is given by the equation N = x2 + 2x + 300, where N is

the number of ants, and x is the height of the anthill in centimetres. How high could the anthill be when there are 850 ants in the colony? 8 The profit, $P, made when a particular make of car is manufactured in t hours is given by the equation

P = 40t2 − 7t − 5000. In order to just ‘break even’, how long must the manufacturing process take?

9 The position of an item x metres from the start of an assembly line is given by x = −4t2 + 20t + 5 where t

is the time in minutes elapsed since the item started on the line. When is the item at position x = 0 metres on the assembly line? Chapter 2 • Quadratic functions

61

10 A golf ball is hit from the bottom of a bunker as shown

at right. The height, h metres, of the ball above the ground is given by h = 5.2t − 1.5t2 − 1, where t seconds is the time the ball has been in flight. a How deep is the bunker? b When is the ball first level with the top of the bunker? c At which times is the ball at a height of 3 metres? 11 Debbi plans to pave sections of her backyard, as shown at right. She has a total of 12 square metres of pavers. a Find the value of x if she wishes to use her entire supply of 5 pavers. She then finds that the value of x (found above) will need to be rounded either up or down to a multiple of 0.5 m. Calculate the effect this will have on her existing supply of pavers if she rounds: b up c down. 12 The surface area (in m2) of cement transport containers made by a certain company is given by 4π r2 + 24π r, where r is the radius of the container. If the surface area of a particular container is 60 m2, determine its radius. 13 The Gateway Arch in St Louis, Missouri, was designed by Eero Saarinen in 1948 but not completed until 1964. The line of the arch can be represented by the quadratic function 2 (x − 95)2 + 190 where y is the height y = − 95 of the arch at a horizontal distance x along the base (ground level) from one side of the arch. Both x and y are in metres. a Re-express this function in an expanded form. b How wide is the arch at its base? c This is the tallest arch in the world. How high is the Gateway Arch?

2H

h

xm

4m

xm

Sloppo Cement Co.

r

The discriminant

You may have found on occasions that no solutions (or ‘roots’) can be found for a quadratic equation. If you were using the quadratic formula at the time, you would have found the trouble started when you tried to evaluate the square root part of the formula. The expression under the square root sign is called ‘the discriminant’. The discriminant is used to determine how many roots of an equation exist and is denoted by the upper case Greek letter ‘delta’ (Δ). −

b ± b 2 − 4 ac Δ = b2 − 4ac 2a The quadratic formula The discriminant If Δ < 0 (that is, negative), then no real solutions exist, as we cannot yet evaluate the square root of a negative value. The word ‘real’ is used to describe numbers we can deal with at present. The set of real numbers includes positives, negatives, fractions, decimals, surds, rationals (numbers that may be expressed as a ratio — for example 49 ) and irrationals. (In further studies of maths, you will learn about a way of dealing with square roots of negative numbers using what are known as ‘imaginary numbers’.) If Δ > 0, then the discriminant can be evaluated, and its square root can be added and subtracted in the quadratic formula (see above) to produce two real solutions. If Δ = 0, then there is nothing to add or subtract in the quadratic formula, and so there is only one real solution to evaluate.

62 Maths Quest 11 Mathematical Methods CAS

x=

Worked example 17

Find the value of the discriminant for the equation 3x2 − 5x + 2 = 0. think

Write

1

Write the expression. Note the value of a, b and c.

3x2 − 5x + 2 = 0 a = 3, b = −5, c = 2

2

Write the formula for Δ.

Δ = b2 − 4ac

3

Substitute for a, b and c.

Δ = (−5)2 − 4(3)(2)

4

Evaluate.

Δ = 25 − 24 Δ=1

Worked example 18

How many real solutions are there to the equation −7x2 + 3x − 1 = 0? think

Write

1

Write the expression. Note the value of a, b and c.

−7x2

2

Write the formula for Δ.

Δ = b2 − 4ac

3

Substitute for a, b and c.

Δ = (3)2 − 4(−7)(−1)

4

Simplify.

Δ = 9 − 28 = −19

5

Comment on the number of solutions.

Since Δ < 0, there are no real solutions.

+ 3x − 1 = 0 a = −7, b = 3, c = −1

Worked example 19

For what values of k does −x2 + 2kx − 9 = 0 have: i two distinct solutions? ii one solution? iii no solutions? think

tUtorial eles-1409 Worked example 19

Write

1

Write the expression. Note the value of a, b and c.

−x2

2

Write the formula for Δ.

Δ = b2 − 4ac

3

Substitute for a, b and c.

Δ = (2k)2 − 4(−1)(−9)

4

Simplify.

Δ = 4k2 − 36

5

Consider case i, two distinct solutions. Note: |k| means the ‘size’ or ‘absolute value’ of k without regard to + or −. Since k2 is always positive, we require k > 3 (values such as 3.1, 4, 5 etc.) or k < −3 (values such as −3.1, −4, −5 etc.) if k2 is to be > 9.

iii Two distinct solutions, require Δ > 0.

+ 2kx − 9 = 0 a = −1, b = 2k, c = −9

So

4k2 − 36 > 0 4k2 > 36 k2 > 9 |k| > 3 k > 3 or k < −3

Chapter 2 • Quadratic functions

63

6

7

Consider case ii, one solution. The ‘size’ of k must be = 3. (That is, 3 or −3 are our only choices.)

iii One solution, require Δ = 0.

Consider case iii, no solutions. The ‘size’ of k must be < 3 (values such as −2.9, −2, −1, 0, 1, 2, 2.9 etc.). An alternative method is shown in the next example.

iii No solutions, require Δ < 0.

4k2 − 36 = 0 4k2 = 36 k2 = 9 |k| = 3 k = 3 or k = −3 k = ±3

So

4k2 − 36 < 0 4k2 < 36 k2 < 9 |k| < 3 −3 < k < 3

So

Worked example 20

For which values of k does 2x2 + 2kx + (k + 4) = 0 have: i two solutions? ii one solution? iii no solutions? think

Write/draW

1

Write the expression. Note the value of a, b and c.

2x2 + 2kx + (k + 4) = 0 a = 2, b = 2k, c = k + 4

2

Write the formula for Δ.

Δ = b2 − 4ac

3

Substitute for a, b and c.

Δ = (2k)2 − 4(2)(k + 4)

4

Simplify and factorise.

Δ = 4k2 − 8(k + 4) = 4k2 − 8k − 32 = 4(k2 − 2k − 8) = 4(k + 2)(k − 4)

5

Since Δ is a more complicated expression than those in the previous example, a graph of Δ versus k (Δ on the vertical axis, k on the horizontal axis) is useful. (Recall how you sketched quadratic graphs in previous work, or see the next section.)

Δ>0

Δ

−2

4

−32 6

7

8

64

Δ>0 Δ=0

Δ<0

For case i, we require Δ > 0 (quadrants 1 and 2) From the graph, Δ > 0 when k < −2 or k > 4.

iii Two solutions, Δ > 0.

For case ii, we require Δ = 0 (k-intercepts) From the graph, Δ > 0 when k = −2 or k = 4.

iii One solution, Δ = 0.

For case iii, we require Δ < 0 (quadrants 3 and 4) From the graph, Δ < 0 when −2 < k < 4. Note: This method involving sketching a graph of Δ may be used as an alternative to the method shown in the previous example.

iii No solutions, Δ < 0.

Maths Quest 11 Mathematical Methods CAS

So k < −2 or k > 4.

So k = −2 or k = 4. So −2 < k < 4.

k

Worked example 21

Find an expression for the discriminant of the equation x2 + (2k +1)x + 4k = 0. Use this result to: a find exact and approximate values of k for which the equation has no solution b explain the result obtained in part a through substitution of appropriate k-values into the equation. think

Write

1

Identify the coefficients required.

x2 + (2k + 1)x + 4k = 0 a = 1, b = 2k + 1, c = 4k

2

Write the formula for Δ.

Δ = b2 − 4ac

3

Substitute for a, b and c.

Δ = (2k + 1)2 − 4 × 1 × 4k

4

Simplify.

Δ = 4k2 + 4k + 1 − 16k = 4k2 − 12k + 1

a 1 For the original equation to have no solution,

we require Δ < 0.

a 4k2 − 12k + 1 < 0

12 ± ( − 12)2 − 4 × 4 × 1 2×4 12 ± 144 − 16 = 8 12 ± 128 = 8

2

Solve the left-hand side using the quadratic formula.

k=

3

Simplify.

k=

12 ± 8 2 8

k = 32 ± 2 4

Interpret the result for k.

If 32 − 2 < k < 32 + 2 , then Δ < 0 and the equation has no solutions.

5

Evaluate approximate values for k.

If 0.086 < k < 2.914, then Δ < 0 and the equation has no solutions.

b 1 A selection of values for k will produce Δ < 0,

Δ = 0 and Δ > 0. Find a k-value that gives Δ > 0.

b Δ = 4k2 − 12k + 1

Let k = −2 Δ = 4(−2)2 − 12 × −2 + 1 Δ = 16 + 24 + 1 Δ = 41 Δ > 0, so there are two solutions.

2

Find a k-value that gives Δ < 0. (Pick a value in the range from step 4 of part a.)

Let k = 1 Δ = 4(1)2 − 12 × 1 + 1 Δ = 16 + 24 + 1 Δ = −7 Δ < 0, so there are no solutions.

3

From step 3 of part a, we know the k-values for which Δ = 0.

3 If k = 2 ± 2 , Δ = 0, so there is one solution.

4

Evaluate the result.

If 3 − 2 < k < 3 + 2 , there are no solutions 2 2 to the original equation.

Chapter 2 • Quadratic functions

65

exercise 2h

the discriminant

1 We17 Find the value of the discriminant in each case. a x2 + 9x + 2 = 0 b x2 − 4x − 1 = 0 2 d 2x − 3x + 10 = 0 e −3x2 + x + 3 = 0 2 g x + 15x = 1 h 9 − 7x = 4x2

c 5x2 + 6x − 7 = 0 f −x2 − 2x − 6 = 0 i −3x2 = 5

2 We18 How many real solutions are there to the following equations?

Do not actually work out any solutions. diGital doC doc-9718 Calculating the discriminant

b −x2 + 4x + 4 = 0 e −2x2 − 8x − 8 = 0 h 6 − 6x = x2

a 5x2 + x + 2 = 0 d 3x2 + 6x + 3 = 0 g 5x2 = 2 − x

c 3x2 − 3x + 1 = 0 f 9 − x2 + x = 0 i 12x = 9x2 + 4

3 We19 Find an expression for the discriminant of each of the following. a x2 + ax + 1 = 0 b ax2 + 2x + 3 = 0 c x2 + 6x + a = 0 d ax2 + bx + 1 = 0 e mx2 + 2mx + 1 = 0 f x2 + (m + 1)x + 3 = 0 2 2 g x − mx − (m + 4) = 0 h (k − 1)x − kx + 2 = 0 4 mC Which values of m below both result in the expression x2 − mx + 5 = 0 having two

distinct solutions? a −20 and 3

B 0 and 20

C 4 and 5

5 We19, 20 For what values of k does each equation have: i two distinct solutions? ii one solution? a x2 + kx + 4 = 0 b x2 − 4x + k = 0 2 d kx − 18x + 20 = 0 e x2 − 4x + (k + 1) = 0 2 g 4kx + 12kx + 9k = 0 h (k + 4)x2 + 10x + 5 = 0

d −6 and −5

iii c f i

e 0 and 5

no solutions? x2 + 4kx + 4 = 0 6x2 + 4kx + (k + 3) = 0 (k − 1)x2 − (k + 1)x + 2 = 0

6 We21 Assume k = −1 in each of questions 5 a–i. Use substitution and your previous answers to question 5 to complete the following table. (There is no need to actually solve the equations.)

Question

Equation after substitution k = −1

Number of solutions

Reasoning

a b c d e f g h i

Graphs of quadratic functions as power functions (turning point form) 2i

Quadratic functions are also power functions. Power functions are functions of the form f (x) = xn, n ∈ R. The value of the power, n, determines the type of function. When n = 1, f (x) = x, the function is linear. When n = 2, f (x) = x2, the function is quadratic. Other power functions will be discussed later. Under a sequence of transformations of f (x) = xn, n ∈ R, the general form of a power function is f (x) = a(x − b)n + c (where a, b, c, and n ∈R). All linear and quadratic polynomials are also linear and quadratic power functions, because all linear and quadratic functions are transformations of f (x) = x and f (x) = x2 respectively. When a quadratic function is written in turning point form, it is written in power form. For example, the quadratic function y = x2 + 4x + 6 can also be represented as the power function y = (x + 2)2 + 2. 66

Maths Quest 11 Mathematical Methods CAS

In previous years’ work, you will have discovered the following connections between a quadratic function in turning point (TP) form, and its graph. Summary of the sequence of transformations y = a(x − b)2 + c a is the dilation factor from the x-axis. If a > 1, the graph gets thinner or narrower. If 0 < a < 1, the graph gets wider. If a < 0, the graph is turned upside down (or, reflected in the x-axis).

x-coordinate of turning point (horizontal shift)

y

y-coordinate of turning point (vertical shift) Minimum value of y for ∪ shape graphs Maximum value of y for ∩ shape graphs

a positive, increasing

y x Turning point (b, c) x

a negative, increasingly negative

Worked Example 22

For the graph of y = −3(x + 2)2 − 1: i state the turning point coordinates ii describe the width of the graph as ‘same’, ‘thinner’ or ‘wider’ compared to the width of y = x2 iii state whether the graph is a minimum (∪) or maximum (∩) type, and state the maximum or minimum value of y iv find the y-intercept v sketch the graph (x-intercepts are not required). Think

Write/draw

y = −3(x + 2)2 − 1

1

Write the equation. Compare y = −3(x + 2)2 − 1 with y = a(x − b)2 + c a = −3, b = −2, c = −1.

2

TP x-coordinate: b = −2. TP y-coordinate: c = −1

i TP (−2, −1)

3

y dilation factor = −3 (magnitude greater than 1 means it is thinner than y = x2)

ii Thinner

4

a is negative, so the graph is ‘upside down’ or a maximum shape.

iii The graph is a maximum (∩) type.

5

Maximum y-value is c, which equals −1.

The maximum value of y is −1.

6

The y-intercept can be found by substituting x = 0 into the original expression and simplifying.

iv y-intercept: y = −3(0 + 2)2 − 1

Use the information above to sketch the graph.

v

7

= −3(2)2 − 1 = −13

y x

(−2, −1)

−13

Chapter 2 • Quadratic functions 67

If a quadratic function is not in power form or turning point form, it must be manipulated in order to answer questions like those posed in the previous example. To do this, we use the method of completing the square as demonstrated in the following example.

Worked example 23

Convert the function y = x2 − 6x + 1 to power form. State the coordinates of the turning point and the maximum or minimum value of y. think

Write

1

Write the equation. Ensure the x2 coefficient is 1. It is. (If not, divide the equation by whatever will change the x2 coefficient to 1.)

y = x2 − 6x + 1

2

Halve the x-coefficient and square it. Add and subtract this value after the x term.

y = x2 − 6x + 9 − 9 + 1

3

Group terms to make a perfect square.

y = (x2 − 6x + 9) − 8

4

Factorise the perfect square part.

y = (x − 3)2 − 8

5

Compare with y = a(x − b)2 + c.

a = 1, b = 3, c = −8

6

The turning point is (b, c).

The turning point is (3, −8).

7

a is positive, so there is a minimum value of y = −8 when x = 3.

The minimum value of y is −8.

Worked example 24

Identify the coordinates of the turning point of the graph of y = 2x2 + x − 4 using algebraic methods. think

Write

y = 2x2 + x − 4 1 = 2[ x 2 + 2 x − 2]

1

Write the rule for the function. Begin the process of completing the square by first taking out 2 as the common factor.

2

Half of shown.

3

Continue on with completing the square, within the square [ ] brackets.

32 1 = 2[( x + 14 )2 − 16 − 16 ]

4

Multiply the 2 through the square brackets, leaving it as a factor of the curved brackets.

= 2( x + 14 )2 − 33 8

5

Read the coordinates of the turning point from this form. Write your answer.

1 2

1 , so add and subtract this as is 14 , and ( 14 )2 = 16

1 1 = 2[( x 2 + 12 x + 16 ) − 2 − 16 ]

33 = 2[( x + 14 )2 − 16 ]

The turning point has coordinates − − ( 14 , 33 8 ).

Graphs of quadratic functions as power functions (turning point form) exercise 2i

1 We22 State the turning point coordinates for each of the following. a y = (x − 5)2 b y = (x + 7)2 + 3 2 c y = (x − 2) − 7 d y = (x − 1)2 + 8 e y = 2(x + 3)2 − 4 f y = −6(x − 2)2 + 2 68

Maths Quest 11 Mathematical Methods CAS

2 We22 For each of the following: i state the turning point coordinates ii describe the width of the graph as same width, thinner or wider than y = x2 iii state whether the graph is a minimum (∪) or maximum (∩) type, and state the maximum or

minimum value of y iv find the y-intercept v sketch the graph (x-intercepts are not required). a y = (x − 4)2 + 2 c y = (x + 5)2 − 8

b y = (x − 2)2 − 3 d y = (x − 1)2 − 1

e y = 2(x + 5)2 − 9

f

g y = 65 (x + 1)2 − 12

h y = −7(x − 3)2 + 3

i

y = −(x − 8)2 − 9

j

y = 13 (x − 4)2 + 4 y = (1 − x)2 + 20

3 Sketch graphs having the following properties, but do not show intercepts. a Turning point (2, 5), dilation factor of 3 from the x-axis b Turning point (−1, 3), dilation factor of 1 from the x-axis c Turning point (0, −4), dilation factor of 2 from the x-axis d Turning point (6, 0), dilation factor of −1 from the x-axis e Turning point (7, −7), dilation factor of −4 from the x-axis f Turning point (0, 2), dilation factor of 12 from the x-axis 4 Determine the function rules of the following graphs, given that they are all of the power function form

y = (x − b)2 + c (that is, dilation factors all equal to 1). a

y

b

(−1, 5)

y

c

y

4

8 (2, 4)

x

x

−1

x

(2, −5) d (−3, 1)

e

y

y

y 7

x −8

f

(6, 6) x

x

−30

(−3, −2)

Find the y-intercepts of graphs with the following properties.

5

a Turning point (1, 4), dilation factor of 2 from the x-axis b Turning point (−2, −3), dilation factor of −1 from the x-axis 6 We23 Convert each of the following to power form, and state the coordinates of the turning point and

the maximum or minimum value of y. y = x2 − 4x + 9 y = x2 − 12x + 37 y = x2 − 4x − 5 y = x2 + 18x + 0 y = 3x2 + 12x + 15

a c e g i

b d f h

y = x2 − 6x + 17 y = x2 + 8x + 13 y = x2 + 7 y = 2x2 − 12x + 22

7 We24 For each of the following quadratic functions: i use by-hand algebraic methods to find the exact coordinates of the turning point ii use a CAS calculator to verify this algebra and find decimal equivalents for these coordinates. a y = 3x2 − 2x + 1 b y = −2x2 − 5x − 4 c y = 6x2 − 3x − 4 Chapter 2 • Quadratic functions

69

8 A section of perspex for a parabolic window is drawn on a

coordinate grid as shown at right. What equation should be programmed into the automatic glass cutting machine, using the grid system on the diagram?

8 6 4 2 0

2

4

6

8

What is the effect on the graph of increasing k in the equation y = k(x − 2)2 + 1? The graph is: a raised B lowered C thinned d widened e translated left F translated right 10 mC What is the effect on the graph of increasing k in the equation y = (x + k)2? The graph is: a raised B lowered C thinned d widened e translated left F translated right 11 mC What is the effect on the graph of increasing k in the equation y = (x + 1)2 − k? The graph is: a raised B lowered C thinned d widened e translated left F translated right 9 mC

diGital doC doc-9719 WorkSHEET 2.2

12 For the graph y = M(x + h)2 + n, state: a the x-coordinate of the turning point b the y-coordinate of the turning point c the dilation factor from the x-axis d the coefficient of x2 in the expanded version e the coefficient of x in the expanded version f the constant term in the expanded version.

Graphs of quadratic functions (intercepts method) 2J

In this section we will consider graphs of quadratic functions of the form y = ax2 + bx + c. When we talk about sketching a graph, we mean drawing a diagram showing the main features — not a true scale graph showing every point plotted accurately using a computer package or other means. To sketch a quadratic graph, the following features should generally be apparent or labelled.

1 the y-intercept

y

The y-intercept is found by substituting x = 0 into the rule and solving the equation for the corresponding y-value. That is, it is the c in the general form y = ax2 + bx + c. All quadratics have a y-intercept.

y-intercept

c x

2 any x-intercepts if they exist Any x-intercepts are found by substituting y = 0 into the rule and solving the equation for the corresponding x-value. Let’s say the rule factorises to y = (x − d)(x − e). Then substituting y = 0 into the rule gives the equation 0 = (x − d)(x − e). For this to be true, we must have x = d or x = e.

70

Maths Quest 11 Mathematical Methods CAS

y x-intercepts c d

e

x

3 The turning point coordinates Consider the general quadratic function, and complete the square as follows: y = ax 2 + bx + c

y

Axis of symmetry x-intercepts

y-intercept

b c = a x 2 + x + a a

x

b b 2 b 2 c = a x 2 + x + − + 2a 2a a a

Turning point b b2 — — (− 2a , c − 4a )

b 2 b2 c = a x + − 2 + a 2a 4a b 2 b2 +c = a x + − 2a 4a b 2 b2 = a x + + c − 2a 4a You may recognise this form as turning point form. In this case, the coordinates of the turning point are b2 −b 2a , c − 4 a Using xt and yt for the coordinates of the turning point, we have −

b b2 and yt = c − 2a 4a If a quadratic function has two x-intercepts, the x-coordinate of the turning point may be found by averaging them. xt =

y

d

d+e —— 2

e

x

If there is only one x-intercept, then it is the x‑coordinate of the turning point. y

d

x

The y-coordinate of the turning point may be found using the general turning point form above, by completing the square from scratch or by substituting the x-coordinate into the original equation. Recall also the two main types of parabolas: y

y

x

Minimum parabola y = ax2 . . . (a > 0)

x

Maximum parabola y = ax2 . . . (a < 0) Chapter 2 • Quadratic functions 71

Worked Example 25

Sketch the graphs of the following, showing all intercepts and the turning point in each case. a y = x2 − 4x − 32 b y = x2 + 10x + 25 c y = −2x2 + 11x − 15 Think

a 1 Write the rule.

Write/draw

a y = x2 − 4x − 32

2

Find the y-intercept (when x = 0).

If x = 0, y = −32

3

Factorise before finding x-intercepts.

y = (x + 4)(x − 8)

4

Find the x-intercepts (when y = 0).

If y = 0, 0 = (x + 4)(x − 8) x = −4 or x = 8

5

Find the x-coordinate of the turning point (TP).

, b = −4, a = 1 2a − ( − 4) = 2(1) = 2

6

Substitute into the original rule to find the y-coordinate of the TP. Alternatively, use b2 yt = c − . 4a Write the turning point coordinates.

7 8

TP xt =

−b

yt = 22 − 4(2) − 32 = 4 − 8 − 32 = −36 TP (2, −36)

Combine the above information and sketch the graph.

y

−4

8

x

(2, −36) b 1 Write the rule.

b y = x2 + 10x + 25

2

Find the y-intercept (when x = 0).

If x = 0, y = 25

3

Factorise before finding x-intercepts.

y = (x + 5)(x + 5) = (x + 5)2

4

Find the x-intercept (when y = 0).

If y = 0, 0 = (x + 5)2 x = −5

5

Find the x-coordinate of the turning point. Alternatively, since there is only one x‑intercept (−5), it must be the turning point x-coordinate.

TP

6

Substitute into the original rule to find the y-coordinate of the turning point.

7

Write the turning point coordinates.

8

Combine the information and sketch the graph.

xt =

, b = 10, a = 1 2a − 10 = 2(1) = −5

yt = (−5)2 + 10(−5) + 25 = 25 − 50 + 25 =0 TP (−5, 0) y 25 (−5, 0)

72 Maths Quest 11 Mathematical Methods CAS

−b

x

c 1 Write the rule.

c y = −2x2 + 11x − 15

2

Find the y-intercept (when x = 0).

If x = 0, y = −15

3

Factorise before finding x-intercepts.

y = −(2x2 − 11x + 15) = −(2x − 5)(x − 3)

4

Find the x-intercepts (when y = 0).

If y = 0, 0 = −(2x − 5)(x − 3) 2x − 5 = 0 or x − 3 = 0 2x = 5 or x = 3 5 x = 2 or x = 3

5

Find the x-coordinate of the turning point.

TP xt =

−b

2a

, b = 11, a = −2 − 11

= = 6

Substitute into the original rule to find the y-coordinate of the turning point.

7

Write the turning point coordinates.

8

Sketch the graph, showing all important features.

yt = − 2

11 4

(or 2.75)

( ) + 11( ) − 15 2

11 4

− 242 + 16 − 242 = 16 + 2 = 16 1 = 8 (or

=

11

TP

2(− 2)

11 4

121 − 15 4 484 240 − 16 16

0.125)

1

( 4 , 8 ) y 5– 2

—, 1– ) (11 4 8

3

x

−15

Don’t be put off if asked to sketch a quadratic graph whose equation doesn’t have 3 terms. Such cases are easier to sketch, as the following example shows. Worked example 26

Sketch the graphs of the following equations. a y = x2 − 16 b y = −2x2 + 14x think

a 1 Write the rule.

Write/draW

a y = x2 − 16

2

Find the y-intercept (when x = 0).

If x = 0, y = 02 − 16 = −16

3

Factorise before finding x-intercepts. In this case, recognise a difference of squares.

y = (x + 4)(x − 4)

4

Find the x-intercepts (when y = 0).

If y = 0, So

0 = (x + 4)(x − 4) x = −4 or x = 4 Chapter 2 • Quadratic functions

73

xt =

−b

5

Find the x-coordinate of the turning point.

, b = 0, a = 1 2a 0 xt = 2(1) =0

6

Substitute into the original rule to find the y-coordinate of the turning point.

yt = 02 − 16 = −16

7

Write the turning point coordinates.

8

Sketch the graph, showing all important features.

TP (0, −16) y

−4

x

4

(0, −16) b 1 Write the rule.

b y = −2x2 + 14x

2

Find the y-intercept (when x = 0).

If x = 0, y = −2(0)2 + 14(0) =0

3

Factorise before finding x-intercepts. In this case, use a common factor of −2x.

y = −2x(x − 7)

4

Find the x-intercepts (when y = 0).

If y = 0, So

5

Find the x-coordinate of the turning point.

xt =

= Substitute into the original rule to find the y-coordinate of the turning point. b2 We could also use yt = c − . 4a

Write the turning point coordinates.

8

Sketch the graph.

14 4 7 2

(or 3.5)

( ) + 14 ( ) 2 ( ) + 49

yt = − 2 = = =

7

−

b , b = 14, a = − 2 2a − 14 xt = − 2( 2) =

6

0 = −2x(x − 7) x = 0 or x = 7

−

7 2

49 4

− 49 2 49 2

(

TP

2

7 2

+

98 2

(or 24.5)

7 49 , 2 2

)

y 7– 49 ,— 2 2

0

74

Maths Quest 11 Mathematical Methods CAS

7

x

Worked example 27

Sketch the graphs of the following equations. a y = 3x2 + 6x − 1 b y = −2x2 + x − 7 think

a 1 Write the rule.

Write/draW

a y = 3x2 + 6x − 1

2

Find the y-intercept (when x = 0).

If x = 0, y = −1

3

Factorising before finding x-intercepts (when y = 0) is not easily done, so use the quadratic formula.

If y = 0, 0 = 3x2 + 6x − 1 −b ±

b 2 − 4 ac 2a a = 3, b = 6, c = −1 x=

x= = = = =

−6 ±

62 − 4(3)(− 1) 2(3)

−6 ±

36 + 12 6

−

6 ± 48 6

−

6 ± 16 × 3 6

−

6±4 3 6

−

3±2 3 3 x = 0.155 or =

4

x = −2.155

xt =

−b

, b = 6, a = 3 2a − 6 xt = 2(3)

Find the x-coordinate of the turning point.

= −1 5

Substitute into the original rule to find the y-coordinate of the turning point.

6

Sketch the graph, showing all important features.

yt = 3(−1)2 + 6(−1) − 1 =3−6−1 = −4 TP (−1, −4) y −2.155

0.155 x −1

(−1, −4) b 1 Write the rule. 2

Find the y-intercept (when x = 0).

b y = −2x2 + x − 7

If x = 0, y = −2(0) + (0) − 7 = −7

Chapter 2 • Quadratic functions

75

3

Factorising before finding x-intercepts (when y = 0) is not easily done, so try the quadratic formula.

If y = 0, 0 = −2x2 + x − 7 b 2 − 4 ac 2a a = −2, b = 1, c = −7 −1 ±

x=

−1 ±

−1 ±

cannot be evaluated so there are no x-intercepts.

No x-intercepts

5

Find the x-coordinate of the turning point.

xt =

−

b 2a −

=

1 2( 2)

=

−1 −4

=

1 4

−

(or 0.25)

yt = − 2 ( 14 ) + 14 − 7 2

1 = − 2 ( 16 ) + 14 − 7 −1 8+ −1 = 8 + − 55 = 8

=

7

− 55

−4

4

Substitute into the original rule to find the y-coordinate of the turning point.

12 − 56 −4

=

6

12 − 4(− 2)(− 7) 2(− 2)

=

− 55

−b ±

x=

Sketch the graph, showing all important features.

1 4 2 8

−7 − 56 8

(or − 6.875)

y —) ( 1–4 , − 55 8

x

−7

Worked example 28

Consider the graph of y = x2 − 5x + 9. Use written algebra to: a show that the parabola has no x-intercepts b find how many units the parabola needs to be translated down, or ‘lowered’, so that it has exactly one x-intercept c find how many units the original parabola needs to be lowered so that its x-intercepts are 13 units apart. think

a 1 Write the rule for the function. Any potential

x-intercepts depend on the solutions of the equation 0 = x2 − 5x + 9. 76

Maths Quest 11 Mathematical Methods CAS

Write

a y = x2 − 5x + 9

2

Calculate the discriminant, given a = 1, b = −5 and c = 9.

a = 1, b = −5 and c = 9 Δ = (−5)2 − 4 × 1 × 9 = 25 − 36 = −11

3

The discriminant is negative. Interpret the meaning of this.

The discriminant is negative, so there are no solutions to the quadratic equation 0 = x2 − 5x + 9; that is, the parabola has no x‑intercepts.

b 1 Replace the original equation with a ‘lowered’

one. Change 9 to a variable, c.

b y = x2 − 5x + c

2

For this equation to have one x-intercept, the discriminant, Δ, must equal 0. Calculate Δ.

a = 1, b = −5 Δ = (−5)2 − 4 × 1 × c = 25 − 4c

3

Set Δ = 0 and solve for c.

25 − 4c = 0 c = 6.25

4

Write the new equation.

y = x2 − 5x + 6.25

5

Compare the new equation with the original equation.

The graph is lowered by 9 − 6.25 = 2.75 units.

c 1 Calculate the axis of symmetry of

y = x2 − 5x + 9.

c Axis of symmetry =

b = −5, a = 1 Axis of symmetry =

−b

2a − ( − 5)

2 ×1 5 = 2 = 2.5 The turning point is at x = 2.5.

2

The x-coordinate of the turning point (here, 2.5) is always halfway between any x-intercepts; the required distance between the intercepts is 13 units.

13 ÷ 2 = 6.5 2.5 + 6.5 = 9 and 2.5 − 6.5 = −4 The x-intercepts must be 9 and −4.

3

The x-intercepts of a function y = (x − d)(x − e) are d and e given a dilation factor parallel to the x-axis of 1.

The rule of the function is y = (x + 4)(x − 9).

4

Expand to find the y-intercept.

y = (x + 4)(x − 9) y = x2 − 5x − 36 The y-intercept is −36; the answer is correct as the first two terms are identical to those in the previous functions.

5

The amount of vertical translation will be the difference between the original and final y-intercepts. Answer the question.

9 − (−36) = 45 The original parabola must be lowered by 45 units.

Using the discriminant Since finding x-intercepts for a quadratic graph involves solving a quadratic equation, we can use the discriminant to decide the number of x-intercepts such a graph has. Chapter 2 • Quadratic functions 77

y

y

y x

x x

Δ>0 2 x-intercepts

Δ=0 1 x-intercept

Δ<0 No x-intercepts

Graphs of quadratic functions (intercepts method) exercise 2J

diGital doCS doc-9720 Quadratic graphs — general form doc-9721 Quadratic graphs — turning point form

1

Find the y-intercepts for each of the following. b y = −3x2 + 6x − 4 e y = 5x + 2 − 4x2

a y = x2 + 9x + 2 d y = 6 − x2 a d g j

State the x-intercepts for each of the following. y = (x − 1)(x − 6) b y = (x − 3)(x + 2) y = (x − 4)(x − 5) e y = x(x − 2) y = (x + 5)(5 − x) h y = −x(x + 8) − y = 3(2 − x)(x + 10)

a c e g

State the x-intercepts for each of the following. y = (2x − 5)(x + 1) y = (x − 9)(4x − 9) y = (2x + 3)(4x + 1) y = (Ax + a)(Bx + b)

2

3

4

c y = 4x2 + 2x f y = 1 − x − x2 c y = (x + 5)(x + 1) f y = (3 − x)(4 − x) i y = −(x + 9)2

b y = (−3x + 1)(x + 2) d y = −(x + 1)(6 − 5x) f y = 3x(x + 4)

Use the quadratic formula to find exact values (if possible) for the x-intercepts of: b y = −x2 + 7x − 7 d y = 10 − 3x + 3x2.

a y = 4x2 − 2x + 3 c y = −2x2 − 9x − 1

5 We25a, b Sketch graphs of the following, showing all intercepts and the turning point in each case.

(Hint : Factorise first.) y = x2 − 4x + 3 y = x2 + 6x + 8 y = x2 − 8x + 12 y = x2 + 3x + 2 y = x2 − 11x − 12 y = x2 − 16x + 64

a c e g i k

b d f h j l

y = x2 + 2x + 1 y = x2 + 12x + 35 y = x2 + 2x − 63 y = x2 − 5x + 6 y = x2 + 14x + 49 y = x2 + 8x − 153

6 Sketch the following graphs, showing all intercepts and the turning point in each case. a y = 3x2 + 2x − 8 b y = 5x2 + 18x − 8 c y = 3x2 − 4x − 15 d y = 4x2 − 8x + 3 e y = 8x2 − 10x + 3 f y = 7x2 + 18x − 9 2 2 g y = 15x + 48x + 9 h y = 9x − 2x − 7 i y = 2x2 + x − 28 2 2 j y = 3x + 5x + 2 k y = 2x − 3x − 9 7 We25c Sketch the graphs of the following. a y = −x2 − 8x + 33 b y = −x2 + 2x + 3 − 2 d y = x + 18x − 81 e y = −4x2 + 12x − 5

78

c y = −x2 − 18x − 45 f y = −8x2 − 6x + 5

8 We26 Sketch the graphs of the following. a y = x2 − 25 b y = x2 − 121 2 d y=3−x e y = 2x2 − 18 2 g y = x + 5x h y = x2 − 8x j y = −4x2 − 24x k y = 21x − 3x2

c y = −x2 + 1 f y = −3x2 + 12 i y = 5x2 − 10x

9 We27 Sketch the graphs of the following. a y = x2 + 2x − 7 b y = x2 + 4x + 1 d y = −4x2 + 2x + 3 e y = −x2 − 18x − 1 2 g y = x − 7x − 2 h y = 2x2 + 7x + 4 − 2 j y = 3x + 17 k y = 4x2 − 5x

c y = x2 + 8x + 3 f y = x2 − 3x + 1 i y = 3x2 − 9x − 5

Maths Quest 11 Mathematical Methods CAS

If a and b are positive numbers, which of the following graphs could be that of y = (ax + 1)(x + b)?

10 mC a

B

y

C

y

y

x x x y

d

e

y

x x

If k and p are positive numbers, which of the following could be the graph of y = x2 − 2kx + p?

11 mC

y

a

B

y

x

x

y

d

e

x

12

y

C

x

y x

Without sketching, determine how many x-intercepts each of the following graphs have. b y = −3x2 − 4x + 8 d y = 4x2 + x + 9 f y = 4x2 + 72x + 324

a y = x2 + 37x + 208 c y = −9x2 + 78x − 169 e y = −x2 − 12x − 35

13 Consider the value of the discriminant in deciding which graph matches which equation below. a y = 5x2 + x + 1 b y = 6x2 − 2x − 1 c y = −7x2 − x + 2 d y = −4x2 + 8x − 4 a

y

B

y x

x

C

d

y

y x

x 14 We28b Use a CAS calculator and written algebra to find how many units and in which direction the

graph of y = 2x2 − 3x − 2 should be translated vertically so that it has exactly one x-intercept. Also, state the function rule for this new parabola in both expanded and turning point forms.

15 We28c Use a CAS calculator and written algebra to find how many units and in which direction the

graph of y = −3 (x + 2)2 − 4 should be translated vertically so that its x-intercepts are exactly 4 units apart. Also, state the function rule for this new parabola in both expanded and turning point forms.

Chapter 2 • Quadratic functions

79

Using technology to solve quadratic equations 2k

Quadratic expressions, equations and functions are linked closely, as you have previously seen. Study the following table carefully. Classification

Characteristics

Examples

Quadratic expression

Degree 2 polynomial, no equals sign Can be written in expanded and/or factorised form

4x2 − 3.4x + 2 (x − 7)(x + 3) x2 + 5x − 6

Quadratic equation

One variable, one equals sign 0, 1 or 2 solutions exist

4x2 − 3.4x + 2 = 19 (x − 7)(x + 3) = 0 x2 + 5x − 6 = −4

Quadratic function

Two variables, one equals sign in the rule Can be sketched, as it is a set of ordered pairs Sketch is a parabola, with 0, 1 or 2 x-intercepts

y = 4x2 − 3.4x + 2 y = (x − 7)(x + 3) y = x2 + 5x − 6

Note: The solutions (also known as the roots) of a quadratic equation (say, 3x2 − 4.3x − 1.68 = 0) are identical to the x-intercepts of its related parabola (here, the sketch of y = 3x2 − 4.3x − 1.68). They are also known as the zeros of the related expression (here, 3x2 − 4.3x − 1.68). Can you see why the solutions of x2 + 5x − 6 = −4 become the x-intercepts of the parabola given by y = x2 + 5x − 2? Worked example 29

Consider the expressions x2 − 8x + 12, x2 − 10x + 21, and x2 − 12x + 32. Use written algebra and/or a CAS calculator to: a find the zeros and factors of the expressions b find the turning points of their related parabolas c use the patterns seen in these answers to predict the next three and previous two quadratic expressions, along with the features of their related parabolas. Now consider a general quadratic function (variable x) whose graph is an upright parabola with a dilation factor from the x-axis of 1. Its x-intercepts are j and k. d Find the rule for this quadratic function. e Find the function rule for a second parabola that has been translated 3 units to the left of the original parabola. f Verify your results for d and e (algebraically and graphically) by letting j = −5 and k = 2. think

a 1 Factorise x2 − 8x + 12.

a x2 − 8x + 12 = (x − 2)(x − 6)

Set the factorised expression equal to 0 and solve.

(x − 2)(x − 6) = 0 x−2=0 or x=2 or

3

Factorise x2 − 10x + 21.

x2 − 10x + 21 = (x − 3)(x − 6)

4

Set the factorised expression equal to 0 and solve.

(x − 3)(x − 7) = 0 x−3=0 or x=3 or

5

Factorise x2 − 12x + 32.

x2 − 12x + 32 = (x − 4)(x − 8)

6

Set the factorised expression equal to 0 and solve.

(x − 4)(x − 8) = 0 x−4=0 or x=4 or

2

80

Write/draW

Maths Quest 11 Mathematical Methods CAS

x−6=0 x=6

x−7=0 x=7

x−8=0 x=8

b 1 The x-value of the turning point is halfway

between the roots (zeros) of the associated parabola.

2+6 =4 2 For x2 − 10x + 21, 3+ 7 TP x-value = =5 2 For x2 − 12x + 32, 4+8 TP x-value = =6 2

TP x-value =

2

Find the y-values of each turning point by substituting the x-values in the corresponding functions.

For x2 − 8x + 12, 42 − 8 × 4 + 12 = −4 For x2 − 10x + 21, 52 − 10 × 5 + 21 = −4 For x2 − 12x + 32, 62 − 12 × 6 + 32 = −4

3

State the turning points of each parabola.

The turning points of these three parabolas are (4, −4), (5, −4) and (6, −4).

c 1 Predict the next three expressions based upon the patterns from parts a and b. 2

Predict the previous two expressions.

3

The intercepts and turning points are moving to the right (horizontally) by 1 unit for each step in the progression. This means that only the x-coordinates change. Also, the coefficient of the x term in the expanded expression is the negative sum of the zeros, and the constant term is the product of the zeros.

d 1 Assign variables to the x-values of the zeros

c (x − 5)(x − 9)

(x − 6)(x − 10) (x − 7)(x − 11) (x − 1)(x − 5) (x − 0)(x − 4) Expression

Zeros

TP

Factors

x2 − 4x

0, 4

(2, −4)

(x)(x − 4)

x2 − 6x + 5

1, 5

(3, −4)

(x − 1)(x − 5)

x − 8x + 12

2, 6

(4,

−4)

(x − 2)(x − 6)

x2 − 10x + 21

3, 7

(5, −4)

(x − 3)(x − 7)

x2 − 12x + 32

4, 8

(6, −4)

(x − 4)(x − 8)

x2 − 14x + 45

5, 9

(7, −4)

(x − 5)(x − 9)

x2 − 16x + 60

6, 10

(8, −4) (x − 6)(x − 10)

x2 − 18x + 77

7, 11

(9, −4) (x − 7)(x − 11)

2

d Let j and k be the x-values of the zeros

of each quadratic function.

(x-intercepts) of the quadratic function.

2

State the general function.

y = (x − j)(x − k)

3

Expand.

y = x2 − ( j + k)x + jk

e 1 When the parabola moves 3 units to the left,

3 is subtracted from both zeros (j and k).

f

b For x2 − 8x + 12,

e The zeros are now j − 3 and k − 3.

2

State the new function.

y = (x − ( j − 3))(x − (k − 3))

3

Expand.

y = x2 − ( j + k − 6)x + ( j − 3)(k − 3)

1

Given j = −5 and k = 2, state the quadratic function.

2

Translate this function 3 units to the left by subtracting 3 from j and k.

f y = (x + 5)(x − 2)

y = x2 + 3x − 10

j − 3 = −5 − 3 = −8 k − 3 = 2 − 3 = −1 Chapter 2 • Quadratic functions

81

3

State the new function.

4

Sketch the graphs of y = x2 + 3x − 10 and y = x2 + 9x + 8 on the same set of axes.

y = (x + 8)(x + 1) y = x2 + 9x + 8 y

−8

−5

(TP)

5

Observe the features of the two parabolas.

−1

y = x2 + 9x + 8 y = x2 + 3x − 10

2

x

(TP)

The turning points and zeros have been translated 3 units to the left. The second parabola has been translated 3 units to the left from the first parabola. The answers are verified.

Using technology to solve quadratic equations exercise 2k

1 Use a CAS calculator to find all solutions of each of the following. a x2 + 4x + 1 = 0 b x2 = 11x + 2 − 2 c x + 4x = 2 d x2 = 8x − 8 2 e x + 12x + 9 = 0 f −3x2 + 5x − 1 = 0

diGital doC doc-9722 Simultaneous, quadratic and linear equations

2 Find all roots of the quadratics below using a CAS calculator. a 2.3x2 + 0.7x − 0.59 = 0 b −0.811x2 − 5.2x − 3.1 = 0 c −3.97x2 + 17x + 8.05 = 0 d 5.18x2 = 2.66x + 9 e 0.006x2 + 0.923x + 0.361 = 0 f x2 + 500x = 47 3 The distance, d, of a comet from one of the moons of Jupiter is given by the equation

d = 47.9t2 + 0.03t − 908.7, where t is the number of hours since the comet was first discovered on 28 June 2001. At what value of t will the comet reach this moon?

4 The number of marine organisms, N, in a marine research organisation’s testing tank is found to follow

the equation (or model) N = −0.0751h2 + 0.69h + 200, where h is the number of hours since the tank was supplied with nutrient and stocked with 200 organisms. How long after being fed could the colony survive without further food before none were left?

82

Maths Quest 11 Mathematical Methods CAS

5 A diver follows a parabolic path from the diving board to the water, given by the function below. What

is the horizontal distance travelled by the diver from leaving the diving board to entering the water? (h represents the height of the diver above the water for a distance, d, from the diving board in the equation shown. Both h and d are in metres.)

h = −0.5d2 + 2d + 5

6 We29 As part of a Year 11 Maths extension activity, Harry and Christine were each asked by their

teacher to come up with a quadratic expression whose zeros had a sum of 10.25 (or as close to that as possible). The coefficient of the linear term had to be −5. When they returned to class the next day, Harry announced his expression was 0.48x2 − 5x − 2.3, and Christine said hers was 0.49x2 − 5x + 1.9. Who was closest, and by how much? 7 One of the solutions of the equation 10x2 + 11x = k (where k is a constant) is −1.6. a Find the value of k. b Find the other solution to the original quadratic equation.

Simultaneous quadratic and linear equations 2l

In previous studies you have dealt with pairs of simultaneous linear equations and solved these using algebra. The solution could also be represented graphically. The same is true when we have one linear and one quadratic equation as a pair of simultaneous equations. Consider the following pair of simultaneous equations: y = x2 + x − 2 and y = 3x + 1 If x = 3 is substituted into the first equation, y = 32 + 3 − 2 = 10 is obtained. If x = 3 is substituted into the second equation, y = 3 × 3 + 1 = 10 is obtained. That is, the coordinate pair (3, 10) ‘fits’ both equations, so it is a solution. If x = −1 is substituted into the first equation, y = (−1)2 + (−1) − 2 = −2 is obtained. If x = −1 is substituted into the second equation, y = 3(−1) + 1 = −2 is obtained. That is, the coordinate pair (−1, −2) is also a solution. To illustrate the situation graphically, the related linear and quadratic functions must be sketched on the same set of axes. As shown below, there are three possible relationships: • no points in common (hence, no solutions) • one point in common (hence, one solution) • two points in common (hence, two solutions).

interaCtiVitY int-0261 Simultaneous quadratic and linear equations

Chapter 2 • Quadratic functions

83

y

y

x

y

One solution

No solution

x x

Two solutions

When a linear equation and a quadratic equation are solved simultaneously, a new quadratic equation is formed, as you will see in the following examples. The number of solutions relates to the discriminant (positive = 2 solutions; negative = no solutions; zero = one solution) of the ‘new’ quadratic. Worked Example 30

a Solve the system of equations y = x2 + x − 2 and y = 3x + 1. b Illustrate the solution using a sketch graph. The turning point of the quadratic graph is not

required.

Think

a 1 Write the original equations and label them.

Write/draw

a y = x2 + x − 2

[1] [2]

y = 3x + 1

2

Equate [1] and [2]. Put RHS [1] = RHS [2].

3

Collect terms on the side that makes the x2 term positive.

4

Combine like terms.

x2 − 2x − 3 = 0

5

Factorise if possible.

(x − 3)(x + 1) = 0

6

Solve for x.

7

Substitute x-values into the linear equation [2] to find the corresponding y-values.

If x = 3, y = 3(3) + 1 = 10 If x = −1, y = 3(−1) + 1 = −2

8

Write the solution coordinates.

Solutions: (3, 10) and (−1, −2)

b 1 Find intercepts for a sketch of the linear graph.

2

Find intercepts for a sketch of the quadratic graph. (The turning point is not required here.)

3

Sketch the linear and quadratic graphs on the same axes, and include the points of intersection found in part a.

x2 + x − 2 = 3x + 1 x2 + x − 2 − 3x − 1 = 0

x = 3 or x = −1

b For y = 3x + 1,

if x = 0, y = 1 if y = 0, 0 = 3x + 1 −1 = 3x −1 x = 3

For y = x2 + x − 2, if x = 0, y = −2 if y = 0, 0 = x2 + x − 2 so 0 = (x + 2)(x − 1) and x = −2 or x = 1 y (3, 10)

− 1–3 −2 (−1, −2)

84 Maths Quest 11 Mathematical Methods CAS

1 1 −2

x

Worked example 31

Solve the equations y = −0.5x2 − 4x + 2 and 11x + 2y = 6. think 1

Write the two equations and label them.

2

Solve [2] for y.

Write

y = −0.5x2 − 4x + 2 11x + 2y = 6

[1] [2]

2y = −11x + 6 y = −5.5x + 3 −0.5x2

[3]

− 4x + 2 = −5.5x + 3

3

Set [1] = [3].

4

Simplify.

5

Multiply by −2.

6

Factorise and state the solutions.

7

Find the corresponding y-values and substitute in [3].

When x = 2, y = −5.5(2) + 3 = −8 When x = 1, y = −5.5 + 3 = −2.5

8

Write the coordinates of the two points of intersection.

The solutions are (2, −8) and (1, −2.5).

−0.5x2

+ 1.5x − 1 = 0

x2 − 3x + 2 = 0 (x − 2)(x − 1) = 0 x = 2 or x = 1

Worked example 32

a Solve the system of equations y = x2 − 8x + 12 and y = −6x + 11. b Illustrate the solution using a sketch graph. The turning point of the quadratic graph

is not required.

think

a 1 Write the original equations and label them.

Write/draW

a y = x2 − 8x + 12

[1] [2]

y = −6x + 11

x2 − 8x + 12 = −6x + 11

2

Put RHS [1] = RHS [2].

3

Collect terms on the side that makes the x2 term positive.

4

Combine like terms.

x2 − 2x + 1 = 0

5

Factorise if possible.

(x − 1)(x − 1) = 0

6

Solve for x.

7

Substitute x-values into the linear equation [2] to find the corresponding y-value.

If x = 1,

8

Write the solution coordinates.

Solution: (1, 5) only

b 1 Find intercepts for a sketch of the linear graph.

x2 − 8x + 12 + 6x − 11 = 0

x = 1 only y = −6x + 11 y = −6(1) + 11 y = −6 + 11 y=5

b For y = −6x + 11,

if x = 0, if y = 0,

y = 11 0 = −6x + 11 6x = 11 5 x = 11 6 or 1 6 Chapter 2 • Quadratic functions

85

2

Find intercepts for a sketch of the quadratic graph. (The turning point is not required here.)

3

Sketch the linear and quadratic graphs on the same axes, and include the points of intersection found in part a.

For y = x2 − 8x + 12, if x = 0, y = 12 if y = 0, 0 = x2 − 8x + 12 so 0 = (x − 6)(x − 2) and x = 6 or x = 2 y 12 11

Tangent point (1, 5) 11 — 6

2

x

6

Worked example 33

a Solve the system of equations y = −x2 + 3x + 18 and y = 4x + 22. b Illustrate the solution using a sketch graph. The turning point of the quadratic graph is not

required.

think

a 1 Write the original equations and label them.

Write/draW

a y = −x2 + 3x + 18 −x2

2

Put RHS [1] = RHS [2].

3

Collect terms on the side that makes the x2 term positive.

4

Combine like terms.

5

Try to factorise. No solution using the Null Factor Law is apparent. Check the value of the discriminant Δ of the quadratic in step 4.

(x )(x ) = 0? Δ = b2 − 4ac = (1)2 − 4(1)(4) = 1 − 16 = −15

6

Since Δ < 0, there is no solution.

Δ < 0, ∴ no solution; the graphs don’t meet.

b 1 Find intercepts for a sketch of the linear graph.

+ 3x + 18 = 4x + 22 0 = x2 − 3x − 18 + 4x + 22 x2 + x + 4 = 0

b For y = 4x + 22,

if x = 0, if y = 0,

y = 22 0 = 4x + 22 −22 = 4x x= x=

2

86

[1] [2]

y = 4x + 22

Find intercepts for a sketch of the quadratic graph. (The turning point is not required here.)

Maths Quest 11 Mathematical Methods CAS

− 22 4 − 11 2

1

or −5 2

For y = −x2 + 3x + 18, if x = 0, y = 18 if y = 0, 0 = −x2 + 3x + 18 x2 − 3x − 18 = 0 so (x − 6)(x + 3) = 0 and x = 6 or x = −3

3

y

Sketch the linear and quadratic graphs on the same axes. Note that the graphs do not intersect, indicating no solution.

22 18

— − 11 2

−3

6

x

Worked example 34

The graphs with equations y = x2 + 4 x + 33 and y = mx + 24 intersect once only. Find the possible values of m. think

Write

1

Write and label the equations.

y = x2 + 4x + 33 y = mx + 24

2

Put equation [1] = equation [2] and form a new quadratic equation [3].

x2 + 4x + 33 = mx + 24 + 4x − mx + 9 = 0 x2 + (4 − m)x + 9 = 0 [3]

3

For one solution only, we require Δ = 0.

Δ = (4 − m)2 − 4(1)(9) = 16 − 8m + m2 − 36 = m2 − 8m − 20 = 0 for one solution only.

4

Factorise and solve for m.

(m − 10)(m + 2) = 0 m = 10 or m = −2

5

State the rules for the two straight lines.

The two lines are y = 10x + 24 and y = −2x + 24.

[1] [2]

x2

Simultaneous quadratic and linear equations exercise 2l

1 We30, 31, 32, 33 For each of the following systems of equations: i solve to find any solution coordinates ii illustrate the solution (or lack of solution) using a sketch graph. The turning point of the quadratic

graph is not required. You may use a CAS calculator to verify solutions. a y = x2 + 6x + 5 and y = 11x − 1 c y = x2 + 9x + 14 and y = 3x + 5 e y = x2 − 2x − 3 and y = x − 6 g y = x2 + 5x − 36 and y = 15x − 61 i y = x2 − 2x − 24 and y = 4x + 3 k y = −x2 + 4x + 21 and y = x + 11 m y = −x2 + 4x + 12 and y = 9x + 16 o y = −x2 − 4x + 5 and y = −4x + 9

b y = x2 + 5x − 6 and y = 8x − 8 d y = x2 − 7x + 10 and y = −11x + 6 f h j l n p

y = x2 + 11x + 28 and y = 10x + 40 y = x2 − 6x − 16 and y = −4x − 17 y = x2 − 7x + 10 and y = −4x + 6 y = −x2 + 14x − 48 and y = 13x − 54 y = x2 + 7x + 12 and y = 20 y = x2 − 4x + 4 and y = 8x − 32

2 State how many points of intersection exist with each of the following pairs of simultaneous equations. a y = x2 + x − 6 and y = −9x − 31 b y = x2 − 16 and y = 6x + 11 c y = −x2 + 3x + 4 and y = −7x + 25 d y = x2 − 6x + 5 and y = 2x − 12 Chapter 2 • Quadratic functions

87

3 WE34 The graphs of equations y = x2 − 14x + 49 and y = mx + 48 intersect once only, at x = 1. Find

the value of m. 4 The system of equations y = x2 − 4x + c and y = −7x + 8 has two solutions, one at x = 1 and another at x = −4. Find the value of c. 5 The graphs of y = x2 + bx − 14 and y = −9x + c intersect at (−1, −8) and (−3, 10). Find the values of b and c. 6 Using a CAS calculator, find the points of intersection, in exact form, for each of the following systems of equations. a y = x2 + 5x − 3 and y = 2x − 2 b y = 2x2 + 7x − 8 and y = 3x + 4 c y = −3x2 − 2x + 12 and y = −4x + 8 d y = −5x2 − 2x − 5 and y = 43 x − 5 y

7 An engineer’s plans for a proposed road through a

mountain are shown at right. At what heights above sea level will the entrance and exit to the tunnel be, given the equations of the mountain profile and road path as shown on the plan?

y=−

x–2 +5 2

y = x–4 +2 Proposed road

Entrance of tunnel

x

8 A graphic designer draws a logo involving a parabola ‘sitting’ in a

y

V shape on a set of axes as shown at right. Find the equation of the parabola, given it is of the form y = kx2, and find the points of intersection of the V with the parabola. −1 −2 9 Use a CAS calculator to find the values of a in exact form such that the linear equation y =

and the quadratic equation y =

88 Maths Quest 11 Mathematical Methods CAS

−3 8

x 2 + 3 x − 5 do not intersect.

Sea level

y = kx 2

1

ax +a 2

x

10 Use a CAS calculator to help answer the following questions.

A pole 11 metres high is firmly secured to the ground. A parabolic arch is to be attached to the pole 3 metres above the ground as shown in the diagram below. A metal rod will run directly from the top of the pole straight to the ground so that it just touches the arch in one position only. Let y be the height above the ground in metres and x be the distance along the ground from the foot of the pole in metres. y (0, 11)

(0, 3) 0

A

B

x

Give all answers in exact form unless told otherwise. −1 a a The arch is to be modelled by the quadratic equation y = 8 x 2 + 4 x + 3, where a is a non-zero positive constant. Find where the arch meets the ground in terms of a, that is, find the x-intercept. b If it is decided that the distance from the foot of the pole to the point A is 12 metres, find the value of the constant a. c If the equation of the metal rod is modelled by the linear function y = 11 + bx, find the value of b. (Hint: Consider the discriminant.) d Find the coordinates of the point where the arch touches the metal rod. e Find the maximum height of the arch correct to 2 decimal places. f Find the distance from point A to point B both in exact form and correct to 2 decimal places.

Chapter 2 • Quadratic functions 89

Summary polynomials

1. A polynomial in x, sometimes denoted by P(x), is an expression containing only non-negative whole powers of x. 2. The degree of the polynomial is given by the highest power of the variable x. 3. The general form of a degree n polynomial is: P(x) = an x n + an − 1xn − 1 + . . . + a2 x 2 + a1x + a0 where n is a positive whole number and an, an − 1, . . ., a2, a1 and a0 are coefficients.

expanding quadratic expressions

• To expand: 2 1. First term × everything in the second brackets, then 1 2. Second term × everything in the second brackets. (4x + 9)(2x − 3) • Perfect squares (ax + b)2 = a2x2 + 2abx + b2 3 4 • Difference of squares (ax + b)(ax − b) = a2x2 − b2 • Expand brackets first, then multiply if there is an ‘external’ factor in expressions like k(ax + b)(cx + d).

Factorising quadratic expressions

• • • •

Look for a common factor first. General quadratics: Write down (_x )(_x ) and try factors of the constant term. Perfect squares: (ax + b)2 = a2x2 + 2abx + b2 Difference of squares: ( a x + b )( a x − b ) = ax 2 − b

Factorising by completing the square

• • • •

Use when whole number factors are not apparent. Halve and square the x-coefficient, then add and subtract this new term. Form a perfect square from three of the terms. Continue to factorise using a difference of squares.

Solving quadratic equations — null Factor law

• Factorise. • Set each factor equal to zero. • Solve two mini-equations.

Solving quadratic equations — completing the square

• Halve and square the x-coefficient. • Add and subtract the new term. • Form a perfect square and solve for x.

the quadratic formula

• If ax2 + bx + c = 0, then solutions may be found using x =

the discriminant

• • • •

If ax2 + bx + c = 0, then the discriminant Δ = b2 − 4ac. If Δ < 0, no real solutions exist. If Δ > 0, there are two real solutions. If Δ = 0, there is only one real solution.

Graphs of quadratic functions as power functions (turning point form)

• • • • •

Turning point form: y = a(x − b)2 + c. The turning point is at (b, c) and the dilation factor is a. If a > 0, the graph is a positive (∪) shape. If a < 0, the graph is a negative (∩) shape. To convert to turning point form, complete the square.

90

Maths Quest 11 Mathematical Methods CAS

−b ±

b 2 − 4 ac . 2a

y

(b, c) x

Graphs of quadratic functions (intercepts method)

• To sketch a quadratic graph of the form y = ax2 + bx + c: 1. Find the y-intercept (when x = 0) 2. Factorise if possible, and find the x-intercepts (when y = 0) It may be easier to use the quadratic y formula in some cases. If Δ < 0, there are no x-intercepts. If Δ = 0, one intercept only. y-intercept If Δ > 0, two intercepts. 3. Find the x-coordinate of the turning point −b using xt = or by completing the 2a square.

Axis of symmetry x-intercepts x

Turning point b b2 — — (− 2a , c − 4a )

4. Find the y-coordinate of the turning point by substituting xt into the equation for y, or by using b2 or by completing the square. yt = c − 4a 5. Combine all the information and sketch it. Functions with a positive x2 coefficient are ∪ shaped, and those with negative x2 coefficients are ∩ shaped. If the information you have gathered doesn’t seem to fit, check for calculation errors. Simultaneous quadratic and linear equations

• If given equations of the form y = ax2 + bx + c and y = mx + k: 1. Set ax2 + bx + c = mx + k. 2. Rearrange to form a new quadratic equation Ax2 + Bx + C = 0. 3. Solve to find any x-coordinates of intersection. 4. Substitute any x-coordinates into the linear equation to find the corresponding y-coordinates. 5. State the solutions (or state that there are none).

Chapter 2 • Quadratic functions

91

Chapter review S h ort anS Wer

1 Expand: a (7x + 8)(7x − 8) 2 Factorise: a 25x2 + 110x + 121

b (2x − 9)2

c (5 x − 3)(5 x + 3)

d (4x − 9)(2x − 13)

b 6x2 + 37x + 6

c 12x2 − 37x + 21

d 36x2 − 49

3 Factorise x2 − 6x − 14. 4 Solve the following. a (4x + 1)(3x − 9) = 0

b 3x2 − 40x − 75 = 0

5 Solve: a 4x2 − 5 = 0

b x2 − 14x + 42 = 0

6 Solve x2 + 6x − 2 = 0. 7 Use the quadratic formula to solve 5x2 − 7x + 1 = 0. 8 Evaluate the discriminant for 4x2 − 3x + 9 = 0. 9 Find the value(s) of k for which the equation 2x2 + 3kx + 6 = 0 has: a no solution b one solution c two solutions. 10 Sketch the following, showing the turning point and y-intercept in each case (x-intercepts not required). a y = −4(x + 1)2 + 2 b y = 6x2 − 1 11 Convert y = x2 + 8x + 3 to turning point form. 12 Sketch the following. a y = (x − 6)(x − 14)

b y = x2 + 2x − 80

c y = 3x2 − 26x + 48

d y = −2x2 − 5x − 3

13 Find any points of intersection of the line y = −2x − 6 and the parabola y = x2 + 7x + 12. 14 Solve 4kx2 − 5x + 3k = 0 for x. m U lt ip l e C h oiCe

1 Which of the following is the expansion of (3x − 7)(x + 5)? a 3x2 + 8x − 35 d 4x2 − 2x − 2

B 3x2 − 7x − 35 e 4x − 2

2 Which of the following is a perfect square quadratic expression? a x2 + 10x + 16 d (x + 6)(x − 6)

B x2 − 8x + 16 e (x2 + 7)2

3 The expression 9x2 − 64 is an example of: a a perfect square d a quadratic term

B a difference of squares e a factorised expression

4 Which of the following is equivalent to 36x2 − 49? a (6x + 7)2 d (36x + 1)(x − 49)

B (6x − 7)2 e (9x + 7)(4x − 7)

5 The expression (x − 2)2 + 8 can be factorised to:

C 3x2 − 2x − 35

C x2 − 25

C a negative quadratic

C (6x + 7)(6x − 7)

a (x + 6)2

B (x + 6)(x − 10)

d ( x − 2 + 2 2)( x − 2 + 2 2)

e The expression cannot be factorised using real numbers.

C (x − 2 + 4)(x − 2 − 4)

6 Which of the following gives all solutions of the equation (2x − 7)(x + 4) = 0? a x = 7, −4 d x=

−7 2

B x = −2, −1

,4

e

x=

C x = 2, 1

7 − , 4 2

7 Which of the following represent the full solution to x2 − 6x + 9 = 0? a x = −3 d x = −3, −6

B x = −3, +3 e No real solution

C x=3

B x = 5+

C x = 7+ 5

8 A solution to (x − 5)2 − 7 = 0 is: a x = −5− d x=

92

−7+

7 5

Maths Quest 11 Mathematical Methods CAS

e

x=

−5+

7 7

9 If the equation 3x2 = 9 + x is to be solved using the quadratic formula, which set of values should be

substituted into the formula? a a = 3, b = −1 and c = 0 1 d a = 1, b = 3 and c = 9

B a = 3, b = −1 and c = −9 e a = 9, b = 1 and c =

C a = 3, b = 9 and c = 1

−3

10 The quadratic formula, which can be used to solve equations of the type ax2 + bx + c = 0, is given by: a x= d x=

−b ±

− b2

b 2 − 4 ac 2a

± b − 4 ac 2a

B x=

e

x=

b ± b 2 − 4 ac 2a −b

C x = −b ±

b 2 − 4 ac 2a

± b − 4 ac 2a

11 The value of the discriminant in 2x2 + 3x + 4 = 0 is: a −23 d 32

B −7 e 41

C 25

12 For a quadratic equation to have at least one real solution, the discriminant could be: a negative d either positive or negative

B zero e a perfect square

13 The turning point of the graph of y = 5(x + 2)2 − 1 is at: a (1, 2) d (−2, 1)

B (1, −2) e (−2, −1)

C positive

C (2, −1)

14 If y = −3(x − 2)2 + 8, the maximum value of y is: a −3 C 2 e 20

B −4 d 8

y

15 The graph shown could be for the rule: a B C d e

y = + 5x − 30 y = x2 − 11x + 30 y = x2 + 11x + 30 y = −x2 + 11x + 30 y = −x2 + 11x − 30 x2

30 −6

−5

x

16 The turning point of y = x2 + bx + 40 is at x = −7. The value of b is: a −14

B −7

7 2

d 7

C

e 14

17 Using a CAS calculator, the approximate solutions to y = −x2 + x + 4 are: a B C d e

x = 0.5 and y = 4.25 x = 0 and y = 4 x = 2.56 and x = 0 x = −1.56 and x = 2.56 x = 1.56 and x = 0

18 The system of 2 simultaneous equations represented by the graphs

at right has: a no solutions for x B one solution for x C one positive and one negative solution for x d two negative solutions for x e insufficient information for us to find a solution

y

x

19 A solution to the system of equations y = 2x − 1 and y = x2 + 8x + 8 is: a (−3, −7) C (0, 8) e (2, 28)

B (0, −1) d (3, 5)

Chapter 2 • Quadratic functions

93

e x tended r e Sp onS e

1 A 100 m length of steel cable is threaded through a series of posts in order to construct

w

a fence around a paddock with four straight sides as shown at right. a Write an equation that links l and w. b Rearrange the equation in part a to write an expression for l in terms of w. l c Write an expression for the area of the paddock in terms of w. d Plot a graph of area against w. Plot area on the vertical axis and w on the horizontal axis. On the graph, label all intercepts and the turning point. e What is the maximum area of the paddock? f What values of l and w give the maximum area of the paddock? g Comment on the relationship between l and w, and state what type of shape the paddock is. 2 MacBurger’s restaurants have employed a mathematician to y design a new logo based on an ‘M’ made up of two parabolas Parabola 1 Parabola 2 as shown in the sketch at right. The parabolas both have the form y = −2x2 + bx + c. h a Give the coordinates of two points on each parabola. b Find the equation of each parabola. (−2, 0) (2, 0) x c Find the height, h, of the sign. d Which domain of x-values should each graph be restricted to so that the parabolas form the logo shown above? e Determine the equations of two parabolas that may be used to form the ‘W’ shape below, given they are both of the form y = x2 + bx + c. y 9

(−3, 0)

x

(3, 0)

3 Find the equation of a parabola that goes through the points listed in each case.

(Hint: Assume equations of the form y = ax2 + bx + c and form simultaneous equations by substituting coordinate values.) a (0, 1), (1, 0) and (2, 3) b (0, −1), (1, 4) and (2, 15) c (0, 5), (−1, 11) and (1, −3) d Find the equations of two parabolas that form a path similar to the one below depicting a proposed water slide, given the x2 coefficient of each curve is 1 or −1. y (8, 8) (4, 4) x

4 a Using written algebra, sketch the graphs of the quadratic equations y = x2 − 6x + 8 and diGital doC doc-9723 Test Yourself Chapter 2

94

y = 3x2 + 5x − 28, showing intercepts and turning points. b Why was one equation easier to sketch than the other? c Determine another quadratic function whose graph has integer intercepts and turning point coordinates. How can this be ensured?

Maths Quest 11 Mathematical Methods CAS

ICT activities Chapter opener diGital doC • 10 Quick Questions doc-9710: Warm up with ten quick questions on quadratic functions (page 39)

2a

polynomials

diGital doC • History of mathematics doc-9711: Learn about the life of Galois, a 19th century mathematician (page 40)

2B

expanding quadratic expressions

diGital doCS • SkillSHEET 2.1 doc-9712: Practise expanding perfect squares (page 43) • SkillSHEET 2.2 doc-9713: Practise expanding differences of squares (page 43)

2e

Solving quadratic equations — null Factor law

tUtorial • We 10 eles-1407: Watch how to apply quadratic algebra skills to determine the dimensions of a triangle (page 51) diGital doCS • doc-9714: Apply the Null Factor Law to solve quadratic equations (page 53) • WorkSHEET 2.1 doc-9715: Expanding brackets and factorising quadratic expressions (page 54)

2F Solving quadratic equations — completing the square tUtorial • We 12 eles-1408: Watch how to solve a quadratic equation giving solutions in exact form (page 55) diGital doCS • SkillSHEET 2.3 doc-9716: Practise solving equations in the complete square form (page 57) • SkillSHEET 2.4 doc-9717: Practise simplifying surds (page 57)

2h

the discriminant

tUtorial • We 19 eles-1409: Watch how to find the possible values of an unknown constant in a quadratic equation with zero, one and two solutions (page 63)

diGital doC • doc-9718: Observe the discriminant and the number of solutions to a quadratic equation (page 66)

2i Graphs of quadratic functions as power functions (turning point form) diGital doC • WorkSHEET 2.2 doc-9719: Solve quadratic equations, understanding the discriminant and sketching parabolas (page 70)

2J Graphs of quadratic functions (intercepts method) diGital doCS • doc-9720: Observe axial intercepts of the graph of a quadratic in general form (page 78) • doc-9721: Observe the turning point of the graph of a quadratic in turning point form (page 78)

2k Using technology to solve quadratic equations diGital doC • doc-9722: Calculate the solutions to linear and quadratic simultaneous equations (page 82)

2l Simultaneous quadratic and linear equations interaCtiVitY • Simultaneous quadratic and linear equations int-0261: Consolidate your understanding of how to solve simultaneous quadratic and linear equations (page 83)

Chapter review diGital doC • Test Yourself doc-9723: Take the end-of-chapter test to test your progress (page 94)

To access eBookPLUS activities, log on to www.jacplus.com.au

Chapter 2 • Quadratic functions

95

Answers CHAPTER 2 QUadratiC FUnCtionS exercise 2a

1 a d 2 a d 3 a b c 4 a d 5 a c e 6 a b c 7 a b c d

polynomials 3 b1 c 2 6 e 5 x bt c x x e u Polynomial 1b Polynomial 1c Polynomial 1a 5 N; bP c N; −3 x x − N; k 2 e N; 2x 3 bx −3 d5 Coefficient of the quadratic term 1 − 19 −2x2 + x + 2 6 t 0.9937; she is swimming 0.9937 m/s at 0.2 s. 1.818 m/s

exercise 2B expanding quadratic expressions 1 a 2x2 + 16x + 30 b 12x2 − 5x − 3 2 c 25x − 25x − 14 d 48x2 − 26x + 3 e 7x2 + 19x − 36 f x2 + 21x + 90 g −18x2 + 51x − 26 h x2 − 11x + 30 i 9x2 − 82x + 9 j 4x2 + 9x − 63 2 a 4x2 + 12x + 9 b 9x2 − 30x + 25 c 36x2 + 12x + 1 d 49x2 − 84x + 36 2 e x − 16x + 64 f x2 + 26x + 169 g 4x2 − 36x + 81 h 4x2 + 36x + 81 2 i 16 − 24x + 9x j 36 − 12x + x2 2 3 a 4x − 36 b 9x2 − 25 c 36x2 − 1 d 4x2 − 81 2 e 121x − 9 f x2 − 144 g x2 − 36 h 49 − 4x2 i 1 − x2 j 25x2 − 1 2 4 a 6x + 46x + 60 b 6x2 − 3x − 84 c −8x2 + 4x + 40 d 24x2 − 66x + 27 e 96x2 − 56x + 8 f −70x2 + 35x + 210 2 g 4x − 44x + 112 h 35x2 − 90x + 40 i 2ax2 − 3ax − 54a j 2bx2 − 32b 5 a 36x2 − 169 b 60x2 − x − 10 c −6x2 + 27x + 168 d 9x2 + 66x + 121 e 4x2 − 196 f 36x2 − 60x + 25 2 g x + 7x − 144 h −x2 + 16 i −100x2 + 120x − 36

j

20x2 − 245

2x2 − 7x − 34 12x2 + 25x − 55 4x2 − 24x − 62 g x2 − 3

6 a c e i 7 a b c d

96

4x2

+ 9x +

7 2

18x2 − 14x + 5 −2x2 − 18x − 159 23x2 − 114 h 12x2 + 2 x − 2 5 j 15x2 − 24x − b d f

4a2 + 8a + 2ab + 4b 3x2 − 10y2 + xy 49c2 − 56c + 12 u2 + 8uv + 16v2

3

e f g h

12r2 + 28rs − 5s2 9u2 − 4t2 2h2 − 128k2 −3m2 + 36mn − 108n2

exercise 2C

Factorising quadratic

expressions 1 a 5xy2(xy + 4) b −2ax(4x − 7) 2 c 7(p q − 3p + 1) d 11r2s(2r2s2 + 1) e 5(x2 + 2x + 15) f −3(x2 + 6x − 2) 2 a (x + 9)(x + 7) b (x − 11)(x − 6) c 2(x + 3)(x + 8) d −3(x − 2)(x + 4) e (8 − x)(x + 12) f (3x + 5)(2x − 7) g (2x − 9)(3x − 5) h −2(4x + 1)(5x − 7) 3 a (2x + 3)(2x − 3) b (9x + 5)(9x − 5) c 2(x + 4)(x − 4) d 18(x + 3)(x − 3) e (x + 3)(x − 1) f 3(x − 1)(3x − 5) g 2(x + 10)(4 − x) h −15(x − 1)(5x − 11) 4 a (x + 7)2 b (5x − 1)2 c 2(x + 3)2 d 3(x − 2)2 e −2(6x − 1)2 f (x + 3)2 2 g (2x − 5) h (2x − 3 2)2 5 a x(x + 6) b (x + 1)(x − 7) c −(2x + 1)(2x + 13) d (3x + 7)(3x − 11) e 2(x − 1)(x + 3) f 3x(8 − x) g 6(5x + 1)(15x + 17) h −(x − 5)(x − 7) i 3(x + 2)(10 − 3x) 6 a (x + 11)(x + 12) b (3x − 4)(2x − 5) c 2(2x − 1)(x + 3) d (9x − 31)(12x − 53) 7 a 3(x − 9)(x + 1) b (5x + 1)2 c (x − 14)2 d (x − 14)2 e 3 x ( 2 x − 2) f 5(2x + 1)(6x + 1) g 5x(12x − 1) h 9(1 − xy)(1 + xy) i 7x(5x − 4) j −2(6x + 7)(x − 7) 8 a C b E c A 9 k = 5, m = −4 exercise 2d

Factorising by completing

e

9 65 9 65 x + 2 + 2 x + 2 − 2

f

11 145 11 145 x + 2 + 2 x + 2 − 2

No real factors 5 17 5 17 x+ − h x+ + 2 2 2 2 g

i

1 13 1 13 x + 2 + 2 x + 2 − 2 3 E 4 E 5 B 6 C 7 a 3( x − 3 + 10)( x − 3 − 10) j

b c d

1 a

2 a

6, −12

b 4,

d

5, −11

e 3, −3

g

2 −2 , 5 5

h 0, 5

3 a

5, 9

b

d

d

( x + 1 + 2 2)( x + 1 − 2 2)

e

( x + 4 + 3)( x + 4 − 3)

f

( x − 2 + 5)( x − 2 − 5)

g

( x − 6 + 17)( x − 6 − 17)

h

( x − 1 + 6)( x − 1 − 6)

i

( x − 4 + 6)( x − 4 − 6)

j

( x − 3 + 13)( x − 3 − 13)

1 5 1 5 x − 2 + 2 x − 2 − 2

Maths Quest 11 Mathematical Methods CAS

6

−1

h 0, 1

j

d

1 b −1,

6 − , 6 7

( x + 3 + 2)( x + 3 − 2)

7 41 7 41 x − 2 + 2 x − 2 − 2

2

g

c

c

−3

−6,

e

g

5 37 5 37 x + 2 + 2 x + 2 − 2

− 1)(x + 3) or 5(1 − x)(x + 3) − x + 2)

−4(3x2

−12 − 2 , 7 9

( x + 5 + 5)( x + 5 − 5)

b

−5(x

d

b

3 5 3 5 x + 2 + 2 x + 2 − 2

5 17 5 17 2 x + + x+ − 2 2 2 2

exercise 2e Solving quadratic equations — null Factor law

the square 1 a ( x + 2 + 7)( x + 2 − 7)

2 a

13 173 13 173 x − 2 + 2 x − 2 − 2

1 −1 , 6 6 9 −9 4

,

4

e

2

,3

−1 2

−7 7 , 2 3 − −5, 1 3

h 0,

8 5

c 5, −2 f −7, 23

−5 c −2, 4

f

c f

1 4

−1 4 3 2

i 2,

−5 2

5 1 , 3 2

4 D 5 E 6 D 7 Length = 8 cm, width = 5 cm 8 a The area is doubled. b

Length = 20 2 + 33 + 1 cm Width = 20 2 + 33 − 1 cm

9 4 hours 10 a 30 11 a 42 b 16 12

9 2

b 13

or 4.5 kg

13 a 3.7 s b 2 5 14 a x = 1.998 using fixed point iteration.

(The correct answer is x = 2.)

b x = −0.562 15 a 0.228 b 1.140 c 0.268 d 0.863 16 Answers will vary. 17 Answers will vary. 18 Answers will vary.

exercise 2F Solving quadratic equations — completing the square

1 a

−4

c e

5± 7

2 a c e 3 a

b −6 ±

± 15

−2 ±

6

−3 ±

13

7 ± 29 2 − 11 ±

d 3±

5

f 3±

3

−5 ±

b

2

9 ± 89 2

f −2, 3

2 2± 5

b 3±

c

9 ± 73 2

e

3 ± 19

g

17

2

d

105

33

d

11

65

7±5 2

2 15 ± 201 f 2 h No solutions

4 a

± 14

b± 6

c

±2 3

d ±3

5 6 7 8

2

Solutions do not exist.

exercise 2G

the quadratic formula 1 a 1, 4, −3 b 1, −7, 9 − c 1, 4, 9 d −6, −3, 7 1 − e , 7, 5 f 1, 2, 1 2

4, −12, 9 7, −21, 4 −4, −5 8,

h j b

−3

−2,

0, 1 3, 0, 2 2, 8 −

−5

b

b − 4 ± 11

c

d − 5 ± 13 f 6, 1

9 113 ± 2 2

g

2± 6

h

i

3 17 ± 4 4

3 j ± 2 2

k l 4 a

x=

±

3k 2

+1 +1

k k + 1 ± k 2 − 30 k + 1 x= 4k −2 ±

7; k =

−1 2

8± 2 b ; k = 31 31

2 a d g 3 a b c d e f g h 4 D 5 a

1 4

d 3,

2

13 ± 2 2 −3 5 c ± 2 2 e 3± 7

3 a

exercise 2h

g

Solutions do not exist. 21 b x =2± 6 3 − c , 1

c

13

d

a

g i 2 a

12

1 a

E C D

d

7 8 9 10

7

−7 ±

6.317, −0.317 b 1.854, −4.854 d No real solutions 3.637, −0.137 f No real solutions 0.123, −8.123 1.472, −7.472 b 1.121, −3.121 − d 7.162, 0.838 0.225, −1.775 3.851, 0.649 f 7.772, −0.772 No real solutions h No real solutions 22.5 cm 11.3 hours 5.2 minutes a 1 metre b At t = 0.2 seconds c At t = 1.2 and 2.3 seconds a 1.2 m b She will need to buy an extra 5.25 m2 of pavers. c She will have 3 m2 of pavers left over. 0.711 m −2 2 a y= x + 4x 95 b 190 m c 190 m

5 a c e 6 a c e g

the discriminant 73 b 20 −71 e 37 229 h 193 0 b 2 1 e 1 2 h 2 a2 − 4 4 − 12a 36 − 4a b2 − 4a 4m2 − 4m m2 + 2m − 11 m2 + 4m + 16 k2 − 8k + 8 i iii i iii i iii

k > 4 or k < −4 44 k > 1 or k < −1 −1 < k < 1 −

6

Question

c 176 f −20 i c f i

k < 4.05 ii k > 4.05 e k<3 ii k>3 f k > 3 or ii k < −1.5 iii −1.5 < k < 3 ii g i No values of k iii No values of k h i k<1 ii iii k > 1 i i k < 3 or k > 3 ii iii No values of k 6 (See table bottom of page) d

v

60 0 2 1

k = ±4

ii

k=4

ii

k = ±1

Equation after substitution k = −1

k = 4.05 k=3 k = −1.5 or k = 3 All values of k k=1 k=3

exercise 2i Graphs of quadratic functions as power functions (turning point form) 1 a (5, 0) b (−7, 3) − d (1, 8) c (2, 7) f (2, 2) e (−3, −4) 2 a i (4, 2) ii Same iii Minimum y = 2 iv 18

−

ii

i iii i iii i

y

(0, 18)

(4, 2) x

b

−3)

i (2, iii Minimum y = −3 v y

ii Same iv 1

1 x

(2, −3)

Number of solutions

Reasoning Δ = k2 − 16 = −15

a

x2

−x+4=0

0

b

x2 − 4x − 1 = 0

2

Δ = 16 − 4k = 20

c

x2

− 4x + 4 = 0

1

Δ = 16k2 − 16 = 0

− 18x + 20 = 0

2

Δ = 324 − 80k = 404

x2 − 4x = 0

2

Δ = 12 − 4k = 16

− 4x + 2 = 0

0

Δ = 16k2 − 24k − 72 = −32

d e

−x2

f

6x2

g

−4x2

− 12x − 9 = 0

1

Δ = 144k2 − 144k2 =0

h

3x2 + 10x + 5 = 0

2

Δ = 20 − 20k = 40

2

Δ = k2 − 6k + 9 = 16

i

−2x2

+2=0

Chapter 2 • Quadratic functions

97

c

i ii iii iv v

(−5, −8) Same Minimum y = −8 17

h

i (3, 3) iii Maximum y = 3 y v

ii Thinner iv −60

y = (x − 2)2 + 4 y = −(x + 1)2 + 5 y = (x − 2)2 − 5 y = −(x + 3)2 + 1 y = −(x − 6)2 + 6 y = (x + 3)2 − 2 5 a 6 b −7 6 a (2, 5) minimum y = 5 b (3, 8) minimum y = 8 c (6, 1) minimum y = 1 d (−4, −3) minimum y = −3 e (2, −9) minimum y = −9 f (0, 7) minimum y = 7 g (−9, −81) minimum y = −81 h (3, 4) minimum y = 4 i (−2, 3) minimum y = 3 1 2 7 a i( , ) ii (0.33, 0.67) 3 3 4 a b c d e f

(3, 3)

x

y 17 −60

i

x (−5, −8)

d

i ii iii iv v

(1, −1) Same Minimum y = −1 0 y

−73

j

(1, −1)

i ii iii iv v

ii Same iv −73

x

(8, −9)

x

e

i (8, −9) iii Maximum y = −9 y v

(−5, −9) Thinner Minimum y = −9 41

i ii iii iv v

8 9 10 11 12

(1, 20) Same Minimum y = 20 21 y

y

(1, 20)

x

3 a

y

y

b

− − i ( 5 , 7) 4 8

ii (−1.25, −0.875)

c

− i ( 1 , 35 ) 4 8

ii (0.25, −4.375)

y = −(x − 5)2 + 9 C E B b n a −h d M e 2Mh

3 a d x

g

(−5, −9)

f

i (4, 4)

iv

(−1, 3)

28 3

x

x

c

d

y

c x

28 — 3

i

d 5 a

x

x

g

b

y (6, 0)

(4, 4)

(−1, −12)

iii Minimum y = −12 y v

iv

− 54 5

e

f

y

−1, 6 5 −a −b

e

7 2

±

−9 4

±

5

98

Maths Quest 11 Mathematical Methods CAS

2

,

4

73 4

No x-intercepts y

b

x

y

(7, −7)

— − 54

(−1, −12)

1 3 − 3 −1

21 2

1 3 (2, −1)

y

x

x

b , −2

3

(0, −4)

ii Thinner

5 − , 1 2

, A B 4 a No x-intercepts

(2, 5)

ii Wider

iii Minimum y = 4 v y

c M f Mh2 + n

exercise 2J Graphs of quadratic functions (intercepts method) 1 a 2 b −4 c 0 e 2 f 1 d 6 2 a 1, 6 b 3, −2 c −5, −1 d 4, 5 e 0, 2 f 3, 4 g −5, 5 h 0, −8 i −9 j 2, −10

21

41

b

1

(0, 2) x

(−1, 0)

x

c 9,

9 4

f 0, −4

y

c

y

i

d

8

y

3 −1

x

12

−12 −4 −2 (−3, −1)

1– 2

x — , − 169 —– ) (11 2

4

y

j

y

d

x

3– 2

(1, −1)

e

y 3

35

49

k

1– 2

x

(−7, 0)

x

3– 4

( 5– , − 1– ) 8

8

y

f

−7 −5 (−6, −1)

y 3– 7

x −3

64

x

−9

e

y x

(8, 0) 12

–— ) (− 9– , − 144

y

l

7

−17 2

x

9

x

6

7

y

g

(4, −4) 9

f

y

−153

(−4, −169)

−9

x

7

y

6 a

− 1– 5

−3

x

–— ) (− 8– , − 147 5

(−1, −64)

−63

−2

y

g

4

c

y 6

—) ( 1– , − 64 9

i

—– ) (− 9– , − 121 5

h

−7

x

y

y

3

−4 3

2

4

7– 2

x

x

−15 3 ( 5– , − 1– )

9

5

− 5–

2

x

1

9

−8

x

−1

− 7–

2– 5

−4

(− 3– , − 1– ) 2

−8

b

y

h

y

2 −2

x

4– 3

—) (− 1– , − 25 3 3

5

−28

x —) ( 2– , − 49 3 3

–—) (− 1– , − 225 4

8

Chapter 2 • Quadratic functions

99

y

j

y

e

y

f

( 3– , 4)

12

2

2 1– 2

−5

x

5– 2

− 2– 3

−1

x

−2

f

1 ) (− 5– , − — 6 12

—) (− 3– , 49 8

8

y

g 5

y

k

x

2

y

2

4

−5

x

0

—) (− 5– , − 25 2 4

y

8 a

5 x

−5 −9

x

1– 2

− 5–

3 x

− 3–

y

h

—) ( 3– , − 81 8

4

7 a

(−4, 49)

0

x

8

y (4, −16)

−25 33

i

y

b −11 −11

3

y

11 x 0

x

x

2

(1, −5) y

b

(1, 4)

−1

3

j

−121

3

y

c

x

y

(−3, 36)

1 −6

1 x

−1

x

0

y

c

k

(−9, 36) −15

d

x

−3

y

–—) ( 7– , 147 2

y

4

3

−45 3 x

− 3

0

d

y

y

e

(9, 0) x

−3

−81

100

Maths Quest 11 Mathematical Methods CAS

y

9 a 3

−18

x

x

7

−3.828

1.828

−7 (−1, −8)

x

y

b

−0.479

1

−0.268 −3.732

y

i

exercise 2l Simultaneous quadratic and linear equations

3.479 x

y

1 a

(3, 32)

−5

x

(2, 21) —) ( 3– , − 47

(−2, −3)

2

y

c

j −0.395

5

y

1 — 11

−1

−3 −2

(0, 17)

3 −7.606

4

x

x

y

b

(2, 8) 2.380 x

−2.380

k

(−4, −13)

d

−6

y

−6

y

—) ( 1– , 13 4 4

5– 4

0

3

−0.651

1.151 x y

(−9, 80)

−17.944

−0.056

−1 x

10 D 12 a 2 b 2 c 1 d 0 e 2 f 1 13 a C b A c D d B 14

f

25 8

x

−8

x

—) ( 5– , − 25 8

e

(1, 0)

y

c

16

11 B

14 5

−2

−7

x

(−3, −4)

d

y

(−2, 28)

3

units; up; y = 2x2 − 3x + 9 ; y = 2( x − 4 )2 8

15 16 units; up; y = −3x2 − 12x;

y = −3(x + 2)2 + 12

y

10 6

exercise 2k

1 0.382

2.618 ( 3– , − 5– ) 2

x

4

y

g

7.275 x

−0.275 −2 —) ( 7– , −57 2

4

y

h

4

−2.781

−0.719

—) (− 7– , − 17 4 8

x

Using technology to solve quadratic equations 1 a −3.732, −0.268 b −0.179, 11.179 c 3.414, 0.586 d −8.899, 0.899 e −11.196, −0.804 f 0.232, 1.434 2 a 0.377, −0.681 b −5.747, −0.665 c 4.712, −0.430 d −1.086, 1.600 e −153.441, −0.392 f 0.094, −500.094 3 4.355 h 4 56.4 h 5 5.7 m 6 Christine; her sum was 0.046 less than 10.25, whereas Harry’s was 0.167 more than 10.25. 7 a k=8 b x = 0.5

6 — 11

2

e

x

5

y

−1 −6

3

6

x

y

f

(3, 70)

40 28 (−4, 0) −7

−4

x

Chapter 2 • Quadratic functions

101

y

g

y

m

16

−2 — − 16 9

x

4

6

x

6.13 metres |AB| = 2.67 metres (approx) = 2 2 m (exact)

49x2 − 64 b 25x2 − 3 d (5x + 11)2 b (3x − 7)(4x − 3) d 3 (x − 3 + 23)(x − 3 −

8 x

(1, −21)

4 a 3,

x

−4 −3

−16

5 a ±

y

o

7 y 9– 4

(9, 39) −5

p

1

8

b 15,

4 5 2

b 7±

−4 3 3

b k=±

y

c k>

x

(6, 16)

4 3 3

10 a

or k <

(−1, 2)

−4 3 3

y

x

3 4 5

6 3– 2

2

5

x

11

(5, 16)

y

7 x

54 — 13

6 8 (3, −15)

−48

−54

(−2, −80)

102

d 0

b

y

b

x

− 3 − 13 − , 5 − 13 and 2 − 3 + 13 − , 5 + 13 2

−3

l

c 2

4 b = −5, c = −17

6 a

21 (−2, 9)

b2

−12

y

k

1

−2

x

2 4 (2, 0)

y

2 a

7

4 3 3

4

10

3

4 3 3

−24

j

−5

7 29 ± 10 10 −135

9 a

x

3 6

−1

4x2 − 36x + 81 8x2 − 70x + 117 (x + 6)(6x + 1) (6x + 7)(6x − 7) 23)

6 − 3 ± 11

9

− 3–4

e

d (8, 5)

4

1 a c 2 a c

12

−2

−4 (−3, −9)

b=

Short anSWer (1, 20)

y

i

c

Chapter reVieW

y

n

b a=5

−3

3

(−8, 20)

−17

x = a + a 2 + 24

(−4, −20)

−61

– −17 4

10 a

f

−36

h

3 − 3 < a <3 3 + 9

(−1, 7)

(5, 14)

−9

9

(0, −1)

11 y = (x + 4)2 − 13 y 12 a

(− 1 − 7, 1 − 3 7) and

84

(− 1 + 7, 1 + 3 7) c

1 − 13 4(5 + 13) 3 , and 3 1 + 13 4(5 − 13) 3 , 3

x

d

− 11

, 20

6 14 (10, −16)

b

y

− 433

and (0, −5) 80

8 x

−10

7 1.322 km and 2.553 km 8 y=

x2 , (2, 2) and (−2, 2) 2

Maths Quest 11 Mathematical Methods CAS

x

−80 (−1, −81)

c

y

7 10 13 16 19

48 8– 3 ( 13 —, 3

d

6

—) − 25 3

− 3–2

−1

8 11 14 17

B A D D

9 12 15 18

B C C D

extended reSponSe

x

1 a b c d

y

( − 5– , 1– ) 4 8

C A E E A

100 = 2l + 2w l = 50 − w A = −w2 + 50w Area

d e 3 a b c d 4 a

13

14 x =

0) and 5±

2 4 (3, −1)

6)

8k 2B 5E

3 B 6 E

(50, 0)

w

Area = 625 m2 l = 25 m, w = 25 m l = w: the paddock is a square. Parabola 1: (−2, 0) and (0, 0), parabola 2: (2, 0) and (0, 0) b y = −2x2 − 4x, y = −2x2 + 4x c h = 2 units

e f g 2 a

25 − 48 k 2

mUltiple ChoiCe

1A 4C

y

8

x

(0, 0)

(−6,

≤ x ≤ 0, 0 ≤ x ≤ 2 y = x2 + 6x + 9, y = x2 − 6x + 9 y = 2x2 − 3x + 1 y = 3x2 + 2x − 1 y = −x2 − 7x + 5 y = −x2 + 5x, y = x2 − 11x + 32

(25, 625)

−3

(−3,

−2

x

The turning point of the first graph has whole number (integer) coordinates. c The x-coordinate of the turning point − b , so is a whole number if b is a is 2a multiple of 2a. b

Chapter 2 • Quadratic functions

103

ChapTer 3

Cubic and quartic functions diGiTal doC doc-9724 10 Quick Questions

ChapTer ConTenTS 3a Expanding 3B Long division of polynomials 3C Polynomial values 3d The remainder and factor theorems 3e Factorising polynomials 3F Sum and difference of two cubes 3G Solving polynomial equations 3h Cubic graphs — intercepts method 3i Quartic graphs — intercepts method 3J Graphs of cubic functions in power function form 3k Domain, range, maximums and minimums 3l Modelling using technology 3m Finite differences

polynomials of degree 3 and 4 This chapter will deal mainly with polynomials of degree 3 (cubics). The general equation of a cubic polynomial is P(x) = a3x3 + a2x2 + a1x + a0, more commonly written as y = ax3 + bx2 + cx + d. Degree 4 polynomials (quartics) will also be considered. The general equation of a quartic polynomial is P(x) = a4 x4 + a3 x3 + a2 x2 + a1 x + a0, more commonly written as y = ax4 + bx3 + cx2 + dx + e.

3a

expanding

If we expand three linear factors, for example, (x + 1)(x + 2)(x − 7), we get a cubic polynomial (a polynomial of degree 3) as the following worked example shows. Worked example 1

Expand: a x(x + 2)(x − 3)

b (x − 1)(x + 5)(x + 2).

Think

a 1 Write the expression. 2

Expand two linear factors and simplify.

3

Multiply by the remaining factor.

b 1 Write the expression. 2

Expand two linear factors and simplify.

3

Multiply by the remaining factor and simplify.

WriTe

a x(x + 2)(x − 3)

= x(x2 − 3x + 2x − 6) = x(x2 − x − 6) = x3 − x2 − 6x

b (x − 1)(x + 5)(x + 2)

= (x − 1)(x2 + 2x + 5x + 10) = (x − 1)(x2 + 7x + 10) = x3 + 7x2 + 10x − x2 − 7x − 10 = x3 + 6x2 + 3x − 10

ChapTer 3 • Cubic and quartic functions

105

Note: Just as there is a shortcut for expanding perfect squares, there is also a shortcut for expanding cubes. We can find the shortcut by expanding (a + b)3 as usual. (a + b)3 = (a + b)(a + b)(a + b) = (a + b)(a2 + 2ab + b2) = a3 + 2a2b + ab2 + a2b + 2ab2 + b3 (a + b)3 = a3 + 3a2b + 3ab2 + b3 = a3 + 3a2b + 3ab2 + b3 (a − b)3 = a3 − 3a2b + 3ab2 − b3 Similarly, (a − b)3 = a3 − 3a2b + 3ab2 − b3. Worked example 2

Expand the perfect cube (x − 4)3 using the appropriate rule. Think

WriTe

1

Use the rule (a − b)3 = a3 − 3a2b + 3ab2 + b3. In this case a is x and b is 4.

(x − 4)3 = x3 − 3 × x2 × 4 + 3 × x × 42 − 43

2

Simplify.

(x − 4)3 = x3 − 12x2 + 48x − 64

Worked example 3

Expand −2x(x + 5)(x − 12). Think

WriTe −2x(x

+ 5)(x – 12)

1

Write the expression.

2

Expand the two linear factors and simplify.

= −2x(x2 − 12x + 5x − 60) = −2x(x2 − 7x − 60)

3

Multiply the remaining factor. No simplification is needed.

= −2x3 + 14x2 + 120

exercise 3a

expanding

1 We1a Expand each of the following. a x(x + 6)(x + 1) b x(x − 9)(x + 2) e −3x(x − 4)(x + 4) f 5x(x + 8)(x + 2) i (5x)(−6x)(x + 9) j −7x(x + 4)2

c x(x − 3)(x + 11) g x2(x + 4)

2 We1b Expand each of the following. a (x + 7)(x + 2)(x + 3) b (x − 2)(x + 4)(x − 5) d (x − 1)(x − 2)(x − 3) e (x + 6)(x − 1)(x + 1) g (x + 11)(x + 5)(x − 12) h (x + 5)(x − 1)2 j (x + 1)(x − 1)(x + 1) 3

d 2x(x + 2)(x + 3) h −2x2(7 − x)

c (x − 1)(x − 4)(x + 8) f (x − 7)(x + 7)(x + 5) i (x + 2)(x − 7)2

Expand each of the following. a (x − 2)(x + 7)(x + 8) d (5x + 3)(2x − 3)(x − 4) g −9x(1 − 2x)(3x + 8) j 2(7 + 2x)(x + 3)(x + 4)

b (x + 5)(3x − 1)(x + 4) e (1 − 6x)(x + 7)(x + 5) h (6x + 5)(2x − 7)2

c (4x − 1)(x + 3)(x − 3) f 3x(7x − 4)(x − 4) i (3 − 4x)(2 − x)(5x + 9)

4 We2 Expand the following using the appropriate rule for expanding cubes. a (x + 2)3 b (x + 5)3 c (x − 1)3 3 3 d (x − 3) e (2x − 6) f (3x + 4)3 5 We3 Expand each of the following. a (x + 5)(x − 11)(x + 2) b 3x(x + 6)(x − 1) d 106

(−x

+ 5)(x −

12)2

Maths Quest 11 Mathematical Methods CAS

e

−x(x

+

10)2

c f

6(x − 5)(−x + 15)(x + 8) + 13)3

−(x

3B

long division of polynomials

The reverse of expanding is factorising (expressing a polynomial as a product of its linear factors). Before learning how to factorise cubics, you must be familiar with long division of polynomials. You may remember in earlier levels doing long division questions. Consider 745 ÷ 3, or 3 745 The process used is as follows. 3 into 7 goes 2 times. Write 2 at the top. 2 × 3 = 6. Write down the 6. Subtract to get 1.

2 3 745 −6 14

Bring down the 4 to form 14. 3 into 14 goes 4 times. Write 4 at the top. 4 × 3 = 12. Write down the 12. Subtract to get 2.

Divisor

Bring down the 5 to form 25. 3 into 25 goes 8 times. Write 8 at the top. 8 × 3 = 24. Write down the 24. Subtract to get 1. Answer: 745 ÷ 3 = 248 remainder 1

248 3 745

Quotient Dividend

−6 14 − 12 25 − 24 1

Remainder

The same process can be used to divide polynomials by polynomial factors. Consider (x3 + 2x2 − 13x + 10) ÷ (x − 3)

or x − 3 x3 + 2x2 − 13x + 10

x into x3 goes x2 times (consider only the leading terms). Write x2 at the top. x2 × (x − 3) = x3 − 3x2 Write down the x3 − 3x2.

x−3

x3

x2 + 2x2 − 13x + 10

− (x3 − 3x2) 5x2 − 13x

Subtract. (x3 − x3 = 0, 2x2 − −3x2 = 5x2) Bring down the −13x. x into 5x2 goes 5x times. Write + 5x at the top. 5x × (x − 3) =

5x2

− 15x

Divisor

x2 + 5x + 2 + 2x2 − 13x + 10

Quotient Dividend

− (x3 − 3x2) 5x2 − 13x − (5x2 − 15x) 2x + 10 − (2x − 6) 16

Remainder

x−3

Write down the 5x2 − 15x. Subtract. Note: 5x2 − 5x2 = 0, −13x − −15x = +2x Bring down the 10.

x3

x into 2x goes 2 times. Write + 2 at the top. 2 × (x − 3) = 2x − 6 Write down the 2x − 6. Subtract to get 16. Answer: (x3 + 2x2 − 13x + 10) ÷ (x − 3) = x2 + 5x + 2 remainder 16

ChapTer 3 • Cubic and quartic functions

107

Worked example 4

Perform the following long divisions and state the quotient and remainder. a (2x3 + 6x2 − 3x + 2) ÷ (x − 6) b (x3 − 7x + 1) ÷ (x + 5) Think

WriTe

a 1 Write the question in long division format. a 2

Perform the long division process.

3

Write down the quotient and remainder.

b 1 Write the question in long division

x2

Perform the long division process.

3

Write down the quotient and remainder.

2 2 + 18xx + 105 Q 2x x −6 + 6x2 − 3x + 2 3 − (2 (2x − 12x 12 2) 18x2 − 3x − (18x2 − 108x) 105x + 2 − (105x − 630) 632 R The quotient is 2x2 + 18x + 105; the remainder is 632. 2 3 2x

b

format. Note that there is no term in this equation. Include 0x2 as a ‘place holder’. 2

TUTorial eles-1412 Worked example 4

x2 − 5x + 18 Q x + 5 + 0x2 − 7x + 1 − (x3 + 5x2) −5x2 − 7x − − ( 5x2 − 25x) 18x + 1 − (18x + 90) −89 R The quotient is x2 − 5x + 18; the remainder is −89. x3

Worked example 5

Find the quotient and remainder when x4 − 3x3 + 2x2 − 8 is divided by x + 2. Think 1

Write the question in long division format. Include 0x as a ‘place holder’.

2

Divide x into x4 and write the result above.

3

Multiply x3 by x + 2 and write the result underneath.

4

Subtract and then bring down the next term.

5

Continue to perform the long division process (as you did for cubic polynomials).

6

Write down the quotient and remainder.

WriTe

x3 − 5x2 + 12x − 24 x + 2 − 3x3 + 2x2 + 0x − 8 − (x4 + 2x3) −5x3 + 2x2 − − ( 5x3 − 10x2) 12x2 + 0x − (12x2 + 24x) −24x − 8 − (−24x − 48) 40 x4

The quotient is x3 − 5x2 + 12x − 24. The remainder is 40.

Worked example 6

Calculate the quotient and remainder when x3 − 4x2 − 7x − 5 is divided by x − 1. Think

108

1

Write the question in long division format.

2

Perform the long division process.

Maths Quest 11 Mathematical Methods CAS

WriTe

x2 − 3x − 10 x − 1 − 4x2 − 7x − 5 − (x3 − x2) −3x2 − 7x − (−3x2 + 3x) −10x − 5 − (−10x + 10) −15 x3

3

The quotient is x2 – 3x – 10. The remainder is –15.

Write the quotient and the remainder.

exercise 3B

long division of polynomials

1 We4a Perform the following long divisions, and state the quotient and remainder. 3 2 a (x + 6x + 3x + 1) ÷ (x + 3) b (x3 + 4x2 + 3x + 4) ÷ (x + 2) c (x3 + x2 + x + 3) ÷ (x + 1) 3 2 3 2 d (x + x + 4x + 1) ÷ (x + 2) e (x + 2x − 5x − 9) ÷ (x − 2) f (x3 + x2 − 9x − 5) ÷ (x − 2) 3 2 3 2 g (x − 5x + 3x − 8) ÷ (x − 3) h (x − 9x + 2x − 1) ÷ (x − 5) i 3x3 − x2 + 6x + 5, x + 2 3 2 3 2 j 4x − 4x + 10x − 4, x + 1 k 2x − 7x + 9x + 1, x − 2 l 2x3 + 8x2 − 9x − 1, x + 4 2

Divide the first polynomial by the second, and state the quotient and remainder. b 6x3 + 23x2 + 2x − 31, 3x + 4 c 8x3 + 6x2 − 39x − 13, 2x + 5 3 2 e 3x + 5x − 16x − 23, 3x + 2 f 9x3 − 6x2 − 5x + 9, 3x − 4

a 6x3 − 7x2 + 4x + 4, 2x − 1 d 2x3 − 15x2 + 34x − 13, 2x − 7

State the quotient and remainder for each of the following.

3 a c

− x3

− 6 x 2 − 7 x − 16 x +1

− 2x3

b

+ 9 x 2 + 17 x + 15 2x + 1

d

− 3x 3

+ 7 x 2 + 10 x − 15 x−3

4 x 3 − 20 x 2 + 23 x − 2 −2 x+3

4 We4b State the quotient and remainder for each of the following. a (x3 − 3x + 1) ÷ (x + 1) b (x3 + 2x2 − 7) ÷ (x + 2) − 3 d ( x − 7x + 8) ÷ (x − 1) e (5x2 + 13x + 1) ÷ (x + 3) − 3 g ( 2x − x + 2) ÷ (x − 2) h (−4x3 + 6x2 + 2x) ÷ (2x + 1)

c (x3 − 5x2 + 2x) ÷ (x − 4) f (2x3 + 8x2 − 4) ÷ (x + 5)

5 We5 Find the quotient and remainder for each of the following. a (x4 + x3 + 3x2 – 7x) ÷ (x – 1) b (x4 – 13x2 + 36) ÷ (x – 2) c (6x4 – x3 + 2x2 – 4x) ÷ (x – 3) 6 We6 Calculate the quotient and remainder for each of the following. a (x3 + 9x2 + 11x + 25) ÷ (x + 15) b (2x3 − 18x2 + 5x − 9) ÷ (x − 31) 3 c (12x + 32x − 9) ÷ (3x + 4) d (18x4 + 3x3 + 45) ÷ (2x + 7)

3C

polynomial values

Consider the polynomial P(x) = x3 − 5x2 + x + 1. The value of the polynomial when x = 3 is denoted by P(3) and is found by substituting x = 3 into the equation in place of x. That is, P(3) = (3)3 − 5(3)2 + (3) + 1 = 27 − 5(9) + 3 + 1 = 27 − 45 + 4 = −14.

Worked example 7

If P(x) = 2x3 + x2 − 3x − 4, determine: a P(1) b P(−2) Think

a 1 Write the expression. 2

Replace x with 1.

3

Simplify.

c P(a)

d P(x + 1).

WriTe

a P(x) = 2x3 + x2 − 3x − 4

P(1) = 2(1)3 + (1)2 − 3(1) − 4 =2+1−3−4 = −4 ChapTer 3 • Cubic and quartic functions

109

b 1 Write the expression. 2

Replace x with −2.

3

Simplify.

P(x) = 2x3 + x2 − 3x − 4 P(−2) = 2(−2)3 + (−2)2 − 3(−2) − 4 = 2(−8) + (4) + 6 − 4 = −16 + 4 + 6 − 4 = −10

c 1 Write the expression. 2

b

c P(x) = 2x3 + x2 − 3x − 4

P(a) = 2a3 + a2 − 3a − 4

Replace x with a. No further simplification is possible.

d 1 Write the expression. 2

Replace x with (x + 1).

3

Expand the right-hand side and collect like terms. Use the rules for expanding cubics and quadratics.

d

P(x) = 2x3 + x2 − 3x − 4 P(x + 1) = 2(x + 1)3 + (x + 1)2 − 3(x + 1) − 4 = 2(x3 + 3x2 + 3x + 1) + x2 + 2x + 1 − 3x − 3 − 4 = 2x3 + 6x2 + 6x + 2 + x2 − x − 6 = 2x3 + 7x2 + 5x − 4

Worked example 8

Determine the following, considering P(x) = 16x4 + 3x3 − 22x + 17. a P(−14) b P(y + 7) Think

WriTe

a 1 Write the expression.

a

P(x) = 16x4 + 3x3 – 22x + 17

2

Replace x with −14.

P(−14) = 16(–14)4 + 3(−14)3 – 22(−14) + 17

3

Simplify.

P(−14) = 614 656 – 8232 + 308 + 17 = 606 749

b 1 Write the expression.

b

P(x) = 16x4 + 3x3 – 22x + 17

2

Replace x with y + 7.

3

Expand each term.

16(y + 7)4 = 16y4 + 448y3 + 4704y2 + 21 952y + 38 416 3(y + 7)3 = 3y3 + 63y2 + 441y + 1029 −22(y + 7) = −22y − 154 17 = 17

4

Collect like terms.

P(y + 7) = 16y4 + 451y3 + 4767y2 + 22 371y + 39 308

exercise 3C

P(y + 7) = 16(y + 7)4 + 3(y + 7)3 − 22(y + 7) + 17

polynomial values

1 We7 If P(x) = 2x3 − 3x2 + 2x + 10, determine the following. a P(0) b P(1) d P(3) e P(−1) − h P(a) g P( 3) j P(x + 2) k P(x − 3)

c f i l

2 We8 Determine the following considering P(x) = 16x4 + 3x3 − 22x + 17. b P(102) a P(−11) c P(2x + 9) d P(x3 + 2) 110

Maths Quest 11 Mathematical Methods CAS

P(2) P(−2) P(2b) P(−4y)

3

Copy the following table. Column Column Column Column Column Column Column Column Column 1 2 3 4 5 6 7 8 9

P(x)

P(1)

P(2)

P(−1)

P(−2)

Rem. Rem. Rem. Rem. when when when when divided divided divided divided by by by by (x − 1) (x − 2) (x + 1) (x + 2)

a b c d

Complete columns 2 to 5 of the table for each of the following polynomials. a P(x) = x3 + x2 + x + 1 b P(x) = x3 + 2x2 + 5x + 2 3 2 c P(x) = x − x + 4x − 1 d P(x) = x3 − 4x2 − 7x + 3 4 Find the remainder when each polynomial in question 2 is divided by (x − 1) and complete column 6 of the table. 5 Find the remainder when each polynomial in question 2 is divided by (x − 2) and complete column 7 of

diGiTal doC doc-9725 Cubic valuer

the table. 6 Find the remainder when each polynomial in question 2 is divided by (x + 1) and complete column 8 of

the table. 7 Find the remainder when each polynomial in question 2 is divided by (x + 2) and complete column 9 of

the table. 8

Copy and complete the following sentences, using your answers to questions 3 to 7 to find the pattern. a A quick way of finding the remainder when P(x) is divided by (x + 8) is to calculate b A quick way of finding the remainder when P(x) is divided by (x − 7) is to calculate c A quick way of finding the remainder when P(x) is divided by (x − a) is to calculate

3d The remainder The remainder theorem

. . .

diGiTal doC doc-9726 WorkSHEET 3.1

and factor theorems

In the previous exercise, you may have noticed that: The remainder when P(x) is divided by (x − a) is equal to P(a). That is, R = P(a). This is called the remainder theorem. We could have derived this result as follows. If 13 is divided by 4, the quotient is 3, and the remainder is 1. That is, 13 ÷ 4 = 3 + 1 4

and

13 = 4 × 3 + 1. Similarly, if P(x) = is 15. That is,

x3

+

x2

+ x + 1 is divided by (x − 2), the quotient is x2 + 3x + 7 and the remainder

15 and x−2 (x3 + x2 + x + 1) = (x2 + 3x + 7)(x − 2) + 15.

(x3 + x2 + x + 1) ÷ (x − 2) = x2 + 3x + 7 +

In general, if P(x) is divided by (x − a), the quotient is Q(x), and the remainder is R, we can write R P(x) ÷ (x − a) = Q(x) + and ( x − a) P(x) = (x − a)Q(x) + R. ChapTer 3 • Cubic and quartic functions

111

Substituting x = a into this last expression yields P(a) = (a − a)Q(x) + R = 0 × Q(x) + R = R as before. Worked Example 9

Without actually dividing, find the remainder when x3 − 7x2 − 2x + 4 is divided by: a x − 3 b x + 6. Think

a

Write

a Let P(x) = x3 − 7x2 − 2x + 4

1

Name the polynomial.

2

The remainder when P(x) is divided by (x − 3) is equal to P(3).

b The remainder when P(x) is divided by

(x + 6) is equal to P(−6).

R = P(3) = 33 − 7(3)2 − 2(3) + 4 = 27 − 7(9) − 6 + 4 = 27 − 63 − 6 + 4 = −38 b R = P(−6)

= (−6)3 − 7(−6)2 − 2(−6) + 4 = −216 − 7(36) + 12 + 4 = −216 − 252 + 12 + 4 = −452

Worked Example 10

The remainder when x3 + kx2 + x − 2 is divided by (x − 2) is equal to 20. Find the value of k. Think

Write

1

Name the polynomial.

2

The remainder when P(x) is divided by (x − 2) is equal to P(2).

3

We are given R = 20. Put 8 + 4k = 20.

4

Solve for k.

Let P(x) = x3 + kx2 + x − 2. R = P(2) = 23 + k(2)2 + 2 − 2 = 8 + 4k Since R = 20. 8 + 4k = 20 4k = 12 k = 3

The factor theorem The remainder when 12 is divided by 4 is zero, since 4 is a factor of 12. Similarly, if the remainder (R) when P(x) is divided by (x − a) is zero, then (x − a) must be a factor of P(x). Since R = P(a), all we need to do is to find a value of a that makes P(a) = 0, and we can say that (x − a) is a factor. If P(a) = 0, then (x − a) is a factor of P(x). This is called the factor theorem. Imagine P(x) could be factorised as follows: P(x) = (x − a)Q(x), where Q(x) is ‘the other’ factor of P(x). Then we have P(a) = (a − a)Q(a) = 0 × Q(a) = 0. So if P(a) = 0, (x − a) is a factor. 112 Maths Quest 11 Mathematical Methods CAS

Worked example 11

Apply the factor theorem to determine which of the following is a factor of x4 − 4x3 − 43x2 + 58x + 240. a (x + 2) b (x − 1) Think

WriTe

a 1 Name the polynomial.

a Let P(x) = x4 − 4x3 − 43x2 + 58x + 240

2

To find the remainder when P(x) is divided by (x − a), find P(a).

P(−2) = (−2)4 − 4(−2)3 − 43(−2)2 + 58(−2) + 240 = 16 − 4(−8) − 43(4) − 116 + 240 = 16 + 32 − 172 − 116 + 240 =0

3

State the answer.

As P(−2) = 0, the remainder when P(x) is divided by (x + 2) is zero; therefore, (x + 2) is a factor.

b 1 To find the remainder when P(x) is

divided by (x − a), find P(a).

2

State the answer.

exercise 3d

b P(1) = (1)4 − 4(1)3 − 43(1)2 + 58(1) + 240

= 1 − 4 − 43 + 58 + 240 = 252

As P(1) = 252, the remainder when P(x) is divided by (x − 1) is 252; therefore, (x − 1) is not a factor.

The remainder and factor theorems

1 We9 Without actually dividing, find the remainder when x3 + 3x2 − 10x − 24 is divided by: a x−1 b x+2 c x−3 d x+5 e x−0 f x−k g x+n h x + 3c.

diGiTal doC doc-9725 Cubic valuer

2 Find the remainder when the first polynomial is divided by the second without performing

long division. a x3 + 2x2 + 3x + 4, x − 3 c x3 + 3x2 − 3x + 1, x + 2 e 2x3 + 3x2 + 6x + 3, x + 5 g x3 + x2 + 8, x − 5 i −x3 + 8, x + 3

b d f h j

x3 − 4x2 + 2x − 1, x + 1 x3 − x2 − 4x − 5, x − 1 −3x3 − 2x2 + x + 6, x + 1 x3 − 3x2 − 2, x − 2 x3 + 2x2, x − 7

The remainder when x3 + kx + 1 is divided by (x + 2) is −19. Find the value of k. The remainder when x3 + 2x2 + mx + 5 is divided by (x − 2) is 27. Find the value of m. The remainder when x3 − 3x2 + 2x + n is divided by (x − 1) is 1. Find the value of n. The remainder when ax3 + 4x2 − 2x + 1 is divided by (x − 3) is −23. Find the value of a. The remainder when x3 − bx2 − 2x + 1 is divided by (x + 1) is 0. Find the value of b. The remainder when −4x2 + 2x + 7 is divided by (x − c) is −5. Find a possible whole number value of c. g The remainder when x2 − 3x + 1 is divided by (x + d) is 11. Find the possible values of d. h The remainder when x3 + ax2 + bx + 1 is divided by (x − 5) is −14. When the cubic polynomial is divided by (x + 1), the remainder is −2. Find a and b.

3a b c d e f

We10

4 We11 Apply the factor theorem to determine which of the following are factors of x3 + 2x2 − 11x − 12. a (x − 1) b (x − 3) c (x + 1) d (x + 2) 5

Prove that each of the following are linear factors of x3 + 4x2 − 11x − 30 by substituting values into the cubic function: (x + 2), (x − 3), (x + 5).

6 Use the factor theorem to show that the first polynomial is exactly divisible by the second (that is, the

second polynomial is a factor of the first). a x3 + 5x2 + 2x − 8, x − 1 c x3 − 7x2 + 4x + 12, x − 2 e x3 + 3x2 − 9x − 27, x + 3 g −2x3 + 9x2 − x − 12, x − 4

b x3 − 7x2 − x + 7, x − 7 d x3 + 2x2 − 9x − 18, x + 2 f −x3 + x2 + 9x − 9, x − 1 h 3x3 + 22x2 + 37x + 10, x + 5 ChapTer 3 • Cubic and quartic functions

113

7 mC a When x3 + 2x2 − 5x − 5 is divided by (x + 2), the remainder is: a −5

B −2

C 0

d 2

e 5

a (x − 1)

B (x − 2)

C (x + 3)

d (x − 5)

e (x + 4)

a −3

B −2

C −1

d 0

e 1

d (x + 5)

e (x + 7)

b Which of the following is a factor of 2x3 + 15x2 + 22x − 15?

c When x3 − 13x2 + 48x − 36 is divided by (x − 1), the remainder is: d Which of the following is a factor of a (x − 2)

diGiTal doC doc-9727 SkillSHEET 3.1 reviewing the discriminant

B (x + 2)

x3

− 5x2 − 22x + 56? C (x − 5)

8 Find one factor of each of the following cubic polynomials. a x3 − 3x2 + 3x − 1 b x3 − 7x2 + 16x − 12 3 2 c x + x − 8x − 12 d x3 + 3x2 − 34x − 120 9 Consider the polynomial P(x) = 6x3 + 7x2 − x − 2. a Determine:

i P(−1)

ii P

( 12 )

iii P

( ) −2 3

i Factorise P(x) as the product of (x + 1) and a quadratic factor. ii Further factorise so P(x) is written as the product of three linear factors. c Explain how the other two linear factors relate to what you found in parts b and c. d Copy and complete the following: In general if (ax + b) is a factor, then P(…) = 0. b

3e Factorising Using long division

polynomials

Once one factor of a polynomial has been found (using the factor theorem as in the previous section), long division may be used to find other factors. Worked example 12

Use long division to factorise x3 − 19 x + 30. Think

WriTe

1

Name the polynomial. Note: There is no x2 term, so include 0x2.

P(x) = x3 − 19x + 30 P(x) = x3 + 0x2 − 19x + 30

2

Look at the last term in P(x), which is 30. This suggests it is worth trying P(5) or P(−5). Try P(−5). P(−5) = 0, so (x + 5) is a factor.

P(−5) = (−5)3 − 19 × (−5) + 30 = −125 + 95 + 30 =0 Therefore (x + 5) is a factor.

3

Divide (x + 5) into P(x) using long division to find a quadratic factor.

x2 − 5x + 6 x +5 + 0x2 − 19x + 30 3 − (x + 5x2) −5x2 − 19x −(−5x2 − 25x) 6x + 30 −(6x + 30) 0

4

Write P(x) as a product of the two factors found so far.

P(x) = (x + 5)(x2 − 5x + 6)

5

Factorise the second bracket if possible.

P(x) = (x + 5)(x − 2)(x − 3)

TUTorial eles-1413 Worked example 12

x3

Note: In this example, P(x) may have been factorised without long division by finding all three values of x that make P(x) = 0, and hence three factors, then checking that the three factors multiply to give P(x).

Using short division The process of long division can take a lot of time (and space). One short division method is shown here; it may take a little longer to understand, but it is quicker than long division once mastered. 114

Maths Quest 11 Mathematical Methods CAS

Consider P(x) = x3 + 2x2 − 13x + 10. Using the factor theorem, we can find that (x − 1) is a factor of P(x). So, P(x) = (x − 1)(ax2 + bx + c). Actually, we know more than this: as P(x) begins with x3 and ends with +10, we could write: P(x) = (x − 1)(x2 + bx − 10) Imagine expanding this version of P(x). Our x2 terms give −1x2 + bx2. Since P(x) = x3 + 2x2 − 13x + 10, we need +2x2. That is, we need −1x2 + 3x2. To get this, the bx must be 3x, as when x in the first bracket is multiplied by 3x in the second bracket, +3x2 results. That is, we have deduced P(x) = (x − 1)(x2 + 3x − 10). Factorising the second bracket gives P(x) = (x − 1)(x + 5)(x − 2) Worked example 13

Factorise the following, using short division where possible. a x3 − 5x2 − 2 x + 24 b x4 + x3 − 13 x2 − 25 x − 12 Think

a 1 Name the polynomial. 2

Look for a value of x such that P(x) = 0. Try P(−2). P(−2) does equal 0, so (x + 2) is a factor.

WriTe

a Let P(x) = x3 − 5x2 − 2x + 24

P(−2) = (−2)3 − 5 × (−2)2 − 2 × (−2) + 24 = −8 − 20 + 4 + 24 = −28 + 28 =0 So (x + 2) is a factor.

3

Write the original polynomial as the found factor multiplied by ax2 + bx + c. The first term in the brackets must be x2, and the last term must be 12.

P(x) = x3 − 5x2 − 2x + 24 P(x) = (x + 2)(ax2 + bx + c) = (x + 2)(x2 + bx + 12)

4

Imagine the expansion of the expression in step 3. We have 2x2, and require −5x2. We need an extra −7x2. So b = −7.

2x2 + bx2 = −5x2 b = −7 P(x) = (x + 2)(x2 − 7x + 12)

5

Factorise the second bracket if possible.

P(x) = (x + 2)(x − 3)(x − 4)

b 1 Name the polynomial.

b Let P(x) = x4 + x3 − 13x2 − 25x − 12

2

Look for a value of x such that P(x) = 0. Try P(−1).

P(−1) = 0 So (x + 1) is a factor.

3

It is difficult to factorise a quartic using short division, so we will use long division here.

x3 + 0x2 − 13x − 12 x + 1 + x3 − 13x2 − 25x − 12 − (x4 + x3) 0 − 13x2 − 25x − (−13x2 − 13x) −12x − 12 − − ( 12x − 12) 0

4

Name the cubic factor and try to find another factor using the factor theorem.

Let Q(x) = x3 − 13x − 12. Q(−3) = 0 So (x + 3) is a factor.

5

Factorise the cubic using short division.

Q(x) = x3 − 13x − 12 = (x + 3)(x2 − 3x − 4)

x4

ChapTer 3 • Cubic and quartic functions

115

6

Factorise the quadratic if possible.

Q(x) = (x + 3)(x − 4)(x + 1)

7

Write the original polynomial in factorised form.

P(x) = x4 + x3 − 13x2 − 25x − 12 = (x + 1)(x + 3)(x − 4)(x + 1) = (x + 1)2(x + 3)(x − 4)

Worked example 14

Factorise the expression 2 x3 − 9 x2 − 2 x + 24. Think

WriTe

Name the polynomial.

2

Look for values of x such that P(x) = 0.

x = 1: P(1) = 2 – 9 – 2 + 24 ≠ 0 x = 2: P(2) = 16 – 36 – 8 + 24 = 0 So x – 2 is a factor.

3

Write P(x) as the product of x – 2 and an unknown quadratic.

P(x) = (x – 2)(ax2 + bx + c)

4

Consider the x3 term (2x3). This must equal ax3.

(x)ax2 = ax3 = 2x3 Therefore a = 2.

5

Consider the constant term (24). This must equal –2c.

(–2)c = –2c = 2c Therefore c = –12.

6

Rewrite P(x).

P(x) = (x – 2)(2x2 + bx – 12)

7

Consider the x2 term from step 6. This must equal –9x2 from the original cubic.

+ bx2 = –9x2 bx2 = –5x2 Therefore b = –5.

8

Consider the x term from step 6. This must equal –2x from the original cubic. This confirms step 7.

–2bx

9

Write P(x).

P(x) = (x – 2)(2x2 – 5x – 12)

Factorise the quadratic term.

P(x) = (x – 2)(2x + 3)(x – 4)

10

exercise 3e 1 We12

–4x2

– 12x = –22 –2bx = 10x b = –5

Factorising polynomials

Use long division to factorise each dividend.

a x + 1 x3 + 10x2 + 27x + 18

b x + 2 x3 + 8x2 + 17x + 10

c x + 9 x3 + 12x2 + 29x + 18

d x + 1 x3 + 8x2 + 19x + 12

e x + 3 x3 + 14x2 + 61x + 84

f

x + 7 x3 + 12x2 + 41x + 42

g x + 2 x3 + 4x2 + 5x + 2

h x + 3 x3 + 7x2 + 16x + 12

i

x + 5 x3 + 14x2 + 65x + 100

k x x3 + 7x2 + 12x

l

x + 5 x3 + 10x2 + 25x

j

x x3 + 13x2 + 40x

m x + 1 x3 + 6x2 + 5x

116

P(x) = 2x3 – 9x2 – 2x + 24

1

Maths Quest 11 Mathematical Methods CAS

n x + 6 x3 + 6x2

2 We12,13 Factorise the following as fully as possible. a x3 + x2 − x − 1 b x3 − 2x2 − x + 2 3 2 d x + x − 8x − 12 e x3 + 9x2 + 24x + 16 3 2 g x + 2x − x − 2 h x3 − 7x − 6 3 2 j x +x +x+6 k x3 + 8x2 + 17x + 10 3 2 m x − x − 8x + 12 n x3 + 9x2 − 12x − 160 4 3 2 p x + 3x − 6x − 28x − 24 q x4 + 6x3 + 8x2 − 6x − 9

c f i l o r

x3 + 7x2 + 11x + 5 x3 − 5x2 − 4x + 20 x3 + 3x2 − 4 x3 + x2 − 9x − 9 x4 + 4x3 + 3x2 − 4x − 4 x4 − 5x3 − 17x2 + 21x

diGiTal doC doc-9728 polynomials zero search

3 We14 Use a CAS calculator to factorise the following as fully as possible. a 3x3 − x2 − 10x b 4x3 + 2x2 − 2x c 3x3 − 6x2 − 24x − 3 2 3 2 d 2x − 12x − 18x e 6x − 6x f −x3 − 7x2 − 12x − 3 2 − 3 2 g x − 3x + x + 3 h 2x + 10x − 12x i −6x3 − 5x2 + 12x − 4 − 3 2 − 5 4 3 2 j 5x + 24x − 36x + 16 k x − x + 21x + 49x − 8x − 60 l 24x4 − 53x3 − 71x2 + 152x + 20 4 a d g j

Factorise the following as fully as possible. 2x3 + 5x2 − x − 6 b 3x3 + 14x2 + 7x − 4 3 2 4x + 35x + 84x + 45 e 5x3 + 9x2 + 3x − 1 3 2 4x + 16x + 21x + 9 h 6x3 − 23x2 + 26x − 8 3 2 7x + 12x − 60x + 16 k 2x4 − x3 − 11x2 − 11x − 3

3F

c f i l

3x3 + 2x2 − 12x − 8 x3 + x2 + 4 10x3 + 19x2 − 94x − 40 6x4 + 11x3 − 22x2 − x + 6

Sum and difference of two cubes

Two special cases of cubic polynomials, called ‘sum of cubes’and ‘difference of cubes’, are discussed in this section. There are shortcuts for factorising such cubic expressions. Examples of each are shown in the table below. Sum of cubes x + 3

23

Difference of cubes x3 − 27

125 + 64b3

x3 − 81y3 1000

x3y3 + 1

w6 − 1

(2x + 1)3 + 8

216 − (uv)3

Consider the following expansions. (a + b)(a2 − ab + b2) and (a − b)(a2 + ab + b2) 3 2 2 2 2 3 = a − a b + ab + ba − ab + b = a3 + a2b + ab2 − ba2 − ab2 − b3 3 2 2 2 2 3 = a − a b + ab + a b − ab + b = a3 + a2b + ab2 − a2b − ab2 − b3 3 3 =a +b = a3 − b3 These expansions show that: a3 + b3 = (a + b)(a2 − ab + b2) and a3 − b3 = (a − b)(a2 + ab + b2). That is, we have two formulas that may be used to factorise sums and differences of cubes. Worked example 15

Factorise the following using the sum or difference of cubes formula. a x3 − 1000y3 b 2(x + 6)3 + 16 Think

a 1 Write the expression.

TUTorial eles-1414 Worked example 15

WriTe

a x3 − 1000y3

= x3 − (10y)3

2

Recognise a difference of cubes.

3

Identify a and b for use with the formula a3 − b3 = (a − b)(a2 + ab + b2).

4

Use the formula to factorise.

= (x − 10y)[x2 + x(10y) + (10y)2]

5

Simplify.

= (x − 10y)(x2 + 10xy + 100y2)

a = x, b = 10y

ChapTer 3 • Cubic and quartic functions

117

b 2(x + 6)3 + 16

b 1 Write the expression.

= 2[(x + 6)3 + 8] = 2[(x + 6)3 + 23]

2

Take out a common factor of 2 to produce a sum of cubes.

3

Identify a and b for use with the formula a3 + b3 = (a + b)(a2 − ab + b2).

4

Apply the sum of cubes formula.

= 2[(x + 6) + 2][(x + 6)2 − (x + 6)(2) + 22]

5

Simplify.

= 2(x + 8)(x2 + 12x + 36 − 2x − 12 + 4) = 2(x + 8)(x2 + 10x + 28)

exercise 3F

a = (x + 6), b = 2

Sum and difference of two cubes

1 Identify a and b (as used in the above sum and difference of cubes expressions) in each of the following

(do not factorise). a x3 + 63

b 8y3 + z3

d 1 − 64h3

e

2 We15a a x3 − 125

g j

+ 8p3

729 s3t3

+

− e3g3

f

t3 −

u3 216

Factorise the following using the sum or difference of cubes formula. b j3 + k3 c y3 − 8

d 27x3 + y3

x3

1 3 8c

c (x + 5)3 + 27

e

64t3 − 216u3

h 27r3 − 1

f

x3 − 1

i

(3k)3 − 1 8

g6

Factorise the following. a (a − + a3 b (x + 2)3 − 8 c (2x + 3)3 + 1 3 3 3 3 d (w − 5) − w e (2m + p) + (3m − p) f 27x3 − (x + 3)3 3 3 3 3 g (2y + 7) + (y − 2) h (3x + y) + (x − 4y) i (2 − 4p)3 − (p + 1)3 3 3 6 9 j (5x − 9) − (7 − x) k x +y l 2x3 − 54 3 2 3 m 3a + 3 n 6(x + 1) + 162 4 When mx3 − ny3 is fully factorised it gives (3x − y)(9x2 + 3xy + y2). What are the values of m and n? 5 a Write 3x3 + my3 in the form a3 + b3. b Identify the values of a and b. c Factorise using the rule for sum of cubes. 3 We15b

1)3

3G Solving polynomial equations making x the subject of the equation Cubic equations of the form a(x − b)3 + c = 0 may be solved by isolating x as follows. a(x − b)3 = −c −c (x − b)3 = a −

c a

x=b+

3

x−b=

3

−

c a

Unlike a square root, a cube root can be only positive or negative, not both; for example, 3 − 8

118

+= −2, 3 8 = 2.

Maths Quest 11 Mathematical Methods CAS

Worked example 16

Solve 3(x + 2)3 + 192 = 0 by making x the subject. Think

WriTe

3(x + 2)3 + 192 = 0

1

Write the equation.

2

Subtract 192 from both sides.

3

Divide both sides by 3.

4

Take the cube root of both sides.

5

Subtract 2 from both sides and simplify.

3(x + 2)3 = −192 (x + 2)3 = −64 x + 2 = −4 x = −4 − 2 = −6

Factorising to solve polynomial equations The Null Factor Law applies to cubic and quartic equations just as it does for quadratics. If P(x) = (x − a)(x − b)(x − c) = 0, then the solutions are x = a, x = b and x = c. If P(x) = k(lx − a)(mx − b)(nx − c) = 0, solutions are found by solving the following equations: lx − a = 0, mx − b = 0 and nx − c = 0 Worked example 17

Solve each of the following equations. a x4 = 16x2 b 2x3 − 11x2 + 18x − 9 = 0 Think

a 1 Write the equation.

WriTe

a

x4 = 16x2 x4 − 16x2 = 0

2

Rearrange so that all terms are on the left.

3

Take out a common factor of x.

4

Factorise the brackets using a difference of squares.

x2(x + 4)(x − 4) = 0

5

Use the Null Factor Law to solve.

x = 0, x + 4 = 0 or x − 4 = 0 so x = 0, x = −4 or x = 4

b 1 Name the polynomial.

x2(x2 − 16) = 0

b Let P(x) = 2x3 − 11x2 + 18x − 9.

2

Use the factor theorem to find a factor (search for a value a such that P(a) = 0). Consider factors of the constant term (that is, factors of 9 such as 1, 3). The simplest value to try is 1.

P(1) = 2 − 11 + 18 − 9 =0 So (x − 1) is a factor.

3

Use long or short division to find the other factors of P(x).

2 2 − 9x + 9 2x x − 1 2x 2 3 − 11x2 − 18x − 9 22x3 − 2x 2 2 −9x2 + 18x −9x2 + 9x 9x − 9 9x − 9 0

4

Factorise the brackets.

P(x) = (x − 1)(2x2 − 9x + 9) P(x) = (x − 1)(2x − 3)(x − 3)

5

Consider the factorised equation to solve.

For (x − 1)(2x − 3)(x − 3) = 0

6

Use the Null Factor Law to solve.

x − 1 = 0, 2x − 3 = 0 or x − 3 = 0 3

so x = 1, x = 2 or x = 3

ChapTer 3 • Cubic and quartic functions

119

Polynomial equations can also be solved using a CAS calculator, which is useful when the solutions are not rational. The solutions can be found by using the solve function on a calculator page, or by constructing a graph. Worked example 18

Use a CAS calculator to solve x3 + 3 x2 − 8 x − 5 = 0. Think

WriTe

1

Use the CAS calculator to solve the equation.

Enter solve (x3 + 3x2 − 8x − 5 = 0, x) into the CAS calculator.

2

Read the results supplied by the calculator.

x = −4.524 04 or x = −0.536 396 or x = 2.060 43

3

Write the answer.

The solutions of x3 + 3x2 − 8x − 5 = 0 are x = −4.52, −0.54 and 2.06, rounded to 2 decimal places.

exercise 3G 1 We16

Solving polynomial equations

Solve the following equations by making x the subject.

a 2(x − − 250 = 0 c (x − 4)3 − 1000 = 0 e −2(x − 5)3 − 2 = 0 g (2x + 3)3 − 27 = 0

b 3(x + 2)3 + 81 = 0 d (x + 7)3 − 8 = 0 f −(x + 3)3 + 1 = 0 h 4(3x − 1)3 + 500 = 0

1)3

i

2 3

(x + 1)3 + 18 = 0

j

k (x − 5)3 = 343

1 2

(5 − x)3 − 32 = 0

l 4−

4 5

(x + 8)3 = 104

Find all solutions of the following equations.

2 a c e g i k m

(x − 1)(x − 2)(x − 5) = 0 (x − 5)(x + 2)(x − 9) (x − 1) = 0 (3x + 12)(x − 4)(x + 4) = 0 (x + 5)(x − 8)2 = 0 x2(x + 1)(x − 1) = 0 (6 − x)2 (2 + x) = 0 x(5x − 6)(2x + 3) (6 − 7x) = 0

3 We17a a d g j m

b d f h j l n

(x + 3)(x + 4)(x + 7) = 0 (2x − 4)(x + 1)(x − 3) (x + 2) = 0 (2 − x)(x + 2)(1 − x)(1 + x) = 0 (x − 1)3 = 0 −3x(x − 9)3 = 0 x2(2x + 7) = 0 (3 − 4x)2(5x − 1) = 0

Solve the following equations.

x3 − 4x = 0 −3x4 + 81 = 0 −4x3 + 8x = 0 x4 − 5x3 + 6x2 = 0 9x2 = 20x + x3

b e h k n

x3 − 16x = 0 x3 + 5x2 = 0 12x3 + 3x2 = 0 x3 − 8x2 + 16x = 0 x3 + 6x = 4x2

c f i l

2x4 − 50x2 = 0 x3 − 2x2 = 0 4x3 − 20x4 = 0 x3 + 6x2 = 7x

4 We17b Use the Null Factor Law to solve the following equations. a x3 − x2 − 16x + 16 = 0 b x3 − 6x2 − x + 30 = 0 3 2 c x − x − 25x + 25 = 0 d x3 + 4x2 − 4x − 16 = 0 3 2 e x − 4x + x + 6 = 0 f x3 − 4x2 − 7x + 10 = 0 4 3 2 − g x − 3x − 7x + 15x = 18 h x4 + 2x3 − 13x2 + 10x 3 2 i 2x + 15x + 19x + 6 = 0 j −4x3 + 16x2 − 9x − 9 = 0 − 3 2 k 2x − 9x − 7x + 6 = 0 l 2x3 + 4x2 − 2x − 4 = 0 5 We18 Use a CAS calculator to find all solutions to the following equations. a x3 − 17x2 − 56x + 1153 = 0 b x3 + 12x2 − 49x − 588 = 0 − 3 2 c x + 17x + 65x − 1521 = 0 d x3 – 48x2 + 768x − 4096 = 0 120

Maths Quest 11 Mathematical Methods CAS

e x3 + 6x2 − 6x + 2 = 0 g 3x2 + 2x + 1 = x3 i x4 + 4x3 − 5x2 + 15 = 0

f x3 – 14x2 – 4x + 13 = 0 h −2x3 – 3x2 + 2x + 0.5 = 0 j 2x4 + 25x3 + 17x = 9

6 mC Which of the following is a solution to x3 − 7x2 + 2x + 40? a −5 d 1

B −4 e 2

C −2

7 mC A solution of x3 − 9x2 + 15x + 25 = 0 is x = 5. How many other (distinct) solutions are there? a 0 d 3

3h

B 1 e 4

C 2

diGiTal doC doc-9729 WorkSHEET 3.2

Cubic graphs — intercepts method

A good sketch graph of a function shows: 1. x- and y-intercepts 2. the behaviour of the function at extreme values of x, that is, as x approaches infinity (x → +∞) and as x approaches negative infinity (x → −∞) 3. the general location of turning points. The graphs below show the two main types of cubic graph. y

y Turning points

x

x

A positive cubic

A negative cubic

For positive cubic graphs, as positive values of x become larger and larger, y-values also become larger. For negative cubic graphs, as positive values of x become larger and y Point of inflection larger, y-values become smaller. Sometimes, instead of two turning points, there is a point of inflection, where the graph changes from a decreasing gradient to zero to an increasing gradient (or vice versa). When this occurs, there is only one x-intercept. Consider the general factorised cubic f (x) = (x − a)(x − b)(x − c). x The Null Factor Law tells us that f (x) = 0 when x = a or x = b or x = c. The y-intercept occurs when x = 0; that is, the y-intercept is f (0) = (0 − a)(0 − b)(0 − c) = −abc y

c

b

a

Cubic with a point of inflection

x

−abc

ChapTer 3 • Cubic and quartic functions

121

Worked Example 19

Sketch graphs of the following, showing all intercepts. a y = (x − 2)(x − 3)(x + 5) b y = (x − 6)2 (4 − x) c y = (x − 2)3 Think

Write/Draw

a 1 Note that the function is already

factorised and that the graph is a positive cubic.

a y = (x − 2)(x − 3)(x + 5)

2

The y-intercept occurs where x = 0. Substitute x = 0 into the equation.

y-intercept: if x = 0, y = (−2)(−3)(5) = 30 Point: (0, 30)

3

Use the Null Factor Law to find the x-intercepts. (Make each bracket equal to 0 and solve a mini-equation.)

x-intercepts: if y = 0, x − 2 = 0, x − 3 = 0 or x + 5 = 0 x = 2, x = 3 or x = −5 Points: (2, 0), (3, 0), (−5, 0)

4

Combine information from the above steps to sketch the graph.

y

30 −5 b 1 The graph is a negative cubic −x

(the in the last factor produces a negative x3 coefficient if the RHS is expanded).

2

3

x

b y = (x − 6)2(4 − x)

2

Substitute x = 0 to find the y-intercept.

y-intercept: if x = 0, y = (−6)2(4) = 144 Point: (0, 144)

3

Use the Null Factor Law to find the x-intercepts. (Make each bracket equal to 0 and solve a mini-equation.)

x-intercepts: if y = 0, x − 6 = 0 or 4 − x = 0 x = 6 or x=4 Points: (6, 0), (4, 0)

4

Combine all information and sketch the graph. Note the skimming of the x-axis indicative of a repeated factor, in this case the (x − 6)2 part of the expression.

y 144

4 c 1 Positive cubic.

2

Substitute x = 0 to find the y-intercept.

122 Maths Quest 11 Mathematical Methods CAS

6

x

c y = (x − 2)3

y-intercept: if x = 0, y = (−2)3 = −8

3

Use the Null Factor Law to find the x-intercept. (Make each bracket equal to 0 and solve a mini-equation.)

4

Combine all information and sketch the graph. The cubed factor, (x − 2), indicates a point of inflection and only one x-intercept.

x-intercept: if y = 0, x−2=0 x=2 y

x

2 −8

If a cubic function is not in the form f (x) = (x − a)(x − b)(x − c), we may try to factorise to find the x-intercepts. We can use the factor theorem and division of polynomials to achieve this. Worked example 20

Sketch the graph of y = 2x3 − 11x2 + 7x + 20, showing all intercepts. Think 1

Write the equation and name the polynomial P(x).

2

Note the graph is a positive cubic.

3

Let x = 0 to find the y-intercept. Note: All terms involving x are equal to zero.

4

Factorise P(x) to find x-intercepts. (x − 1) is not a factor.

TUTorial eles-1415 Worked example 20

WriTe/draW

y = P(x) = 2x3 − 11x2 + 7x + 20

y-intercept: if x = 0, y = 20 Point: (0, 20) P(1) = 2 × 13 − 11 × 12 + 7 × 1 + 20 = 2 − 11 + 7 + 20 = 18 ≠0 P(−1) = 2 × (−1)3 − 11 × (−1)2 + 7 × (−1) + 20 = −2 − 11 − 7 + 20 =0 So (x + 1) is a factor.

5

Use long or short division to factorise P(x). Here, short division has been used.

P(x) = (x + 1)(2x2 − 13x + 20) = (x + 1)(2x − 5)(x − 4)

6

Write down the x-intercepts (determined by making each bracket equal to 0 and solving for x).

x-intercepts: if y = 0, x = −1, 52 , 4

Points: (−1, 0), ( 52 , 0), (4, 0) 7

Use all available information to sketch the graph.

y 20

−1

5– 2

4

x

ChapTer 3 • Cubic and quartic functions

123

Worked example 21

Sketch the graph of y = 3x3 + x2 − 2x + 5 using a CAS calculator. Find all intercepts and stationary points. Think

WriTe/draW

1

Enter the rule for the function into the CAS calculator.

f1(x) = 3x3 + x2 − 2x + 5

2

Determine the y-intercept (when x = 0).

f1(0) = 3(0)3 + (0)2 − 2(0) + 5 =5 The y-intercept is (0, 5).

3

Use the CAS calculator to help sketch the graph.

A

y

(0, 5) B

x

(−1.51, 0)

4

Analyse the function with the CAS calculator to find zeros (x-intercepts). In this case there is only one x-intercept.

The x-intercept is (−1.51, 0).

5

Analyse the function with the CAS calculator to find the local maximum, point A.

The local maximum (A) is (−0.62, 5.9).

6

Analyse the function with the CAS calculator to find the local minimum, point B.

The local minimum (B) is (0.36, 4.6).

exercise 3h

Cubic graphs — intercepts method

1 We19 Sketch the following, showing all intercepts. a y = (x − 1)(x − 2)(x − 3) c y = (x + 8)(x − 11)(x + 1) e y = (4x − 3)(2x + 1)(x − 4) 2

Sketch the following.

a y = (2 − x)(x + 5)(x + 3) c y = x(x + 1)(x − 2) e y = 4x2(x + 8)

b y = (x + 8)(x − 8)(2x + 3) d y = 3(x + 1)(x + 10)(x + 5) f y = (6x − 1)2(x + 7)

Sketch each of the following as fully as possible.

3 We20 a c e g

b y = (x + 6)(x + 1)(x − 7) d y = (2x − 5)(x + 4)(x − 3) f y = (x − 3)2(x − 6)

y = + 2x2 − x − 2 y = x3 + 7x2 + 14x + 8 y = x3 − 3x2 − 25x − 21 y = 6x3 − 17x2 + 6x + 8 x3

b d f h

y = x3 + 6x2 + 11x + 6 y = x3 − x2 − 14x + 24 y = 3x3 + 17x2 + 28x + 12 y = −2x3 − 18x2

4 We21 Sketch the following using a CAS calculator. Find the intercepts and the coordinates

of all turning points, correct to 2 decimal places. y = −x3 − 8x2 − 5x + 14 y = −x3 + 8x2 + 13x − 140 y = 3x3 − 3x2 − 15x − 9 y = 7x3 + 29x2 + 32x + 4

a b c d 124

Maths Quest 11 Mathematical Methods CAS

5 mC Which of the following is a reasonable sketch of y = (x + 2)(x − 3)(2x + 1)? y

a

−3

1 – 2

−2

− 1– 2

−2

2 x

y

d

y

B

y 3

C

−2

y

e

1– 2

3 x

3 x

2

y

6 mC The graph shown on the right could be that of: a y= + 2) C y = (x − 2)(x + 2)2 e y = (x − 2)3

x2(x

x

− 1– 2

3 x

− 1– 2

diGiTal doC doc-9730 Cubic graphs — general form

B y = (x + d y = (x − 2)2(x + 2)

2)3

−2

2

x

−8 7 mC The graph below has the equation: y

−3

−1

2

x

−6 a y = (x + 1)(x + 2)(x + 3) d y = (x + 1)(x + 2)(x − 3)

B y = (x + 1)(x − 2)(x + 3) e y = (x − 1)(x − 2)(x − 3)

C y = (x − 1)(x + 2)(x + 3)

8 mC If a, b and c are positive numbers, the equation of the graph shown below is: a y = (x − a)(x − b)(x − c) d y = (x + a)(x + b)(x + c)

B y = (x + a)(x − b)(x + c) e y = (x − a)(x + b)(x − c)

C y = (x + a)(x + b)(x − c)

y

−b

c

a

x

9 mC Which of the following has only two distinct x-intercepts when graphed? a y = x(x + 1)(x + 2) d y = (x + 1)(x + 2)2

10 a b c d

B y = (x + 1)(x + 2)(x + 3) e y = x(x + 1)(x − 1)

C y = x3

Sketch the graph of y = x3 − x2 + 3x + 5 showing all intercepts. Factorise y = x3 − x2 + 3x + 5 expressing your answer in the form of y = (x + a)(x2 + bx + c). Hence, show that y = x3 − x2 + 3x + 5 has only one real solution. (Hint: Consider the discriminant.) If y = (x + a)(x2 + kbx + c) where k is a constant, find the values of k such that the cubic has: i two real solutions ii three real solutions. Note: a, b and c are the same values from part b. ChapTer 3 • Cubic and quartic functions

125

3I

Quartic graphs — intercepts method

Quartic functions are polynomial functions of degree 4. The graphs shown below are the main types of quartic graphs. y y = x4

y = (x − a)(x − b)(x − c)(x − d) y

y = (x − a)3(x − b) y

(2, 16) 0

2

x

a repeated factor (x − a)3

b

x a

b

d x

c

Negative quartics are reflected across the x-axis. Consider the general factorised quartic, f (x) = (x – a)(x – b)(x – c)(x – d). As for the cubic functions, the Null Factor Law tells us that f (x) = 0 (that is, an x-intercept occurs) when x = a, x = b, x = c or x = d. The y-intercept occurs when x = 0, therefore the y-intercept is f (0) = (0 – a)(0 – b)(0 – c)(0 – d) = abcd

y abcd

a

b

c

d x

Worked Example 22

Sketch the following graphs, showing all intercepts. a y = (x − 2)(x − 1)(x + 1)(x + 3) b y = (x − 3)2(x + 1)(x + 5) c y = (2x − 1)(x + 1)3 Think

a 1 The y-intercept occurs when x = 0.

Write/Draw

a y = (x − 2)(x − 1)(x + 1)(x + 3)

y = (−2)(−1)(1)(3) =6 Point: (0, 6)

2

Use the Null Factor Law to find the x-intercepts. (Make each bracket equal to 0 and solve a mini-equation.)

x-intercepts: if y = 0, x − 2 = 0, x − 1 = 0, x + 1 = 0 or x + 3 = 0 x = 2, x = 1, x = −1 or x = −3 Points: (2, 0), (1, 0), (−1, 0), (−3, 0)

3

The graph has a positive x4 coefficient, so large positive values for x result in large positive values for y.

Shape:

4

Combine information from steps 1 to 3 to sketch the graph.

y

6 −3

b 1 The y-intercept occurs when x = 0.

2

Use the Null Factor Law to find the x-intercepts. (Make each bracket equal to 0 and solve a mini-equation.)

126 Maths Quest 11 Mathematical Methods CAS

−1 0

1

2

x

b y = (x − 3)2(x + 1)(x + 5)

y = (−3)2(1)(5) = 45 Point: (0, 45)

x-intercepts: if y = 0, x − 3 = 0, x + 1 = 0 or x + 5 = 0 x = 3, x = −1 or x = −5 Points: (3, 0), (−1, 0), (−5, 0)

3

The graph has a positive x4 coefficient. (Large positive values for x result in large positive values for y.)

4

Combine all the information above to sketch the graph. Note that it touches the x-axis where there are repeated squared factors, (x − 3)2.

Shape:

y 45 −1 0

−5

c 1 The y-intercept occurs when x = 0.

2

Use the Null Factor Law to find the x-intercepts. (Make each bracket equal to 0 and solve a mini-equation.)

x

3

c y = (2x − 1)(x + 1)3

y = (−1)(1)3 = −1 Point: (0, −1)

x-intercepts: if y = 0, 2x − 1 = 0 or x + 1 = 0 x = 12 or x = −1 Points: (1, 0) and (−1, 0) 2

x4

3

The graph has a positive coefficient. (Large positive values for x result in large positive values for y.)

4

Combine all the information from above to sketch the graph. Note that the graph has a point of inflection where it crosses the x-axis with repeated cubic factors (x + 1)3.

Shape:

y

0

−1

1– 2

x

−1

Follow the instructions in worked example 21 to draw quartic graphs using a CAS calculator.

exercise 3i

Quartic graphs — intercepts method

1 We22 Sketch the following graphs, showing all intercepts. a y = (x − 3)(x − 2)(x + 1)(x + 2) b y = (x − 2)2(x + 1)(x + 2) 3 c y = (x + 5)(x − 1) d y = (x − 1)4 − e y = x(2x − 1)(x – 3)(x + 3) f y = (x − 2)2(x + 1)2 4 g y = (1 − 3x) h y = (x + 5)3(1 − x) 2 Compare your answers to question 1 to those found using a CAS calculator. 3 If the graph of y = (x + a)(x + 3)(x + 1)(x − 3) has four distinct x-intercepts and has a y-intercept at

(0, 45), find the value of a. 4 A quartic graph has only two x-intercepts, at x = a and x = b, and a y-intercept at the point (0, 81).

If a = −b: a find a possible equation for the quartic graph b sketch the graph, labelling all intercepts.

5 mC Which of the following has two distinct x-intercepts when graphed? a B C d e

y = x(x + 7)(x − 7)(x + 2) y = x2(x + 3)(x − 3) y = x3(x + 27) y = x(x + 7)2(x − 7) y = x4

ChapTer 3 • Cubic and quartic functions

127

y

6 MC If a, b and c are positive numbers, the equation of the graph

shown is: A y = (x + a)2(x + b)(x + c) C y = (x + a)2(x − b)(x + c) E y = (x − a)2(x + b)(x − c)

B y = (x − a)2(x − b)(x − c) D y = (x − a)(x + b)(x – c)

−b

−abc

a

c

x

7 MC For the graph of the quadratic equation y = (x − 2)(x + 1)(x + 3)2,

the y-intercept occurs at: A 6 b −6 C −12 D −18 E 18 8 Use a CAS calculator to help you sketch the following quartics, showing all intercepts and turning points correct to 2 decimal places. a y = 2x4 + x3 − 5x2 + 7x + 2 b y = 3x4 − 9x3 − 8x2 + 12x + 9 c y = −x4 − 2x3 + 5x2 + 4x d y = x4 − 5x3 − 45x2 + 8x + 120 e y = 3x4 − 10x2 − 3 f y = −8x4 − 10x3 + 120x2 + 15x + 358

Graphs of cubic functions in power function form 3J

Remember the power form or turning point form for quadratic graphs y = a(x − b)2 + c, which was related to transformations of the basic parabola? The same understanding of transformations can be used to sketch cubic functions. Cubic functions can be power functions. Power functions are functions of the form f (x) = xn, n ∈ R. The value of the power, n, determines the type of function. When n = 1, f (x) = x, and the function is linear. When n = 2, f (x) = x2, and the function is quadratic. When n = 3, f (x) = x3, and the function is cubic. When n = 4, f (x) = x4, and the function is quartic. Other power functions will be discussed later. Under a sequence of transformations of f (x) = xn, n ∈ R, the general form of a power function is f (x) = a(x − b)n + c (where a, b, c and n ∈ R). Although all linear and quadratic polynomials are also linear and quadratic power functions, this is not the case for cubic functions (or quartic functions). For example, a cubic power function in the form of f (x) = a(x − b)n + c has exactly one x-intercept and one stationary point of inflection. A cubic polynomial in the form f (x) = ax3 − bx2 + cx + d can have one, two or three x-intercepts and is therefore not always a power function. For example, the cubic function y = 2(x − 3)3 + 1 is a polynomial and a power function. It is the graph of y = x3 under a sequence of transformations. A cubic power function has a stationary point of inflection at (b, c). A stationary point of inflection is where a graph ‘levels off’ to have a zero gradient at one point with the same sign gradient either side. y

y = x3

y = −x3 y

x

x

Stationary point of inflection

Summary of transformations y = a(x − b)3 + c

y (b, c)

Dilation factor from the x-axis (y-stretch) 128 Maths Quest 11 Mathematical Methods CAS

x-translation

y-translation

b

c

Stationary point of inflection x

The effect of a is illustrated below. y

y = 2x3 y = x3 y = 1–2 x3

y = −2x3 y y = −x3 y = − 1–2 x3

x

x

Positive a

Negative a

intercepts Intercepts may be found by substituting x = 0 (to find the y-intercepts) and y = 0 (to find the x-intercepts) into the equation. Worked example 23

Sketch the graph of each of the following, showing the stationary point of inflection and intercepts. a y = 3(x − 2)3 + 3 b y = −2x3 + 54 c y = −2(1 − 2x)3 − 16 Think

a 1 Compare the equation with y = a(x − b)3 + c, which

WriTe/draW

a y = 3(x − 2)3 + 3

has a stationary point of inflection (b, c). 2

Note the values that match, namely a = 3, b = 2 and c = 3. State the stationary point of inflection (b, c).

Stationary point of inflection (2, 3)

3

Find the y-intercept.

If x = 0,

4

Find the x-intercept. (Note that the cube root of −1 is −1.)

If y = 0,

5

Note that the equation is for a positive cubic.

Shape:

6

Sketch, showing the stationary point of inflection and intercepts.

y = 3(0 − 2)3 + 3 y = 3(−8) + 3 y = −21 0 = 3(x − 2)3 + 3 − 2)3 = 3 (x − 2)3 = −1 x − 2 = −1 x=1

−3(x

y

(2, 3)

1

2

x

−21 b 1 Manipulate into y = a(x − b)3 + c form. 2

Note the graph is a negative cubic with stationary point of inflection (0, 54).

b y = −2x3 + 54

y = −2(x − 0)3 + 54

Stationary point of inflection (0, 54)

ChapTer 3 • Cubic and quartic functions

129

3

Find the y-intercept.

If x = 0,

4

Find the x-intercept.

If y = 0,

5

Sketch, showing the stationary point of inflection and intercepts.

y = −2(0) + 54 y = 54 0 = −2x3 + 54 2x3 = 54 x3 = 27 x=3

y (0, 54)

x

3

c 1 Manipulate into y = a(x − b)3 + c form.

c y = −2(1 − 2x)3 − 16

= −2[−2(x − 1)]3 − 16 2

= −2[−8(x − 12)3] − 16 = 16(x − 12)3 − 16 2

Note the graph is a positive cubic with stationary point 1 of inflection ( 2 , −16).

Stationary point of inflection ( 12 , −16)

3

Find the y-intercept.

If x = 0,

−

1

y = 16( 2 )3 − 16 = −2 − 16 = −18

4

Find the x-intercept. Use the result from step 1.

1

0 = 16( x − 2 )3 − 16

If y = 0,

1

16 = 16( x − 2 )3 1

1 = ( x − 2 )3 1=x − 3 2

5

Sketch, showing stationary point of inflection and intercepts.

=x

y

3– 2

−18

x

( 1–2, −16)

Graphs of cubic functions in power function form exercise 3J 1

130

Without sketching graphs for each of the following, state: i the dilation factor ii the coordinates of the stationary point of inflection. a y = 2(x − 1)3 + 3 b y = 3(x + 5)3 − 2 − 3 c y = 2(x − 6) − 8 d y = −7(x + 4)3 + 1 3 e y = (x − 9) + 4 f y = x3 − 7

Maths Quest 11 Mathematical Methods CAS

1 2

1

g y = −(x + 1)3 − 1

y= k y= i

−1 4 (x −1 3 x 3

h y = 2 (x + 2)3

− 3)3 + 2

j l

y = 4x3 y = −2x3 − 2

Sketch the graph of each of the following, showing the stationary point of inflection and intercepts. a y = 2(x − 2)3 + 2 b y = −3(x + 3)3 + 81 3 c y = 4(x − 4) − 32 d y = −5(x − 1)3 + 5 − 3 e y= x −8 f y = x3 − 1 1 3 g y = (x + 2) + 27 h y = (x + 5)3 − 32 2

2 We23

i

y=

k y= m y=

−1 3 3 (x − 3) − 9 1 (x + 2)3 + 25 5 5x3

−1 4

j

y=

(x + 1)3 + 2

l

y = −2x3

n y = 3x3 − 3

3 Sketch the following, showing the stationary point of inflection. Intercepts are not required. Use a

CAS calculator to verify answers. a y = (4 − x)3 + 1 c y = 2(4x − 1)3

b y = 3(5 − x)3 − 3 d y = 5(3 − 2x)3 + 1

y = − 17 (3 − 4x)3 − 2 h y = (9 − 5x)3 − 7 j y = −2(5 − 2x)3 − 1

2

e y = 5 (1 − x)3

f

g y = −(4 − x)3 + 3 i

diGiTal doC doc-9731 Cubic graphs — basic form

y = 83 (6 − x)3 + 4

4 mC The basic cubic graph y = x3 undergoes a dilation factor of 6 from the x-axis and is translated

right 4 units and down 3 units. The equation for this graph is:

a y = 6(x − 4)3 − 3 C y = −6(x − 3)3 − 4 e y = −4(x + 3)3 + 6

B y = 3(x − 4)3 − 6 d y = 4(x + 6)3 + 3

5 mC The graph of y = 5(2 − x)3 + 9 has a stationary point of inflection at: a (5, 2) d (2, −9)

B (5, 9) e (2, 9)

C (−2, 9)

6 Suggest a possible equation for each of the following, given that each is a cubic with a dilation factor of

1 or −1 from the x-axis.

a

y

y

b (1, 5)

(−2, 2) x

x y

c

y

d

(−3, 4)

(−3, 0)

x

x

7 Write an equation for a cubic with: a a dilation factor of 4 from the x-axis and a stationary point of inflection at (2, 3) b a dilation factor of –2 from the x-axis and a stationary point of inflection at (−5, 1) c a dilation factor of 1 from the x-axis and a stationary point of inflection at (1, −2) d a dilation factor of

4 −1 2

from the x-axis and a stationary point of inflection at (0, 4).

diGiTal doC doc-9732 Investigation Graphs of the form y = a (x − b )n + c

ChapTer 3 • Cubic and quartic functions

131

domain, range, maximums and minimums 3k

Range = [−8, 10]

The domain of a function is the set of x-coordinates of points on its graph. The range is the set of y-coordinates of points on the graph. Normally, the domain and range of a cubic function are the set of all real numbers, or R for short, as such graphs extend indefinitely in both positive and negative axis directions. The domain and range of a restricted cubic function may be a smaller set of numbers. Actual maximum (within given domain) y Local maximum 10

−5

3

x

−8 Domain = [−5, 3]

Range = [−8, 10]

The restricted graph has a domain of x-values between −5 and 3, denoted [−5, 3]. y The range is [−8, 10]. 10 Square brackets are used to indicate that an end value is included. We show this with a small coloured-in circle on the graph. If an end value is not included, a curved bracket is used. We show such points on a graph using a ‘hollow’ circle. −5

3

x

−8 Domain = [−5, 3]

Function notation

When we wish to convey information about the domain of a function, the following notation may be used: f : [−4, 1] → R, where f (x) = (x − 1)(x + 2)(x + 4)

}

} The ‘name’ of the function

The domain The co-domain. The range is within this set.

The rule for the function

Note: The range is not necessarily equal to R; the range is within R. Worked example 24

For the function f: [−4, 1] → R where f (x) = (x − 1)(x + 2)(x + 4), sketch the graph of f (x), showing intercepts and the coordinates of any local maximum or local minimum, and state the range. Think

132

WriTe/draW

1

Determine the y-intercept by setting x = 0.

f (x) = (x − 1)(x + 2)(x + 4) f (0) = (0 − 1)(0 + 2)(0 + 4) = −8 The y-intercept is (0, −8).

2

Determine the x-intercept by setting y = 0. Use the Null Factor Law.

0 = (x − 1)(x + 2)(x + 4) x − 1 = 0 or x + 2 = 0 or x + 4 = 0 x=1 x = −2 x = −4 − The x-intercepts are (1, 0), ( 2, 0) and (−4, 0).

Maths Quest 11 Mathematical Methods CAS

3

Enter the rule into a CAS calculator (including the restricted domain).

4

Sketch the graph over the restricted domain.

f (x) = (x − 1)(x + 2)(x + 4) | −4 ≤ x ≤ 1 y A (−2, 0)

(1, 0) x

(−4, 0)

B 5

Use the CAS calculator to determine the maximum at point A.

The maximum (A) is (−3.12, 4.06).

6

Use the CAS calculator to determine the minimum at point B.

The minimum (B) is (−0.21, −8.21).

7

Use the graph and the local maximum and minimum to determine the range.

The range is [−8.21, 4.06].

y

The maximum and minimum of a graph within a certain domain are not necessarily the values of the local maximum or minimum. Sometimes an extreme value is simply the y-coordinate of an end point of a graph.

Absolute maximum

Local maximum

Local minimum x Absolute minimum Worked example 25

Sketch f: [0, 7) → R where f (x) = x(x − 5)2, showing intercepts, end points, and the local maximum and minimum, and state the range. Think 1

Determine the y-intercept by setting x = 0.

2

Determine the x-intercept by setting y = 0. Use the Null Factor Law.

3

Calculate the value of the end points of the restricted domain.

4

Use a CAS calculator to sketch the graph over the restricted domain. Note the right end point (7, 28) should be shown as a hollow circle.

WriTe/draW

f (x) = x(x − 5)2 f (0) = 0(0 − 5)2 =0 The y-intercept is (0, 0). 0 = x(x − 5)2 x = 0 or (x − 5)2 = 0 → x = 5 The x-intercepts are (0, 0) and (5, 0). f (0) = 0 f (7) = 7(7 − 5)2 = 7 × 22 = 28 y (7, 28) A

(0, 0)

(0, 5)

x

ChapTer 3 • Cubic and quartic functions

133

5

Use the CAS calculator to determine the local maximum at point A.

The local maximum is (1.67, 18.52).

6

Use the graph to determine the range. Note the use of a curved bracket to indicate that the end value is not included in the range.

The range is [0, 28).

Domain and range will be discussed in more detail in chapter 4. A method of finding maximums and minimums without a calculator will be covered in the study of calculus later in this book.

domain, range, maximums and minimums exercise 3k

A graphics calculator is required for this exercise. State the domain and range of the sections of graph shown in each case.

1

y

a diGiTal doC doc-9733 SkillSHEET 3.2 interval notation

y

b

(−2, 5)

(1, 7)

(−5, 2)

(−2, 3)

(4, 0) x

x

(2, −2) c

(4, −2) d

y

(6, 10)

y

(−4, 2) x

(4, 1) x

(2, −3)

(−2, −2)

(−3, −5) y

e

(5, 5)

y

f

(2, 0)

(−1, 2)

(−3, 0)

x

(3, 0)

x

(−3, −8)

(4, −5)

(−1, −9) 2 We24,25 For each of the following, sketch the graph (showing local maximums and minimums, and

intercepts) and state the range. a f: [−1, 4] → R where f (x) = (x − 3)(x − 4)(x + 1) b f: [−5, 1] → R where f (x) = (x + 2)(x + 5)(x − 1) c f: [1, 3) → R where f (x) = (x − 2)2(x − 1) d f: (−3, 0] → R where f (x) = (x + 3)(x + 1)2 e f: [−8, 2) → R where f (x) = (2x − 3)(x + 1)(x + 7) f f: [0, 4] → R where f (x) = x3 − 6x2 + 9x − 4 g f: [−4, −1.442] → R where f (x) = x3 + 6x2 + 11x + 6 h f: (−2, 2.1) → R where f (x) = x3 + 2x2 − 5x − 6 i f: [−3, 5.1] → R where f (x) = −x3 + x2 + 17x + 15 1 j f: (−3, 2 ) → R where f (x) = 3x3 + 5x2 − 19x − 21 3 134

Maths Quest 11 Mathematical Methods CAS

3 mC The range of the function shown at right is: a

[−4.061,

18]

y (−0.786, 8.209)

B [−4.061, 8.209]

(4, 18)

6

C (0, 18] d (−2, 3] e (−2, 4]

x 1 3 (2.120, −4.061)

−2 y

4 mC Point A on the curve is: a B C d e

an intercept a local minimum an absolute minimum a local maximum an absolute maximum

A x y

5 A roller-coaster ride is modelled by the function

f (x) = 0.001(x − 10)(x + 20)(x − 40). a What is the height above ground level of the track at x = 50? b How far apart vertically are points A and B?

−20

20 A

Ground level 40

x

B

6 The course of a river as marked on a map follows the curve defined by the function

f (x) = 1.5x3 − 2.7x2 + x − 1. Find the coordinates of the southernmost point on the river between x = 0 and x = 2.

y

N

5 −1.5

1.5 −5

x diGiTal doC doc-9734 Investigation modelling the path of a roller-coaster

ChapTer 3 • Cubic and quartic functions

135

3l inTeraCTiViTY int-0262 modelling data using polynomials

modelling using technology

Scientists, economists, doctors and biologists often wish to find an equation that closely matches, or ‘models’, a set of data. For example, the wombat population of a particular island may vary as recorded in the following table. Year of study (x) Wombat population (W)

0

1

2

3

4

5

6

7

8

9

10

59

62

69

83

81

76

70

66

52

49

41

Wombat population

The graph below shows these data, with a possible model for the wombat population superimposed. 90 80 70 60 50 40 30 20 y = 0.0888x3 − 2.4598x2 + 14.196x + 55.063 10 0

2

4 6 Year

8 10

We will examine polynomial models up to degree 3, that is, models of the form: y y = a3x3 + a2x2 + a1x + a0 35 where a0, a1, a2 and a3 are constants. 30 Several technological options are available to assist in 25 Sum of squares of these obtaining models for data, including a CAS calculator, 20 lengths is minimised. spreadsheets and computer algebra systems such as 15 Mathcad. Many of these applications use a method 10 y = 2.6636x + 7.3182 5 involving minimising the sum of the squares of the vertical distances of the data points from the graph of the 0 2 4 6 8 10 12 x function — this is known as the least squares method.

modelling using a calculator A CAS calculator can be used to find a model or regression for a set of data. The following example employs cubic regression, but the general approach is the same for all types of regression. Worked example 26

Fit a cubic model to the following data using a CAS calculator. Write the equation and draw a rough sketch of the graph. 0 1 2 3 4 5 6 7 8 9 10 x 627 545 580 528 436 318 238 229 134 169 139 y Think 1 2 3

WriTe/draW

Use the CAS calculator to perform a cubic Enter the x- and y-values into two columns in regression. This fits a best-fit cubic to the given data. a spreadsheet on the CAS calculator. Write the equation. y = 1.37x3 + 19.14x2 + 8.55x + 607.48 Use the CAS calculator to sketch the graph over the even domain.

y 600

0

136

TUTorial eles-1416 Worked example 26

Maths Quest 11 Mathematical Methods CAS

10

x

modelling using technology

exercise 3l

Use a CAS calculator or other technology to answer the questions in this exercise. 1 We26 Find a linear model for each of the following sets of data, and draw a rough sketch of the graph. a

x

0

1

2

3

4

5

6

7

8

9

10

y

−30

0

5

−9

28

29

50

68

73

77

84

x

0

1

2

3

4

5

6

7

8

9

10

y

−15

−12

−26

−27

−12

−20

−39

−46

−50

−40

−67

c

x y

0 11

1 8

2 9

3 14

4 19

5 18

6 29

7 29

8 28

9 32

10 39

d

x y

0 53

1 44

2 39

3 42

4 35

5 32

6 30

7 29

8 23

9 27

10 19

b

diGiTal doC doc-9735 modelling

2 Find a quadratic model for each of the following sets of data, and draw a rough sketch of the graph. a

x y

0 19

1 4

2 48

3 60

4 36

5 88

6 126

7 116

8 159

9 168

10 122

b

x y

0 65

1 33

2 80

3 12

4 50

5 248

6 228

7 252

8 496

9 439

10 694

c

x y

0 −14

1 16

2 32

3 36

4 37

5 51

6 57

7 56

8 55

9 54

10 56

0

1

2

3

4

5

6

7

8

9

10

70

−27

2

−148

−327

−447

−639

−733

−910

−1204

d

x y

92

3 Find the cubic model for each of the following sets of data, and draw a rough sketch of the graph. a

x y

0 627

1 545

2 580

3 528

4 436

5 318

6 238

7 229

8 134

9 169

10 139

b

x y

0 21

1 28

2 91

3 182

4 81

5 203

6 345

7 397

8 730

9 873

10 1205

c

x

0

1

2

3

4

5

6

7

8

9

10

35

182

400

631

7 755

8 800

9 868

10 854

d

y

55

84

64

29

10

4

−17

x y

0 45

1 26

2 109

3 201

4 399

5 466

6 621

4 For the following data set, find and sketch: a a linear model b a quadratic model

x y

0 537

1 681

2 536

3 624

4 632

5 763

c a cubic model.

6 686

7 885

8 1090

9 1230

10 1451

5 Which of the models in question 4 fits best? 6 Use the model from question 1a to predict the value of y when x = 20. 7 Use the model from question 2a to predict the value of y to the nearest unit when x = 5.5. 8 Use the model from question 3a to predict the value of y to the nearest unit when x = 12. ChapTer 3 • Cubic and quartic functions

137

9 The value of shares in the company Mathsco is plotted by a sharemarket analyst over a

12-month period as shown. 0 J

Month Share price

1 J

2 A

3 S

4 O

5 N

6 D

7 J

8 F

9 M

10 A

11 M

0.50 0.58 0.53 0.76 1.00 1.50 1.55 2.20 3.06 3.83 4.79 4.40

a Find and sketch a quadratic model for the data. b Use your model to predict the share price 2 months later. c Give reasons why such a prediction may not be accurate. 10 The population of a colony of yellow-bellied sap-suckers on an isolated island is studied

over a number of years. The population at the start of each year is shown in the table below. Year Population

0

1

2

3

4

5

6

7

8

9

10

250

270

310

375

410

395

335

290

290

320

325

Find and sketch a cubic model for the population, and use it to estimate the population at the start of year 11.

3M

Finite differences

If pairs of data values in a set obey a polynomial equation, that equation or model may be found using the method of finite differences. Consider a difference table for a general polynomial of the form y = a3x3 + a2x2 + a1x + a0. We begin the difference table by evaluating the polynomial for x values of 0, 1, 2 etc. The differences between successive y-values (see table) are called the first differences. The differences between successive first differences are called second differences. The differences between successive second differences called the third differences. We will call the first shaded cell (nearest the top of the table) stepped cell 1, the second shaded cell stepped cell 2 and so on. x

y (= a3 x3 + a2 x2 + a1x + a0)

0

a0

First differences

Second differences

Third differences

a3 + a2 + a1 1

a3 + a2 + a1 + a0

6a3 + 2a2 7a3 + 3a2 + a1

2

8a3 + 4a2 + 2a1 + a0

6a3 12a3 + 2a2

19a3 + 5a2 + a1 3

27a3 + 9a2 + 3a1 + a0

6a3 18a3 + 2a2

37a3 + 7a2 + a1 4

64a3 + 16a2 + 4a1 + a0

6a3 24a3 + 2a2

61a3 + 9a2 + a1 5

125a3 + 25a2 + 5a1 + a0

If a3 ≠ 0, the above polynomial equation represents a cubic model, and the third differences are identical (all equal to 6a3). If a3 = 0, a2 ≠ 0 and the polynomial reduces to y = a2x2 + a1x + a0, that is, a quadratic model, and the second differences become identical (all equal to 2a2). 138 Maths Quest 11 Mathematical Methods CAS

If a3 = 0 and a2 = 0, the polynomial becomes y = a1x + a0, that is, a linear model, and the first differences are identical (all equal to a1). 1. Stepped cell 1 = a0 2. Stepped cell 2 = a1 + a2 + a3 3. Stepped cell 3 = 2a2 + 6a3 4. Stepped cell 4 = 6a3 Worked Example 27

Complete a finite difference table based on the data below, and use it to determine the equation for y in terms of x. x

0

1

2

3

4

5

y

−1

0

7

20

39

64

Think 1

2

3

Place the data in columns as shown, allowing space for 3 difference columns. Calculate the first differences and place them in the next column. The first differences are not constant, so we need to find the second differences. Calculate these and place them in the next column. The second differences are constant, so our table is complete. Showing the third differences is optional. The curve is a quadratic.

Write

x

y

0

−1

Differences 1st

2nd

3rd

1 1

0

6 7

2

0

7

6 13

3

0

20

6 19

4

39

5

64

0 6

25

4

Recall the stepped cell equations, and equate them to the shaded cells as shown: Stepped cell 1 = a0 Stepped cell 2 = a1 + a2 + a3 Stepped cell 3 = 2a2 + 6a3 Stepped cell 4 = 6a3 Here, * is used to denote solved values.

a0 = −1* [1] a1 + a2 + a3 = 1 [2] 2a2 + 6a3 = 6 [3] 6a3 = 0 [4]

5

[1] gives a0 = −1 and [4] gives a3 = 0. Substitute this information into [2] and [3].

a3 = 0* Sub a3 = 0 into [2]: a1 + a2 + 0 = 1 a1 + a2 = 1 Sub a3 = 0 into [3]: 2a2 + 6 × 0 = 6 2a2 = 6 a2 = 3* Sub a2 = 3 into [5]: So

[5]

6

Substitute a2 = 0 into [5] to find a1.

a1 + 3 = 1 a1 = −2*

7

Substitute your values for a0, a1, a2 and a3 into the equation y = a3x3 + a2x2 + a1x + a0.

y = a3x3 + a2x2 + a1x + a0 becomes y = (0)x3 + (3)x2 + (−2)x + (−1) y = 3x2 − 2x − 1

Chapter 3 • Cubic and quartic functions 139

The stepped equations work only if the finite differences table begins with x = 0 and x increases in steps of 1. It may be necessary on occasions to adjust the table to achieve this, as the following example shows. Worked Example 28

Complete a finite difference table based on these data and use it to determine the equation for y in terms of x. Think 1

2

3

4

Construct a difference table, leaving room for the x = 0 row. Calculate and fill in the first differences where possible. Note the first differences are constant, so the last two columns are optional. The relationship is linear. Working ‘backwards’, the first stepped cell must be −11 in order for the difference between it and the next cell to be 8.

1

x y

2 5

−3

3 13

4 21

5 29

Write

x

y

0

−11

Differences 1st

2nd

3rd

8 1

−3

0 8

2

5

0 0

8 3

13

0 0

8 4

21

0 0

8 5 5

Recall the stepped cell equations, and equate them to the shaded cells as shown: Stepped cell 1 = a0 Stepped cell 2 = a1 + a2 + a3 Stepped cell 3 = 2a2 + 6a3 Stepped cell 4 = 6a3 Here, an asterisk (*) is used to denote solved values.

6

[1] gives a0 = −11, [4] yields a3 = 0, and hence [3] yields a2 = 0. Substitute this information into [2].

7

Use the asterisked values to build the equation y = a3x3 + a2x2 + a1x + a0.

29

a0 = −11* [1] a1 + a2 + a3 = 8 [2] 2a2 + 6a3 = 0 [3] 6a3 = 0 [4]

a3 = 0* and a2 = 0* Sub a2 = 0 and a3 = 0 into [2]: a1 + 0 + 0 = 8 a1 = 8* y = a3x3 + a2x2 + a1x + a0 becomes y = (0)x3 + (0)x2 + (8)x + (−11) y = 8x − 11

So

Using simultaneous equations to find a polynomial model The method of fitting a polynomial to a set of data using finite differences requires the data to be sequential. Often this is not the case. Simultaneous equations can be used to find a polynomial model when the data are not sequential. 140 Maths Quest 11 Mathematical Methods CAS

The number of simultaneous equations required to find the rule of a degree n polynomial is n + 1. For example, to find a quadratic model, 2 + 1 = 3 points are required as a quadratic is a degree 2 polynomial. Each of the points are substituted into the general equation of the quadratic polynomial, y = ax2 + bx + c, to generate 3 simultaneous equations. These can be solved using elimination or by using a CAS calculator. Worked example 29

Using simultaneous equations, find a quadratic model for the points (−2, −9), (3, 1) and (1, 9). Think 1

Write down the general rule of a quadratic.

2

Substitute each point into the general equation to get three simultaneous equations.

3

WriTe

y = ax2 + bx + c a(−2)2 + b(−2) + c = −9 ∴ 4a − 2b + c = −9 a(3)2 + b(3) + c = 1 ∴ 9a + 3b + c = 1 a(1)2 + b(1) + c = 9 ∴a+b+c=9

Solve equations [1], [2] and [3] using elimination. Equation [1] − [2] to eliminate c. Equation [2] − [3] to eliminate c. Equation [4] × 8 Equation [5] × 5 Add equations [6] and [7] to eliminate a and solve for b. Substitute b = 4 into equation [4] to find a.

−5a

− 5b = −10 8a + 2b = −8 −40a − 40b = −80 40a + 10b = −40 −30b = −120 ∴b=4 −5a − 5(4) = −10 −5a = 10 ∴ a = −2 −2 + 4 + c = 9 ∴c=7

Substitute a = −2 and b = 4 into equation [3] to find c. 4

Write the rule.

[1] [2] [3] [4] [5] [6] [7]

y = −2x2 + 4x + 7

Worked example 30

Using simultaneous equations, find a cubic model for the points (−2, −10), (1, 2), (3, −20) and (6, 22). Use a CAS calculator to solve the simultaneous equations. Think

WriTe

1

Write the general rule for a cubic.

y = ax3 + bx2 + cx + d

2

Substitute each point into the general equation to get 4 simultaneous equations.

a(−2)3 + b(−2)2 + c(−2) + d = −10 −8a + 4b − 2c + d = −10 3 a(1) + b(1)2 + c(1) + d = 2 a+b+c+d=2 a(3)3 + b(3)2 + c(3) + d = −20 27a + 9b + 3c + d = −20 3 a(6) + b(6)2 + c(6) + d = 22 216a + 36b + 6c + d = 22

[1] [2] [3] [4]

3

Use the CAS calculator’s solve feature to determine a, b, c and d.

The values are a = 1, b = −5, c = −4 and d = 10.

4

Write the rule.

y = x3 − 5x2 − 4x + 10

ChapTer 3 • Cubic and quartic functions

141

exercise 3m

Finite differences

For each of the following, complete a finite difference table based on the data below and use it to determine the equation of y in terms of x.

1 We27,28 a

b

c

d

x

0

1

2

3

4

5

y

6

17

28

39

50

61

x

0

1

2

3

4

5

y

100

74

48

22

−4

−30

x

0

1

2

3

4

5

y

−4

7

32

71

124

191

x y

e

f

g

h

i

j

k

l

2

1

2

3

4

5

1

−9

−13

−11

−3

x

0

1

2

3

4

5

y

−7

−10

−1

26

77

158

x

0

1

2

3

4

5

y

16

17

20

31

56

101

x

0

1

2

3

4

5

y

−23

−11

5

25

49

77

x

1

2

3

4

5

y

28

27

32

31

12

x

0

1

2

3

4

5

y

−27

−18

−9

0

9

18

x y

1

2

3

4

5

−7

−3

−3

−7

−15

x

0

1

2

3

4

5

y

−66

−5

58

99

94

19

x

0

1

2

3

4

5

y

43

35

27

19

11

3

Triangular numbers may be illustrated as shown at right. If x is the number of dots on the base of each diagram, and y is the total number of dots: a complete the table below b find an equation linking x and y x (base dots)

0

1

2

3

4

5

y (total dots) c find the total number of cans in the supermarket display shown at right using the equation found in b, and check your answer by counting the cans.

142

Maths Quest 11 Mathematical Methods CAS

x=1

x=2

x=3

x=4

3

The diagonals in polygons of various types are shown at right in red. Find the relationship between the number of dots (x) and the number of diagonals (n). (Hint: Continue patterns in a difference table so that it is completed back to x = 0.)

4

If n is the number of different squares that can be found within a square grid of edge length x, find an equation for n in terms of x and use this equation to find the number of different squares on a chessboard.

x=1 x=2 n=1 n=5 5 Find a linear model for the following sets of points. a (2, 1) (−1, −23)

x=3 n=?

b (−4, 6) (8, −3)

6 We29 Using simultaneous equations, find a quadratic model for the following sets of points. a (−2, −13) (6, −37) (−4, −57) b (−1, 4) (1, −2) (4, 19) − c (4, 8) (0, 8) ( 4, 24) d (−5, −360) (−2, −96) (6, −272) 7 We30 Using simultaneous equations, find a cubic model for the following sets of points. Use a CAS

calculator to solve the simultaneous equations.

a (−6, 3) (−3, −27) (3, −33) (2, 3) c (4, −10) (6, 90) (8, 302) (−2, 2)

b (−2, −39) (1, 6) (4, 141) (−3, −118) d (−1, −4) (1, −8) (4, −314) (0, −6)

8 Using simultaneous equations, find a quartic model for the following sets of points. Use a

CAS calculator to solve the simultaneous equations. a (1, 2) (−3, 354) (4, 1313) (2, 79) (−1, −2) b (−4, 73) (0, 1) (2, −11) (−2, 13) (6, −707)

ChapTer 3 • Cubic and quartic functions

143

Summary expanding

• When expanding three linear factors: 1. expand two factors first, then multiply the result by the remaining linear factor 2. collect like terms at each stage 3. (x + 2)3 may be written as (x + 2)(x + 2)(x + 2) 4. (a + b)3 = a3 + 3a2b + 3ab2 + b3 (a − b)3 = a3 − 3a2b + 3ab2 − b3

long division of polynomials

• Long division of polynomials is similar to long division with numbers. • The highest power term is the main one considered at each stage. • The key steps are: 1. How many? 2. Multiply and write the result underneath. 3. Subtract. 4. Bring down the next term. 5. Repeat until no variables remain to be divided. 6. State the quotient and the remainder.

polynomial values

• P(a) means the value of P(x) when x is replaced by a and the polynomial is evaluated.

The remainder and factor theorems

• The remainder R = P(a) when P(x) is divided by x − a. • If P(a) = 0, then (x − a) is a factor of P(x).

Factorising polynomials

• To factorise a polynomial: 1. let P(x) = the given polynomial 2. use the factor theorem to find a linear factor 3. use long or short division to find another factor 4. repeat steps 2 and 3, or factorise by inspection if possible. Alternatively, use the factor function on a calculator.

Sum and difference of two cubes

• a3 + b3 = (a + b)(a2 − ab + b2) • a3 − b3 = (a − b)(a2 + ab + b2)

Solving polynomial equations

• To solve a polynomial equation: 1. rewrite the equation so it equals zero 2. factorise the polynomial as much as possible 3. let each linear factor equal zero and solve for x in each case.

Cubic graphs — intercepts method

y • To sketch a cubic function of the form f (x) = Ax3 + Bx2 + Cx + D: 1. determine if the expression is a positive or negative cubic (that is, 20 if A is positive or negative) 2. find the y-intercept (let x = 0) 3. factorise if necessary or possible; for example, obtain an expression in the form f (x) = (x − a)(x − b)(x − c) 4. find the x-intercepts (let factors of f (x) equal 0) −1 5. use all available information to sketch the graph.

Quartic graphs — intercepts method

Maths Quest 11 Mathematical Methods CAS

4

x

y

• To sketch a quartic function in the form f (x) = (x – a)(x – b)(x – c) (x – d): 1. find the y-intercept (f (0) = abcd) 2. find the x-intercepts (let factors of f (x) = 0) 3. use all available information to sketch the graph.

840

−8

144

5– 2

−5

0

3

7

x

Graphs of cubic functions in power function form

• A cubic function in the form y = a(x − b)3 + c has a y-dilation of a and a stationary point of inflection at (b, c). y

y

a>0

a<0

(b, c)

(b, c) x

x

Positive a

Negative a

domain, range, maximums and minimums

• The absolute maximum or minimum is either the y-value at a local maximum or minimum, or the y-value at an end of the domain.

modelling using technology

• An equation to model data may be obtained using a calculator. 1. Enter data as lists. 2. Find the regression equation (linear, quadratic etc.).

Finite differences

• To use the method of finite differences: 1. Set up a table as shown and find differences by subtracting successive values (value – previous value). Circle or shade the stepped cells. x

y

0

Stepped cell 1

First differences

Second differences

Third differences

Stepped cell 2 1

Stepped cell 3 Stepped cell 4

2

3

Etc. Etc.

4

Previous value Value – previous value

5

Value

2. Use the following equations to determine the polynomial model’s coefficients. 1. Stepped cell 1 = a0 2. Stepped cell 2 = a1 + a2 + a3 The equation of the polynomial model is 3. Stepped cell 3 = 2a2 + 6a3 y = a3x3 + a2x2+ a1x + a0. 4. Stepped cell 4 = 6a3

}

ChapTer 3 • Cubic and quartic functions

145

Chapter review S h orT anS Wer

1 Expand: a (x − 2)2(x + 10) c (x − 7)3

b (x + 6)(x − 1)(x + 5) d (5 − 2x)(1 + x)(x + 2).

2 Find the quotient and remainder when the first polynomial is divided by the second in each case. a x3 + 2x2 − 16x − 3, x + 2 b x3 + 3x2 − 13x − 7, x − 3 c −x3 + x2 + 4x − 7, x + 1 3 If P(x) = −3x3 + 2x2 + x − 4, find: a P(1)

b P(−4)

+ − 29x − 42.

4 Without dividing, find the remainder when 5 Show that x + 3 is a factor of

x3

−

2x2

c P(2a).

x3

3x2

− 16x + 5 is divided by x − 1.

6 Factorise x3 + 4x2 − 100x − 400. 7 Factorise: a 1 − 125x3

b (x − 2)3 + (x + 3)3.

8 Solve: a 5(x + 5)3 + 5 = 0

b (2x + 1)2 (x − 3)2 = 0

c x3 − 9x2 + 26x − 24 = 0.

9 Sketch: a y = x(x − 2)(x + 11)

b y = x3 + 6x2 − 15x + 8

c y = −2x3 + x2.

b y = (2x – 1)(x + 1)(x + 4)2

c y = −x(x + 5)3.

10 Sketch: a y = x(x – 7)(x – 2)(x + 4) 11 Sketch y =

−1 8

(x + 1)3 + 8.

12 Find the range of f : [−6, 3] → R, where f (x) = (x + 1)(2 − x)(x + 5). 13 Complete a finite difference table, and use it to determine the equation for y in terms of x for the

following data set. x

0

1

2

3

4

5

y

8

7

8

17

40

83

14 The following series of diagrams show the maximum number of regions

produced by drawing chords in a circle. x=0 x=1 x=2 x=3 Find a relationship between the number of chords (x) and the maximum r = 1 r = 2 r = 4 r = 7 number of regions (r). 15 Using simultaneous equations, find a cubic model for the points (−1, −10), (2, −4), (−3, −104) and (0, −2). mUlTip l e Ch oiCe

1 The expansion of (x + 5)(x + 1)(x − 6) is:

2

a x3 − 30

B x3 − 6x2 + 5x − 6

d x3 − 31x − 30

e x3 + 5x2 − 36x − 30

x3

+

5x2

+ 3x − 9 is the expansion of:

a (x + 3)3

C x3 + 12x2 − 31x + 30

B x(x + 3)(x − 3)

C (x − 1)(x + 3)2

d (x − 1)(x + 1)(x + 3)

e (x + 1)(x + 2)(x − 3) Questions 3 and 4 refer to the following long division.

x2 + x + 2 x + 4)x3 + 5x2 + 6x − 1 x3 + 4x2 x2 + 6x x2 + 4x 2 −1 2x 2 +8 2x −9 3 The quotient is: a −9

146

B 4

Maths Quest 11 Mathematical Methods CAS

C x+4

d x2 + x + 2

e x3 + 5x2 + 6x − 1

4 The remainder is: a −9

B 2

C 4

d 2x − 1

e 2x + 8

d 7

e 35

d 7

e 8

d (x + 3)

e (2x − 1)

5 If P(x) = x3 − 3x2 + 7x + 1, then P(−2) equals: a −34

B −33

C −9

− 7x is divided by x − 1 is: C 6 3 7 Which of the following is a factor of x − 3x2 − 18x + 40? a (x − 4) B (x − 2) C (x + 1) 6 The remainder when a −6

8

9

x3

+

x3

B 1

6x2

− 15x + 8 factorises to:

a (x − 1)2(x + 8)

B (x + 1)2(x + 8)

d (x + 1)(x + 2)(x + 4)

e (x − 1)(x + 2)(x + 4)

−

64x3

y3

C (x + 2)3

factorises to:

a (4x − y)(16x2 + 4xy + y2)

B (4x − y)(16x2 − 4xy + y2)

d (4x + y)(16x2 − 8xy + y2)

e (4x + y)(16x2 − 4xy + y2)

C (4x − y)(16x2 + 8xy + y2)

10 Which of the following is the solution to −(x − 4)3 − 2 = 6? a −6

B −2

C 2

d 4

e 6

11 Which of the following is a solution to (x − 11)(3x + 5)(7 − 3x)(2x + 5) = 0? a −11

B

−3

3

d 3

12 The equation for this graph could be: a B C d e

7

5

C 7

5

e 3

y

y = (x − 5)(x + 1)(x + 3) y = (x − 3)(x − 1)(x + 5) y = (x − 3)(x + 1)(x + 5) y = (3x − 1)(x + 1)(x − 5) y = (5 − x)(1 + x)(3 + x)

−3

−1

5

x

y

13 The equation for the graph shown below could be: a B C d e

y = (x – 3)2(x + 3)2 y = (x – 3)(x + 3)3 y = (x – 3)3(x + 3) y = (x – 3)4 y = (x + 3)4

0

−3

3

x

−81

14 Which of the following shows the graph of y = −2(x + 5)3 − 12? y y a B x

C

y x

(−5, 12)

(−5, −12)

(5, −12) x

d

y

y

e

(5, 12) x

(−5, −12)

x

ChapTer 3 • Cubic and quartic functions

147

Questions 15 and 16 refer to the following graph (below). y (−7, 252) (4.813, 60.370) −5

2

7 x

(−2.147, −108.222)

15 The domain of the graph is: a [−108.222, 252)

B (−2.147, 4.183]

d (−7, 7]

e (−6, 7]

16 The range of the graph is: a [−108.222, 252)

B [−2.147, 4.183]

d [7, 252)

e [0, 252)

C (−108.222, 60.370]

C [−108.222, 60.370]

17 The data below obey which type of relationship?

x

0

1

2

3

4

5

y

0

4

16

66

208

520

a Linear

B Quadratic

C Cubic

d Quartic

e None of the above

18 Which of the following points lies on the curve of the quadratic model that fits the points (1, 0), (0, −7)

and (2, 11)? a (3, −4)

e x T ended r e SponS e

C (−2, −7)

B (0, 7)

d (−1, −10)

e (−1, −14)

1 For P(x) = 5x3 − 3x2 − 6x − 22, find P(3) and P(−x). 2 Find the value of m if x + 3 is a factor of 2x3 − 15x2 + mx − 21. 3 Factorise x3 − 2x2 − 9x + 18. Sketch the graph of f (x) = x3 − 2x2 − 9x +18. 4 Factorise (3x − 2)3 + (x + 5)3. 5 Determine the x- and y-intercepts of the cubic graph y = (2 − 3x)(4x + 1)(2x − 7). Hence, sketch the

graph. 6 The graph y = x3 has been moved parallel to the x-axis 5 units to the left and moved upwards 2 units

from the x-axis. What is the equation of the translated graph and what are the coordinates of the point of inflection? Sketch the translated graph. 7 Sketch the graph of y = (x + 2)2(x − 3)(x − 4), showing all intercepts. 8 The polynomial P(x) = x3 + ax2 + bx + 54 is exactly divisible by x − 9 and also exactly divisible by x − 6. a Find the values of a and b. b Find the third factor. c Hence, sketch the graph of the polynomial y = x3 + ax2 + bx + 54. 9 Factorise x3 − 2x2 − 3x + 6 over the real number field. Sketch the graph of y = x3 − 2x2 − 3x + 6. 10 Find the points of intersection between y = x3 − x2 − 19x − 13 and 3x + y − 7 = 0. 11 Use the method of finite differences to fit a polynomial model to the following data.

x y

148

0 4

1 16

Maths Quest 11 Mathematical Methods CAS

2 25

3 30

4 30

12 A diagram of a proposed waterslide based on a cubic function appears below.

Find: a the height, h1, of the top of the slide b the coordinates of point A (where the slide enters the water) c the length, L, of the ladder d the height, h2, of the ‘mini-hump’ to the nearest centimetre. y

y = −0.008(x3 − 30x2 + 285x − 900)

L

h1 h2

(−5, 0)

A

(8, 0.22)

x

13 An innovative local council decides to put a map of the district on a website. Part of the map involves

two key features — the Cubic River and the Linear Highway. A mathematically able website designer has found the following equations for these features: Cubic River: y = x3 + x2 − 4x − 4 Linear Highway: y = 5x + 5. a Sketch the river and highway, showing x- and y-axis intercepts. b Find the coordinates of the points of intersection of the highway and the river. c A fun-run organiser wishes to arrange checkpoints at the closest points of intersection. Find the distance between the proposed checkpoints. 14 A cubic function in the form f (x) = ax3 + bx2 + cx + d has the following values. x y

0 42

1 36

2 20

3

4

5

0

−18

−28

Use finite differences to find the values of a, b, c and d. State one factor of f (x), giving your reasoning. Using long or short division, factorise f (x). Sketch the graph of f (x), labelling all intercepts. 15 The height (in centimetres) of a wave above a 1-metre pole is measured over an interval of 8 seconds. The wave’s height above the 1-metre pole has been found to approximate the function H1 = t3 − 13t2 + 48t. a Find the initial height of the wave above the 1-metre pole. b Using a CAS calculator, sketch the function and find the local maximum and minimum height of the wave. c The height of a later wave above the 1-metre pole is found to approximate the function H2 = t3 − 14t2 + 53t − 40. Show at what times the height of this wave is exactly the same height as the pole. a b c d

diGiTal doC doc-9737 Test Yourself Chapter 3

ChapTer 3 • Cubic and quartic functions

149

ICT activities Chapter opener diGiTal doC • 10 Quick Questions doc-9724: Warm up with ten quick questions on cubic and quartic functions (page 105)

3B

long division of polynomials

TUTorial • We4 eles-1412: Watch how to perform long division of polynomials (page 108)

3C

polynomial values

diGiTal doCS • doc-9725: Investigate solutions to cubic equations (page 111) • WorkSHEET 3.1 doc-9726: Review the discriminant (page 111)

3d

The remainder and factor theorems

diGiTal doCS • doc-9725: Investigate solutions to cubic equations (page 113) • SkillSHEET 3.1 doc-9727: Practise calculating and using the discriminant (page 114)

3e

Factorising polynomials

diGiTal doC • doc-9728: Investigate zeros of cubics (page 117) TUTorial • We12 eles-1413: Use long division to factorise a cubic (page 114)

3F

Sum and difference of two cubes

TUTorial • We15 eles-1414: Watch how to factorise expressions using the sum or difference of two cubes formulae (page 117)

3G

Solving polynomial equations

diGiTal doC • WorkSHEET 3.2 doc-9729: Factorising cubics and quartics using long division, applying the Null Factor Law to determine x-intercepts and sketching cubics and quartics (page 121)

150

Maths Quest 11 Mathematical Methods CAS

3h

Cubic graphs — intercepts method

TUTorial • We20 eles-1415: Sketch the graph of a cubic showing axial intercepts (page 123) diGiTal doC • doc-9730: Investigate the effect of changing coefficients of cubics in general form on its graph (page 125)

3J Graphs of cubic functions in power function form diGiTal doCS • doc-9731: Investigate the graphs of cubic functions in power form (page 131) • Investigation doc-9732: Graphs of the form y = a(x − b)n + c (page 131)

3k

domain, range, maximums and minimums

diGiTal doCS • SkillSHEET 3.2 doc-9733: Practise expressing intervals using varying notation (page 134) • Investigation doc-9734: Modelling the path of a roller-coaster (page 135)

3l

modelling using technology

inTeraCTiViTY • Modelling data using polynomials int-0262: Use the interactivity to consolidate your understanding of how to fit a polynomial model to data (page 136) TUTorial • We26 eles-1416: Watch how to fit a cubic model to a set of data using a CAS calculator (page 136) diGiTal doCS • doc-9735: Investigate the best model for a set of data (page 137)

Chapter review diGiTal doC • Test Yourself doc-9737: Take the end-of-chapter test to test your progress (page 149)

To access eBookPLUS activities, log on to www.jacplus.com.au

Answers CHAPTER 3 CUBiC and QUarTiC FUnCTionS exercise 3a

1 a b c d e f g h i j 2 a b c d e f g h i j 3 a b c d e f g h i j 4 a b c d e f 5 a b c d e f

expanding x3 + 7x2 + 6x x3 − 7x2 − 18x x3 + 8x2 − 33x 2x3 + 10x2 + 12x 48x − 3x3 5x3 + 50x2 + 80x x3 + 4x2 2x3 − 14x2 − 30x3 − 270x2 −7x3 − 56x2 − 112x x3 + 12x2 + 41x + 42 x3 − 3x2 − 18x + 40 x3 + 3x2 − 36x + 32 x3 − 6x2 + 11x − 6 x3 + 6x2 − x − 6 x3 + 5x2 − 49x − 245 x3 + 4x2 − 137x − 660 x3 + 3x2 − 9x + 5 x3 − 12x2 + 21x + 98 x3 + x2 − x − 1 x3 + 13x2 + 26x − 112 3x3 + 26x2 + 51x − 20 4x3 − x2 − 36x + 9 10x3 − 49x2 + 27x + 36 −6x3 − 71x2 − 198x + 35 21x3 − 96x2 + 48x 54x3 + 117x2 − 72x 24x3 − 148x2 + 154x + 245 20x3 − 19x2 − 69x + 54 4x3 + 42x2 + 146x + 168 x3 + 6x2 + 12x + 8 x3 + 15x2 + 75x + 125 x3 − 3x2 + 3x − 1 x3 − 9x2 + 27x − 27 8x3 − 72x2 + 216x − 216 27x3 + 108x2 + 144x + 64 x3 − 4x2 − 67x − 110 3x3 + 15x2 − 18x −6x3 + 72x2 + 510x − 3600 − 3 x + 29x2 − 264x + 720 −x3 − 20x2 − 100x −x3 − 39x2 − 507x − 2197

exercise 3B

1 a b c d e f g h i j k l 2 a b

long division of polynomials 2 x + 3x − 6, 19 x2 + 2x − 1, 6 x2 + 1, 2 x2 − x + 6, −11 x2 + 4x + 3, −3 x2 + 3x − 3, −11 x2 − 2x − 3, −17 x2 − 4x − 18, −91 3x2 − 7x + 20, −35 4x2 − 8x + 18, −22 2x2 − 3x + 3, 7 2x2 − 9, 35 3x2 − 2x + 1, 5 2x2 + 5x − 6, −7

3

4

5

6

4x2 − 7x − 2, −3 x2 − 4x + 3, 8 x2 + x − 6, −11 3x2 + 2x + 1, 13 −x2 − 5x − 2, −14 − 2 3x − 2x + 4, −3 −x2 + 5x + 6, 9 −2x2 + 7x − 1, 1 x2 − x − 2, 3 x2, −7 x2 − x − 2, −8 −x2 − x − 8, 0 5x − 2, 7 2x2 − 2x + 10, −54 −2x2 − 4x − 9, −16 −2x2 + 4x − 1, 1 x3 + 2x2 + 5x – 2, −2 x3 + 2x2 – 9x – 18, 0 6x3 + 17x2 + 53x + 155, 465 The quotient is x2 − 6x + 101; the remainder is −1490. b The quotient is 2x2 + 44x + 1369; the remainder is 42 430. 16 x 160 c The quotient is 4x2 − + ; − 721 3 9 the remainder is 9 .

exercise 3d The remainder and factor theorems b 0 c 0 1 a −30 d −24 e −24 f k3 + 3k2 − 10k − 24 g −n3 + 3n2 + 10n − 24 h −27c3 + 27c2 + 30c − 24 2 a 58 c 11 b −8 f 6 d −9 e −202 g 158 i 35 h −6 j 441 3 a 6 b 3 c 1 e 2 f 2 d −2 g −5, 2 h a = −5, b = −3 4 a Not a factor b Is a factor c Is a factor d Not a factor 5 Show P(−2) = 0, P(3) = 0 and P(−5) = 0. 6 For example, for a show P (1) = 0. 7 a E b C c D d A 8 a (x − 1) b (x − 3) or (x − 2) c (x − 3) or (x + 2) d (x − 6) or (x + 4) or (x + 5) 9 a i 0 ii 0 iii 0 b i (x + 1)(6x2 + x − 2) ii (x + 1)(2x − 1)(3x + 2)

c d e f a b c d a b c d e f g h a b c a

d The quotient is the remainder is exercise 3C

1 a 10 d 43 g h i j k l 2 a b c

−77

9x3

−

30x2

+ 105x −

5235 2 .

c P 1 = 0 and (2x − 1) is a factor. 2

735 2 ;

1

2x − 1 = 0 gives x = 2 . −2 P( 3 )

= 0 and (3x + 2) is a factor.

3x + 2 = 0 givesx =

polynomial values b 11 c 18 e 3 f −22

−2

3.

d In general if (ax + b) is a factor, then

− b = 0. P a

2a3 − 3a2 + 2a + 10 16b3 − 12b2 + 4b + 10 2x3 + 9x2 + 14x + 18 2x3 − 21x2 + 74x − 77 −128y3 − 48y2 − 8y + 10 P(−11) = 230 522 P(102) = 1 735 072 853 P(2x + 9) = 256x4 + 4632x3 + 31 428x2 + 94 726x + 106 982 d P(x3 + 2) = 16x12 + 131x9 + 402x6 + 526x3 + 253

exercise 3e

1 a b c d e f g h i j

3–7

Factorising polynomials (x + 1)(x + 3)(x + 6) (x + 1)(x + 2)(x + 5) (x + 1)(x + 2)(x + 9) (x + 1)(x + 3)(x + 4) (x + 3)(x + 4)(x + 7) (x + 2)(x + 3)(x + 7) (x + 1)2(x + 2) (x + 2)2(x + 3) (x + 4)(x + 5)2 x(x + 5)(x + 8)

Column Column Column Column Column Column Column Column Column 1 2 3 4 5 6 7 8 9

P (x)

P (1)

P (2)

P (−1)

P (−2)

Rem when divided by (x − 1)

a

4

15

0

−5

4

15

0

−5

b

10

28

−2

−8

10

28

−2

−8

c

3

11

−7

−21

3

11

−7

−21

d

−7

−19

5

−7

−7

−19

5

−7

8 a

P (−8)

b P (7)

Rem when divided by (x − 2)

Rem when divided by (x + 1)

Rem when divided by (x + 2)

c P (a)

ChapTer 3 • Cubic and quartic functions

151

k x(x + 3)(x + 4) l x(x + 5)2 m x(x + 1)(x + 5) n x2(x + 6) 2 a (x − 1)(x + 1)2 b (x − 2)(x − 1)(x + 1) c (x + 1)2(x + 5) d (x − 3)(x + 2)2 e (x + 1)(x + 4)2 f (x − 5)(x − 2)(x + 2) g (x − 1)(x + 1)(x + 2) h (x − 3)(x + 1)(x + 2) i (x − 1)(x + 2)2 j (x + 2)(x2 − x + 3) k (x + 1)(x + 2)(x + 5) l (x − 3)(x + 1)(x + 3) m (x − 2)2(x + 3) n (x − 4)(x + 5)(x + 8) o (x − 1)(x + 1)(x + 2)2 p (x − 3)(x + 2)3 q (x − 1)(x + 1) (x + 3)2 r x(x − 1)(x − 7)(x + 3) 3 a x(x − 2)(3x + 5) b 2x(x + 1)(2x − 1) c 3x(x − 4)(x + 2) d −2x(x + 3)2 e 6x2(x − 1) f −x(x + 4)(x + 3) g −(x − 1)(x + 1)(x + 3) h −2x(x − 3)(x − 2) i −(x + 2)(2x − 1)(3x − 2) j −(x − 2)2(5x − 4) k −(x − 1)(x + 3)(x − 5)(x + 2)2 l (x − 2)2(3x + 5)(8x + 1) 4 a (2x + 3)(x − 1)(x + 2) b (3x − 1)(x + 1)(x + 4) c (3x + 2)(x − 2)(x + 2) d (4x + 3)(x + 3)(x + 5) e (5x − 1)(x + 1)2 f (x + 2)(x2 − x + 2) g (x + 1)(2x + 3)2 h (x − 2)(2x − 1)(3x − 4) i (x + 4)(2x − 5)(5x + 2) j (7x − 2)(x − 2)(x + 4) k (x − 3)(x + 1)2(2x + 1) l (x − 1)(x + 3)(2x + 1) (3x − 2) exercise 3F

Sum and difference of

two cubes 1 a a = x, b = 6 b a = 2y, b = z c a = x + 5, b = 3 d a = 1, b = 4h c u e a= f a = t, b = , b = eg 2 6 2 a (x − 5)(x2 + 5x + 25) b ( j + k)( j2 − jk + k2) c (y − 2)( y2 + 2y + 4) d (3x + y)(9x2 − 3xy + y2) e 8(2t − 3u)(4t2 + 6tu + 9u2) f (x − 1)(x2 + x + 1) 2 x x − 2 xp + 4 p2 g + 2 p 9 81 9

3 a (2a − 1)(a2 − a + 1) b x(x2 + 6x + 12) c 2(x + 2)(4x2 + 10x + 7) d −5(3w2 − 15w + 25) e 5m(7m2 − 3mp + 3p2) f (2x − 3)(13x2 + 15x + 9) g (3y + 5)(3y2 + 21y + 67) h (4x − 3y)(7x2 + 9xy + 21y2) i (1 − 5p)(13p2 − 16p + 7) j 2(3x − 8)(21x2 − 60x + 67) k (x2 + y3)(x4 − x2y3 + y6) l 2(x − 3)(x2 + 3x + 9) m 3(a + 1)(a2 − a + 1) n 6(x2 + 4)(x4 − x2 + 7) 4 m = 27, n = 1

exercise 3G

152

−6 y

b −6

7 x

−1 −42

b d f

g 0

h

i −4

c e g

−2,

k

−

−4,

2, 6

−3

l 6 6

3 a −2, 0, 2 −3,

3

−5 −2 −4

y

d

3

60

2, 0, 2

m 0, 4, 5 4 a −4, 1, 4 −4, −2,

2

g −2, −1, 3 −1 2

,

3 2,

3

−

5– 2

−2, −1,

−7

2,

−4

e

y 12 − 1–

0

3 3– 4

4

2

x

x

n 5, 4 b −4, 0, 4 e

j 0, 2, 3

j

x

1 3

m 2 , 0, 7 , 5

d

11

−5

3, −4, −7 2, 3 −2, −1, 1, 2 1 j 0, 9

b d f h

1, 5, 9 4 −5, 8 i −1, 0, 1

g

−1 −88

l −13

2 a 1, 2, 5

−

−8

j 1

k 12

d

y

c

Solving polynomial

equations 1 a 6 c 14 e 4

6 C 7 B

3k

x

1 2 3

c (3 3x + 3 m y) ((3 3 x) 2 − 3 3mxy + ( 3 my)2 )

j (st + g2)(s2t 2 − stg2 + g4)

1

y

1 a

b a = 3 3x , b = 3 m y

1

2 + i 3k − 9 k + 2 2 4

Cubic graphs — intercepts

5 a 3x3 + my3 = ( 3 3 x )3 + ( 3 m y)3

5 a b c d e f g h i j

h (3r − 1)(9r2 + 3r + 1)

exercise 3h

method

h

−5,

c −5, 0, 5

0

f 0, 2

,0 4

i 0, 5

−1

k 0, 4

l −7, 0, 1

n 0 b −2, 3, 5

c −5, 1, 5

e

−1,

2, 3

h −5, 0, 1, 2 k −3, −2,

1 2

−1

2,

6 x

3

f −2, 1, 5 i −6,

y

f

1

−54 −1

y 30

2 a

l −2, −1, 1

−

8.002, 9.008, 15.994 12, −7, 7 − 9, 13 16 − 6.91 14.22, −1.071, 0.854 3.627 0.646, −1.948, −0.199 −4.894, −1.325 −12.556, 0.418 −

Maths Quest 11 Mathematical Methods CAS

−5

−3

2

x

8

x

y

b −8

− 3– 2

−192

y

e

y

c −1

−3 −1 −21

x

2

5D 8 E

x

7

6 C 9 D

10 a

7 B

y

5 y

f

y

d

−1

x

12

150

−10

−1

−5

x

−3 −2

−1

4 3

2

−8

x

x

2

y

h

x

−9

y

f

factor, Δ = (−2)2 − 4 × 1 × 5 = 4 − 20 = −16. ∴ As Δ < 0, the quadratic factor has no real solutions, so y = x3 − x2 + 3x + 5 has only one real solution at x = −1. d Find the discriminant of the quadratic factor, Δ = (−2k)2 − 4 × 1 × 5 = 4k2 − 20. i For 2 real solutions, the quadratic factor must have 1 solution. ∴ Δ = 0. Solve 4k2 − 20 = 0. ∴ If k = ± 5 , the cubic will have 2 real solutions. ii For 3 real solutions, the quadratic factor must have 2 solutions. ∴ Δ > 0. Solve 4k2 − 20 > 0.

x

y 8

g

y

e

b y = (x + 1)(x2 − 2x + 5) c Find the discriminant for the quadratic

−2 3

7 −7

x

1– 6

4 a

y (−0.33, 14.81) 14

y

3a

−7

−2

−1

−2

exercise 3i

1 a

12 (−5, −36)

−2

b y

b

y

x

1

Quartic graphs — intercepts

method

x

1

∴ If k < − 5 or k > 5 , the cubic will have 3 real solutions.

0 −2 −1

y (6.04, 10.02) −4 5 7 x

b

2 3

y

6

8 −2 −1 0

−3

−2

−140 (−0.71, −144.84)

−1

x

c

y

c

−1

3

−1

x

d

−2

1

y

0

e

x

(3, 16) x

1 y

−1 7

(−2, 0)

2 3

x

y

(1.67, −28.44)

4 24

−4

0 1 −5

−9

y

d

x

y −5

x

d −2

2

c

y

(−1, 0)

8

−4

x

(−0.76, −6.64)

x

−3

0

1– 2

3 x

ChapTer 3 • Cubic and quartic functions

153

6 E 7 D 8 a

y

f

4

i i y

−1

2

−2.23 −0.24 2

x

0

1– 3

h

x

b

2 a

ii (0, −2) y (2, 2)

(0.46, 12.09) −14 9

125

1.30

3.36 x

(−0.83, 0.10)

(−3, 81) (2.61, −34.98)

3 a = −5 4 a There are many possible quartic

equations, including y = −(x + 3)(x – 3)3 or y = −x4 + 81 or y = −(x − 3)(x + 3)3. b y = −(x + 3)(x – 3)3

y c (−2.38, 13.68)

x (1.22, 6.47)

−3.18 −0.68 0

y

81

d

1.86

c

0

3

y (0.09, 120.35) 9.51

−288

10

(1, 5)

y

e

2 y

−1.90 (0, −3) −3

81

1.90

x

0

−2

− 3)(x +

3

(−3.22, 1027.74)y

−4.75

y

y

f

358

3)3

x

−8

(−0.06, 357.53) (2.35, 682.19)

x

x

y

e

(−1.29, −11.33) (1.29, −11.33)

−3

y

d

f

x

x

(6.95, −1343.39)

y = −x4 + 81

6 (4, −32)

(−3.28, −98.19)

x

y

x

120 1.62 −4.33 −1.79 −3

y

b

0 1 x

−5

x

1

y

y

y=

ii (0, 0)

3

(−1.55, −13.04)

1

−(x

ii (0, 0)

−1

l i −2

x

y

g

ii (3, 2)

4

j i 4 k i

0

−1

3.74

1

x

x

−1 81

−3

5 C

154

0

3

x

exercise 3J Graphs of cubic functions in power function form 1 a i 2 ii (1, 3) b i 3 ii (−5, −2) c i −2 ii (6, −8) d i −7 ii (−4, 1) e i 1 ii (9, 4) f i 1 ii (0, −7) g i −1 ii (−1, −1) 1 h i ii (−2, 0) 2

Maths Quest 11 Mathematical Methods CAS

y

g

(−2, 27)

−5

35

x

y

h

b

1

y

y

j

30 –2 x

x

4

−1

( 5– , −1) 2

(5, −3)

x

(−5, −32) y

i

y

c x

x

( 1– , 0) 4

(3, −9) y

j (−1, 2)

y

d

1

c y = 4 (x − 1)3 − 2

1 3–4 1

d y=

( 3– , 1) 2 x

x

y

k (−2, 25)

y

e

26 3–5

(1, 0) x x

−7

4 A 5 E 6 a y = (x − 1)3 + 5 b y = (x + 2)3 + 2 c y = −(x + 3)3 + 4 d y = −(x + 3)3 7 a y = 4(x − 2)3 + 3 b y = −2(x + 5)3 + 1

−1

3 2x

+4

exercise 3k domain, range, maximums and minimums 1 a Domain [−5, 4], range [−2, 5] b Domain [−2, 4], range [−2, 7] c Domain [−4, 2], range [−5, 2] d Domain [−2, 6), range [−2, 10) e Domain (−3, 5), range [0, 5) f Domain (−3, 4], range [−9, 0]

2 a

y 12

(0.472, 13.129)

y

l

y

f

y

4 (3.528, −1.128)

( 3– , −2) 4

(1, −2)

m

−1

x

x

x

Range [−1.128, 13.129] y

b (−3.732, 10.392) (1, 5)

y

g

(4, 3)

−5 x

x

1 x

−2

(−0.268, −10.392)

−10

Range [−10.392, 10.392] y

n

c

y

h x

1

y

x

−3

(9–5 , −7)

(1.333, 0.148) 1

3 a

y

i

(4, 1) 5

Range [0, 2)

(3, 2)

y 3

d

y

x

2

Range [0, 3]

(−2.333, 1.185)

(6, 4) x x

−3

−1

x

ChapTer 3 • Cubic and quartic functions

155

e (−4.689, 105.526)

y

−7

b 3– (2, 27) 2

−1

x −21 (0.355, −22.822)

y 0 −10 −20 −30 −40 −50 −60

600 500 400 300 200 100 0

x

2 4 6 8 10

y = −4.6273x − 9.0455

1

−4

4

40 30 20 10 0

x

d

(3, −4)

g (−2.577, 0.385) y x −2 (−1.422, −0.385)

−3

2 a (−4, −6)

Range [−6, 0.385] h (−2, 4) y (2.1, 1.581) −1 −6

b x

2

(0.786, −8.209)

Range [−8.209, 4) y

i

(2.737, 48.517)

−3 −1 (−2.070, −7.035)

5 (5.1, −4.94)

−3

0 2 4 6 8 10

−21

Range [−32, 13.169]

3 A 5 a 28 m 6 (0.971, −1.201) exercise 3l

1 a

156

4 B b 20.785 m

modelling using technology

y y = 11.436x − 23.091 100 80 60 40 20 0 −20 2 4 6 8 10 x

2 4 6 8 10

y y = 12.953x2 − 47.679x + 613.66

0

c

0 2 4 6 8 10

b

y y = 1.5056x3 − 6.6906x2 + 35.465x + 20.615 1200 1000 800 600 400 200 0 0 2 4 6 8 10 x

Maths Quest 11 Mathematical Methods CAS

x

5 The cubic model 6 205.629 7 99 8 330 9 a y

x

y y = 1.3747x3 − 19.138x2 + 8.5536x + 607.48 600 500 400 300 200 100 0 0 2 4 6 8 10 x

x

0 2 4 6 8 10

y y = 10.006x3 − 2.1369x2 + 9.8652x + 577.45 1400 1200 1000 800 600 400 200 0

x

y = −11.124x2 − 15.655x + 79.329

x

0 2 4 6 8 10

1400 1200 1000 800 600 400 200

6 5 4 3 2 1 0

y −200 −400 −600 −800 −1000 −1200

3 a

b

y y = −1.1014x2 + 16.75x − 5.5664

0

(1, −32)

x

x

y y = 81.855x + 419.36 1400 1200 1000 800 600 400 200 0

y y = 8.2459x2 − 21.777x + 56.371 800 600 400 200 0 0 2 4 6 8 10 x

2 4 6 8 10

0 2 4 6 8 10

4a

y y = −0.5291x2 + 20.8x + 0.5175 250 200 150 100 50 0 0 2 4 6 8 10 x

d

2.333 x

x

0 2 4 6 8 10

50 40 30 20 10 0 −10

y −1

0

c

x

Range [−7.035, 48.517] j (−2.111, 13.169)

300

y y = −2.8818x + 48.318 50 40 30 20 10 0

Range [−4, 0]

600

Share price ($)

y

f

x

900

y y = 3.0273x + 6.3182

2 4 6 8 10 12 Time (months)

x

b $6.78 c Anything can happen with prices in the

sharemarket. The last data point may be the beginning of a share price crash!

10

y y = 1.048x3 − 19.309x2 + 98.296x + 220.7 Population

c

2 4 6 8 10

d y y = −2.2988x3 + 32.575x2 −14.094x + 30.154

(−8, −133)

Range [−133, 105.526]

y y = 2.5307x3 − 23.29x2 + 38.808x + 59.448

c

400 300 200 100 0

0 2 4 6 8 10 Year

x

Year 11 estimate: 360 birds exercise 3m

1 a b c d e

Finite differences y = 11x + 6 y = −26x + 100 y = 7x2 + 4x − 4 y = 3x2 − 19x + 17 y = x3 + 3x2 − 7x − 7

f y = x3 − 2x2 + 2x + 16 g y = 2x + 10x − 23

exTended reSponSe

y

9a

2

1 P(3) = 68

P(−x) = −5x3 − 3x2 + 6x − 22

h y = −2x3 + 15x2 − 32x + 47

2 m = −70 3 Factors: (x − 3)(x + 3)(x − 2)

i y = 9x − 27

j y = −2x2 + 10x − 15

y

k y = −4x3 + 13x2 + 52x − 66 l y= 2 a

−8x

−11

+ 43

x (base dots)

0

1

2

3

y (total dots)

0

1

3

6 10 15

b y=

1 2

4

b

5

18

y

8

1

−8

x (dots)

0

n (diagonals) 0 1

−1 −1

−1 2 1 , ,3 4 3 2 y-intercepts: y = −14

5 x-intercepts: x =

0 2 5 9

1

1

a chessboard

b y = 4x + 3

c y= d y=

d y=

2

7

x

− 1 0 2 3 1– 4 3 2

x

−14 y

b

1 2 x − 2x + 8 2 −10x2 + 18x –

3 3 1 x + 2 x 2 − 12 x 4 −5x3 + 3x – 6

0

−4

−5 2 2 x + 7x + 11 2x2 − 3x − 1

6 Equation: y = (x + 5)3 + 2

20 7 a y = −x3 − 5x2 + 8x + 15 b y = 3x3 − 4x2 + 2x + 5 c y=

y

10 a

−3

b y=

x

1– 2

+ 2 x 2 + 6 x, 204 squares on

5 a y = 8x − 15 6 a y=

}

y

3

1 3 x 3

x

2 3

4 (4x + 3)(7x2 − 15x + 39)

y

n = 2 x2 − 2 x 4 n=

x

1

c

1 2 3 4 5 6

0

−3

x2 + 2 x

c 28 3

x

2

−4

Point of inflection: (−5, 2)

1– x 2

−1 0

y

−16

− 18

(−5, 2) y

c

2

8 a y = 5x4 + x3 − 2x2 + x − 3

x

0

−5

−1

b y = 4 x4 − 2x3 + x2 + 2x + 1

0

−5

x

ChapTer reVieW ShorT anSWer

1 a x3 + 6x2 − 36x + 40 b x3 + 10x2 + 19x − 30 c x3 − 21x2 + 147x − 343 d −2x3 − x2 + 11x + 10 2 a x2 − 16, 29 b x2 + 6x + 5, 8 c −x2 + 2x + 2, −9 3 a −4 b 216 c −24a3 + 8a2 + 2a − 4 4 −7 5 Show P (−3) = 0. 6 (x − 10)(x + 4)(x + 10) 7 a (1 − 5x)(1 + 5x + 25x2) b (2x + 1)(x2 + x + 19) 8 a −6 b

−1

2,

3

c 2, 3, 4

y

7 y

11

7

(−1, 8)

48

7 –8

3

−2 0

x

12 [−32, 40] 13 y = x3 – 2x2 + 8 1

3

x

4

8 a a = −14, b = 39 b The third factor is x + 1. y c

1

14 r = 2 x 2 + 2 x + 1 15 y = 2x3 – 5x2 + x – 2 mUlTiple ChoiCe

1D 4A 7B 10 C 13 B 16 A

2C 5B 8A 11 E 14 A 17 D

54

3D 6A 9A 12 E 15 D 18 D

−10

6 9

x

ChapTer 3 • Cubic and quartic functions

157

9 y = ( x − 2)( x −

3)( x + 3)

y

c 8.766 m d 58 cm 13 a

(3, 20)

y

6

14 a a = 1, b = −8, c = 1, d = 42 b (x – 3), f (3) = 0 c (x – 7)(x – 3)(x + 2) d y 42

0− 3

− 3

2

x

−3

10 Points of intersection are (−2, 13) and (5, −8).

158

−2

−1

(−3, −10)

− 3

79 x − x2 + x + 4 6 6 12 a 7.2 m b (15, 0) 11 y =

−2

Linear Highway: 5 y = 5x + 5 0 −4

1

3

x

Cubic River: y = x3 + x2 − 4x − 4

b (−3, −10), (−1, 0), (3, 20) c 10.198 km

Maths Quest 11 Mathematical Methods CAS

2

0

3

7

x

15 a Initial height = 0 cm b y 54.5185 (2 2–3 , 54.5185) (8, 64)

64

(6, 36)

36 0

22–3

c 1 s, 5 s and 8 s

6

8 x

Exam practice 1 CHAPTERS 1–3 1 Determine the exact values for x for which x2 − 5 = 3x.

[−3,

3 marks

2 Let f : 1] → R, where f (x) = + − 4x − 3. a Show that (x − 1) is a linear factor. b When factorised, f (x) = (x − 1)(x + 3)(ax + b). Determine the values of a and b. c Hence, sketch the graph of f (x).

2x3

5x2

2 marks

S ho rT a n S W er 25 minutes

3 marks 3 marks

3 The line y = ax + b passes through the point (2, 1) and is parallel to the line y − 4x + 3 = 0. Determine

the values of a and b. [−1,

2 marks

4 Let f : 3] → R, f (x) = − 4x − 3. a Express f in the form f (x) = a(x − b)2 + c. Hence, state the coordinates of the turning point. b State the range for f. c Determine the exact values of the x-intercepts. d Sketch f on a set of axes. Label all key features.

2x2

3 marks 2 marks 3 marks 3 marks

1 A straight line passes through the points (2, 4) and (−1, −5). Its equation would be: a y=6−x

B y=x+2

C 3y = x + 10

e y = −3x + 10

d y = 3x − 2

d

B (3x + 4)(x − 2)

− 13 13 or 10 10

e

−1

2

or

C

−4

3

12 minutes

each question is worth one mark.

2 The exact values of x for which 3(2x − 1)2 + 2(2x − 1) − 8 = 0 are: a (2x + 1)(6x − 7)

m U lTip l e C ho iC e

or 2

7 6

3 If (x − a) is a linear factor of the function f (x) = x3 + 2x2 − 11x − 12, then the possible value for a is: a −3

B −1

C 0

d 1

e 4

4 A cubic function has the following axis intercepts: x-intercepts 1, 3 and −4, and y-intercept 24. The

equation that would best describe this function would be: B (x + 1)(x + 3)(x − 4) + 24 e 2(x + 1)(x + 3)(x − 4)

a 24(x + 1)(x + 3)(x − 4) d 2(x − 1)(x − 3)(x + 4)

C (x − 1)(x − 3)(x + 4) + 24

5 Three linear functions are defined as:

L1: 4y − 3x − 8 = 0 L2: 3y + 4x + 1 = 0 L3: 3y − 4x − 2 = 0 Which one of the following statements is correct? a L1 and L2 are parallel. B L1 and L3 are parallel. d L1 and L3 are perpendicular. e L1, L2 and L3 are parallel.

6

C L1 and L2 are perpendicular.

y

−a

b

x

Which one of the following rules best describes the graph above? a y = (x + a)2(b − x) B y = (x − a)2(x − b) d y = (x − b)2(x + a) e y = (x − a)2(x − b)

C y = (x + a)(x − b)

7 A polynomial of degree 3 passes through the origin and has x-intercepts at −2 and 3. The equation for

this polynomial would best be described by which one of the following? a y = (x − 2)(x + 3) B y = (x + 2)(x − 3) C y = x(x − 2)(x + 3) d y = x(x + 2)(x − 3) e y = (x − 1)(x − 2)(x + 3) 8 A linear function L(x) = ax + 1 intersects the quadratic function Q(x) = 2x2 − x − 1 at the point (−0.5, 0).

The coordinates of the second point of intersection would be closest to: a (−1, 1) B (−1, 2) C (−1, 4) d (2, 3) e (2, 5) Exam practice 1

159

e xT ended r e S p onS e 40 minutes

1 Points A (−2, 1), B (1, 5) and C (5, 2) are vertices on a triangle. a Show that the triangle is an isosceles triangle. b Determine the equation of the perpendicular bisector of the line AC. c Show that the perpendicular bisector found in part b passes through vertex B.

3 marks 4 marks 2 marks

2 Seng and Victor are training for a 1000-metre race. The distance in metres they each ran was recorded

in 1-minute intervals. Victor was delayed for 1 minute because he was tying up his shoelaces when he was 110 m from the starting line. The individual distances are recorded in the table below. Time (minutes) Seng’s distance (m) Victor’s distance (m)

0 0 −110

1 200 0

2 380 200

3 680 560

4 880 875

5 1040 1050

a The relationship between distance and time for Victor can be modelled using a quartic function.

diGiTal doC doc-10161 Solutions exam practice 1

Determine this quartic function, Qv. Write your function in terms of Qv (distance in metres) and t (minutes). Express coefficients correct to 2 decimal places. 2 marks b Seng’s distance and time relationship can be modelled using a cubic function. Determine this cubic function, Cs. Write your function in terms of Cs (distance in metres) and t (minutes). Express coefficients correct to 2 decimal places. 2 marks c i State the domain of Qv. 1 mark ii State the domain of Cs. 1 mark iii Sketch both Qv and Cs on the same set of axes. 3 marks d Using your functions Qv and Cs, determine the time first Victor passes Seng. Write your answer in minutes and seconds. 2 marks 3 A children’s playground in the shape of a rectangle is to be constructed at a local park. The longer side is 5 metres longer than the shorter side. a If x is the shorter side, write down an expression for the longer side in terms of x. 1 mark b Write down an equation for the area, A(x), of the playground in terms of x. 2 marks c If the maximum area of the playground is 150 m2, determine the y dimensions of the playground. 3 marks A miniature bike path will be built within the playground. It will travel from the gate to the seesaws, slides and swings. The bike path can be modelled by the cubic function B(x) = x3 − 6x2 +10x, where x is the horizontal distance, in metres, and B(x) is the vertical distance, in metres, from the gate. d Assuming that the slides are on the path, determine the vertical distance the slides are from the gate if their measurement is 3 metres horizontally from the gate. 2 marks e If the seesaws are on the bike path, show that the seesaws are located 1 metre horizontally and 5 metres vertically from the gate. 2 marks f Determine the shortest distance, in metres, between the seesaws and the slides. Write your answer in exact form. 2 marks 0

160

Maths Quest 11 Mathematical Methods CAS

x

Chapter 4

Relations, functions and transformations Chapter ContentS 4a 4b 4C 4d 4e 4F 4G 4h 4i 4J

Set notation Relations and graphs Domain and range Types of relations (including functions) Power functions (hyperbola, truncus and square root function) Function notation Special types of function (including hybrid functions) Inverse relations and functions Circles Functions and modelling

4a

diGital doC doc-9738 10 Quick Questions

Set notation

Set notation is used in mathematics in the same way as symbols that are used to represent language statements.

definitions 1. A set is a collection of things. 2. The symbol {. . .} refers to a set. 3. Anything contained in a set, that is, a member of a set, is referred to as an element of the set. (a) The symbol ∈ means ‘is an element of’, for example, 6 ∈ {2, 4, 6, 8, 10}. (b) The symbol ∉ means ‘is not an element of’, for example, 1 ∉ {2, 4, 6, 8, 10}. Chapter 4 • Relations, functions and transformations

161

4. A capital letter is often used to refer to a particular set of things. 5. The symbol ⊂ means ‘is a subset of’, so, if B ⊂ A, then all of the elements of set B are contained in set A. For example, {2, 4} ⊂ {2, 4, 6, 8, 10}. 6. The symbol ⊄ means ‘is not a subset (or is not contained in)’, for example {2, 3} ⊄ {2, 4, 6, 8, 10}. 7. The symbol ∩ means ‘intersection’, so, A ∩ B is the set of elements common to sets A and B. For example, {1, 2, 3} ∩ {2, 4, 6} is {2}. 8. The symbol ∪ means ‘union’, so, A ∪ B is the set of all elements belonging to either set A or B or both. For example, {1, 2, 3} ∪ {2, 4, 6} is {1, 2, 3, 4, 6}. 9. The symbol A \ B denotes all of the elements of A that are not elements of B. 10. The symbol ∅ means the null set. It implies that there is nothing in the set, or that the set is empty. Worked exaMple 1

If A = {1, 2, 4, 8, 16, 32}, B = {1, 2} and C = {1, 2, 3, 4}, find: a A∩B b A∪C c A\B d {3, 4} ∩ B e whether or not: i 8 ∈ A ii B ⊂ A iii C ⊂ A. think

Write

a The elements that A and B have in common are 1 and 2.

a {1, 2}

b The elements that belong to either A or C are 1, 2, 3, 4, 8, 16 and 32.

b {1, 2, 3, 4, 8, 16, 32}

c The elements of A that are not elements of B are 4, 8, 16 and 32.

c {4, 8, 16, 32}

d {3, 4} and B have no common elements.

d {∅}

e iii 8 is an element of A. ii All elements of B belong to A. iii 3 is an element of C but not A.

e i Yes. 8 ∈ A ii Yes. B ⊂ A iii No. C ⊄ A

Sets of numbers Certain letters are reserved for important sets that arise frequently in the study of mathematics. 1. R is the set of real numbers, that is, any number you can think of. 2. N is the set of natural numbers, that is, {1, 2, 3, 4, 5, . . .}. 3. Z is the set of integers, that is, {. . ., −3, −2, −1, 0, 1, 2, 3, . . .}. a 4. Q is the set of rational numbers (that is, numbers that can be expressed as fractions in the form b where a and b are integers and b ≠ 0). 5. Q′ is the set of numbers that are not rational (that is, cannot be expressed as a ratio of two whole numbers). These numbers are called irrational, for example, π, 3, etc. Note that N ⊂ Z ⊂ Q ⊂ R, that is, R

Q'

Q 1– 2 3– 4 2– 3

Z N 0 −1 −2 −3 ...

exercise 4a

3 π ...

1 2 3 4 ...

7– 5 33 — ... 51

Set notation

1 We1 If A = {2, 4, 6, 8, 10, 12, 14}, B = {1, 3, 5, 7, 9, 11, 13}, C = {4, 5, 6, 7} and D = {6, 7, 8}, find: a A∩B b A∩C c A∩C∩D d A∪B f A\C g C \ D. e C∪D 162

Maths Quest 11 Mathematical Methods CAS

2 If A = {−3, −2, −1, 0, 1, 2, 3}, B = {0, 1, 2, 3} and C = {−3, 2, 3, 4}, find: b A \ B a A ∩ B ∩ C c A \ (B ∪ C) d A \ (B ∩ C) e A ∪ C. 3 If F = {a, e, i, o, u}, G = {a, b, c, d, e, f, g, h, i} and H = {b, c, d, f, g, h}, find: c G \ H a F ∩ G ∩ H b G ∩ H d H \ F e (F ∪ H) \ G. 4 MC Given that A ⊂ B, then A ∩ B is equivalent to: b ∅ e A

A B d A ∪ B

c {1, 2}

5 MC Given that C ⊂ B ⊂ A, then it follows that: a A ∪ B ∪ C is equivalent to: A B

B C

C A

D A ∪ B

E B ∪ C

A B

B ∅

C C

D A ∩ B

E B \ C

b (A \ B) ∩ C is equivalent to:

6 Answer true (T) or false (F) to each of the following statements relating to the number sets N, Z, Q

and R. a 7 ∈ R d 5 ∈ Q g {5, 10, 15, 20} ⊂ N j Q ⊂ N

4B

b −4 ∈ N e 1.5 ∈ Z

c 6.4217 ∈ Q f {5, 10, 15, 20} ⊂ Z

h Z \ N = {. . ., −3, −2, −1} k Q ∩ Z = ∅

i Z ∩ N = N l (Z ∪ Q) ⊂ R

Relations and graphs

A relation is a set of ‘ordered pairs’ of values or ‘variables’. Consider the following relation. The cost of hiring a trailer depends on the number of hours for which it is hired. The table below outlines this relation. Number of hours of hire Cost ($)

3 50

4 60

5 70

6 80

7 90

8 100

Cost of trailer hire ($)

Since the cost depends upon the number of hours, the cost is said to be the dependent variable, and the number of hours is called the independent variable. The information in the table can be represented by a graph, which usually gives a better indication of how two variables are related. When graphing a relation, the independent variable is displayed on the horizontal (or x) axis and the dependent variable is displayed on the vertical (or y) axis. So we can plot the set of points {(3, 50), (4, 60), (5, 70), (6, 80), (7, 90), (8, 100)}. The points are called (x, y) ordered pairs, where x is the first element and y is the second element. y 100 90 80 70 60 50 40 0 1 2 3 4 5 6 7 8 x Number of hours

This graph clearly shows that the cost increases as the number of hours of hire increases. The relation appears to be linear. That is, a straight line could be drawn that passes through every point. However, the dots are not joined as the relation involves ‘integer-valued’ numbers of hours and not minutes or seconds. The number of hours can be referred to as a ‘discrete dependent variable.’ Discrete variables include names and numbers of things; that is, things that can be counted (values are natural numbers or integers). Chapter 4 • Relations, functions and transformations 163

Some variables are referred to as continuous variables. Continuous variables include height, weight and volume; that is, things that can be measured (values are real numbers). If a relationship exists between the variables, we may try to find a rule and then write this rule in mathematical terms. In our example, the relationship appears to be that for each extra hour of hire the cost increases by $10 after an initial cost of $20. Cost = 10 × number of hours + 20 Using x and y terms, this is written as y = 10x + 20 Worked Example 2

Sketch the graph by plotting selected x-values for the following relations and state whether each is discrete or continuous. a y = x2, where x ∈ {1, 2, 3, 4} b y = 2x + 1, where x ∈ R Think

a 1 Use the rule to calculate y and state the

ordered pairs by letting x = 1, 2, 3 and 4.

2

Plot the points (1, 1), (2, 4), (3, 9) and (4, 16) on a set of axes.

Write/draw

a When x = 1, y = 12

= 1 x = 2, y = 22 = 4 x = 3, y = 32 = 9 x = 4, y = 42 = 16

(1, 1) (2, 4) (3, 9) (4, 16)

y 16 12 8 4 0

3

Do not join the points as x is a discrete variable (whole numbers only).

b 1 Use the rule to calculate y. Select values of x,

say x = 0, 1 and 2 (or find the intercepts). State the ordered pairs.

2

Plot the points (0, 1), (1, 3) and (2, 5) on a set of axes.

1

2

Join the points with a straight line, continuing in both directions as x is a continuous variable (any real number).

164 Maths Quest 11 Mathematical Methods CAS

4

x

It is a discrete relation as x can be only whole number values. b When x = 0, y = 2(0) + 1

= 1 x = 1, y = 2(1) + 1 = 3 x = 2, y = 2(2) + 1 = 5

(0, 1) (1, 3) (2, 5)

y y = 2x + 1 5 4 3 2 1 −2 −1 −10 −2 −3

3

3

1

2

x

It is a continuous relation as x can be any real number.

Worked Example 3

The pulse rate of an athlete, R beats per minute, t minutes after the athlete finishes a workout, is shown in the table below. 0 180

t R

2 150

4 100

6 80

8 70

a Plot the points on a graph. b Estimate the athlete’s pulse rate after 3 minutes.

Think

Write/draw

a 1 Draw a set of axes with t on the horizontal axis

a&b

and R on the vertical axis because heart rate is dependent on the time.

R (beats/min)

2

180 160 140 120 100 80 60 40 20

Plot the points given in the table.

b 1 Join the points with a smooth curve as t (time)

is a continuous variable. 2

Construct a vertical line up from t = 3 until it touches the curve.

3

From this point draw a horizontal line back to the vertical axis.

4

Estimate the value of R where this line touches the axis.

0 1 2 3 4 5 6 7 8 t (min)

When t = 3, the pulse rate is approximately 125 beats per minute.

Note: In any defined domain, for example, −3 ≤ x ≤ 3, the variable (x) is assumed continuous in that domain unless otherwise stated.

Exercise 4B

Relations and graphs

Questions 1, 2, and 3 refer to the following information. A particular relation is described by the following ordered pairs: {(0, 4), (1, 3), (2, 2), (3, 1)}. 1 MC The graph of this relation is represented by: A y B y 4 4

C

y 4

3

3

3

2

2

2

1

1

1

0 D y 4

1

2

3

4 x

0 E y 4

3

3

2

2

1

1

0

1

2

3

4

x

0

1

2

3

4 x

1

2

3

4 x

0

1

2

3

4 x

Chapter 4 • Relations, functions and transformations 165

2 MC The elements of the dependent variable are: A {1, 2, 3, 4} d {0, 1, 2, 3}

b {1, 2, 3} E {1, 2}

c {0, 1, 2, 3, 4}

3 MC The rule for the relation is correctly described by: A y = 4 − x, x ∈ R c y = 4 − x, x ∈ N E y = 4 − x, x ∈ {0, 1, 2, 3}

b y = x − 4, x ∈ N d y = x − 4, x ∈ Z

4 MC During one week, the number of people travelling on a particular train, at a certain time,

Number of people

D

M T W T F

0

M T W T F

M T W T F

Number of people

0 E

0

M T W T F

Number of people

0 C

B

Number of people

A

Number of people

progressively increases from Monday through to Friday. Which graph below best represents this information?

0

M T W T F

5 State whether each of the following relations has discrete (D) or continuous (C) variables. a {(−4, 4), (−3, 2), (−2, 0), (−1, −2), (0, 0), (1, 2), (2, 4)} b The relation that shows the air pressure at any time of the day y y c d

0

x

x

0

e The relation that shows the number of student absences per day during term 3 at your school f The relation describing the weight of a child from age 3 months to one year 6 WE2 Sketch the graph representing each of the following relations, and state whether each is discrete

or continuous. a

Day Cost of petrol (c/L)

b c d e f

Mon

Tues

Wed

Thur

Fri

Sat

Sun

68

67.1

66.5

64.9

67

68.5

70

{(0, 0), (1, 1), (2, 4), (3, 9)} y = −x2, where x ∈ {−2, −1, 0, 1, 2} y = x − 2, where x ∈ R y = 2x + 3, where x ∈ Z y = x2 + 2, where −2 ≤ x ≤ 2

166 Maths Quest 11 Mathematical Methods CAS

7 WE3 The table at right shows the temperature of a cup

t (min) 0 2 4 6 8 of coffee, T °C, t minutes after it is poured. 80 64 54 48 44 T (°C) a Plot the points on a graph. b Join the points with a smooth curve. c Explain why this can be done. d Use the graph to determine how long it takes the coffee to reach half of its initial temperature.

8 A salesperson in a computer store is paid a base salary of $300 per week plus $40 commission for each

computer she sells. If n is the number of computers she sells per week and P dollars is the total amount she earns per week, then: a copy and complete the table below. n

0

1

2

3

4

5

6

P b plot the information on a graph. c explain why the points cannot be joined together. 9 The speed of an aircraft, V km/h, t seconds after it starts to accelerate down the runway, is shown in the

following table. t

0

1

2

3

4

5

V

0

30

80

150

240

350

a Plot a graph that represents the information shown in the table. b Use the graph to estimate the speed after: i 2.5 s ii 4.8 s. 10 The cost, C dollars, of taking n students on an excursion to the zoo is $50 plus $6 per student. a Complete a table using 15 ≤ n ≤ 25. b Plot these points on a graph. c Explain why the dots can or cannot be joined.

4C Domain and Domain and range

range

A relation can be described by: 1. a listed set of ordered pairs 2. a graph 3. a rule. The set of all first elements of a set of ordered pairs is known as the domain, and the set of all second elements of a set of ordered pairs is known as the range. Alternatively, the domain is the set of independent values, and the range is the set of dependent values. If a relation is described by a rule, it should also specify the domain. For example: 1. the relation {(x, y): y = 2x, x ∈ {1, 2, 3}} describes the set of ordered pairs {(1, 2), (2, 4), (3, 6)} 2. the domain is the set X = {1, 2, 3}, which is given 3. the range is the set Y = {2, 4, 6}, and can be found by applying the rule y = 2x to the domain values. If the domain of a relation is not specifically stated, it is assumed to consist of all real numbers for which the rule has meaning. This is referred to as the implied domain of a relation. For example: {(x, y): y = x3} has the implied domain R. {(x, y): y = x } has the implied domain x ≥ 0.

Interval notation If a and b are real numbers and a < b, then the following intervals are defined with an accompanying number line. (a, b) implies a < x < b or (a, b] implies a < x ≤ b or a

b

x

a

b

x

Chapter 4 • Relations, functions and transformations 167

(a, ∞) implies x > a or

[a, ∞) implies x ≥ a or x

a

(−∞, b) implies x < b or

(−∞, b] implies x ≤ b or x

b

x

a

b

[a, b) implies a ≤ x < b or

x

[a, b] implies a ≤ x ≤ b or x

x

a b A closed circle indicates that the number is included and an open circle indicates that the number is not included. a

b

Worked Example 4

Describe each of the following subsets of the real numbers using interval notation. a b c −4

0

x

2

−3

5 x

0

Think

0 1

3

Write

a The interval is x < 2 (2 is not included).

a (−∞, 2)

b The interval is −3 ≤ x < 5 (−3 is included).

b [−3, 5)

c The interval is both 1 ≤ x < 3 and x ≥ 5 (1 is included,

c [1, 3) ∪ [5, ∞)

3 is not). Worked Example 5

Illustrate the following number intervals on a number line. a (−2, 10] b [1, ∞) Think

Write/draw

a The interval is −2 < x ≤ 10 (−2 is not included, 10 is).

a −2

b The interval is x ≥ 1 (1 is included).

10 x

0

b 0

x

1

Worked Example 6

State the domain and range of each of the following relations. a {(1, 2), (2, 5), (3, 8), (4, 11)} b Weight (kg) 10 15 20 25 30 Cost per kg ($) 3.5 3.2 3.0 2.8 2.7 c

d

y

y 4

−4 0

168 Maths Quest 11 Mathematical Methods CAS

0

x −4

4

x

5

x

think

Write

a Domain = {1, 2, 3, 4}

a 1 The domain is the set of first elements of the

ordered pairs. 2

Range = {2, 5, 8, 11}

The range is the set of second elements of the ordered pairs.

b 1 The domain is the set of independent values in the

table, that is, the weight values. 2

Range = {2.7, 2.8, 3.0, 3.2, 3.5}

The range is the set of dependent values in the table, that is, the cost values.

c Domain = R

c 1 The domain is the set of values that the graph

covers horizontally. 2

Range = [0, ∞)

The range is the set of values that the graph covers vertically.

d Domain = [−4, 4]

d 1 The domain is the set of values that the graph

covers horizontally. 2

b Domain = {10, 15, 20, 25, 30}

Range = [−4, 4]

The range is the set of values that the graph covers vertically.

Worked exaMple 7

For each relation given, sketch its graph and state the domain and range using interval notation. a {(x, y): y = x − 1} b {(x, y): y = x2 − 4, x ∈ [0, 4]} think

a 1 The rule has meaning for x ≥ 1 because

tUtorial eles-1461 Worked example 7

Write/draW

a

if x < 1, y = negative number. 2

3

Calculate the value of y when x = 1, 2, 3, 4 and 5, and state the coordinate points.

Plot the points on a set of axes.

When x = 1, y = 0 =0 x = 2, y = 1 =1 x = 3, y = 2 x = 4, y = 3 x = 5, y = 4 =2 y

(2, 1) (3, 2) (4, 3) (5, 2)

y= x−1

Join the points with a smooth curve starting from x = 1, extending it beyond the last point. Since no domain is given we can assume x ∈ R (continuous).

2

5

Place a closed circle on the point (1, 0) and put an arrow on the other end of the curve.

−1

6

The domain is the set of values covered horizontally by the graph, or implied by the rule.

Domain = [1, ∞)

7

The range is the set of values covered vertically by the graph.

Range = [0, ∞)

4

(1, 0).

1 0

1

2

3

4 5

x

Chapter 4 • Relations, functions and transformations

169

b 1 Calculate the value of y when x = 0, 1, 2, 3 and 4,

b When x = 0, y = 02 − 4

= −4 x = 1, y = 12 − 4 = −3 x = 2, y = 22 − 4 =0 x = 3, y = 32 − 4 =5 x = 4, y = 42 − 4 = 12

say, as the domain is [0, 4]. State the coordinate points.

2

Plot these points on a set of axes.

3

Join the dots with a smooth curve from x = 0 to x = 4.

4

Place a closed circle on the points (0, −4) and (4, 12).

(0, −4) (1, −3) (2, 0) (3, 5) (4, 12)

y y = x2 − 4, x ∈ [0, 4] 12 10 8 6 4 2 0 −2 −4

1 2

3 4

5

The domain is the set of values covered by the graph horizontally.

Domain = [0, 4]

6

The range is the set of values covered by the graph vertically.

Range = [−4, 12]

x

Verify that the graphs are correct using a graphics calculator or other technology.

domain and range

exercise 4C

1 We4 Describe each of the following subsets of the real numbers using interval notation. b a −2

0

1

0

c

5

d −3

0

−8

4

e

0

9

f −1

0

0

g

1

h −5

−2

0

−3

3

0 1 2

2 We5 Illustrate each of the following number intervals on a number line. a [−6, 2) b (−9, −3) − c ( ∞, 2] d [5, ∞) e (1, 10] f (2, 7) g (−∞, −2) ∪ [1, 3) h [−8, 0) ∪ (2, 6] i R \ [1, 4] j R \ (−1, 5) k R \ (0, 2] l R \ [−2, 1) 3 Describe each of the following sets using interval notation. a {x: −4 ≤ x < 2} b {x: −3 < x ≤ 1} −1 c {y: −1 < y < 3} d {y:

{x: x > 3} R R \ {1} R \ {x: 2 ≤ x ≤ 3}

Maths Quest 11 Mathematical Methods CAS

2

f h j l

2

{x: x ≤ −3} R+ ∪ {0} R \ {−2} R \ {x: −2 < x < 0}

4

4 MC Consider the set described by R \ {x: 1 ≤ x < 2}. a It is represented on a number line as: a

b

0 1 2

0 1 2 C

d

0 1 2

0 1 2

e

0 1 2 b It is written in interval notation as: a b C d e

(−∞, 1) ∪ (2, ∞) (−∞, −1) ∪ [2, ∞) (−∞, 1) ∪ (2, ∞] (−∞, 1] ∪ (2, ∞) (−∞, 1) ∪ [2, ∞)

y 4

5 MC The domain of the relation graphed at right is: a b C d e

[−4, 4] (−4, 7) [−1, 7] (−4, 4) (−1, 7)

−1 0

diGital doC doc-9739 SkillSHEET 4.1 domain and range

7 x

3

−4 6 MC The range of the relation {(x, y): y = 2x + 5, x ∈ [−1, 4]} is: a [7, 13] C [3, ∞) e R \ (7, 13)

b [3, 13] d R

7 We6a, b State i the domain and ii the range of each of the following relations. a {(3, 8), (4, 10), (5, 12), (6, 14), (7, 16)} b {(1.1, 2), (1.3, 1.8), (1.5, 1.6), (1.7, 1.4)} c

Time (min)

Distance (m) d

Day Cost ($)

3

4

5

6

110

130

150

170

Monday

Tuesday

Wednesday

Thursday

Friday

25

35

30

35

30

e y = 5x − 2, where x is an integer greater than 2 and less than 6 f y = x2 − 1, x ∈ R 8 We6c, d State the domain and range of each of the following relations. Use a CAS calculator to view

more of each graph if required. a

b

y 2 −3

d

y

0 1

0

x

y= x−1

0 e

c

y y = 2ex 2

y 4

y 2

x

2

y = 4e−x

x

−2 f

0

2x

y 0

x

−3 0

x

Chapter 4 • Relations, functions and transformations

171

y

g

y= 0

diGital doC doc-9740 WorkSHEET 4.1

1– x

y

h

y

i

1

x

0

x

0 −2

x

9 We7 For each relation given, sketch its graph and state the domain and range using interval notation. a {(x, y): y = 2 − x2} b {(x, y): y = x3 + 1, x ∈ [−2, 2]} c {(x, y): y = x2 + 3x + 2} d {(x, y): y = x2 − 4, x ∈ [−2, 1]} e {(x, y): y = 2x − 5, x ∈ [−1, 4)} f {(x, y): y = 2x2 − x − 6}

Verify that the graphs are correct with a CAS calculator.

diGital doC doc-9741 Investigation interesting relations

10 State the implied domain for each relation defined by the following rules. a y = 10 − x b y=3 x −

16 − x 2 c y= 2 d y=x +3 1 x f y = 10 − 7x2 e y=

types of relations (including functions) 4d

one-to-one relations

y

A one-to-one relation exists if, for any x-value, there is only one corresponding y-value and vice versa. For example:

0

x

{(1, 1), (2, 2), (3, 3), (4, 4)}

one-to-many relations y

A one-to-many relation exists if there is more than one y-value for any x-value but for any y-value there is only one x-value. For example:

x

0

{(1, 1), (1, 2), (2, 3), (3, 4)}

Many-to-one relations A many-to-one relation exists if there is more than one x-value for any y-value but for any x-value there is only one y-value. For example:

y

{(−1, 1), (0, 1), (1, 2)} 0

Many-to-many relations A many-to-many relation exists if there is more than one x-value for any y-value and vice versa. For example: {(0, −1), (0, 1), (1, 0), (−1, 0)}

172

Maths Quest 11 Mathematical Methods CAS

y

y

0

x

x

0

x

Worked exaMple 8

What type of relation does each graph represent? a

b

y

0

c

y

x

y

think

x

0

x

0

tUtorial eles-1462 Worked example 8

Write

a 1 For some x-values there is more than one

a One-to-many relation.

y-value. A line through some x-values shows that two y-values are available: y x = −1

x

0

2

For any y-value there is only one x-value. A line through any y-value shows that only one x-value is available: y y=1 x

0

b 1 For any x-value there is only one y-value. 2

For any y-value there is only one x-value.

c 1 For any x-value there is only one y-value. 2

b One-to-one relation.

c Many-to-one relation.

For some y-values there is more than one x-value.

Functions Relations that are one-to-one or many-to-one are called functions. That is, a function is a relation where for any x-value there is at most one y-value. For example: y

0

y

x

0

x

Vertical line test A function is determined from a graph if a vertical line drawn anywhere on the graph cannot intersect with the curve more than once. Chapter 4 • Relations, functions and transformations

173

Worked Example 9

State whether or not each of the following relations are functions. a {(−2, 1), (−1, 0), (0, −1), (1, −2)} y b c y

x

0

x

0 Think

Write

a For each x-value there is only one y‑value. (Or, a plot of

a Function

the points would pass the vertical line test.) b It is possible for a vertical line to intersect with the curve

b Not a function

more than once. c It is not possible for any vertical line to intersect with

c Function

the curve more than once.

Exercise 4D

Types of relations (including functions)

1 WE8 What type of relation does each graph represent? y y a b

0

d

h

y

x

k

y

0

x

0

y

x

y

x

0

l

x

x

0

i

y

x

0

f

y

0

y

x

y

0

x

0

j

0

e

y

0

g

x

c

x

y

0

x

2 WE9 Use the vertical line test to determine which of the relations in question 1 are functions. 174 Maths Quest 11 Mathematical Methods CAS

3 MC Which of the following relations is not a function? a {(5, 8), (6, 9), (7, 9), (8, 10), (9, 12)} y C y2 = x b

d y = 8x − 3

x

0

y

0

4 MC Consider the relation y ≥ x + 1. a The graph that represents this relation is: a y

b

x

y 1

1

d

y

x

−1 0

x

−1 0

C

e

y 1

1 0 e

0

1

x

x

1

Note: The shaded side indicates the region not required.

y

1 −1 0

x

b This relation is:

a one-to-one C many-to-one e a function

b one-to-many d many-to-many

c The domain and range are respectively:

R and R+ R and R R and R− R+ and R R− and R 5 Which of the following relations are functions? State the domain and range for each function. a {(0, 2), (0, 3), (1, 3), (2, 4), (3, 5)} b {(−3, −2), (−1, −1), (0, 1), (1, 3), (2, −2)} − − − − d {(1, 2), (1, 0), (2, 1), (3, 2), (4, 3)} c {(3, 1), (4, 1), (5, 1), (6, 1)} e {(x, y): y = 2, x ∈ R} f {(x, y): x = −3, y ∈ Z} g y = 1 − 2x h y>x+2 2 2 i x + y = 25 j y = x + 1, x ≥ − 1 3 k y=x +x l x = y2 + 1 a b C d e

power functions (hyperbola, truncus and square root function) 4e

Power functions are functions of the form f (x) = xn, n ∈ R. The value of the power, n, determines the type of function. We saw earlier that when n = 1, f (x) = x, and the function is linear. When n = 2, f (x) = x2, and the function is quadratic. When n = 3, f (x) = x3, and the function is cubic. When n = 4, f (x) = x4, and the function is quartic.

interaCtiVitY int-0263 domain and range

Chapter 4 • Relations, functions and transformations

175

Other power functions are: − • when n = −1, f (x) = x 1, and the power function produces the graph of a hyperbola − − • when n = 2, f (x) = x 2, and the power function produces the graph of a truncus 1 • when n = 2 , f ( x ) = x , and the function is the square root function. Under a sequence of transformations of f (x) = xn, n ∈ R, the general form of a power function is f (x) = a(x − b)n + c (where a, b, c and n ∈ R). 1 2

The hyperbola

1

• The graph shown is called a hyperbola, and is given by the equation y = x . − • This can also be represented as the power function y = x 1. y

y=0 0

x

x=0

The graph exhibits asymptotic behaviour. • That is, as x becomes very large, the graph approaches the x‑axis but never touches it. As x becomes very small (approaches 0), the graph approaches the y‑axis, but never touches it. • So the line x = 0 (the y‑axis) is a vertical asymptote, and the line y = 0 (the x‑axis) is the horizontal asymptote. • Both the domain and the range of the function are all real numbers, except 0; that is, R \ {0}. 1 • The graph of y = can be subject to a number of transformations. x a − • Consider y = + c or y = a( x − b) 1 + c. x−b y

Dilation

3 2 1

• The value a is a dilation factor. It dilates the graph from the x‑axis. 3 • For example, the graph of y = compared to the basic graph of x 1 y = is shown at right. x

0 −3 −2 −1 −1

y= (1, 3)

(1, 1)

1 2 3

3– x

y=

1– x

x y=0

−2 −3

x=0

Reflection

• If a is negative, the graph of the basic hyperbola is reflected in the x‑axis. • If x is replaced with −x, the graph of the basic hyperbola is reflected in the y‑axis. 1 1 and y = − are reflections of each other across the y‑axis. • For example, the graphs of y = x +1 x+1 y 2 1 0 −2 −1 −1 −2

y y = x 1+ 1

1 2

3 2 1 x

0 −2 −1 −1 1

2

x

−2

y = −x1+ 1 x = −1

176 Maths Quest 11 Mathematical Methods CAS

x=1

translation horizontal translation

• The value b translates the graph b units horizontally, that is, parallel to the x-axis. If b > 0, the graph is translated to the right, and if b < 0, the graph is translated to the left. 1 is a basic hyperbola translated one unit to the left, since • For example, the graph with equation y = x +1 b = −1. This graph has a vertical asymptote of x = −1 and domain R \ {−1}, and a horizontal asymptote of y = 0. 1 , with a vertical asymptote • If a basic hyperbola is translated one unit to the right, it becomes y = x −1 of x = 1 and domain R \ {1}. • Hence, the equation of the vertical asymptote is x = b and the domain is R \ {b}. The horizontal asymptote and the range remain the same, that is, x = 0 and R \ {0} respectively. Vertical translation

• The value c translates the graph c units vertically, that is, parallel to the y-axis. If c > 0, the graph is translated upward, and if c < 0, the graph is translated c units downward. 1 • The graph with equation y = − 1 is a basic hyperbola translated one unit down. This graph has a x horizontal asymptote of y = −1, a range of R \ {−1} and a vertical asymptote of x = 0. 1 • If a basic hyperbola is translated two units up, it becomes y = + 2, with a horizontal asymptote of x y = 2 and a range of R \ {2}. Hence, the equation of the horizontal asymptote is y = c, with a vertical asymptote of x = 0, and the range is R \ {c}. y y=c

y=

a x−b

+c

c 0

b

x

x=b

a − + c or y = a(x − b) 1 + c shows the combination of these transformations. The graph of y = x−b Worked exaMple 10

3 Sketch the graph of y = + 2, clearly showing the intercepts with the axes and the position of +1 x the asymptotes. think

Write/draW

a Compare the given equation with y = + c, x−b and state the values of a, b and c.

a = 3, b = −1, c = 2

2

Write a short statement about the effects of 1 a, b and c on the graph of y = . x

The graph of y =

3

Write the equations of the asymptotes. The horizontal asymptote is at y = c. The vertical asymptote is at x = b.

Asymptotes: x = −1; y = 2

4

Find the value of the y-intercept by letting x = 0.

y-intercept: x = 0

1

1 is dilated by the factor of 3 x from the x-axis (a = 3), translated 1 unit to the left (b = −1) and 2 units up (c = 2).

3 +2 0 +1 =3+2 =5

y=

Point (0, 5)

Chapter 4 • Relations, functions and transformations

177

5

Find the value of the x-intercept by making y = 0.

x-intercept: y = 0

3 +2 x +1 3 − 2= x +1 − 2(x + 1) = 3 −2x − 2 = 3 −2x = 5 0=

x= Point 6

To sketch the graph: (a) draw the set of axes and label them (b) use dotted lines to draw the asymptotes The asymptotes are x = −1 and y = 2. (c) mark the intercepts with the axes −5 The intercepts are y = 5 and x = 2 . (d) treat the asymptotes as your new set of axes, sketch the graph of the hyperbola. (As a is positive, the graph is not reflected in the x-axis.)

(

−5

2

,0

−5 2

) y

y= 6 (0, 5) 4 2 0 −3 −2 −1 −2 (− 5 , 2

0)

3 x+1

+2

y=2 x

1 2 3

−4 −6

x = −1

the truncus • The graph shown is known as a truncus. The equation of the graph is given by: 1 y= 2 x −

• This can also be represented as the power function y = x 2. • The function is undefined for x = 0. Hence, the equation of the vertical asymptote is x = 0 and the domain of the function is R \{0}. • We can also observe that the graph approaches the x-axis very closely but never touches it. So y = 0 is the horizontal asymptote. • Since the whole graph of the truncus is above the x-axis, its range is R+ (that is, all positive real numbers). • Similarly to the graphs of the functions, discussed in the previous 1 sections, the graph of y = 2 can undergo various transformations. x a + c or y = a(a − b)−2 + c. • Consider the general formula y = ( x − b) 2

dilation • The value a is the dilation factor. It dilates the graph from the x-axis. The dilation factor does not affect the domain, range or asymptotes. 4 • Consider the graph of y = 2 . x

y y= y=0

1 x2

x

0 x=0

y 6 4 2 −3 −2 −1 0

y= 1 2 3

4 x2

x y=0

x=0

reflection • If a is negative, the graph of a basic truncus is reflected in the x-axis. The range becomes R− (that is, all negative real numbers). • If x is replaced with −x, the graph of the basic truncus is reflected in the y-axis. The effect of this reflection cannot be seen in the basic graph, but it becomes more obvious if the graph has been translated horizontally first. 178

Maths Quest 11 Mathematical Methods CAS

1 1 and y = − are reflections across the y-axis. 2 ( x − 2)2 ( x − 2) − The vertical asymptote changes from x = 2 to x = 2, and the domain changes from R \{2} to R \{−2}.

• For example, the graphs of y =

y

y 3 2 1 0 −2 −1 −1

y=

1

2

1 (x − 2)2

y=

x

1 (−x − 2)2

x

−3 −2 −1 0

−2

x = −2 x=2

translation horizontal translation

• The value b translates the graph b units horizontally. If b > 0, the graph is translated to the right, and if b < 0, the graph is translated left. 1 • For example, the graph of the equation y = results from translating a basic truncus 3 units to ( x − 3)2 the right. The vertical asymptote is x = 3 and the domain is R \ {3}. 1 , where the vertical asymptote • If a basic truncus is translated 2 units to the left, it becomes y = ( 2)2 x + is x = −2 and the domain is R \ {−2}. • Hence, the equation of the vertical asymptote is x = b, and the domain is R \ {b}. The range is still R+, and the equation of the horizontal asymptote is y = 0. Vertical translation

• The value c translates the graph c units vertically. If c > 0, the graph is translated upward, and if c < 0, the graph is translated c units downward. 1 • For example, the graph with equation y = 2 + 1 results when a basic truncus is translated 1 unit x upward. The horizontal asymptote is y = 1, and the range is (1, ∞). 1 • If a basic truncus is translated 1 unit down, it becomes y = 2 − 1, with y = −1 as the horizontal x asymptote and (−1, ∞) as the range. • Hence, the equation of the horizontal asymptote is y = c, and the range is (c, ∞). Note: If a is positive (see graph below), the whole graph of the truncus is above the line y = c (the horizontal asymptote), and hence its range is y > c, or (c, ∞). y • If a is negative, the whole graph is below its horizontal asymptote, y= − and therefore the range is y < c, or ( ∞, c). a + c or y = a(x − b)−2 + c shows the • The graph of y = ( x − b) 2 c combination of these transformations. 0

b

a (x − b)2

+c

y=c x

x=b

Worked exaMple 11

1 + 3, clearly showing the position of the asymptotes and the ( x − 2) 2 intercepts with the axes. Sketch the graph of y = think

Write

a +c ( x − b) 2

1

Write the general formula for the truncus.

y=

2

Identify the values of a, b and c.

a = 1, b = 2, c = 3

Chapter 4 • Relations, functions and transformations

179

1 The graph of y = 2 is translated 2 units to the right and 3 units up. x

3

Write a short statement about the 1 transformations the graph of y = 2 should undergo in order to be changed x into the one in question.

4

Write the equations of the asymptotes (y = c and x = b).

Asymptotes: x = 2 and y = 3

5

Find the x-intercept. As you cannot get the square root of a negative number, there is no solution and therefore no x-intercepts. An inspection of the equation of the graph would also have revealed this.

x-intercept: y = 0 1 0= +3 ( x − 2)2 1 = −3 ( x − 2)2 ( x − 2)2 =

−1 3 −1

( x − 2) = ±

3

y-intercept: x = 0 1 y= +3 (0 − 2)2 1 = +3 4 13 = 4

6

Find the y-intercept.

7

To sketch the graph: (a) draw the set of axes and label them (b) use dotted lines to draw asymptotes (c) mark the x- and y-intercepts (d) treating the asymptotes as the new set of axes, draw the basic truncus curve (e) make sure it intersects the axes in the right places.

y (0,

13 ) 4

6 4 2

−2 −1 0

y=3 1 2 3 4x x=2

the square root function • • • • •

The square root function is given by y = x . This can be written as the power function y = x . The function is defined for x ≥ 0; that is, the domain is R+ ∪ {0}, or [0, ∞). As can be seen from the graph, the range of the square root function is also R+ ∪ {0}, or [0, ∞). Throughout this section we will refer to the graph of y = x as ‘the basic square root curve’. 1 2

y y= x x

0

• Let us now investigate the effects of various transformations on the basic square root curve. • Consider the function y = a x − b + c or y = a( x − b) + c. 1 2

dilation • The value a is a dilation factor; it dilates the graph from the x-axis. The domain is still [0, ∞). 180

Maths Quest 11 Mathematical Methods CAS

Reflection • If a is negative, the graph of a basic square root curve is reflected in the x‑axis. The range becomes (−∞, 0]. The domain is still [0, ∞). • If x is replaced with −x, the graph is reflected in the y‑axis. For example, the graphs with equations y = x and y = − x are reflected across the y‑axis. • The domain becomes (−∞, 0] and the range is [0, ∞). y

y y= x

0

1 2

y = −x x

−2 −1 0

1 2

x

Translation Horizontal translation

• The value h translates the graph horizontally. If b > 0, the graph is translated to the right, and if b < 0, the graph is translated to the left. • The graph with the equation y = x − 1 results when the basic curve is translated 1 unit to the right. This translated graph has domain [1, ∞) and range [0, ∞). • If the basic curve is translated 3 units to the left, it becomes y = x + 3 and has domain [−3, ∞) and range [0, ∞). y y= x+3 2 1

3

−3 −2 −1 0

1 2 3

x

• The domain of a square root function after a translation is given by [b, ∞). Vertical translation

• The value c translates the graph vertically. If c > 0, the graph is translated vertically up, and if c < 0, the graph is translated vertically down. • If y = x is translated 3 units vertically up, the graph obtained is y = x + 3, with domain [0, ∞) and range [3, ∞). y 5 4 3 2 1

y= x+3

0 −2 −1 −1

1 2 3 4x

• If the basic curve is translated 2 units down, it becomes y = x − 2, with domain [0, ∞) and range [−2, ∞). • The range of the square root function is [c, ∞) for a > 0. • The graph of y = a x − b + c or y = a( x − b) + c shows the combination of these transformations. 1 2

y y=a b−x+c

(−b, c)

y=a x+b+c

(b, c) x

Chapter 4 • Relations, functions and transformations 181

Worked Example 12

Sketch the graph of y = 3 x − 1 + 2, clearly marking intercepts and the end points. Think

Write/draw

1

Write the equation.

y = 3 x − 1 + 2,

2

Write the coordinates of the end point.

End point: (1, 2)

3

State the shape of the graph.

Shape:

4

Inspection of the equation reveals that there is no y-intercept.

There is no x-intercept.

5

Inspection of the equation reveals that there is no y-intercept.

There is no y-intercept.

6

To help sketch the graph, determine the coordinates of a second point. Let x = 3.

x = 3: y = 3 3−1 + 2

7

.

=3× 2+2 Point: (3, 3 2 + 2)

Sketch the graph by plotting the end point, showing the second point, and drawing the curve so that it starts at the end point and passes through the second point.

y 7 6 5 4 3 2 1 0 −2 −1 −1

y=3 x−1+2 (3, 3 2 + 2)

(1, 2) 1 2 3 4x

Power functions (hyperbola, truncus and square root function)

Exercise 4E

1 State the dilation factor and the vertical and horizontal translations or reflections for each of the

following. Write a short statement about the effects each has on the basic graph of that function. 3 +1 a y= x−5 2 b y = 2 +1 x 1 −2 c y= 4−x d y= 2 x+4 e y= f

1 (3 − x )2

y = −4 x −5

2 WE10, 11, 12 Sketch a graph for each of the following functions, clearly showing x- and y‑intercepts

and any asymptotes. 3 a f (x) = 2 + 1 x 2 c f (x) = +5 x −1 182 Maths Quest 11 Mathematical Methods CAS

b f (x) =

3− x −1

d f (x) = 2 −

2 ( x − 1)2

3 Assuming the dilation factor is 1, write the equation of the graph. y 9 8 7 6 5 4 y=3 3 5– 2 2 1 1 2 3 4x

−1 0 5– 3

x=2

4F

Function notation

Consider the relation y = 2x, which is a function. The y-values are determined from the x-values, so we say ‘y is a function of x’, which is abbreviated to y = f (x). So, the rule y = 2x can also be written as f (x) = 2x. If x = 1, then y = f (1) If x = 2, then y = f (2) =2×1 =2×2 =2 = 4, and so on.

evaluating functions For a given function y = f (x), the value of y when x = 1 is written as f (1), the value of y when x = 5 is written as f (5) and so on. Worked exaMple 13

If f (x) = x2 − 3, find: a f (1) b f (−2)

c f (a)

d f (2a).

think

a 1 Write the rule. 2

Substitute x = 1 into the rule.

3

Simplify.

b 1 Write the rule. 2

Substitute x = −2 into the rule.

3

Simplify.

c 1 Write the rule. 2

Substitute x = a into the rule.

d 1 Write the rule. 2

Substitute x = 2a into the rule.

3

Simplify the expression if possible.

Write

a f (x) = x2 − 3

f (1) = 12 − 3 =1−3 = −2 b f (x) = x2 − 3

f (−2) = (−2)2 − 3 =4−3 =1 c

f (x) = x2 − 3 f ( a) = a2 − 3

d f (x) = x2 − 3

f (2a) = (2a)2 − 3 = 22a2 − 3 = 4a2 − 3

Chapter 4 • Relations, functions and transformations

183

Fully defining functions To fully define a function: 1. define the domain 2. state the rule. That is, if a function f (x) has domain X, the function may be defined as follows: f : X → Y, f (x) = . . . . . . Domain Co-domain Rule Y is not necessarily the range but is a set that contains the range, called the co‑domain. The co-domain gives the set of possible values that contains y. It is usually R (the set of real numbers). The actual values that y can be — the range — is determined by the rule. When using function notation the domain can be abbreviated as dom f and the range as ran f. For example, the function defined by {(x, y): y = 2x, x ∈ [0, 3]} can be expressed in function notation as f : [0, 3] → R, f (x) = 2x. For this function we can write dom f = [0, 3]. The co-domain = R. Also, ran f = [0, 6] (x = 0 gives y = 0 and x = 3 gives y = 6, which are the minimum and maximum values of y). R

R –5.1

– 7–8

– 2

1– 3

0.6

[0, 3]

3 — – 11 3

f : domain

0

etc. 2– 2 1.2 3 46 2 3 etc. 0 Range

1 2 3

–10

etc. Domain

[0, 6]

Co-domain co-domain , f(x) = rule

The graph of this function is shown below. y 6 5 4 3 2 1 0

f(x)

1

2

3x

The maximal domain of a function is the largest possible set of values of x for which the rule is defined. The letters f, g and h are usually used to name a function, that is, f (x), g(x) and h(x). Note: If a function is referred to by its rule only, then the domain is assumed to be the maximal domain. Worked Example 14

Express the following functions in function notation with maximal domain. 1 a {(x, y): y = x2 − 4} b y = 3x − 4, −2 ≤ x ≤ 5 c y = x Think

Write

a The rule has meaning for all values of x (it is a

quadratic), so the domain of the function is R. b The rule has meaning for all values of x in the

given domain

[−2,

5].

c The rule has meaning for all values of x except 0.

184 Maths Quest 11 Mathematical Methods CAS

a f : R → R, f (x) = x2 − 4 b f : [−2, 5] → R, f (x) = 3x − 4

c f : R \ {0} → R, f (x) =

1 x

Worked exaMple 15

State i the domain, ii the co-domain and iii the range for each of the following functions. 1 a f : R → R, f (x) = 5 − x b g : R+ → R, g(x) = x think

Write/draW

a 1 The domain is given as R. 2

The co-domain is given as R.

3

Use a CAS calculator to obtain the graph of the function, or sketch it.

a i dom f = R ii The co-domain is R. y 5

0

4

From the graph the range is observed to be R.

b 1 The domain is given as R+. 2

The co-domain is given as R.

3

Use a CAS calculator to obtain the graph of the function, or sketch it.

5

iii ran f = R b i dom g = R+ ii The co-domain is R. y

1 0 4

The range is observed from the graph to be R+.

x f(x)

g(x) x

1

iii ran g = R+

Worked exaMple 16

State i the maximal domain and ii the range for the function defined by the rule: 1 . a y = x+1 b y = x+2 think

Write/draW

a 1 The rule has meaning for all x if x + 1 ≥ 0

(that is, contents of

tUtorial eles-1463 Worked example 16

a Require x + 1 ≥ 0

are positive).

2

Solve this inequation.

So x ≥ −1

3

State the maximal domain.

i Maximal domain = [−1, ∞)

4

To confirm the maximal domain and see the range, use a CAS calculator or other technology to obtain the graph of the function, or sketch it by plotting selected points.

y y= x+1 1 (−1, 0) 0

5

The range is observed from the graph to be [0, ∞).

x

ii Range = [0, ∞)

Chapter 4 • Relations, functions and transformations

185

b 1 The rule exists for all x, except when x + 2 = 0. 2

Therefore x ≠ −2.

x ≠ −2

3

State the maximal domain.

i Maximal domain = R \ {−2}

4

Use a CAS calculator to obtain the graph of the function, or sketch it by plotting selected points to confirm the maximal domain and see the range.

5

y 1 y = —— x+2

1 −2 −1 0

x

ii Range = R \ {0}

The range is observed from the graph to be R \ {0}.

exercise 4F

Function notation

1 We13 a If f (x) = 3x + 1, find: diGital doCS doc-9742 SkillSHEET 4.2 Substitution doc-9743 SkillSHEET 4.3 transposition of equations

bx+2≠0

b If g(x) =

x + 4 , find: 1 c If g(x) = 4 − , find: x d If f (x) = (x + 3)2, find: 24 e If h (x) = , find: x

i f (0),

ii f (2),

iii f (−2) and

iv f (5).

i g(0),

ii g(−3),

iii g(5) and

iv g(−4),

i g(1),

ii g 1 ,

iii g

i f (0),

ii f (−2),

iii f (1) and

iv f (a).

i h (2),

ii h (4),

iii h (−6) and

iv h (12).

2

− 1 , 2

− 1 , 5

and

iv g

2 Find the value (or values) of x for which each function has the value given. a f (x) = 3x − 4, f (x) = 5 b g(x) = x2 − 2, g(x) = 7

1 , f (x) = 3 x e g(x) = x2 + 3x, g(x) = 4 c f (x) =

10 3 Given that f ( x ) = − x , find: x a f (2) d f (x2)

d h (x) = x2 − 5x + 6, h (x) = 0 f

f (x) =

b f (−5) e f (x + 3)

8 − x , f (x) = 3 c f (2x) f f (x − 1)

4 We14 Express the following functions in function notation with maximal domain. b y=

x−6 2 d y= x +1 f y = x2 + 3x, where x ≥ 2 h y = x2 + x

a {(x, y): y = 4x + 1}

1 x −1 e y = (x + 2)2, where x ∈R+ g y = 8 − x, where x ≤ 0 c y=

5 We15 For each of the following functions, state: i the domain ii the co-domain iii the range. diGital doC doc-9744 Square root graphs

a f : {0, 1, 2, 3} → Z, f (x) = 3x − 7 c f : {2, 4, 6, 8, 10} → N, f (x) =

x 2

e g: R+ → R, g(x) = x2 − 2

d f : (−∞, 0) → R, f (x) = f

6 We16 State i the maximal domain and

186

b g: (0, 10] → R, g(x) =

3 x 1 −x

h: [−3, 3] → R, h(x) = 9 − x 2

ii the range for the function defined by the rule:

a f (x) = 3 − x

b f (x) = 5 x

c y = x3 + 2

d y = 5 − 3x2

e y=

f

Maths Quest 11 Mathematical Methods CAS

x−4

y=

1 x−3

Special types of function (including hybrid functions) 4G

one-to-one functions

As we have already seen, one-to-one relations and many-to-one relations are functions. A one-to-one function has, at most, one y-value for any x-value and vice versa. The graph of a relation is a function if any vertical line crosses the curve at most once. Similarly, a one-to-one function exists if any horizontal line crosses the curve at most once. For example: y

0

y

x

0

A function that is not one-to-one; this passes the vertical line test but not the horizontal line test.

eleSSon eles-0077 hybrid functions

x

A one-to-one function; this passes the vertical line test and the horizontal line test.

Worked exaMple 17

Which of the following functions are one-to-one? a {(0, 1), (1, 2), (2, 3), (3, 1)} b {(2, 3), (3, 5), (4, 7)} c f (x) = 3x think 1

Write/draW

a When x = 0 and x = 3, y = 1.

Check whether each function has, at most, one y-value for any x-vaue and vice versa.

It is not a one-to-one function. b There is only one x-value for each y-value.

2

Sketch the graph of f (x) = 3x. Check whether both a vertical line and a horizontal line crosses only once.

y

c

f(x)

3 0

1

x

It is a one-to-one function. 3

Write a statement to answer the question.

The functions are one-to-one for b and c.

Worked exaMple 18

Which of the following graphs show a one-to-one function? a

b

y

0

x

c

y

0

x

think

If a function is one-to-one, any vertical or horizontal line crosses the graph only once.

y

0

x

Write

Only b is a one-to-one function.

Chapter 4 • Relations, functions and transformations

187

Restriction of functions Restrictions can be placed on a function through its domain. If we have one relation, for example f (x) = x2, we can create several different functions by defining different domains. For example: y

f(x)

x

0

y

y

h(x)

g(x) −1 0

1x

x

0

f : R → R, f (x) = x2 g: [−1, 1] → R, g(x) = x2 h: R+ → R, h(x) = x2 The restriction imposed on the function f to produce the function h has created a one-to-one function.

Worked Example 19

For each function graphed below state two restricted, maximal (largest possible) domains that make the function one-to-one. y y = (x − 2)2 b a y 1 y=— 2

4

0

x

2

Think

a 1 One-to-one functions will be formed if the

x

0

x

Write/draw

a

curve is split into two through the vertical line x = 2.

y

y

4

0 2

State the required domains.

b 1 One-to-one functions will be formed if the

curve is split into two through the line x = 0.

2

x

0

x

2

For the function to be one-to-one, the domain is (−∞, 2] or [2, ∞). b

y

0

y

x

x

0

2

State the required domains.

For the function to be one-to-one, the domain is (−∞, 0) or (0, ∞).

Hybrid functions A hybrid, mixed, or piecewise defined function is a function that has different rules for different subsets of the domain. For example: x + 1, for x ≤ 0 f (x) = 2 is a hybrid function that obeys the rules y = x + 1 if x , for x > 0 x ∈ (−∞, 0] and y = x2 if x ∈ (0, ∞). The graph of f (x) is shown at right. Note that the domains do not overlap. 188 Maths Quest 11 Mathematical Methods CAS

y

f(x)

1 −1 0

x

Worked exaMple 20

x, a Sketch the graph of f ( x ) = x + 1, 5 − x, b State the range of f.

x<0 0≤x<2

tUtorial eles-1464 Worked example 20

x≥2

think

Write/draW

a If x = −1, y = x

a Calculate and plot points as shown.

= −1

1

Sketch the graph of y = x for the domain (−∞, 0).

If x = 0, y = x =0

2

On the same axes sketch the graph of y = x + 1 for the domain [0, 2).

If x = 0, y = x + 1 =1 If x = 2, y = x + 1 =3

3

On the same axes sketch the graph of y = 5 − x for the domain [2, ∞).

If x = 2, y = 5 − x =3 If x = 5, y = 5 − x =0 y 3 y=x+1 2 1

f(x) y=5−x

0 1 2 3 4 5 x y = x−1−1

b ran f = (−∞, 3]

b The range is made up of (or is the union of) two

sections, (−∞, 0) with (−∞, 3].

Special types of function (including hybrid functions) exercise 4G

1 We17 Which of the following functions are one-to-one? Use a CAS calculator or other technology to

obtain the graph of the function where appropriate.

a {(1, −1), (2, 1), (3, 3), (4, 5)} c {(x, y): y = x2 + 1, x ∈ [0, ∞)} e {(x, y): y = 3 − 2x2}

b {(−2, 1), (−1, 0), (0, 2), (1, 1)} d {(x, y): y = 3 − 4x} f f (x) = x3 − 1

g y = x2, x ≤ 0

h g(x) = 1 − x 2

2 We18 Consider the relations below and state: i which of them are functions ii which of them are one-to-one functions. a b y y

0

x

0

c

x

y

0

x

Chapter 4 • Relations, functions and transformations

189

y

d

h

y

j

y

x

l

y

0

x

0

x

y

x

0

x

0

i

0

k

y

x

y

x

0

y

f

0

x

0

g

y

e

x

0

3 WE19 For each function below, state two restricted, maximal domains that make the function

one-to-one. a

−1 0 d

b

y

y

g f (x) = 1 − x2

3 x

−3 e

x

y −3 0

x

2

0

x

(3, 4)

0

c

y

f

y 0

−4

(−1, 4)

y

x

(−2, −2)

0 (1, 0)

4 − x 2 , x ∈ [−2, 2] j f (x) = (x + 3)2 h g(x) =

x

1 , x ∈ R \ {0} x2 4 MC Use the graph of the relation y2 = x − 1, shown below, to answer the following questions. a A one-to-one function can be formed by: y A restricting the domain to R+ B restricting the domain to [1, ∞) C restricting the domain to (1, ∞) x 0 1 D restricting the range to [0, ∞) E restricting the range to R \ {0} i g(x) =

b A rule that describes a one-to-one function derived from the relation y2 = x − 1 is: a y2 = x − 1 b y = ±

x −1

−

x −1

c y = d y =

x −1

e y =

x −1

190 Maths Quest 11 Mathematical Methods CAS

5 MC Consider the following hybrid function.

− x, x < 1 f (x) = x , x ≥ 1

a The graph that correctly represents this function is: a b y y

d

y

1

1 −1

C

0 1

2

x

0 −1

e

y

1

2

1

x

0 −1

x

1

y

1 0 −1

1

x

1

0 −1

x

1

b The range of this hybrid function is:

b R \ {−1} e R+

a R

d [0, ∞)

C (−1, ∞)

6 We20 a Sketch the graph of the following function.

1 x<0 , f (x) = x x + 1, x ≥ 0 b State the range of f. 2 x + 1,

7 a Sketch the graph of the function g( x ) = 2 − x , b State the range of g. c Find: i g(−1) ii g(0) iii g(1).

x≥0 x<0

.

x − 2, x < − 2 8 a Sketch the graph of the function f ( x ) = x 2 − 4, − 2 ≤ x ≤ 2 . x + 2, x > 2 b State the range of z. c Find: i f (−3)

ii f (−2)

iii f (1)

iv f (2)

v f (5).

y 3 2 1

9 Specify the rule for the function represented by the graph at right.

−2 −1 0 −1 −2 10 The graph of the relation {(x, y): x2 + y2 = 1, x ≥ 0} is shown at right.

From this relation, form two one-to-one functions and state the range of each.

f(x)

1

x

1

x

y 1 0

−1 11 a Sketch the graph of the function f : R → z, f (x) = (x − 3)2. b By restricting the domain of f, form two one-to-one functions that have the same rule as f (use the

largest possible domains). 12 a Sketch the graph of the function g : R → R, g(x) = x2 + 2x + 1. b By restricting the domain of g, form two one-to-one functions that have the same rule as g

(use maximal domains).

diGital doC doc-9745 WorkSHEET 4.2

Chapter 4 • Relations, functions and transformations

191

4H

Inverse relations and functions

A relation is a set of ordered pairs that can be graphed or described by a rule. The inverse of a set of ordered pairs is obtained simply by interchanging the x and y elements. So, the inverse of {(1, 5), (2, 6), (3, 7)} is {(5, 1), (6, 2), (7, 3)}. If these points are plotted on a set of axes, it can be seen that when each original point is reflected across the line y = x, the inverse points are obtained. Similarly, if the graph of a function is given, then its inverse function can be sketched by reflecting the original function across the line y = x. Only functions that are one-to-one have inverses.

Worked Example 21

Sketch the graph of the following and then sketch the inverse. a {(3, −6), (4, −4), (5, −2), (6, 0)} b y

c y = x2 for x ≥ 0

x Think

a 1 Plot the points on a set of axes. 2

Interchange the x- and y-values and plot them. Alternatively, reflect the original points across the line y = x. So, (3, −6) → (−6, 3) (4, −4) → (−4, 4) (5, −2) → (−2, 5) (6, 0) → (0, 6)

b 1 Re-draw the given graph. 2

On the same set of axes, plot the line y = x.

3

Sketch a reflection of the original graph across the line y = x.

Write/draw

a

y y=x 6 5 4 3 2 1 0 −6−5−4−3−2−1 −1 1 2 3 4 5 6 x −2 −3 −4 −5 −6

b y

y=x

x c Sketch the original function. The graph is a

c y

parabola with turning point (0, 0). The domain is restricted.

y = x2

y=x

1 0

192 Maths Quest 11 Mathematical Methods CAS

1

x

inverse relations and functions

exercise 4h

1 We21a Sketch the graph of the following and then sketch the inverse. a (1, 7), (2, 5), (3, 3)} b (1, 3), (2, 6), (3, 9)} c (−2, 11), (0, 6), (2, 1)} 2 We21b Sketch the graph of the following and then sketch the inverse. (Assume each set of axes has

the same scale for x and y.) a

y

b

y

x

x c

y

d

y

x x y

e

y

f

x x 3 We21c Sketch the graph of the following and then sketch the inverse. a y = 4x

4i

b y = x2 + 3, x ≥ 0

1

c y = 2x + 1

d y = x3 + 4

Circles

y r

A circle is a many-to-many relation. The rule that defines a circle with its centre at (0, 0) and of radius r is −r

x2 + y2 = r 2 The graph of this circle is shown at right. The vertical-line test clearly verifies that the circle graph is not a function. Solving the equation for y we have y2 = r 2 − x2, so y = r 2 − x 2 or y = − r 2 − x 2 . These two relations represent two semicircles that together make a complete circle: • y = r 2 − x 2 is the ‘upper semicircle’ (above the x-axis).

r x −r

diGital doC doc-9746 Investigation a special relation

• y = − r 2 − x 2 is the ‘lower semicircle’ (below the x-axis). y r −r

y y = r 2− x2

r

x

−r

r x −r y = − r 2 − x 2

Both these relations are many-to-one functions. Chapter 4 • Relations, functions and transformations

193

Worked Example 22

Sketch the graphs of the following relations. a x2 + y2 = 16 b x2 + y2 = 9, 0 ≤ x ≤ 3 c y = 8 − x 2 Think

draw

a 1 This relation is a circle of centre (0, 0) and radius = 16 = 4. 2

On a set of axes mark x- and y‑intercepts of −4 and 4.

3

Draw the circle.

y

a

4 −4

4 x

0 −4

b 1 This relation is part of a circle of centre (0, 0) and radius = 2

Since the domain is [0, 3], on a set of axes mark y-intercepts −3 and 3 and x‑intercept 3.

3

Draw a semicircle on the right-hand side of the y-axis.

−3

3

Draw a semicircle above the x-axis.

y

c

and radius = 8.

On a set of axes mark the x‑intercepts of y‑intercepts of 8 .

3 x

0

c 1 This relation is an ‘upper semicircle’ (as y > 0) of centre (0, 0)

2

y 3

b

9 = 3.

8 −

8 and 8 and − 8

General equation of a circle

y

The general equation of a circle with centre (h, k) and radius r is (x − h)2 + (y − k)2 = r 2. The domain is [h − r, h + r]. The range is [k − r, k + r].

k+r Range

8x

0

(x − h)2 + (y − k)2 = r2

k

(h, k)

k−r 0 h−r

h Domain

Worked Example 23

Sketch the graphs of the following circles. State the domain and range of each. a x2 + (y − 3)2 = 1 b (x + 3)2 + (y + 2)2 = 9 Think

a 1 This circle has centre (0, 3) and radius 1. 2

On a set of axes mark the centre and four points; 1 unit (the radius) left and right of the centre, and 1 unit (the radius) above and below the centre.

Write/draw

a

y 4

x 2 + (y − 3)2 = 1

3 2 −1 0

1

x

3

Draw a circle that passes through these four points.

4

State the domain.

The domain is [−1, 1].

5

State the range.

The range is [2, 4].

194 Maths Quest 11 Mathematical Methods CAS

h+r x

b 1 This circle has centre (−3, −2) and radius 3. 2

3

b

y 1

(−3, 1)

On a set of axes mark the centre and four points; 3 units left and right of the centre, and 3 units above and below the centre.

−6 −5 −4 −3 −2 −1 0 −1 (−6, −2)

Draw a circle that passes through these four points.

(0, −2)

−2

(−3, −2)

1 x

−3 −4 −5

(−3, −5) 4

State the domain.

The domain is [−6, 0].

5

State the range.

The range is [−5, 1].

Note: When using a CAS calculator to plot circle graphs, ensure that the upper and lower values are entered as separate equations on the Graphs & Geometry page; for example, f1(x) = − 16 − x 2 and f2(x) = 16 − x 2 .

exercise 4i

Circles

1 State the equation of each of the circles graphed below. y y a b 3 1 −3

−1 0 −1

3x

0 −3

d

e

y 10 −10

c

5 −5

x

1

10 x

f

y

y

−10

2 2 −2 2

6 x

0

− 6

5 x

0 −5

6

0

diGital doC doc-9747 Circle graphs

y

2 2x

0

−2 2

− 6 h

y

g

3 −3

y −4

3x

0

0

4 x

−4

2 State the domain and range of each circle in question 1. 3 We22 Sketch the graphs of the following relations. a x2 + y2 = 4 b x2 + y2 = 16 d x2 + y2 = 7

c x2 + y2 = 49

e x2 + y2 = 12

f

x2 + y2 =

1 4

4 Sketch the graph of each of the following relations and state whether it is a function or not. a y = ± 81 − x 2

b y=

4 − x2

c y = − 1− x2

d y=

1 9

e y=−

1 4

− x2

g y = ± 10 − x 2

f

− x2

y = 5 − x2

h x2 + y2 = 3,

−

3≤x≤0

Chapter 4 • Relations, functions and transformations

195

5 MC Consider the circle below.

y 2 0

2

4 x

−2 a The equation of the circle is: A x2 + (y − 2)2 = 4 c (x + 2)2 + y2 = 16 e (x + 2)2 + y2 = 4

b The range of the relation is: A R d [2, 4]

b (x − 2)2 + y2 = 16 d (x − 2)2 + y2 = 4

b [−2, 2] e [−2, 1]

6 MC Consider the equation (x + 3)2 + (y − 1)2 = 1. a The graph that represents this relation is: y B y A 4 2 1 1 0 2 3 4 0 x −3 −6 −2 D

y −1 −2

c [0, 4]

2 1 x

2

3

4

−4 −3 −2 y 2

E

0

y

C

x 1 0

−3.5−3−2.5 b The domain of the relation is: A [−3.5, −2.5] D [2, 4]

B E

x

(−4, −2) [−4, −2]

C R

7 WE23 Sketch the graph of the following circles. State the domain and range of each. a x2 + (y + 2)2 = 1 b x2 + (y − 2)2 = 4 c (x − 4)2 + y2 = 9 d (x − 2)2 + (y + 1)2 = 16 e (x + 3)2 + (y + 2)2 = 25 f (x − 3)2 + (y − 2)2 = 9 2 2 9 3 1 g (x + 5) + (y − 4) = 36 h (x − )2 + (y + )2 = 2

8 Express the relation x2 + y2 = 36 as two

functions, and state the largest domain and range of each. 9 Express the relation x2 + (y − 2)2 = 9 as

two functions, and state the largest domain and range of each. 10 Circular ripples are formed when a

water drop hits the surface of a pond. If one ripple is represented by the equation x2 + y2 = 4 and then 3 seconds later by x2 + y2 = 190, where the length of measurements are in centimetres: a find the radius (in cm) of the ripple in each case b calculate how fast the ripple is moving outwards. (State your answers to 1 decimal place.) 196 Maths Quest 11 Mathematical Methods CAS

2

4

0

x

4J

Functions and modelling

When using functions to model rules in real-life situations the domain usually has practical restrictions imposed on it. For example, the area of a circle is determined by the function A(r) = π 2. For a circle to be drawn the radius needs to be a positive number. Hence the domain is (0, ∞) or R+.

Worked exaMple 24

The table describes hire rates for a removal van. a Express the cost as a hybrid function. b Sketch the graph of the function. Hours of hire (h)

Cost ($C)

Up to 3

200

Over 3 up to 5

300

Over 5 up to 8

450

think

Write/draW

a 1 The cost is $200 if 0 < h ≤ 3. 2

The cost is $300 if 3 < h ≤ 5.

3

The cost is $450 if 5 < h ≤ 8.

4

tUtorial eles-1465 Worked example 24

a

200, C (h) = 300, 450,

State the cost function C(h).

b Sketch a graph with 3 horizontal lines over the

appropriate section of the domain.

0

b C ($) 450 400 350 300 250 200 150 100 50 0 1 2 3 4 5 6 7 8 h (hours)

exercise 4J

Functions and modelling

1 We24 The cost of hiring a paper recycling removalist is described in the following table:

Hours of hire

Cost

Up to 1

$40

Over 1 up to 2

$70

Over 2 up to 4

$110

Over 4 up to 6

$160

a State the cost function, $C, in terms of the time, t hours, for hiring up to 6 hours. b Sketch the graph of the function. Chapter 4 • Relations, functions and transformations

197

2 The charge for making a 10-minute STD call on the weekend is listed below.

Distance d (km) Cost $C

Up to 50 km 0.40

50 to 100 km 0.60

100 to 200 km 0.80

200 to 700 km 1.70

Over 700 km 2.00

a State the cost function in terms of the distance. b Sketch the graph of the function. 1 2

3 A car travels at a constant speed of 60 km/h for 1 hours, stops for half an hour, then travels for another

2 hours at a constant speed of 80 km/h until it reaches its destination. a Construct a function that describes the distance travelled by the car, d (km), at time, t hours. b State the domain and range of this function. c Calculate the distance travelled after: i 1 hour ii 3 hours. 4 At a fun park, a motorised toy boat operates for 5 minutes for every dollar coin placed in a meter. The meter will accept a maximum of 120 one-dollar coins. a Write a rule that gives the time of boat operation, B hours, in terms of the number of dollar coins, n. b Sketch the graph of the function and state the domain and range. c How much is in the meter when the boat has operated for 450 minutes? 5 The tax for Australian residents who earn a taxable income between $37 000 and $80 000 is $4650 plus 30 cents for every dollar earned over $37 000. a Write a rule for the tax payable, $T, for a taxable income, $x, where 37 001 ≤ x ≤ 80 000. b Sketch a graph of this function. c Calculate the tax paid on an income of $42 000. 6 The maximum side length of the rectangle shown is 10 metres. (x + 4) m (x − 1) m a Write a function that gives the perimeter, P metres, of the rectangle. b State the domain and range of this function. 7 A rectangular swimming pool is to have a length 4 metres greater than its width. a Write a rule for the area of the pool, A m2, as a function of the width, x metres. b State the domain and range if the maximum side length is 12 metres. 8 Timber increases in value (appreciates) by 2% each year. If a consignment of timber is currently

worth $100 000: a express the value of the timber, P dollars, as a function of time, t, where t is the number of years from now. b what will be the value of the timber in 10 years?

198 Maths Quest 11 Mathematical Methods CAS

9 The number of koalas remaining in a parkland t weeks after a virus strikes is given by the function

96 koalas per hectare. t+3 How many koalas per hectare were there before the virus struck? How many koalas per hectare are there 13 weeks after the virus struck? How long after the virus strikes are there 23 koalas per hectare? Will the virus kill off all the koalas? Explain why.

N (t ) = 15 + a b c d

10 A school concert usually attracts 600 people at a cost of $10 per person. On average, for every $1 rise

in admission price, 50 less people attend the concert. If T is the total amount of takings and n is the number of $1 increases: a write the rule for the function that gives T in terms of n b sketch the graph of T versus n c find the admission price that will give the maximum takings.

Chapter 4 • Relations, functions and transformations 199

Summary Set notation

• {. . .} refers to a set. • ∈ means ‘is an element of’. • ∉ means ‘is not an element of’. • ⊂ means ‘is a subset of’. • ⊄ means ‘is not a subset (or is not contained in)’. • ∩ means ‘intersection with’. • ∪ means ‘union with’. • \ means ‘excluding’. • ∅ refers to ‘the null, or empty set’. • {(a, b), (c, d), . . .} is a set of ordered pairs. • A relation is a set of ordered pairs. • N refers to the set of natural numbers. • Z refers to the set of integers. • Q refers to the set of rational numbers. • R refers to the set of real numbers.

relations and graphs

• The independent variable (domain) is shown on the horizontal axis of a graph. • The dependent variable (domain) is shown on the vertical axis of a graph. • Discrete variables are things that can be counted. • Continuous variables are things that can be measured.

domain and range

• The domain of a relation is the set of first elements of a set of ordered pairs. • The range of a relation is the set of second elements of a set of ordered pairs. • The implied domain of a relation is the set of first element values for which a rule has meaning. • In interval notation a square bracket means that the end point is included in a set of values, whereas a curved bracket means that the end point is not included. (a, b] a

types of relations (including functions)

b

• A function is a relation that does not repeat the first element in any of its ordered pairs. That is, for any x-value there is only one y-value. • The graph of a function cannot be crossed more than once by any vertical line. y

y

0 x

0

Function power functions (hyperbola, truncus and square root function)

200

1

x

Not a function

1. The graph of y = x is called a hyperbola. a + c is the graph of the basic hyperbola, dilated by the factor of a in the • The graph of y = x−b y-direction, translated b units horizontally (to the right if b > 0 or to the left if b < 0) and c units vertically (up if c > 0 or down if c < 0). If a < 0, the graph is reflected in the x-axis. The equations of the asymptotes are: x = b and y = c. The domain of the function is R \ {b} and its range is R \ {c}.

Maths Quest 11 Mathematical Methods CAS

2. The graph of y = 12 is called a truncus. x a • The graph of y = + c is the basic truncus curve, dilated by a factor of a in the y-direction ( x − b) 2 and translated b units along the x-axis (to the right if b > 0 or to the left if b < 0) and c units along the y-axis (up if c > 0 or down if c < 0). If a is negative, the graph is reflected in the x-axis. The vertical asymptote is x = b. The horizontal asymptote is y = c. The domain is R \ {b}. The range is y > c if a > 0, or y < c if a < 0. 3. The graph of the function y = a x − b + c is the graph of y = x , dilated by the factor of a in the y-direction and translated b units along the x-axis and c units along the y-axis. • If a < 0, the basic graph is reflected in the x-axis. • The end point of the graph is (b, c). • The domain is x ≥ b. • The range is y ≥ c for a > 0, or y ≤ c for a < 0. • If y = a b − x + c, the domain is x ≤ b; the graph of y = a x is reflected in the y-axis. Function notation

• f (x) = . . . is used to describe ‘a function of x’. To evaluate the function, for example, when x = 2, find f (2) by replacing each occurrence of x on the RHS with 2. • Functions are completely described if the domain and the rule are given. • Functions are commonly expressed using the notation f : X → Y, f (x) = . . . . . . Domain

Co-domain

Rule

• dom f is an abbreviation for the domain of f ( x). • ran f is an abbreviation for the range of f ( x). • The maximal domain of a function is the largest domain for which the function will remain defined. Special types of function (including hybrid functions)

• A function is one-to-one if for each x-value there is at most one y-value and vice versa. • A one-to-many function may be ‘converted to’ a one-to-one function by restricting the domain. • A hybrid function obeys different rules for different subsets of the domain.

inverse relations and functions

• An inverse relation is obtained by interchanging the x- and y-values of the original relation. • The graph of a function and its inverse are reflections of each other across the line y = x. • Only one-to-one functions have inverses.

Circles

• The general equation of a circle with centre (h, k) and radius r is (x − h)2 + (y − k)2 = r2 • An ‘upper semicircle’ with centre (0, 0) and radius r is y = r 2 − x 2 . • A ‘lower semicircle’ with centre (0, 0) and radius r is y = − r 2 − x 2 .

Functions and modelling

• When using functions to model situations: 1. form an equation involving one variable and sketch a graph 2. use the graph to determine domain and range.

Chapter 4 • Relations, functions and transformations

201

Chapter review S h ort anS Wer

1 The total number of cars that have entered a car park during the first 5 hours after opening is shown in

the table below. Time, t (hours) No. of cars, n

1

2

3

4

5

30

75

180

330

500

a Plot these points on a graph. b Explain why the dots cannot be joined. 1 c Estimate the number of cars in the park 2 hours after the car park opens. 2

2 a Sketch the graph of the relation {(x, y): y = 1 − x2, x ∈ [−3, 3]}. b State the domain and range of this relation. 3 State the implied domains of the following functions. b y=

a y= x

5 x

c y = 12 − x

4 If g( x ) = x + 2, where x ≥ 0, then find: a g ( x2) b the domain and range of g ( x). 5 Determine which of the following relations are functions. a y = 2x2 − 1 b 3x + y = 2 d x2 + y2 = 10 e y3 = x

c x = y2 + 1 f y2 − x2 = 1

6 Express the following rules in full function notation.

1 b y = (2 − x ) x 7 Sketch the graph of the function described below. x ≤ −1 2 − x , − f ( x ) = 3, 1< x < 3 2 x − 5, x ≥3 8 Sketch the graph of each of the following, stating the domain and range. a y=

a y=

−

1− x2

b (x − 2)2 + (y + 1)2 = 9

9 a Sketch the graph of the relation x2 + y2 = 100. b From this relation form two one-to-one functions (with maximal domains) and state the domain

and range of each. 10 A chicken farmer delivers chicken manure according to the following fee schedule:

Less than half a truckload: $50 Half to a full truckload: $75 More than 1 but less than 2 truckloads: $100 Sketch a graph showing this informations. 2 11 Sketch a graph for the equation y = + 1. x+3 12 State the dilation factor, reflections and translations that have occurred to each of the following equations. −3 5 +7 a f (x) = − 2 x − 4 b f (x) = c f (x) = + 2 x−2 x2 M U lt ip l e C h oiCe

1 If A = {−2, −1, 0, 1, 2, 3} and B = {−2, 0, 2, 4, 6}, then A ∪ B is: a {−2, −1, 0, 1, 2, 3, 4, 6} C {−1, 1, 3, 4, 6} e ∅

b {−2, 0, 2} d {−1, 1, 3}

2 Which of the following statements is false? a Z⊂Q d {0, 1, 2, 3} ∈ N

202

Maths Quest 11 Mathematical Methods CAS

b 3.142 ∈ Q e (N ∪ Z) = Z

C π∈R

y 8

3 The rule describing the relation shown is: a y = 2x b y = 2x, x ∈ {1, 2, 3, 4} C y = 2x, x ∈ N

6

x 2 e y = 2x, x ∈ R+

4

d y=

2 0

1

2

3

x

4

4 Which one of the following graphed relations is continuous? y a y b

C

x

0

y

x

0

x

0 y

d

e

y

x

0

x

0 5 The interval shown below is: −5

−1 0 1

a [−5, −1] ∪ [0, 4] d (−5, −1) ∪ (1, 4)

4 x b [−5, −1) ∪ [0, 4] e [−5, −1) ∪ (1, 4]

C (−5, −1) ∪ (1, 4]

6 The set R+ \ {2} is correctly represented on which number line below? a

b

0

2

x

C

0

2

x

0

2

x

d

0

2

x

0

2

x

e

7 The domain of the relation shown below is: a b C d e

R \ {0, 1} R \ {1} R Z \ {1} R+ ∪ R−

y 4 2 0 1

x

8 A relation has the rule y = x + 3, where x ∈ R+. The range of this relation is: a R+ d R

b R+ \ {3} e (3, ∞)

9 The implied domain of the relation described by the rule y = a (5, ∞) d (0, 5)

b R+ e R−

10 The range of the function f ( x ) = 2 4 − x is: a R d [0, ∞)

b R+ e (2, ∞)

C [3, ∞)

1 is: x−5

C [5, ∞)

C R−

Chapter 4 • Relations, functions and transformations

203

y

11 The relation shown is:

one-to-one one-to-many many-to-many many-to-one none of the above 12 Which of the following is not a relation? a y = x2 b x2 + y2 = 3 e {1, 3, 5, 7, 9} d y=5−x a b C d e

0

C {(1, 1), (2, 1), (3, 2), (4, 3)}

13 Which one of the following graphed relations is not a function? y y a b x

0

x

C

y

x

0

0 d

e

y

y

x

0

x

0

x

14 Which of the following rules does not describe a function?

x b y = 2 − 7x C x=5 d y = 10x2 + 3 5 15 Which of the functions listed below is not one-to-one? a {(10, 10), (11, 12), (12, 13)} b {(5, 8), (6, 10), (7, 8), (8, 9)} d {(x, y): y = 5 − 2x} e f (x) = 2 − x3 16 Which of the graphs below represents a one-to-one function? a y=

a

0

d

b

y

x

0

0

e

y

C

y

x

e y = −8 C {(x, y): y = 4x}

y

0

x

y

x 0

x

17 The function with the domain −5 ≤ x ≤ 5, range −8 ≤ y ≤ 17 and rule y = x2 − 8 can be written in

function notation as:

a f : R → R where y = x2 − 8 C f : [−5, 5] → R where f (x) = x2 − 8 e f : [−8, 17] → R where f (x) = x2 − 8

b f : R → R where f (x) = x2 − 8 d f : R → [−8, 17] where f (x) = x2 − 8

18 The function f : {x: x = 0, 1, 2} → R, where f (x) = x − 4, may be expressed as: a {(0, −4), (1, −3), (2, −2)} d {(−1, −5), (1, −3), (2, −2)}

b {0, 1, 2} e {−4, −3, −2}

19 If g (x) = 6 − x + x2, then g (−2) is equal to: a 6 d 12

204

Maths Quest 11 Mathematical Methods CAS

b 8 e 5

C {(0, 4), (1, 3), (2, 2)}

C 0

20 If f (x) = 3x − 5, then f (2x + 1) is equal to: a 6x − 8

b 6x − 5

C 3x − 5

d 3x − 4

e 6x − 2

21 The graph at right is to be restricted to a one-to-one function. A possible restricted

y

domain could be: a [−3, 1) b (1, 3) C (0, ∞) d (3, ∞) e (−3, ∞)

9

−3 −1 0

x + 1, x < 0 0 ≤ x ≤ 2 is represented by which of 22 The hybrid function f ( x ) = x 2 , 2 − x , x > 2 the following graphs? a

b

y 4 1

1

−10 1 2 d

C

y 4

x e

1

x

y 4 1

−10

y 4

3

2

x

2

x

0

2

x

y 4 1

−1 0

2

x

−1 0

23 The inverse of the graph shown below is: y

x

a

b

y

x d

y

x e

x

C

y

y

x

y

x

Chapter 4 • Relations, functions and transformations

205

24 The graph of the circle relation (x − 2)2 + (y + 1)2 = 4 is: y a b

y

2

1

1 −1 0 −1

1

2

3

4

−1 0 −1

5 x

1

2

3

4 x

−2 −3

−2 −3 −4 y

C

d

4 3 2 1 x 0 −3−2−1 −1 1 2 3 4 5 6 7 −2 −3 −4 −5

y 6 5 4 3 2 1 0 x −7 −6−5−4−3−2−1 −1 1 2 3 −2 −3 −4

y 4

e

3 2 1 −5 −4 −3 −2 −1 0 −1

1 x

−2 25 The equation of the circle shown is: y 2 0 1 −2

a (x + 3)2 + y2 = 4 C (x + 3)2 + y2 = 2 e x2 + (y − 3)2 = 4

3

5 x

b (x − 3)2 + y2 = 2 d (x − 3)2 + y2 = 4

The circle with equation (x + 1)2 + (y − 4)2 = 9 applies to questions 26 and 27. 26 The domain is: a [−10, 8] C (−2, 4) e [−4, 2]

27 The range is a [−7, −1] C [1, 7] e [1, 7]

206

Maths Quest 11 Mathematical Methods CAS

b [−2, 4] d [−3, 3]

b [−5, 13] d [−3, 3]

28 A circle has its centre at (4, −2) and a radius of 5. The equation of the circle is: a b C d e

(x − 4)2 + (y + 2)2 = 25 (x − 4)2 + (y + 2)2 = 5 (x + 4)2 + (y − 2)2 = 5 (x + 4)2 + (y − 2)2 = 25 4x2 − 2y2 = 5

29 The graph that best represents the function f : [−2, 2] → R where f (x) = 4 − x 2 is: a b C y y y 2 2 2

−2

2 x

0

−2

e

y

2 x

0 −2

y 4

4

−4

−2

−2

−2 d

2 x

0

4 x

0

−4

4 x

0 −4

−4

30 The table of maths tutoring fees charged by a Year 11 student is as follows:

Hours (h)

Charge (C$)

0

50

2

80

4

100

Which of the following graphs best shows the information in the preceding table? a

c 100

b

50 0 d

C

50 2

4

6 h

0

c 100 50

2

4

6 h

2

4

6 h

0

2

4

6 h

c e 100

c 100

50

50 0

c 100

2

4

6 h

0

1 Consider the diagram shown at right. a Find an expression for the area, A, in terms of x and y. b Find an expression for the perimeter, P. c If the perimeter is 72 cm, express A as a function of x. d What is the domain of A(x)? e Sketch the graph of this function. f Hence find the maximum area.

ex ten d ed r eS p o n S e

10 m xm ym

xm

Chapter 4 • Relations, functions and transformations

207

2 For the graph at right: a state the domain b state the range c find the rule for x ∈ (−∞, −2) d find the rule for x ∈ (−2, 0] e find the rule for x ∈ [0, 3], given it is of the form y = ax2 f determine the rule when x ≥ 3 g describe the relation using hybrid function notation −2 … … of the form f ( x ) = … . …. … . … 3 A function f is defined as follows: f : [−2, a] → R, where f (x) = (x − 1)2 − 4. a Find f (−2), f (−1), f (0), f (1), f (3). b If f (a) = 12, find the value of a. c Sketch the function f, labelling the graph appropriately. d From the graph or otherwise, state the: i domain of f (x) ii range of f (x). 4 A new rectangular penguin enclosure is to have a maximum side length of 8 m. The width is to be twice the length (x). a Draw a diagram of the enclosure and label the sides. b Define a rule that gives the perimeter, P, of the new enclosure. c What is the largest value that x can be? d State the domain and range. e Write in function notation the rule for the perimeter. f Define the function for the area of the enclosure, A(x). g If the maximum area allowed is 18 m2, find the dimensions of the enclosure.

y 18

(3, 18)

4 0

4 x

5 Thomas is looking to connect to a mobile phone service. He has to decide on one of two plans from

Busytone Communications. The details are as follows.

diGital doC doc-9748 Test Yourself Chapter 4

208

Plan

Flag fall

Cost perminute (cents)

A

20

30

B

50

20

a Write a function A for the cost of making a call from plan A. b Write a function B for the cost of making a call from plan B. c What is the cost of a 2-minute call from: i plan A? ii plan B? d What is the length of one call costing $5 from: i plan A? ii plan B? e How long would a call be for the call costs to be the same for both plans? f If Thomas frequently makes calls lasting more than 4 minutes, which plan should he connect with?

Maths Quest 11 Mathematical Methods CAS

ICT activities Chapter opener diGital doC • 10 Quick Questions doc-9738: Warm up with ten quick questions on relations, functions and transformations (page 161)

4C

domain and range

tUtorial • We 7 eles-1461: Watch how to sketch relations and state their domain and range using interval notation (page 169) diGital doCS • SkillSHEET 4.1 doc-9739: Practise stating the domain and range of relations (page 171) • WorkSHEET 4.1 doc-9740: Use set notation and interval notation, recognise appropriate specific number fields, recognise whether graphs are discrete or continuous and state the domain and range of relations (page 172) • Investigation doc-9741: Investigate some interesting relations using a graphing program to sketch them (page 172)

4d

types of relations (including functions)

tUtorial • We 8 eles-1462: Watch a tutorial on identifying the type of relation given a graph (page 173)

4e power functions (hyperbola, truncus and square root function) interaCtiVitY • Domain and range int-0263: Consolidate your understanding of domain and range for relations and functions (page 175)

4F

Function notation

diGital doCS • SkillSHEET 4.2 doc-9742: Practise substituting values into functions (page 186) • SkillSHEET 4.3 doc-9743: Practise transposing equations (page 186) • doc-9744: Investigate the graph of a square root function (page 186)

tUtorial • We 16 eles-1463: Watch a tutorial on stating maximal domain and range for functions with defined rules (page 185)

4G Special types of function (including hybrid functions) eleSSon • Hybrid functions eles-0077: Watch an eLesson to learn more about hybrid functions (page 187) tUtorial • We 20 eles-1464: Watch a tutorial on sketching the graph of a hybrid function and stating its range (page 189) diGital doC • WorkSHEET 4.2 doc-9745: Recognise types of relations and functions, determine maximal domain and range, identify co-domains and sketch graphs of relations (page 191)

4i

Circles

diGital doCS • Investigation doc-9746: Investigate the graph of x2 + y2 = 25 by constructing a table of values and plotting the graph (page 193) • doc-9747: Investigate graphs of circle relations (page 195)

4J

Functions and modelling

tUtorial • We 24 eles-1465: Watch a tutorial on expressing the hire costs of a removal van as a hybrid function and sketching the graph of the function (page 197)

Chapter review diGital doC • Test Yourself doc-9748: Take the end-of-chapter test to test your progress (page 208)

To access eBookPLUS activities, log on to www.jacplus.com.au

Chapter 4 • Relations, functions and transformations

209

Answers CHAPTER 4 relationS, FUnCtionS and tranSForMationS

y 7 6 5 4 3 2 1 −1 0 −2

e

exercise 4a

1 a b c d e f g 2 a b c d e 3 a b c d e 4 E 5 a 6 a e i

Set notation ∅ {4, 6} {6] {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14} {4, 5, 6, 7, 8} {2, 8, 10, 12, 14} {4, 5} {2, 3} {−3, −2, −1} {−2, −1} {−3, −2, −1, 0, 1} {−3, −2, −1, 0, 1, 2, 3, 4} ∅ {b, c, d, f, g, h} {a, e, i} {b, c, d, f, g, h} {o, u}

C T F T

b b f j

exercise 4b

1 2 3 4 5

cT gT kF

bC f C

cC

2 −2 −1 0

x 200 190 180 170 160 150 140 130

80 70 60 50 40 30 20 10

d F h T l T

0 2 4 6 8

t (minutes)

Cost (¢)

M TWT F S S Day

exercise 4C

80 70 60 50 40 30 20 10

y 9

c

n

0

1 a b c d e f g h 2 a

1

2

−2 −1 0 −1

3

x

1

x

2

−3

y=x−2 2

4

5

x

6

0 2

b −9

−3

0

c 0 2 0

5

e 01

10

f 0 2 0 1 2 3 4 5 6 n

0

7

g −2 0 1 3

h −8

0 2

i 01

4

j −1 0 1 2 3 4 5 t (s)

b i Approx. 110 km/h ii Approx. 320 km/h

Maths Quest 11 Mathematical Methods CAS

∞) (−∞, 5) (−3, 4] (−8, 9) (−∞, −1] (1, ∞) (−5, −2] ∪ [3, ∞) (−3, 1) ∪ (2, 4]

d

350 300 250 200 150 100 50

−4

−2

3

c The variables are discrete. 9 a V (km/h)

−2

0

2

b P ($)

y

y

1

domain and range

[−2,

−6

P($) 300 340 380 420 460 500 540 550 500 450 400 350 300 250 200

0 5 10 15 20 25 n

c The variables are discrete.

b T (°C)

d D

C($) 140 146 152 158 164 170 176 182 188 194 200

b C ($)

7 a T (°C)

8 a

1 0

210

2

c Because the variables are continuous. d Approx. 11 minutes

4

d

1

0 2 4 6 8 10 t (minutes)

60

b

x

2

4

70

6 a

1

n 15 16 17 18 19 20 21 22 23 24 25

y 6

f

relations and graphs

B A E C a D e D

B F T F

−2

10 a

k 0 2

l −2 0 1

5

6

3 a [−4, 2)

b (−3, 1]

c (− 1, 3)

d

−

1 1 , 2 2 − ( ∞, −3] [0, ∞) (−∞, −2) ∪ (−2, ∞) (−∞, −2] ∪ [0, ∞) E B

e (3, ∞) f g (−∞, ∞) h i (−∞, 1) ∪ (1, ∞) j k (−∞, 2) ∪ (3, ∞) l 4 a D b 5C 6 7 a i {3, 4, 5, 6, 7} ii {8, 10, 12, 14, 16} b i {1.1, 1.3, 1.5, 1.7} ii {1.4, 1.6, 1.8, 2} c i {3, 4, 5, 6} ii {110, 130, 150, 170} d i {M, T, W, Th, F} ii {25, 30, 35} e i {3, 4, 5} ii {13, 18, 23} f i R ii [−1, ∞) 8 a R, R b R, (0, ∞) c [−2, 2], [0, 2] d [1, ∞), R e R, (0, 4] f R, (−∞, −3] g R\{0}, R\{0} h R, (−∞, 1] i R, R y

9 a

2 0

− 2

2

b

x

y = x3 + 1 x ∈ [−2, 2]

1 0

−2

2

x

−7

Domain = [−2, 2], range = [−7, 9] c

y y = x2 + 3x + 2

2

x

−2 −1 0

−

d

Domain = (−∞, ∞), range = [ 14 , ∞) y

−2 −1 0 −3

−2 −1 0

y=

x2

−1 0

1 2 3 4

x

2

y=5

−6

c [−4, 4] fR

b [0, ∞) e R\{0}

10 a R d R

exercise 4d types of relations (including functions) 1 a One-to-many b Many-to-one c Many-to-one d One-to-one e One-to-one f Many-to-one g Many-to-many h Many-to-one i One-to-one j Many-to-one k Many-to-many l Many-to-one 2 b, c, d, e, f, h, i, j, l 3C 4 a B b D c B 5 b {−3, −1, 0, 1, 2}, {−2, −1, 1, 3} c {3, 4, 5, 6}, {−1} e R, {2} g R, R

j [−1, ∞), [0, ∞) k R, R a, d, f, h, i and l are not functions.

power functions (hyperbola, truncus and square root function) 1 a Dilation factor of 3 away from the x-axis, horizontal translation of 5 units in the positive direction of the x-axis and a vertical translation of 1 unit up. b Dilation factor of 2 away from the x-axis and a vertical translation of 1 unit up. c Reflection in the y-axis, horizontal translation of 4 units in the positive direction of the x-axis and a vertical translation of 2 units down. d Dilation factor of 2 away from the x-axis and a horizontal translation of 4 units in the negative direction of the x-axis. e Horizontal translation of 3 units in the positive direction of the x-axis. f Reflection in the x-axis, dilation factor of 4 away from the x-axis and a vertical translation of 5 units down. 2 a

y

1

b

y=1 x

1 2 Asymptote x =0

2

x=1

exercise 4F

1 a b c d e 2 a c

−2 −1 −1

1 2 3 (3, −1)

x

i i i i i

1 2 3 9 12

Function notation ii 7 iv iii −5 ii 1 iii 3 iv ii 2 iii 6 iv ii 1 iii 16 iv ii 6 iv iii −4

16 0 9 a2 + 6a + 9 2 b −3 or 3 d 2 or 3

3 1 3 −

e 4 or 1 3 a 3 c e 4 a b c d e f g h 5 a

c

f

3−1

x

1 3 y= +3 ( x − 2)

e

x

y=2

1 2

d

y

x

3 5

x=1

y

d

b

−2 −1

Domain = [−1, 4), Range = [−7, 3)

0

(1, 4)

(−1, 4)

−5 −7

y

Domain = (−∞, ∞), Range = [−6 81, ∞)

− 4, x ∈ [−2, 1]

y = 2x − 5, x ∈ [−1, 4)

c

3

−4 y 3

1

x

1

Domain = [−2, 1], range = [−4, 0] e

y = 2x2 − x − 6

exercise 4e

Domain = (−∞, ∞), range = (−∞, 2] y 9

y

f

6 a b c

f −1 b 3

10 5 d 2 − x2 − 2x x x 10 10 − x +1 f − x−3 x −1 x+3 f : R → R, f ( x) = 4x + 1 f : [6, ∞) → R, f ( x) = x − 6 1 f : R\{1} → R, f ( x) = x −1 2 f : (−1, ∞) → R, f ( x) = x +1 f : (0, ∞) → R, f ( x) = (x + 2)2 f : [2, ∞) → R, f ( x) = x2 + 3x f : (−∞, 0] → R, f ( x) = 8 − x f : [0, ∞) → R, f ( x) = x2 + x i {0, 1, 2, 3} ii Z iii {−7, −4, −1, 2} i (0, 10] ii R 3

iii [ 10 , ∞) i {2, 4, 6, 8, 10} iii {1, 2, 3, 4, 5} i (−∞, 0) iii (0, ∞) i R+ iii (−2, ∞) i [−3, 3] iii [0, 3] i R i [0, ∞) i R

ii N ii R ii R ii R ii R ii [0, ∞) ii R

Chapter 4 • Relations, functions and transformations

211

d i R e i [4, ∞) f i (3, ∞)

(−∞, 5] [0, ∞) (0, ∞)

ii ii ii

exercise 4G Special types of function (including hybrid functions) 1 a, c, d, f, g 2 i a, b, c, d, f, h, i, j, k, l ii c, h, i, k 3 a (−∞, −1], [−1, ∞) b (−∞, 2), (2, ∞) c [−3, 0], [0, 3] d (−∞, 3], [3, ∞) e [−4, −2), (−2, 0] f (−∞, −1], [−1, 1] or [1, ∞) g (−∞, 0], [0, ∞) h [−2, 0], [0, 2] i (−∞, 0), (0, ∞) j (−∞, −3], [−3, ∞) 4 a D b E 5 a B b C

6 a

y 2 1

f(x)

b (−∞, 0) ∪ [1, ∞) 7 a y 3 2 1

y

1 −1 0

x

b g : (−∞, −1] → R, g ( x) = x2+ 2x + 1 or

x

g : [−1, ∞) → R, g ( x) = x2+ 2x + 1

exercise 4h inverse relations and functions 1 a {(7, 1), (5, 2), (3, 3)} y

d y

y=x

8 7 6 5 4 3 2 1

x 1 2 3 4 5 6 7 8 x

0

y

e

y

y=x

9 8 7 6 5 4 3 2 1

g(x)

x

0 1 2 3 4 5 6 7 8 9 x

−2 −1 0

1

b [1, ∞) c i 3 ii 1 y 8 a

2

x

y

iii 2

y=x

10 8 6 4 2

1 2 3 x

−3 −2 −1 0

y

f

c {(11, −2), (6, 0), (1, 2)}

5 4 3 2 1

0

−4 −5

x

2

4 6 8 10

x

3 a

(−∞,

b 0] ∪ (4, ∞) c i −5 ii 0 iii −3

y

y

2 a y iv 0

x + 2, x ≤ 0 9 f (x) = 2 x + 1, x > 0 10 f : [0, 1] → R, f ( x) = 1 − x 2 with range [0, 1] or f : [0, 1] → R, f ( x) = − 1 − x 2 with range [−1, 0]. 11 a

c g(x)

b {(3, 1), (6, 2), (9, 3)} x

01 2

−2 −1

y

12 a

y=x

v 7 x

x

b

b y

f(x)

y = 4x

y

y = x2 + 3

y=x

9

0

3

x

3

b f : (−∞, 3] → R, f ( x) = (x − 3)2 or

f : [3, ∞) → R, f ( x) = (x − 3)2

212

Maths Quest 11 Mathematical Methods CAS

x

0

3

x

c y

y

f

y = 1_2 x + 1

−1–2

0 −1–2

y

h

1– 2

3

x

1– 2

x

0

− 3

− 3

1

Not a function

0

x

1

d

y

4 a

y 9

y = x3 + 4 −9

0

9

x

exercise 4i

x

4

b d f

g y = 9 − x2 2 a Both [−3, 3] c Both [−5, 5]

h b d f

−

b

x2 + y2 = 1 x2 + y2 = 100 x2 + y2 = 8

y

−2

6, 6]

0

−2

2

y

−1

2

0

1

x

c

0

x

1– 3

[1, 7] and [−3, 3] d

Function

x

4

e

y

0

− 1–2

7

x

1– 2

x

d

6 x

2

−5

[−2, 6] and [−5, 3] e

Function f

−7

y 3 −2 −1 0

− 1–2 0

7 x

4

−3 0

− 1–3

y 7

−7

y

0 1

−4

c

x

2

3

1– 3

y 4

−4

0

y

d

−2

b

−2

[−2, 2] and [0, 4] Function

x

y 4

Function c

2

2

[−1, 1] and [−3, −1]

x

b

−

y = 16 − x 2 Both [−1, 1] Both [−10, 10] Both [− 2 2, 2 2] h [−4, 4], [−4, 0]

g [−3, 3], [0, 3] 3 a y

0

x

−1 −1 0 1

−3

Circles

1 a x2 + y2 = 9 c x2 + y2 = 25 e x2 + y2 = 6

−2

y

Not a function 0

e Both [

D B C E

7 a

−9 4

5 a b 6 a b

y 3

y −8

5

y

−3

2 x

0 −2

7 − 7

0

g

− 10

−2 3

−7

[−8, 2] and [−7, 3] f

2 3 x

0 − 10

Not a function

y 5

10

2 3 0

x

y

y

−2 3

5

Function

− 7

e

0

− 5

7 x

10 x

2 −1 0

3

6

x

[0, 6] and [−1, 5]

Chapter 4 • Relations, functions and transformations

213

y 10

g

4 a B=

n 12 c $90

b B (hours) 10

4 5 −11

[−11,

1] and

x

0 1 x −2

−5

[−2,

0

10]

0 1–

−1

2

2

60

120

b f : (− ∞, 2] → R, f ( x) =

n

17 550

x

−2 −1 0

− 3–2

8 a

4650.30

−3

[−1, 2] and [−3, 0] 8 y=

36 − domain 6] and range [0, 6] or y = 36 − x 2 ; domain [−6, 6] and range [−6, 0] 9 y = 2 + 9 − x 2 ; domain [−3, 3] and range [2, 5] or y = 2 − 9 − x 2 ; domain [−3, 3] and range [−1, 2] 10 a 2 cm, 13.8 cm b 3.9 cm/s exercise 4J

Functions and modelling

40, 70, 1 a C (t ) = 110, 160,

0 < t ≤1 1< t ≤ 2

c 6 a b 7 a b 8 a b 9 a c d

2

10 a b

b C ($) 110 70 40 1 2 3 4 5 6 t (hours)

Cost ($)

b

−1

0

Distance (km)

60t ,

0 ≤ t ≤ 1.5 1.5 ≤ t ≤ 2 80t − 70, 2 ≤ t ≤ 4

700

300

9 a

y

x2 + y2 = 100

10

10 x

0

(100 − x 2 ) with dom f = [−10, 10], ran f = [0, 10] and f 2 : [−10, 10] R, f ( x) = − (100 − x 2 ) with dom f = [−10, 10], ran f = [−10, 0]

10

200

100 75 50 25 0

100 1 2 3 4 5 t (hours)

1 −3−2−1 0 1 2 3

x

11

1 2 Number of truck loads y

2 1 −5 −4−3−2−1 −1

5 3

1 2 3

x

−3

3 a d (t ) = 90,

b Domain [0, 4]; range [0, 250] c i 60 km ii 170 km

x

Domain = [−1, 5]; range = [−4, 2]

Cost ($)

400

b The number of cars is a discrete variable. c 120 y 2 a 100 200

5

−4

Graph is not continuous as n ∈ N

0

0.80 0.60 0.40

0 (2, −1)

b f 1 : [−10, 10] → R, f ( x) =

500

2.00 1.70

y 2

1 a

100 < d ≤ 200 200 < d ≤ 700 d > 700

x

1

−10

Short anSWer

50 < d ≤ 100

0

Domain = [−1, 1]; range = [−1, 0]

−10

Chapter reVieW

0 < d ≤ 50

No. of cars (n)

0.40, 0.60, 2 a C (d ) = 0.80, 1.70, 2.00,

x

−1

0 1 2 3 4 5 6 7 8 9 101112 n

0

214

−1

6000 5000 4000 3000 2000 1000

160

b

Domain [37 001, 80 000]; range [4650.3, 17 550] $6150 P = 4x + 6 Domain (1, 6]; range (10, 30] A = x2 + 4x Domain (0, 8]; range (0, 96] P = 100 000(1.02)t $121 899 47 b 21 9 weeks 96 No, as t increases approaches zero, t+3 so N approaches 15. T = 6000 + 100n − 50n2 c $11 T

1 2 3 4 5 y

80 000 x ($)

37 001

[−6,

x2 ; −

2− x

y 5 4 3 2 1

7

5 a T = 0.3x − 6450 b T ($)

y

h

3 a Domain = [0, ∞) b Domain = R\{0} c Domain = (− ∞, 12] 4 a x + 2, x ≥ 0 b Domain = [0, ∞); range = [2, ∞) 5 a, b, e 1 6 a f : R \{0} → R, f ( x) =

12 a Dilation factor of 2 away from the −8

y = 1 − x2

b Domain = [−3, 3]; range = [−8, 1]

Maths Quest 11 Mathematical Methods CAS

x-axis, reflection in the y-axis and vertical translation of 4 units in the positive direction of the x-axis.

b Dilation factor of 5 away from the

x-axis and a horizontal translation of 2 units up. c Dilation factor of 3 away from the x-axis, reflection in the y-axis, vertical translation of 2 units in the positive direction of the x-axis and a horizontal translation of 7 units up.

1 4 7 10 13 16 19 22 25 28

A C B D B C D A D B

2 5 8 11 14 17 20 23 26 29

D E E D C C E E E C

extended reSponSe

1 a A = xy + 10y − x2 b P = 2x + 2y + 20 or

P = 2(x + y + 10)

c A = 260 + 16x − 2x2

3 6 9 12 15 18 21 24 27 30

B C A E B A D A C E

(5, 12)

292 260

(−2, 5)

130

(−1, 0) −2

0 MUltiple ChoiCe

f (x) 12

c

d (0, 13) e A (m2)

2 4 6 8 10 12 14 x (m) −

2 a x ∈ ( ∞, 4)\{ 2}

ii Range = [−4, 12]

2x

c y=4 d y = −2x

x

e y = 2x2 f y = −18x + 72

3 a 5, 0, −3, −4, 0 b a=5

(1, −4)

d i Domain = [−2, 5] 4 a

b [0, 18]

4, − 2 x , g f (x) = 2 2x , − 18 x + 72

5 x

1

(0, −3) −4

f 292 m2 −

0

x ∈(− ∞ , − 2) x ∈(− 2, 0] x ∈[0, 3] x ∈[3, 4]

b d e f g 5 a c d e

P(x) = 6x c 0<x≤4 Domain = (0, 4]; Range = (0, 24] P: (0, 4] → R where P(x) = 6x A: (0, 4] → R where A(x) = 2x2 Length = 3, Width = 6 A(x) = 0.2 + 0.3x b B(x) = 0.5 + 0.2x i 80 cents ii 90 cents i 16 min ii 22.5 min 3 min f Plan B

Chapter 4 • Relations, functions and transformations

215

ChaPTeR 5

Exponential and logarithmic functions DiGiTal DoC doc-9749 10 Quick Questions

ChaPTeR ConTenTS 5a 5B 5C 5D 5e 5F 5G 5h

Index laws Negative and rational powers Indicial equations Graphs of exponential functions Logarithms Solving logarithmic equations Logarithmic graphs Applications of exponential and logarithmic functions

introduction Functions in which the independent variable is an index number are called indicial or exponential functions. For example: f (x) = ax where a > 0 and a ≠ 1 is an exponential function. It can be shown that quantities that increase or decrease by a constant percentage in a particular time can be modelled by an exponential function. Exponential functions have applications in science and medicine (for example, decay of radioactive material, or growth of bacteria like those shown at right), and finance (for example, compound interest and reducing balance loans).

5a

index laws

Recall that a number, a, that is multiplied by itself n times can be represented in index notation: a× a × a … ×a= an n lots of a

Index (or power or exponent) Base

where a is the base number and n is the index (or power or exponent). The expression an is read as ‘a to the power of n’ or ‘a to the n’.

multiplication When multiplying two numbers in index form with the same base, add the indices. For example, 23 × 24 = 2 × 2 × 2 × 2 × 2 × 2 × 2 = 27

am × an = am + n

ChaPTeR 5 • Exponential and logarithmic functions

217

Division When dividing two numbers in index form with the same base, subtract the indices. am ÷ an = am − n 2×2×2×2×2×2 = 24 For example, 26 ÷ 22 = 2×2

Raising to a power To raise an indicial expression to a power, multiply the indices. (am)n = am × n = amn For example, (24)3 = 24 × 24 × 24 = 24 + 4 + 4 = 24 × 3 = 212

Raising to the power of zero Any number raised to the power of zero is equal to one. a0 = 1, a ≠ 0 3 3 3 − 3 0 For example 2 ÷2 =2 = 2 [1] or 23 ÷ 23 = (2 × 2 × 2) ÷ (2 × 2 × 2) =8÷8 =1 So 23 ÷ 23 = 1 [2] Using [1] and [2] we have 20 = 1.

Products and quotients Note the following. (ab)n = anbn For example, (2 × 3)4 = (2 × 3) × (2 × 3) × (2 × 3) × (2 × 3) n n a = a = 2 × 3 × 2 × 3 × 2 × 3 × 2 × 3 b bn = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 4 4 =2 ×3 Worked Example 1

Simplify. 6 2 3 5 a 2x3y2 × 4x2y b (2x2y3)2 × xy4 c (3a)5b6 ÷ 9a4b3 d 8 p m × (3 p) m 4 6p m Think

a

b

c

1

Collect ‘plain’ numbers (2 and 4) and terms with the same base.

2

Simplify by multiplying plain numbers and adding powers with the same base. (Note: y = y1.)

1

Remove the bracket by multiplying the powers. (The power of the 2 inside the bracket is 1.)

Write

a 2x3y2 × 4x2y

= 2 × 4 × x3 × x2 × y2 × y

= 8x5y3 b (2x2y3)2 × xy4

= 22 × x4 × y6 × xy4

2

Convert 22 to a plain number (4) first and collect terms with the same base.

= 4 × x4 × x × y6 × y4

3

Simplify by adding powers with the same base.

= 4x5y10

1

Write the quotient as a fraction.

2

Remove the bracket by multiplying the powers.

c (3a)5 b 6 ÷ 9a 4 b 3 =

(3a)5 b6 9a 4 b 3

=

243a5 b6 9a 4 b 3

=

27a5 b6 a 4 b3

3 4

Simplify by first cancelling plain numbers. Complete simplification by subtracting powers with the same base. (Note: a1 = a.)

218 Maths Quest 11 Mathematical Methods CAS

= 27ab3

d 1 Expand the brackets by raising each term to

the power of 3.

6 2 3 5 6 2 3 3 5 d 8 p m × (3 p) m = 8 p m × 3 p m 4 4

6p m

6p m

8 × 27 × p6 × p3 × m 2 × m 5 6 p4 m

2

Convert 33 to 27 and collect like variables.

=

3

Simplify by first reducing the plain numbers, and then reducing the variables by adding the indices for multiplication and subtracting the indices for division.

= 36p6 + 3 − 4m2 + 5 − 1

4

Simplify the indices of each base.

= 36p5m6

WoRkeD examPle 2 3

Simplify

6 a 4 b3 3 a 2 b ÷ . 16 a7 b6 2 a3 b2

Think

WRiTe

6a 4 b3 3a 2 b ÷ 16a 7 b6 2a3 b 2

3

1

Write the expression.

2

Change the division sign to multiplication and replace the second term with its reciprocal (turn the second term upside down).

=

6a 4 b 3 2a 3 b 2 × 16a 7 b6 3a 2 b

3

Remove the brackets by multiplying the powers.

=

6 a 4 b 3 23 a 9 b 6 × 16a 7 b6 33 a6 b3

4

Collect plain numbers and terms with the same base.

=

6 × 8a 4 + 9 − 7 − 6 b 3 + 6 − 6 − 3 16 × 27

5

Cancel plain numbers and apply index laws.

=

a0 b0 9

6

Simplify.

=

1 9

3

WoRkeD examPle 3

Write the following in simplest index notation and evaluate. 95 × 34 a 23 × 162 b 27 3 Think

WRiTe

a 1 Rewrite the bases in terms of their prime factors.

a 23 × 162 = 23 × (2 × 2 × 2 × 2)2

2

Simplify the brackets using index notation.

= 23 × (24)2

3

Remove the brackets by multiplying the powers.

= 23 × 28

4

Simplify by adding the powers.

= 211

5

Evaluate as a basic number.

= 2048 ChaPTeR 5 • Exponential and logarithmic functions

219

b 1 Rewrite the bases in terms of their prime factors.

b

95 × 34 (3 × 3)5 × 34 = 273 (3 × 3 × 3)3

2

Simplify the brackets using index notation.

=

(32 )5 × 34 (33 )3

3

Remove the brackets by multiplying the powers.

=

310 × 34 39

4

Write in simplest index form.

= 35

5

Evaluate as a basic number.

= 243

Complex expressions involving terms with different bases have to be simplified by replacing each base with its prime factors.

WoRkeD examPle 4

Simplify

3 4 n × 18 n + 1 . 63 n − 2

Think

WRiTe

34 n × 18n + 1 34 n × (3 × 3 × 2)n + 1 = 63 n − 2 (2 × 3)3n − 2

1

Rewrite the bases in terms of their prime factors.

2

Simplify the brackets using index notation.

=

34 n × (32 × 21 )n + 1 (2 × 3)3n − 2

3

Remove the brackets by multiplying powers.

=

34 n × 32 n + 2 × 2 n + 1 23n − 2 × 33n − 2

4

Collect terms with the same base by adding the powers in the products and subtracting the powers in the quotients.

= 34n + 2n + 2 − (3n − 2) × 2n + 1 − (3n − 2)

5

Simplify.

= 36n + 2 − 3n + 2 × 2n + 1 − 3n + 2 − = 33n + 4 × 2 2n + 3 = 33n + 4 × 23 − 2n

index laws

exercise 5a

Simplify each of the following.

1 We1a, b a

x2

×

2 We1c a

a7b8

x5

×

b 52 × 57 × (53)3

x3

c (xy)3 × x4y5

d 3m2p5 × (mp2)3 × 2m4p6

c (3x5)y11 ÷ 6x2y2

d p13q10 ÷ ( pq4)2

Simplify each of the following. ÷ a2b5

b 2a12b9 ÷ (2a)3b4

Simplify each of the following. 6 p8 m 4 × 2 p7 m 6 (3 x )2 y 2 × 5 x 6 y3 a b 9 p5 m 2 10 x 7 y

3 We1d

c

14u11v 9 × (3u 2 )3 v 21u6 v 5

b

5k 12 d 6 kd 4 ÷ 3 2 (2 k ) 25( k 2 d 3 )3

4 We2 Simplify each of the following.

15a8 b3 2a3 b ÷ a 9a 4 b5 3ab 2 220

2

Maths Quest 11 Mathematical Methods CAS

d

(5e3 )2 f 4 × 8e 4 f 3 20e f 5

3 p3 m 4 can be simplified to: p1m 2

5 mC a

a 3p2m2

b

6 5

B 3p4m6

C 3p3m8

D 3p3m2

C x30y16

D

C −3a6b

D

6x y x × can be simplified to: x 5 y3 (2 y)2 a 2x5y4

B 3x5y4 2

3ab3 a 2 b ÷ c − is equal to: ab a5 a −a18

B −3a6

3 x 9 y10 2

x n + 1 × y5 × z 4 − n x × y4 − n × z 3 − n n−2

a

×

42

× 8

c 53 × 152 × 32

34 × 272 f 64 × 35 Write the following in simplest index notation and evaluate.

45 27 (625)4 e (53 )5 We4

e

b 94 × 35 ÷ 27

a

9

f

(25)4 (125)3

8 × 52 23 × 10

c

(162 )3 (25 )4

d

272 (32 )3

g

411 ÷ 82 163

h

272 × 81 93 × 35

c

4 n × 7n − 3 × 493n + 1 14 n + 2

Simplify the following.

× 92 n + 1 12 x − 2 × 4 x b 6n − 2 6x − 2 n − 5n + 1 5n − 4 3 n 5 3 × 16 × 9 d e* 5n + 1 + 5n 4 n + 1 × 181 − n × 63 − 2 n *Hint: Factorise the numerator and denominator first. 362 n × 6n + 3 10 mC In simplest index notation, is equal to: 216n − 2 a 216n + 5 B 65n + 1 C 62n + 5 a

e 3a6

b6

( x n y m + 3 )2 x2 y × x n + 2 y3 − m x n − 5 × y5 − 3m

b 37 × 92 × 273 × 81

d 205 × 84 × 125 8

3x 5 2

Write the following in simplest index notation.

7 We3

24

b

e

− 3a15

6 Simplify each of the following. a

e 3

4

2n

5B negative and negative powers

D 69

e 62n + 9

rational powers

Wherever possible, negative index numbers should be expressed as positive index numbers using the simple rule: When an index number is moved from the numerator to denominator or vice versa, the sign of the power changes. 1 a −n = n , a ≠ 0 a This is easily verified as follows: 1 a0 = since a0 = 1 an an = a0 − n using the division rule for indices =a n simplifying the index. a n 1 1 an = n and − n = . In other words, 1 a a 1 A simple way to remember this rule is ‘change the level, change the sign’. −

−

ChaPTeR 5 • Exponential and logarithmic functions

221

WoRkeD examPle 5

Express each of the following with positive index numbers. 5 a 8 4 b x y

−4

−2

× ( x 4 y) − x 3 y3

TUToRial eles-1417 Worked example 5 −

5

Think

WRiTe

a 1 Remove the brackets by raising the denominator

and numerator to the power of −4.

a 5

−4

8

−

4

−

4

=

5 8

84 54

2

Interchange the numerator and denominator, changing the signs of the powers.

=

3

Simplify by expressing as a fraction to the power of 4.

8 = 5 4 b x y

b 1 Remove the brackets by multiplying powers.

−

× ( x 2 y) x 3 y3

2

Collect terms with the same base by adding the powers on the numerator and subtracting the powers on the denominator.

2

4

−

−

5

=

x4 y

−

=x

=

5

−

−

−

−

Rewrite the answer with positive powers.

−

x 6y 7 x 3 y3 6 − ( − 3)

= x 3y 3

−

−

−

=

× x 10 y x 3 y3

2

−

y

−

7−3

10

1 x 3 y10

Rational powers Until now, the indices have all been integers. In theory, an index can be any number. We will confine ourselves to the case of indices that are rational numbers (fractions). 1

a n , where n is a positive integer, is defined as the nth root of a: 1

an = n a For example, we know that a × a = a 1

1

1

a2 × a2 = a2 = a1 =a

but

+

1 2

1

a = a2

Therefore, 1

1

Similarly, 3 a = a 3 , 4 a = a 4 … etc. 1

a n is defined for all a ≥ 0 and n ≠ 0. In general, m an

222

Maths Quest 11 Mathematical Methods CAS

1 = an

m

=

( n a )m = n a m

WoRkeD examPle 6

Evaluate each of the following without a calculator. a

3 16 2

9 b 25

−3

2

Think

WRiTe 3

3

a 16 2 = (24 ) 2

a 1 Rewrite the base number in terms of its

prime factors. 2

Remove the brackets by multiplying the powers.

= 26

3

Evaluate as a basic number.

= 64

b 1 Rewrite the base numbers of the fraction in

terms of their prime factors.

b 9 25

−3

2

32 = 2 5 −

3

−

3

2

Remove the brackets by multiplying the powers.

=

3 5

3

Rewrite with positive powers by interchanging the numerator and denominator.

=

53 33

4

Evaluate the numerator and denominator as basic numbers.

=

125 27

−3

2

WoRkeD examPle 7

Simplify the following, expressing your answer with positive indices. a

7

128 × 4 64

b

3

x 2 y6 ÷ x 3 y 5

Think

a 1 Write the expression.

WRiTe

a

7

128 × 4 64 1

1

2

Write using fractional indices.

= 128 7 × 64 4

3

Write 128 and 64 in index form.

= (27 ) 7 × (26 ) 4

4

Multiply the powers.

= 21 × 2 4

5

Simplify the powers.

= 21 × 2 2

1

1

6

3

5

= 22 b 1 Write the expression.

b

3

x 2 y6 ÷ x 3 y5 1

1

2

Express the roots in index notation.

= ( x 2 y6 ) 3 ÷ ( x 3 y5 ) 2

3

Remove the brackets by multiplying the powers.

= x 3 y2 ÷ x 2 y 2

2

3

5

ChaPTeR 5 • Exponential and logarithmic functions

223

2

4

Collect terms with the same base by subtracting the powers.

= x3

5

Simplify the powers.

=x

6

Rewrite with positive powers.

=

−

3 5 2− 2y 2

−1 6y 2

−5

negative and rational powers

exercise 5B DiGiTal DoC doc-9750 SkillSHEET 5.1 negative and rational powers

Express each of the following with positive index numbers.

1 We5a a 6

−

3

−

4

−

−

−

−

2

−

3

−

−

−

(3 2 )2 × (2 5 ) 1 e (24 ) 2 × (34 ) 3 −

−

−

−

−

−

x 3 y 2 × ( xy 2 ) 3 (2 x 3 )2 × ( y 3 )2 −

f

−

−

−

Evaluate the following without a calculator.

3 We6 1 92 1 16 4 81 3

4 mC a

−

−

−

e

34 c 3 2

2

Simplify each of the following, expressing your answer with positive index numbers. ( x 2 )3 × ( y 4 ) 2 ( − m)2 × m 3 ×2 4 b c x 5 × ( y 2 )3 (p 2) 1 × p 4 2 3

( − 22))3

x5 (x 4 ) ÷ d x 3 (x 2 )

a

−

3 5

b

2 We5b a

1 5 1 x6 y2

1

2

3

b 27 3

c 83

d 814

f

−

3 25 2 16

g

3 81 4

h

− 2 8 3 27

25 × 125 simplifies to:

5 25 6

7

3

11

13

B 56

C 52

D 56

e 56

5 We7 Simplify each of the following, expressing your answer with positive indices.

d g

2

9 × 3 81

a

5 24

−1

×4

2

b

(64 m 6 ) 3 e 4m 2

3

−

( x + 1)2 x +1

5C

c

3

(xy3 ) ÷ ( x 2 y)

4

−2

×8

1

x3 × x6

f

1 x

−

4

h ( y − 4) y − 4

indicial equations 1

We can solve equations of the form x 3 = 2 as follows: 1

( x 3 ) 3 = 23 The left-hand side becomes x, so x = 8. However, when the unknown (or variable) is not a base number but is an index number, a different approach is required. Take the cube of both sides:

method 1: exact solutions without a calculator To attempt to solve index equations exactly, express both sides of the equation to the same base and equate the powers. If am = an, then m = n. 224

Maths Quest 11 Mathematical Methods CAS

Worked Example 8

Find the value of x in each of the following equations. a 3x = 81 b 4x − 1 = 256 c 63x − 1 = 362 x − 3 Think

a 1 Write the equation. 2

Express both sides to the same base.

3

Equate the powers.

b 1 Write the equation.

Write

a

3x = 81 3x = 34 ⇒x=4

b 4x − 1 = 256

2

Express both sides to the same base.

4x − 1 = 44

3

Equate the powers.

⇒x−1=4

4

Solve the linear equation for x by adding one to both sides.

x = 5

c 1 Write the equation.

c 63x − 1 = 362x − 3

2

Express both sides to the same base.

63x − 1 = (62)2x − 3

3

Remove the brackets by multiplying the powers.

63x − 1 = (6)4x − 6

4

Equate the powers.

⇒ 3x − 1 = 4x − 6

5

Subtract 3x from both sides to make x the subject.

−1

6

Add 6 to both sides to solve the equation.

x = 5

=x−6

More complicated equations can be solved using the same technique. Worked Example 9

Solve for n in the following equation. 23n × 16n + 1 = 32 Think

Write

1

Write the equation.

23n × 16n + 1 = 32

2

Express both sides using the same base, 2.

23n × (24)n + 1 = 25

3

Remove the brackets by multiplying the powers.

23n × 24n + 4 = 25

4

Multiply the terms on the left-hand side by adding the powers.

5

Equate the powers.

6

Solve the linear equation for n.

27n + 4 = 25 ⇒ 7n + 4 = 5 7n = 1 n=

1 7

In some cases indicial equations can be expressed in a quadratic form and solved using the Null Factor Law. Look for numbers in index form similar to a2x and ax appearing in different terms. Chapter 5 • Exponential and logarithmic functions 225

WoRkeD examPle 10

Solve for x if 52x − 4(5x) − 5 = 0. Think

WRiTe

5 − 2x

4(5x)

TUToRial eles-1418 Worked example 10

−5=0

1

Write the equation.

2

Rewrite the equation in quadratic form. Note that 52x = (5x)2.

(5x)2 − 4(5x) − 5 = 0

3

Substitute y for 5x.

Let y = 5x

4

Rewrite the equation in terms of y.

⇒ y2 − 4y − 5 = 0.

5

Factorise the left-hand side.

(y − 5)(y + 1) = 0

6

Solve for y using the Null Factor Law.

⇒ y = 5 or y = −1

7

Substitute 5x for y.

5x = 5 or 5x = −1

8

Equate the powers.

⇒ 5x = 51 and 5x = −1

9

State the solution(s).

⇒x=1

(5x = −1 has no solution.)

Note that in step 9, the possible solution 5x = −1 was rejected because there is no value of x for which it will be satisfied. Recall that exponential functions such as 5x are always positive.

method 2: Using a CaS calculator If answers are not exact, the CAS calculator can be used to solve indicial equations. WoRkeD examPle 11

Solve for x given 3

−

2x+1

=

1 . Write your answer correct to 2 decimal places. 45

Think 1

WRiTe

Use the solve feature of the CAS calculator: 1 solve 3 2 x + 1 = 5 , x 4

The result is x = 3.654 65.

Write the answer.

Solving 3

−

2

exercise 5C

1 for x gives x = 3.65, 45 correct to 2 decimal places. −

2x +1

=

indicial equations

1 We8a Find the value of x in each of the following equations. a 2x = 32 b 5x = 625 c 3x = 243 d 10

−

x

1 = 100

e 4

−

x

= 16

f

6x =

1 216

2 We8b Find the value of n in each of the following equations. 3n + 1 a 2 = 64 b 52n + 3 = 25 c 32 − n = 27 d 16n + 3 = 23 3 We8c

f

364n − 3 = 216

b 274 − x = 92x + 1

c 163x + 1 = 128x − 2

3x + 5 = d 9

c 34x × 27x + 3 = 81

d

1 243

Solve for x in each of the following equations.

a 2x × 83x − 1 = 64

226

1 7

Find x in each of the following.

a 42x = 8x − 1 4 We9

5 − 3n = e 49

Maths Quest 11 Mathematical Methods CAS

b 52x × 1253 − x = 25

812 − x = 92 x 27 x + 3

5 We10 Solve for x in each of the following. a 32x − 4(3x) + 3 = 0 b 22x − 6(2x) + 8 = 0 2x x c 3(4 ) = 15(4 ) − 12 d 25x − 30(5x) + 125 = 0 6 mC Consider the indicial equation 32x − 12(3x) + 27 = 0. The equation can be solved by making the

substitution: a y = 3x D y = 2x

B y = 2x e y = 3x

C y = 32x

7 mC The quadratic equation formed by the appropriate substitution in question 6 is: a y2 − 3y + 27 = 0 D y2 − 12y + 27 = 0

B y2 − 11y + 27 = 0 e y2 − 9y + 3 = 0

C y2 + 12y + 27 = 0

8 mC The solutions to the equation in question 7 are x equals: a 2 or 3 D 0 or 1

B 1 or 2 e 0 or 2

C 1 or 3

9 We11 Solve for x. Write your answer correct to 2 decimal places. a 2x = 3 b 3x = 8 c 10x = 45 d 4x = 10 10 mC The nearest solution to the equation 3x = 10 is: a x = 2.5 D x=2

5D

B x = 2.3 e x = 2.1

C x = 1.9

Graphs of exponential functions

Functions of the form f (x) = ax, where a is a positive real number other than 1 and x is a real number, are called exponential functions. In general, there are two basic shapes for exponential graphs: y = ax, a > 1 or y = ax, 0 < a < 1 y

y

y = ax, a > 1

y = ax, 0 < a < 1

1

1 Asymptote y=0

x

0

Asymptote y=0 Decreasing exponential x

0

Increasing exponential

However, in both cases: the y-intercept is (0, 1) the domain is R the asymptote is y = 0 (x-axis) the range is R+. x x Verify the shapes of these graphs by graphing, say y = 2x, y = 3x, y = 1 and y = 1 on a graphics 2 3 calculator. What is the effect of changing a on the steepness of the graph?

• • • •

Reflections of exponential functions The graph of y = a x is obtained by reflecting y = ax in the y-axis. −

y

The graph of y = −ax is obtained by reflecting y = ax in the x-axis. y

y=

ax,

a>1

y = ax, a > 1 1

y=

1 0

a−x,

a>1 Asymptote x y=0

−1

0

x y = −ax, a > 1

Asymptote y=0

ChaPTeR 5 • Exponential and logarithmic functions

227

Horizontal translations of exponential functions The graph of y = a x + b is obtained by translating y = ax: 1. b units to the right if b < 0 2. b units to the left if b > 0. For example, the graph of y = 2x − 3 is obtained by translating y = 2x to the right 3 units. Check this graph using a graphics calculator. Note also that

y y = 2x

3 units

2 1

()

2 x −3 = (2 x )(2 3 ) = 1 2 x , so the effect is identical to that of 8 multiplying by a constant. −

y = 2x − 3

Vertical translations of exponential functions

Asymptote y=0

x

−1 0 1 2 3 4

y

y = 10 x

10

The graph of y = ax + c is obtained by translating y = ax: 1. up by c units if c > 0 2. down by c units if c < 0. Furthermore the equation of the asymptote becomes y = c. For example, the graph of y = 10x − 5 is obtained by translating y = 10x down by 5 units. −5 units The equation of the asymptote is y = −5. −4. The y-intercept is Check this graph using a graphics calculator.

5 y = 10 x − 5 1 −1 −4 −5

1

x

Asymptote y = −5

Dilation from the x-axis The graph of y = Aax (for positive, real values of A) has a dilation factor of A. The graph is stretched along the y-axis, away from the x-axis (as each y-value is being multiplied by the constant A). Consider the graphs below. The y-intercept in each case is equal to A. Also, as A increases, the graph becomes steeper; as A decreases, the graph becomes less steep. The domain, range and asymptotes are the same as for f (x) = ax. y f(x) = 4 × 2x f(x) = 3 × 2x f(x) = 2x 4 3 (0, 3) 2 1 (0, 1) Asymptote x 0 y=0 −3 −2 −1 1 2 3

y

(0, 4)

Dilation from the y-axis

2 (0, 1) (0, 1–2) −1

(1, 2)

f(x) = 2x f(x) =

1– 2

f(x) =

× 2x 1– 4

× 2x

(1, 1) 1–

(0, 1–4) (1, 2 ) Asymptote x 0 y=0 1

1

The graph of y = akx (for k > 0) has a dilation factor of k from the y-axis. The graph is said to be stretched along the x-axis. Consider the graphs at below. The y-intercept is (0, 1) in each case. As k increases, the graph becomes steeper and closer to the y-axis. The domain, range and asymptotes are the same as for f (x) = ax. y 4

f(x) = 22x f(x) = 2x –x f(x) = 22

2 (0, 1) −3 −2 −1

228 Maths Quest 11 Mathematical Methods CAS

01 2 3

x

Asymptote y=0

Worked Example 12

Find the equation of the asymptote and the y-intercept for each of the following functions. Hence, sketch the graph of each function and state its domain and range. a f : R → R, f (x) = 2x + 3 − 5 x

b f : R → R, f ( x ) = 3 × 2 2 Think

a 1 Write the rule.

Write/Draw

a f (x) = 2x + 3 −5

2

The graph is the same as y = 2x translated 3 units left and 5 units down.

3

State the asymptote.

The asymptote is y = −5.

4

Evaluate y when x = 0 to find the y-intercept.

When x = 0, y = 23 − 5 = 3 Therefore, the y-intercept is (0, 3).

5

Locate the y-intercept and asymptote on a set of axes.

6

Sketch the graph of the exponential function using the y-intercept and asymptote as a guide.

y f(x) = 2x +3 − 5 3 x

0

Asymptote y = −5

−5 7

Use the graph to state the domain and range.

The domain is R and the range is (−5, ∞). x

b 1 Write the rule.

b f (x) = 3 × 2 2 0

2

Find the y-intercept by letting x = 0, or recall that the y-intercept is equal to A in f (x) = Aa x.

f (0) = 3 × 2 2 = 3 × 20 = 3×1 =3 The y-intercept is 3.

3

Locate the horizontal asymptote.

The horizontal asymptote is the x-axis.

4

Locate another point on the graph. This is necessary to be able to see the effect of the dilation. Locate the second point by substituting a value for x into the equation and evaluating a corresponding y-value.

f (2) = 3 × 2 2 = 3 × 21 = 3× 2 =6 Another point is (2, 6).

5

Sketch the graph. Note: The asymptote remains at y = 0 as there is no vertical translation.

y

2

–x

f(x) = 3 × 22 6

(2, 6)

(0, 3) 0 6

State the domain and the range.

1

2

Asymptote x y=0

The domain is R and the range is R+.

Chapter 5 • Exponential and logarithmic functions 229

WoRkeD examPle 13

Use a CAS calculator to solve 2x = 15 (correct to 2 decimal places) by finding the intersection of two graphs. Think

WRiTe

1

Enter the left-hand side as a function into the CAS calculator.

f1(x) = 2x

2

Enter the right-hand side as a function into the CAS calculator.

f2(x) = 15

3

Graph the two functions and use the intersection points feature of the CAS calculator. Write the answer.

The graphs intersect at (3.91, 15).

exercise 5D

Graphs of exponential functions

1 Sketch the graph of each of the following on separate axes. (Use a table of values or copy a DiGiTal DoC doc-9751 SkillSHEET 5.2 Substitution in exponential functions

CAS calculator screen). a y = 3x c y = 10x e y=4 x g y = −3 x

b y = 5x− x d y=2 − x f y= 3 h y = 0.5x

−

−

2 Sketch the following graphs, using a table of values or by copying a CAS calculator screen. State the

equation of the asymptote and the y-intercept for each. a y = 2(3x) c

b y = 0.5(4x)

1

y = 4 (2 x )

d

y = 4 1

x

3

Find the equation of the asymptote and the y-intercept for each of the following. Hence, sketch the graph of each and state its domain and range. a f : R → R, f (x) = 2x − 1 b f : R → R, f (x) = 3x + 2 c f : R → R, f (x) = 51 − x d f : R → R, f (x) = 2x + 3 e f : R → R, f (x) = 3x − 3 f f : R → R, f (x) = 2x + 3 − 1 x h f : R → R, f (x) = 102 − x + 5 6 g f : R → R, f (x) = +3

3 We12a

−

4 mC a The rule for the graph at right is: a B C D e

y

y = 3x − 2 y = 3x y = 2x − 3 y = 3x + 2 y = 3x − 1

3 2 1

b The rule for the graph at right is: a B C D e

Asymptote 0 1 2 3 x y=0 y

y = 2x − 3 y = 3x − 2 y = 2x + 1 − 3 y = 2x − 1 + 3 y = 2x − 1 − 3

x

0 −2 −3 −4

(1, −2)

y = −3

5 We12b Sketch the graph of each of the following, stating the domain and range. a f : R → R, f (x) = 22x b f : R → R, f (x) = 23x x c f : R → R, f (x) = 3 × 2 d f : R → R, f (x) = 5 × 32x x

e f : R → R, f ( x ) = 2 × 5 2 230

Maths Quest 11 Mathematical Methods CAS

x

f

f : R → R, f ( x ) = 2 × 2 3

6 Sketch a graph of f (x) = 4 × 23x + 3, stating the domain and range. Compare your answer to that found

using a CAS calculator. 7 We13 Use a CAS calculator to solve the following indicial equations using the intersection of two

graphs. Give answers rounded to 2 decimal places. a 2x = 10 b 10x = 20 c 3x = 12

5e

d 2x = x + 3

e 3x = x + 4

DiGiTal DoC doc-9752 WorkSHEET 5.1

logarithms

The index, power or exponent (x) in the indicial equation y = ax is also known as a logarithm. This means that y = ax can be written in an alternative form: loga (y) = x, which is read as ‘the logarithm of y to the base a is equal to x’. For example, 32 = 9 can be written as log3 (9) = 2. 105 = 100 000 can be written as log10 (100 000) = 5.

Logarithm y = ax Base numeral Base

ax = y is equivalent to x = loga (y).

In general, for a > 0 and a ≠ 1:

Using the indicial equivalent, it is possible to find the exact value of some logarithms. WoRkeD examPle 14

Evaluate the following without a calculator. 1 a log6 (216) b log 2 ( 8 ) Think

WRiTe

a Let x = log6 (216)

a 1 Let x equal the quantity we wish to find. 2

Express the logarithmic equation as an indicial equation.

6x = 216

3

Express both sides of the equation to the same base.

6x = 63

4

Equate the powers.

x=3

b 1 Write the logarithm as a logarithmic equation. 2

Express the logarithmic equation as an indicial equation.

8

1 b Let x = log2

2x =

1 8

= 1

3

2

= (2 1 )3 x 2 =2 3 −

3

Express both sides of the equation to the same base.

4

Equate the powers.

−

x = −3

logarithm laws The index laws can be used to establish corresponding rules for calculations involving logarithms. These rules are summarised in the following table. Name Logarithm of a product

Rule

Restrictions

loga (mn) = loga (m) + loga (n)

m, n > 0 a > 0, a ≠ 1

loga m = loga (m) − loga (n)

m, n > 0 a > 0 and a ≠ 1

Logarithm of a power

loga (m)n = n loga (m)

m>0 a > 0 and a ≠ 1

Logarithm of the base

loga (a) = 1

a > 0 and a ≠ 1

Logarithm of one

loga (1) = 0

a > 0 and a ≠ 1

Logarithm of a quotient

n

ChaPTeR 5 • Exponential and logarithmic functions

231

It is important to remember that each rule works only if the base, a, is the same for each term. The ‘logarithm of a product’ and ‘logarithm of a quotient’ rules formed the basis for the pre-1970s calculation device for multiplication and division — the slide rule. Worked Example 15

Simplify, and evaluate where possible, each of the following without a calculator. a log10 (5) + log10 (4) b log2 (12) + log2 (8) − log2 (3) Think

Write

a 1 Apply the ‘logarithm of a product’ rule. a log10 (5) + log10 (4) = log10 (5 × 4) 2

b 1

= log10 (20)

Simplify. Multiply the base numerals of the logs being added since their bases are the same.

b log2 (12) + log2 (8) − log2 (3) = log2 (12 × 8) − log2 (3)

= log2 (96) − log2 (3)

2

Apply the ‘logarithm of a quotient’ law.

= log2 (96 ÷ 3)

3

Simplify, noting that 32 is a power of 2.

= log2 (32) = log2 (2)5

4

Evaluate using the ‘logarithm of a power’ and ‘logarithm of the base’ laws.

= 5 log2 (2) =5

Worked Example 16

Simplify 3 log2 (5) − 2 log2 (10). Think

Write

3 log2 (5) − 2 log2 (10) = log2 (5)3 − log2 (10)2

1

Express both terms as logarithms of index numbers.

2

Simplify each logarithm.

= log2 (125) − log2 (100)

3

Apply the ‘logarithm of a quotient’ law.

= log2 (125 ÷ 100)

4

Simplify.

= log2

Worked Example 17

Simplify each of the following. log 8 (49) a b 2 log10 (x) + 1 c 5 log10 (x) − 2 log 8 (343) Think

a 1 Express each base numeral as powers

to the same base, 7.

Write

a

log8 (49) log8 (7)2 = log8 (343) log8 (7)3

2

Apply the ‘logarithm of a power’ law.

=

2 log8 (7) 3 log8 (7)

3

Simplify by cancelling out the common factor of log8 (7).

=

2 3

232 Maths Quest 11 Mathematical Methods CAS

5 4

or log2 (1.25)

b 1 Express 2 log10 (x) as log10 (x)2 and

1 as a logarithm to base 10 also.

2

b 2 log10 (x) + 1 = log10 (x)2 + log10 (10)

= log10 (10x2)

Simplify using the ‘logarithm of a product’ law.

c 1 Express 5 log10 (x) as log10 (x)5 and

2 as 2 log10 (10).

c 5 log10 (x) − 2

= log10 (x)5 − 2 log10 (10)

2

Express 2 log10 (10) as log10 (10)2.

= log10 (x)5 − log10 (10)2

3

Simplify using the ‘logarithm of a quotient’ law.

x5 = log10 2 10 x5 = log10 100

WoRkeD examPle 18

Evaluate each of the following expressions, correct to 3 decimal places. a log2 (5) b log7 (8) Think

WRiTe

a & b 1 On a Calculator page, complete the

entry lines as: log2 (5) log7 (8) 2

Write the answer.

a log2 (5) = 2.332 b log7 (8) = 1.069, correct to 3 decimal places.

logarithms

exercise 5e 1

Express the following indicial equations in logarithmic form. b 35 = 243 c 50 = 1 −2 − 1 d 0.01 = 10 e bn = a f 2 4= 16 a 23 = 8

Express the following logarithmic equations in indicial form.

2

= −1

b log10 (1000 000) = 6

c

log2

d log3 (27) = 3

e log5 (625) = 4

f

log2 (128) = 7

g

log3 1 9

−

= 2

h logb (a) = x

3 mC The value of log5 (25) is: a −2

B 5

C 1

D 2

= 512 is: B log3 (512) = 8 e log8 (3) = 512

4 mC When expressed in logarithmic form, a log3 (8) = 512 D log512 (3) = 8

5

1 2

a log4 (16) = 2

e

4

83

C log8 (512) = 3

When expressed in indicial form, log10 (10 000) = 4 is: a = 10 000 B 10 0004 = 10 C 10 00010 = 4 10 000 10 D 10 =4 e 4 = 10 000 mC

104

6 We14

Evaluate each of the following.

a log2 (16)

b log3 (81)

e log10 (1000)

f

log10 (0.000 01)

j

log2 1 64

i

log2 (32)

1 4 1 log3 243 logn (n5)

c log5 (125)

d log2

g log2 (0.25)

h

k log3

(−3)

l

ChaPTeR 5 • Exponential and logarithmic functions

233

7 We15

Simplify the following, and evaluate where possible.

a log2 (8) + log2 (10) d log6 (8) + log6 (7) g log2

1 3

+ log2 (9)

b log3 (7) + log3 (15) e log2 (20) − log2 (5)

c log10 (20) + log10 (5) f log3 (36) − log3 (12)

h log4 (25) + log4

i

1 5

log3 (8) − log3 (2) + log3 (5)

8 We16 Simplify each of the following. a 3 log10 (5) + log10 (2) b 2 log2 (8) + 3 log2 (3) c log5 (12) − 2 log5 (2) d 4 log10 (2) − 2 log10 (8) e g

1 3 1 2

1

log2 (27) − 2 log2 (36)

f

log2 (x − 4) + 3 log2 (x)

log3 (16) + 2 log3 (4)

h 2 log10 (x + 3) − log10 (x − 2)

9 We17a Simplify the following. a

log3 (25) log3 (125)

b

log2 (81) log2 (9)

c

2 log10 (8) log10 (16)

d

3log5 (27) 2 log5 (9)

e

log3 ( x 6 ) log3 ( x 2 )

f

log10 ( x 3 ) log10 ( x )

g

2 log2 ( x + 1)3 log2 ( x + 1) C

log10 ( x ) log10 ( y)

10 mC The expression log10 (xy) is equal to: a log10 (x) × log10 (y) D y log10 (x)

B log10 (x) − log10 (y) e log10 (x) + log10 (y)

11 mC The expression log5 (xy) is equal to: a x log5 (y)

B y log5 (x)

C 5 logx (y)

D log5 (x) + log5 (y)

e 5y

1

12 mC The expression 3 log2 (64) + log2 (5) can be simplified to: a log2 (40)

B 1

13 mC The expression a log4 (x3)

B

C

log2

64 15

320

D log2 (20)

e log2 3

D log4 (x5 − x2)

e log4 (x7)

log4 ( x 5 ) can be simplified to: log4 ( x 2 ) 5

log4 ( x 2 )

C

5 2

14 We17b Express each of the following in simplest form. a log3 (27) + 1 b log4 (16) + 3 d 2 + 3 log10 (x)

c 3 log5 (2) − 2

e 4 log3 (2) − 2 log3 (6) + 2

f

1 2

+ 3 log10 ( x 2 )

15 We18 Evaluate the following, correct to 3 decimal places where appropriate. a log2 (16) b log3 (81) c log10 (0.001) d log2 (9) e log3 (4) + log2 (7)

5F Solving logarithmic logarithms to the base 10

equations

Logarithms to the base 10 are called common logarithms and can be evaluated using the log function on a calculator. Note: The logarithm of a negative number or zero is not defined. Therefore: loga (x) is defined for x > 0, if a > 0. This can be seen more clearly using index notation as follows: • Let n = loga (x.) • Therefore, an = x (indicial equivalent of logarithmic expression). • However, an > 0 for all values of n if a > 0 (positive based exponentials are always positive). • Therefore, x > 0. 234

Maths Quest 11 Mathematical Methods CAS

WoRkeD examPle 19

Find x if log3 (9) = x − 2. Think 1

Write the equation.

2

Simplify the logarithm using the ‘logarithm of a power’ law and the fact that log3 (3) = 1.

3

Solve for x by adding 2 to both sides.

WRiTe

log3 (9) = x − 2 log3 (32) = x – 2 2 log3 (3) = x – 2 2=x–2 x=4

WoRkeD examPle 20

Solve for x if log6 (x) = −2. Think 1

Write the equation.

2

Express in index form.

3

Evaluate the index number.

WRiTe

log6 (x) = −2 Therefore, x = 6 2. −

x= =

1 62 1 36

WoRkeD examPle 21

Solve for x given that 2 log x (25) = 4, x > 0. Think

WRiTe

2 logx (25) = 4

1

Write the equation.

2

Divide both sides by 2.

3

Write as an index equation.

4

Express both sides of the equation to the same base, 5.

x2 = 52

5

Equate the bases. Note that x = −5 is rejected as a solution, because x > 0.

x=5

logx (25) = 2 Therefore, x2 = 25.

WoRkeD examPle 22

Solve for x correct to 3 decimal places, if 2x = 7. Think 1

Write the equation.

2

Take log10 of both sides.

3

Use the ‘logarithm of a power’ law to bring the power, x, to the front of the logarithmic equation.

4

Divide both sides by log10 (2) to get x by itself.

5

Evaluate the logarithms correct to 4 decimal places, at least one more than the answer requires.

6

Solve for x.

WRiTe

2x = 7 log10 (2x) = log10 (7) x log10 (2) = log10 (7) Therefore x =

logg10 ((7) logg10 ((2)

0.8451 0.3010 x = 2.808 x=

ChaPTeR 5 • Exponential and logarithmic functions

235

Therefore, we can state the following rule: If ax = b, then x =

logg a ( b) logg a ( a)

= logg a ( b) This rule applies to any base a, but 10 is the most commonly used base for this solution technique.

exercise 5F

Solving logarithmic equations

Find x in each of the following.

1 We19

1 10

a log2 (4) = x

b log9 (1) = x

e 2 log2 (8) = x

f

log3 (81) = 2x

b e h k

log3 (x) = 2 c log5 (x) = 4 log8 (x) = −1 f log3 (x) = −3 log2 (3x + 1) = 4 i log10 (2x) = 1 log3 (5) − log3 (4) = log3 (x) − log3 (8)

Solve for x. log2 (x) = 3 log10 (x) = 1 log3 (x − 3) = 3 2 log6 (3x) = 1

c log10

=x

d log3

1 9

=x

g log10 (1000) = 2x − 1

2 We20 a d g j

Solve for x given that:

3 We21

a logx (36) = 2 d

− 2 log

1 x 100

b logx (125) = 3

=4

e

g logx + 1 (27) = 3

h

1 log x 2 −

c 3 logx (16) = 6

(64) = 3

log3 x − 1

1 32

f

5 logx (625) = 10

= 5.

4 mC a The solution to the equation log7 (343) = x is: a x=2

B x=3

C x=1

D x=0

e x = −2

a 4096

B 512

C 64

D 2

e

D 1

e 9

b If log8 (x) = 4, then x is equal to: 1

1 2

c Given that log x (3) = 2 , x must be equal to: a 3

B 6

C 81

d The solution to the equation log3 (x) − 2 = log3 (x − 8) is: a x=8

B x=6

C x=9

D x = −4

e x=2

5 We22 Solve the following equations for x, correct to 3 decimal places. a 2x = 11 b 2x = 0.6 c 3x = 1.7 x x − 1 e 0.7 = 3 f 10 = 18 g 22x + 1 = 5 2 − x i 8 = 0.75

d 5x = 8 − h 10 2x = 7

6 mC The nearest solution to the equation 4x = 5 is: a x = 0.86 DiGiTal DoC doc-9753 WorkSHEET 5.2

236

C x = 1.25

7 mC The nearest solution to the equation a x = 0.18

5G DiGiTal DoCS doc-9821 Investigation investigating logarithmic graphs 1 doc-9822 Investigation investigating logarithmic graphs 2 inTeRaCTiViTY int-0264 logarithmic graphs

B x = 1.2 B x = 0.13

0.62x − 1

= 2 is:

C x = −0.18

D x=1

e x = 0.5

D x = −0.71

e x = −0.13

logarithmic graphs

The graphs of y = loga (x) and y = ax are reflections of each other across the line y = x. Functions such as these that are reflections of each other in the line y = x are called inverses of each other. Consider the logarithm loga (ax). This logarithm can be simplified using the log laws.

loga (ax) = x loga (a) =x×1 =x Notice how the logarithm with base a and the exponential with base a have a cancelling effect on one another, demonstrating that they are inverse operations. This is similar to the way that multiplication and division have a cancelling effect. Multiplication and division are also inverse operations of each other.

Maths Quest 11 Mathematical Methods CAS

Consider now the exponential aloga (x). As the logarithm with base a is the inverse operation to the exponential with base a, the expression aloga (x) simplifies to give x. That is, aloga (x) = x. The inverse properties of logarithms and exponentials can be used to plot the graphs of logarithmic functions. Alternatively, a table of values can be used. For example: y = log10 (x) −1

x y

0

undefined undefined

1

2

3

4

0

0.301

0.477

0.602

y Asymptote x=0 f(x) = log10 (x)

The graph of y = loga (x) does not exist for values of x ≤ 0. It is an increasing function. There is a vertical asymptote along the y-axis, and so there are no y-intercepts. The x-intercept for all values of a is always (1, 0). That’s because loga (1) = 0. Another point on the graph is (a, 1). That’s because loga (a) = 1. The domain of the function is R+ and the range is R.

0

x

1

WoRkeD examPle 23

Sketch the graph of f (x) = log2 (x). Think 1

2

WRiTe/DRaW

Realise that f (x) = log2 (x) is the inverse of f (x) = 2x, so these two graphs are reflections of each other across the line y = x. Alternatively, recall the basic shape of the logarithmic graph. Sketch the basic shape on a set of axes.

3

Mark the x-intercept (1, 0).

4

Mark a second point on the graph (a, 1), which in this case is (2, 1).

f(x) = 2x

y

y=x

f(x) = log2 (x)

1

(2, 1) 0 (1, 0)

x

WoRkeD examPle 24

Find the equation of the inverse of f (x) = 2x. Think

WRiTe

1

Write the function using (x, y) notation.

y = 2x

2

Interchange x and y.

x = 2y

3

Take the logarithm to base 2 of both sides.

log2 (x) = log2 (2y)

4

Simplify using loga (mn) = n loga (m).

log2 (x) = y log2 (2)

5

Simplify using loga (a) = 1.

log2 (x) = y

6

State the inverse of f.

exercise 5G

−

f 1(x) = log2 (x)

logarithmic graphs

1 Simplify the following expressions. log (b) a log2 (2x) b 10 10

c 3 × log5 (5y)

2 We23 Sketch the graphs of each of the following. a f (x) = log2 (x) b f (x) = log5 (x) d f (x) = log10 (x) e f (x) = log12 (x)

d 5 × 20 log20 (x) c f (x) = log8 (x) f f (x) = log15 (x) ChaPTeR 5 • Exponential and logarithmic functions

237

3 Compare the steepness of each of the graphs in question 2, and hence explain how changing the base, a,

affects the steepness of a logarithmic graph of the type f (x) = loga (x).

4 We24 Find the equation of the inverse of the following. a f (x) = 3x b f (x) = 4x c f (x) = log5 (x)

Further work on logarithmic graphs is available on your eBookPLUS.

DiGiTal DoCS doc-9754 logarithmic graphs doc-9755 Extension logarithmic graphs

applications of exponential and logarithmic functions 5h

Exponential and logarithmic functions can be used to model many practical situations in science, medicine, engineering and economics. WoRkeD examPle 25

A square sheet of paper that is 0.1 mm thick is repeatedly folded in half. a Find a rule that gives the thickness, T mm, as a function of the number of folds, n. b What is the thickness after 10 folds? c How many folds are required for the thickness to reach 6 cm? Think

a

b

c

238

1

T = 0.1 when n = 0 and doubles with each fold. This doubling implies that the base should be 2.

2

Complete a table of values showing the thickness, T, for values of n from 0 to 5.

3

Determine the rule for T(n). There is a doubling term (2n) and a multiplying constant for the starting thickness (0.1).

4

Compare the rule for T(n) against the table of values in step 2.

1

Substitute n = 10 into the formula for T.

WRiTe

a When n = 0, T = 0.1, and as n increases

by 1, T doubles. n T

0 0.1

1 0.2

2 0.4

T(n) = 0.1(2n)

b When n = 10,

T(10) = 0.1(210) T = 102.4 mm

2

Calculate T.

1

Change 6 cm to millimetres.

2

Substitute T = 60 into the formula.

3

Divide both sides by 0.1.

4

Take log10 of both sides.

log10 (600) = log10 (2n)

5

Use the ‘logarithm of a power’ law to bring the power n to the front of the logarithm.

log10 (600) = n log10 (2)

6

Divide both sides by log10 (2).

n=

7

Evaluate.

n ≈ 9.23

8

Round the answer up to the nearest whole number, as the number of folds are positive integers and if you round down the thickness will not have reached 60 mm.

Therefore, n = 10 folds.

Maths Quest 11 Mathematical Methods CAS

c 6 cm = 60 mm

When T = 60, 60 = 0.1 (2n) 600 = 2n

log10 (600) log10 (2)

3 0.8

4 1.6

5 3.2

WoRkeD examPle 26

The price of gold since 1980, P (dollars per ounce), can be modelled by the function P = 400 + 50 log10 (5t + 1), where t is the number of years since 1980. a Find the price of gold per ounce in 1980. b Find the price of gold in 2006. c In what year will the price pass $550 per ounce?

TUToRial eles-1419 Worked example 26

Think

a 1 State the modelling function.

WRiTe

a P = 400 + 50 log10 (5t + 1)

In 1980, when t = 0,

2

Determine the value of t represented by the year 1980.

3

Substitute t into the modelling function.

P = 400 + 50 log10 [5(0) + 1] = 400 + 50 log10 (1)

4

Evaluate P.

P = 400

b 1 Repeat part a by determining the value of

t represented by the year 2006. 2

b t = 2006 − 1980

= 26

When t = 26, P = 400 + 50 log10 [5(26) + 1] = 400 + 50 log10 (131) = 400 + 105.864 = $505.86

Substitute the value of t into the modelling function and evaluate P.

c 1 Since P = 550, substitute into the modelling

function and solve for t.

c

550 = 400 + 50 log10 (5t + 1)

2

Simplify by isolating the logarithm part of the equation.

150 = 50 log10 (5t + 1) 3 = log10 (5t + 1)

3

Express this equation in its equivalent indicial form.

103 = 5t + 1

4

Solve this equation for t.

5

Convert the result into years.

1000 = 5t + 1 999 = 5t 199.8 = t The price of gold will reach $550 in 1980 + 199.8 = 2180 (approximately).

applications of exponential and logarithmic functions exercise 5h

1 We25 Before a mice plague that lasts 6 months, the population of mice in a country region is

estimated to be 10 000. The mice population doubles every month during the plague. If P represents the mice population and t is the number of months after the plague starts: a express P as a function of t b find the population after: i 3 months ii 6 months c calculate how long it takes the population to reach 100 000 during the plague. ChaPTeR 5 • Exponential and logarithmic functions

239

2 WE26 The population of a town, N, is modelled by the function N = 15 000(20.01t), where t is the

number of years since 1980. Find the population in 1980. Find the population in: i 1985 ii 1990. What is the predicted population in 2015? In what year will the population reach 20 000?

a b c d

3 The weight of a baby, W kg, t weeks after birth can be modelled by W = 3 log10 (8t + 10). a Find the initial weight. b Find the weight after: i 1 week ii 5 weeks iii 10 weeks. c Sketch the graph. d When will the baby reach a weight of 7 kg? 4 If $A is the amount an investment of $P grows to after n years at 5% p.a. using compound interest: a write A as a function of P b use the function from a to find the value of $10 000 after 10 years c calculate how many years it will be until an investment of $10 000 reaches $26 500. 5 The value of a car, $V, decreases according to the function V = 25 000 2 0.1t, where t is the number of

years since the car was purchased. a Find the value of the car when new. b Find the value of the car after 6 years. c In how many years will the car be worth $10 000?

5

6 The temperature, T (°C), of a cooling cup of coffee in a room of

temperature 20 °C can be modelled by T = 90(3 0.05t ) , where t is the number of minutes after it is poured. a Find the initial temperature. b Find the temperature: i 3 minutes after pouring ii 6 minutes after pouring. c How long is it until the temperature reaches half its initial value? −

7 A number of deer, N, are introduced to a reserve. The deer population can be predicted by the model

N = 120(1.1t ), where t is the number of years since introduction. a Find the initial number of deer in the reserve. b Find the number of deer after: i 2 years ii 4 years iii 6 years. c How long does it take the population to treble? d Sketch the graph of N versus t. e Explain why the model is not reliable for an indefinite time period. 8 After a recycling program is introduced, the weight of rubbish disposed of by a household each week is given by W = 80(2 0.015t ), where W is the weight in kg and t is the number of weeks since recycling was introduced. a Find the weight of rubbish disposed of before recycling starts. b Find the weight of rubbish disposed of after recycling has been introduced for: i 10 weeks ii 40 weeks. c How long is it after recycling starts until the weight of rubbish disposed of is half its initial value? d i Will the model be realistic in 10 years time? ii Explain. 9 The number of hectares (N) of forest land destroyed by a fire t hours after it started is given by N = 40 log10 (500t + 1). a Find the amount of land destroyed after: i 1 hour ii 2 hours iii 10 hours. b How long does the fire take to burn out 155 hectares? −

240 Maths Quest 11 Mathematical Methods CAS

10 A discus thrower competes at several competitions during the year. The best distance, d metres,

that he achieves at each consecutive competition is modelled by d = 50 + log10 (15n), where n is the competition number. a Find the distance thrown at the: i 1st competition ii 3rd competition iii 6th competition iv 10th competition. b Sketch the graph of d versus n. c How many competitions does it take for the thrower to reach a distance of 53 metres?

11 The population, P, of a certain fish t months after being introduced to a reservoir is

P = 400(100.08t), 0 ≤ t ≤ 20. After 20 months, fishing is allowed and the population is then modelled by P = 15 000 + 924 log10 [10(t − 19)], t ≥ 20. a Find the initial population. b Find the population after: i 5 months ii 15 months iii 25 months iv 40 months. c How long does it take the population to pass 10 000? 7 12 A ball is dropped from a height of 5 metres and rebounds to 10 of its previous height. a Find the rule that describes the height of the ball (h metres) after n bounces. b Find the height after: i 4 bounces ii 8 bounces. c Sketch the graph of the height of the ball after n bounces. 13 A computer appreciates in value by 10% per year. If the computer costs $5000 when new, find: a the rule describing the value, V, of the computer at any time, t years, after purchase b the value of the computer after 6 years c the number of years it takes to reach double its original value. 14 From the start of 1996, a small mining town has seen a steady increase in population until 2000 as the

price of minerals improved and mining was extended. Year Population (P)

1996 700

1997 750

1998 804

1999 870

2000 925

Let 1996 be t = 0; then 1997 will be t = 1 and so on. Plot P against t. What does the shape of the curve look like? Calculate the ratio of the population in 1997 to the population in 1996. Calculate all the ratios in successive years, and hence estimate the percentage annual increase or growth. To obtain an accurate estimate of population growth, follow these steps. e On the table above, evaluate log10 (P). f Plot log10 (P) against t. Are the points approximately collinear? g Draw a line of best fit and find its gradient and the intercept on the y-axis. h Write the equation for the line. i Show that P = 700(1.07)t. Is this close to your estimate in d? j Use this formula to estimate the population in 2001 and 2002. k When might the population have reached 2000? l In fact there was a downturn in the population as the mine output decreased. From 2000 onwards there was an annual decline of 10% in population. During which year did the population reduce to below 600? a b c d

15 A used car dealership keeps data on the value of the Fraud Atlas (new at the start of 2005) over 5 years.

Year (t) Value (V) a b c d e

2005 45 000

2006 35 500

2007 28 000

2008 22 500

2009 18 000

Let t = 2005 be x = 0; then 2006 will be x = 1 and so on. Plot V against x. What does the shape of the curve look like? Calculate the ratios of values of the car in successive years, e.g. V2006 ÷ V2005 and so on. Estimate the annual rate of depreciation. For a more accurate result, evaluate log10 (V). Chapter 5 • Exponential and logarithmic functions 241

f g h i j k l

Plot log10 (V) against x. Are the points approximately collinear? Draw a line of best fit and find its gradient and the intercept on the y-axis. Write the equation for the line. Show that V = 45 000(0.79)x (or a formula close to it). What is the annual rate of depreciation? Use this formula to estimate the car’s values in 2010 and 2011. When will it reach a value of $7000?

16 Johannes Kepler was a German astronomer born in the 16th century. He used data collected by Tycho

Brahe to formulate an equation or law connecting the period of a planet’s revolution around the sun to the radius of its orbit. The following table contains the data Kepler used. The radius of the orbit is expressed as a proportion of Earth’s orbit (with the period given in days). Planet Mercury Venus Earth Mars Jupiter

Radius (R) of orbit 0.389 0.724 1.000 1.524 5.200

Period (T) 87.77 224.70 365.25 686.98 4332.62

a Plot T against R (using (0,0) too). What does the graph look like?

The graph has the form T = aRb, where a and b are constants. b To find them, find log10 (R) and log10 (T). c Plot log10 (T) against log10 (R) on graph paper. Are the points collinear? d Draw a line of best fit and find its gradient, correct to 2 decimal places. e Read off the intercept on the y-axis and write it as the equivalent logarithm. f Write an equation for the straight line. g By transposition, show that T = 365.25R1.50 (or a formula close to it). h If Saturn’s orbit has a radius of 9.510, find its period using the formula above. The actual period is 10 759.2 days. Why is there a difference in the results? i Kepler’s Law is T 2 = kR3. What is the value of k? 242 Maths Quest 11 Mathematical Methods CAS

Summary am × an = am + n am ÷ an = am − n (am)n = amn a0 = 1 (ab)n = anbn n n 6. a = a b bn • To simplify indicial expressions: – when dealing with questions in the form (expression 1) ÷ (expression 2), replace expression 2 with its reciprocal and change ÷ to × – remove brackets using laws 4, 5 and 6 – collect plain numbers and terms of the same base – simplify using laws 1, 2 and 3.

index laws

1. 2. 3. 4. 5.

negative and rational powers

•a

−n

=

1 ,a≠0 an

1

• an = n a m

1

n • a n = (a n ) m = ( n a ) m = a m

indicial equations

• If am = an, then m = n. • A graphics calculator may be used to solve indicial equations, using the solve function.

Graphs of exponential functions

• • • • •

f ( x) = ax, a > 1; f ( x) = ax, 0 < a < 1 The y-intercept is (0, 1). The asymptote is y = 0 (x-axis). The domain is R. The range is R+. y

y

1 0

f(x) = ax, a > 1 Asymptote x y=0

1

f(x) = ax, 0 < a < 1 Asymptote x y=0

0

y

• Reflections: f ( x) = ax, a > 1; f ( x) = a−x, a > 1

f(x) = ax, a > 1

1 0

f ( x) = ax, a > 1; f ( x) = −ax, a > 1

f(x) = a−x, a > 1 Asymptote x y=0

y f(x) = ax, a > 1 1 −1

0

x

Asymptote y=0

f(x) = −ax, a > 1

ChaPTeR 5 • Exponential and logarithmic functions

243

• Translations f ( x) = ax, a > 1; f ( x) = ax + b, a > 1, b > 0

−

x+b y f(x) = a , a > 1, b > 0

x b f(x) = a , a > 1

b 1 x

0

f ( x) = ax, a > 1; f ( x) = ax + c, a > 1, c > 0

Asymptote y=0

x y f(x) = a + C, a > 1, C > 0

f(x) = ax, a > 1

C

Asymptote y=C C

1 x

0

• Dilations f ( x) = ax, a > 1; f ( x) = Aax, A > 0, a > 1

y

f(x) = Aax, A > 1, a > 1 f(x) = ax, a > 1

A 1 0

f ( x) = ax, a > 1; f ( x) = akx, a > 1, k > 0

y 4

x

Asymptote y=0

f(x) = 22x f(x) = 2x –x f(x) = 22

2 (0, 1) −3 −2 −1

logarithms

244

Asymptote y=0

• If y = ax, then loga ( y) = x, where a = the base, x = the power, index or logarithm, and y = the base numeral. Log laws: • loga (m) + loga (n) = log a (mn) m, n > 0 m • loga (m) − loga (n) = loga n m, n > 0 • loga (mn) = n loga (m) • loga (a) = 1 • loga (1) = 0

Solving logarithmic equations

0 1 2 3 x

m>0

• Logarithmic equations are solved more easily by: 1. simplifying using log laws 2. expressing in index form 3. solving as required.

Maths Quest 11 Mathematical Methods CAS

• If ax = b, then x = • loga (ax) = x

log10 (b) = loga (b) log10 (a)

log ( x ) • a a =x

logarithmic graphs

• The logarithmic function f ( x) = loga (x) is the inverse function of the exponential function f ( x) = ax. y

f(x) = ax, a > 1

y=x f(x) = loga (x), a > 1

1 0 1

• • • •

x

The x-intercept is (1, 0). The asymptote is x = 0. The domain is R+. The range is R.

ChaPTeR 5 • Exponential and logarithmic functions

245

Chapter review S h oRT anS WeR

1 Simplify the following expression with positive indices. −

1

(16 x 6 y10 ) 2 ÷ 3 (27 x 3 y9 ) 2 Solve the following equations. a 4x3 = 500 3 Find the solutions to: a 9x − 1(3x) − 6 = 0 4 For the function with the rule f ( x) = 3x − 2 + 1: a find the y-intercept b state the equation of the asymptote c sketch the graph of f ( x) d state the domain and range.

b 8x + 1 × 22x = 43x − 1 b 2x + 2

−x + 1

= 3.

5 For the function with the rule f ( x) = 3 × 2x − 3: a find the y-intercept b state the equation of the horizontal asymptote c find a second point on the graph d sketch the graph of f ( x) e state the domain and the range of the function. 6 a Evaluate log3

1 . 27

b Express y in terms of x if log10 ( x) + log10 ( y) = 2 log10 (x + 1). 7 Simplify the following. a 3 log4 (5) − 2 log4 (6)

b

2 log5 ( x 2 ) 1 logg5 ( x ) 3

8 Solve each of the following. a log6 (x) = 3

b 2 logx (125) = 6

c log2 (3x + 6) − log2 (5) = 2

9 If y = loge (7x − 6) + 3, then what does x equal? 10 If f (x) = 3x: a sketch the graph of f (x) and label: i the y-intercept ii the equation of the asymptote b sketch the line y = x, use this line to sketch the inverse function g(x) = log3 (x), and label: i the x-intercept ii the equation of the asymptote. 11 The number of bacteria in a culture, N, is given by the exponential function N = 1500(20.4t ), where t is

the number of days. a Find the initial number of bacteria in the culture. b Find the number of bacteria (to the nearest 100) after: i 5 days ii 10 days. c How many days does it take for the number of bacteria to reach 12 000? mUlTiP l e Ch oiCe

1 When simplified, a

2

x 4 y7 7

5m 4 p2 (5m 2 p6 )3 may be simplified to: ÷ 2m 3 p 3m 7 p m2 3m10 3m 2 a 16 B C 47 p 50 p16 2 p32

3 The value of a

246

(2 xy3 )2 3 x 5 y 2 × is equal to: 7x3 4y 3x 4 y7 3 y7 B C 7 x2

1 20

−1 − 2 64 3 5 125

B 5

Maths Quest 11 Mathematical Methods CAS

D

3x 4 y6

e

x y

D

m15 p29

e

m 20 p24 25

is: 4

C 5

5

D 4

e −5

4 If 252 − x = 125, then x is equal to: 1

a 1

C −1

B 2

5 If 42x − 17(4x) + 16 = 0, then x is equal to: a 1 or 16

B 0 or 1

C 2 or 8

6 The rule for the graph below could be:

D 2

e 5

D 1 or 4

e 0 or 2

y

0

x −1 −2

a y = 3x − 2

B y = 3x + 2

D y = −2x

C y = 3x − 2

e y = 2x + 2

Questions 7 to 9 refer to the function defined by the rule y = 2x + 3 − 1. 7 The graph that best represents this function is: a B y y

C

y

7

−3 D

0

−3

x e

y

0 −1

3

8 The domain is: a (3, ∞)

9 The range is: a [−1, ∞)

−3

x

a 3

5

a 2

e R\{−3}

B R

C R+

D (1, ∞)

e (−1, ∞)

B log5 (x) = 250 e log250 (x) = 5

log3 25 is nearest to: log3 5 B 5

C log5 (250) = x

C 0

D 69

e 1

C 2

D 9

e 20

5 (x 4 )

log7 simplifies to: log7 ( x ) 3

5

B

log7 ( x 2 )

14 The solution to log5 (x) = 4 is: a 25

x

D R

B 7

13 The expression

0

C R+

11 The value of log7 (49) + 3 log2 (8) − 4 is: 12 The value of

B 125

C

log7 ( x 4 )

3

B 7

e

C 1

D 625

e 20

C 5

D 14

e

16 If log3 (2x − 1) + log3 (2) = 2, then x is equal to: a 2

B 1

5 8

D 4

15 The value of x if 2 logx (343) = 6 is: a 3

−1

B [−1, ∞)

a logx (5) = 250 D logx (250) = 5

x

y 1

10 When expressed in log form, 5x = 250 becomes:

a 3

x

0

0

C 3

5

D 2

e

7 11 4

ChaPTeR 5 • Exponential and logarithmic functions

247

17 The solution to the equation 43 − 2x = 12 is nearest to: a x = −1 D x = 0.2

B x = 0.35 e x=0

C x = 0.604

y

18 The equation of the graph shown at right is: a y = log16 (x)

(4, 2)

B y = log2 (x) C y = 2 log8 (x)

0

D y = log10 (x)

1

x

4

e y = 2 log10 (x)

19 The inverse of the graph below would be: y

(2, 1) 1 0 a

x B

y

y

x

0 −1

(1, −2) 1

C

D

y

e

x

1

x (−1, −2)

248

Maths Quest 11 Mathematical Methods CAS

1 0

y

−1

y

(−1, 2)

(2, 1) 0

x

(2, −1)

x

−x

1 a For the function f ( x ) = 5 : i find the y-intercept ii find the values f (1) and f (−1) iii find the equation of the asymptote iv sketch the graph of f (x) v state the domain and range. − b For the function g(x), where g(x) = f (x + 3) − 1 and f (x) = 5 x: i state the transformations to change f (x) to g(x) ii state the equation of the asymptote iii sketch the graph of g(x) iv state the domain and range of g(x).

e x Ten D eD ReS P o n S e

2 The number of lions, L, in a wildlife park is given by L = 20 (100.1t ), where t is the number of years

since counting started. At the same time the number of cheetahs, C, is given by C = 25(100.05t ). a Find the number of: i lions ii cheetahs when counting began. b Find the numbers of each after i 1 year ii 18 months. c Which of the animals is the first to reach a population of 40 and by how long? d After how many months are the populations equal, and what is this population? y 3 The graph of the function f : R → R, where f (x) = −A × 102x + 4, is shown at right. a Give the equation of the horizontal asymptote in the form y = c. c b The graph passes through the point (0, 3.8). Use this information to find the value of A. (0, 3.8) c Find the x-intercept, correct to 3 decimal places. d Find the values of: ii f (4). i f (−3) e Find the value of x if f (x) = −5.2. x f State the domain and range of f (x). 4 The temperature, T °C, of a coffee in a ceramic mug at time t minutes − after it is poured is given by T = 60(4 0.05t ) + 20. a Find the initial temperature of the coffee. b Find the temperature of the coffee, correct to one decimal place, at: i 2 minutes after it is poured ii 25 minutes after it is poured. c Sketch the graph of the equation for 0 ≤ t ≤ 50. If the coffee can be comfortably drunk when it is between temperatures of 30 °C and 45 °C, find: d the time available to drink the coffee e the final temperature the coffee will settle to. 5 The number of bacteria (N) in a culture is given by the exponential function N = 12 000(20.125t), where t is the number of days. a Find the initial number of bacteria in the culture. b Find the number of bacteria in the culture after: i 4 days ii 2 weeks. c Find the time taken for the bacteria to reach 32 000. When the bacteria reach a certain number, they are treated with an anti-bacterial serum. The serum − destroys bacteria according to the exponential function D = N0 × 3 0.789t, where D is the number of bacteria remaining after time t and N0 is the number of bacteria present at the time the serum is added. The culture is considered cured when the number of bacteria drops below 1000. d If the bacteria are treated with the serum when their numbers reach 32 000, find the number of days it takes for the culture to be classed as cured. e How much longer would it take the culture to be cured if the serum is applied after 6 weeks?

DiGiTal DoC doc-9756 Test Yourself Chapter 5

ChaPTeR 5 • Exponential and logarithmic functions

249

ICT activities Chapter opener DiGiTal DoC • 10 Quick Questions doc-9749: Warm up with ten quick questions on exponential and logarithmic functions (page 217)

5B

negative and rational powers

TUToRial • We5 eles-1417: Watch a tutorial on writing exponential expressions with positive indices (page 222) DiGiTal DoC • SkillSHEET 5.1 doc-9750: Practise working with negative and rational powers (page 224)

5C

indicial equations

TUToRial • We10 eles-1418: Watch a tutorial on solving an indicial equation by using substitution (page 226)

5D

Graphs of exponential functions

DiGiTal DoCS • SkillSHEET 5.2 doc-9751: Practise substituting values into exponential functions (page 230) • WorkSHEET 5.1 doc-9752: Write expressions with their simplest index notation, solve indicial equations and sketch graphs of exponentials (page 231)

5F

Solving logarithmic equations

DiGiTal DoC • WorkSHEET 5.2 doc-9753: Simplify logarithmic expressions and solve logarithmic equations (page 236)

250

Maths Quest 11 Mathematical Methods CAS

5G

logarithmic graphs

inTeRaCTiViTY • Logarithmic graphs int-0264: Consolidate your understanding of logarithmic graphs and their features (page 236) DiGiTal DoCS • Investigation doc-9821: Consolidate your understanding of logarithmic graphs and their features (page 236) • Investigation doc-9822: Investigate advanced logarithmic graphs and their features (page 236)

5h applications of exponential and logarithmic functions DiGiTal DoCS • doc-9754: Investigate graphs of exponential functions (page 238) • Extension doc-9755: Practise sketching logarithmic graphs and identifying rules of functions (page 238) TUToRial • We26 eles-1419: Watch a tutorial on calculating the price of gold, where the price is modelled by a logarithmic function (page 239)

Chapter review DiGiTal DoC • Test Yourself doc-9756: Take the end-of-chapter test to test your progress (page 249)

To access eBookPLUS activities, log on to www.jacplus.com.au

Answers CHAPTER 5 exPonenTial anD loGaRiThmiC FUnCTionS exercise 5a

index laws

1 a x10 c x7y8

b

x 3 y9 c 2 4 p10 m8 3a 3

c

a9b5 4

b

9 xy 4 2

125k 11d 6 24 c B b x5y6m − 1

5a A 6 a x3yn + 1z

bE

7 a 211

b 324

c 55 × 34

e

f 2

8a 8

3 24 b 59 049

d 1

e 5

g 16

h 3

e

−2

c 16 1

d x10 3 a 3

1 xy 2

e 213 × 38 b3

d 27

e

1

2 3

p2 m 1 f 4 x 6 y2 c 4 c

f

h 4

4E 5 a

d

7 33

b

1

5 x6

e 64m10

7 24

g (x

125 64

9

g 27

3 + 1) 2

7 D

h

c

exercise 5D

Graphs of exponential

1 a

y

1 a 5 c 5

e −2 2 a 5 d

3 −9 4

e

2 11 6

b

x

y=3

b

y

y = 0.5(4x)

2

0

x

1

y 5

y = 5x

1

y = 0, (0, 0.5) c

y

y = 1–4 (2x)

1 0 y 10

d

1

4– 3

0

1

1

y

x

−1 0 y

y = 4−x

4

0

f

y = 2x − 1

1– 2

4

x

b y = 0, (0, 9)

1 0

−1

x

y = 0, (0, 4) 1 3 a y = 0, (0, 2 ) dom = R, ran = (0, ∞)

2

e

x

y y = 4( 1–3 )x

x

1

1 4)

y = 0, (0,

y

y = 2−x

0

x

y = 10x

1 0

d

1

x

dom = R, ran = (0, ∞) y

y = 3x + 2

y 0

9

x

1

−3

0

x

c y = 0, (0, 5)

dom = R, ran = (0, ∞)

g

y −1 −1 0

c −1 9 8

x

1– 2 1– 4

−1

f

x

1

y = 0, (0, 2)

0.5

0

2 x3

indicial equations b4 d2 f −3 b

2

1

3 ( y − 4) 2

−1

6

3

1 y2

f x2

y = 2(3x)

0

y = −3x exercise 5C

y

8 B

x

1

2 a

b 1.89 d 1.66

negative and rational powers 212 52 b 2 c 316 3 b

0

b 1 or 2 d 1 or 2

functions

c

2 a −22

1 1– 2

−1 11

d

b 23x − 2 d 25n − 6 × 39n − 3

3

1 1 a 3 6

b 7

5

10 E exercise 5B

y = 0.5x

2

f 5

1

9 a 22 × 33n + 4 c 2n − 2 × 76n − 3

5 9 10 −5 7

y

10 E

b

d 222 × 58

− 18

6A 9 a 1.58 c 1.65

d 10e9f 2

15 4

b 7 −5 d

5 a 0 or 1 c 0 or 1

d p11q2

c 18u11v5 4a

4 a

b 518 d 6m9p17

2 a a5b3

c

h

10

3 a −3

−3 y = −3−x

y = 51 − x y x 5 0

x

ChaPTeR 5 • Exponential and logarithmic functions

251

c dom = R, ran = R+

d y = 3, (0, 4)

dom = R, ran = (3, ∞)

y

y y = 2x + 3

4

2 a 42 = 16

f(x) = 3 × 2x

e 5 = 625 −2

g 3

3

e y=

−2)

(0, dom = R, ran = (−3, ∞) y

0

d g j 7 a

x

1

c e g i 8 a

d dom = R, ran = R+

y = 3x − 3

y x

0

(1, 45)

45

f(x) = 5 × 32x

−2 −3

y 5

7

0

y = 2x + 3 − 1 0

e

x

−1

x

1

–x

−

6 log2 (80) log10 (100) = 2 log2 (4) = 2 log2 (3) log3 (20) log10 (250)

10

(2, 10)

4 3 x

y

dom = R, ran = (5, ∞)

f(x) = 2 ×

5

b E

y

6

3

f(x) = 4 × 23x + 3

(1, 4)

y 8

(1, 8)

f(x) = 23x

− 1–3

7 a 3.32 c 2.26 e 1.56,

0

x

x

d log10 (100x3) 6 f log10 ( 10 x )

−3.99

e log b (a) = n

1

g 30

i 5

k 10

j

6 2 B 3.459 1.292 0.661

b 5 f 25 b A

Maths Quest 11 Mathematical Methods CAS

( )

b

c 4 g 2 c E

b e h

−0.737 −

3.080

−0.423

h5

d 10 h1 dC c 0.483 f 2.255 i 2.138

7C

logarithmic graphs b b c 3y

y

1 0

d 2.44, −2.86

f log2 1 = − 4 16

c −3

6 3

b 1.30

13 C

b log4 (1024) = 5

f 27

2 a

d 4

Solving logarithmic equations b 0 c −1 d −2 f 2 g 2 b 9 c 625 d 10

1 a x

logarithms 1 a log 2 (8) = 3 b log 3 (243) = 5 c log 5 (1) = 0 d log 10 (0.01) = −2

1

12 D

b 4 e 4.070

exercise 5G y=3

exercise 5e 1

f 6 11 B

6 B (− 1–3, 5) 7 5 3

9

3

c 2 g 6

e 8

3 a e 4 a 5 a d g

x

y

b dom = R, ran = R+

( x + 3)2 x − 2

b 2

1 a 2 e 6 2 a 8

(3, 4)

0

x

x

h log10

1

4 2

1

g log3 (64)

exercise 5F –x 23

y = 102 − x + 5

f(x) = 22x

f log2 [x3(x − 4)]

15 a 4 d 3.170

2 x

f dom = R, ran = R+

h y = 5, (0, 105)

4 a A 5 a dom = R, ran = R+

b log2 (1728)

e log3 (4)

0

0 1

Undefined log3 (105) log6 (56) f log3 (3) = 1 h log4 (5)

e log2 = −1 2

8

2

f −5 i 5 l 5

−5

1

c log5 25

y=6 +3

0

h k b d

5A c 3

d log10 4

14 a log3 (81) = 4

f(x) = 2 × 52

−x

252

2

−2

10 E

y

0

−

2

y

dom = R, ran = (3, ∞)

0

4 C b 4 e 3

9 a 3 e 3

dom = R, ran = R+

g y = 3, (0, 4)

4 3 2 1

h bx = a

1

dom = R, ran = (−1, ∞)

y

=

c log5 (3)

f y = −1, (0, 7)

15

f 27 = 128

1 9

3D 6 a 4

3

x

0 −3,

d 33 = 27

4

(1, 6)

6

b 106 = 1 000 000

1 − c 2 1= 2

f(x) = log2(x)

(2, 1)

x

2

y

1 0

(5, 1) 5

f(x) = log5(x) x

d 5x

c

e The population will reach a limit at

y

some stage. 8 a 80 kg b i 72.1 kg c 67 weeks

(8, 1) f(x) = log8(x)

1 0

x

8

d i d

f

x

10

(12, 1) f(x) = log12(x) x

12

(15, 1) f(x) = log15(x) 15

rubbish will be disposed of in 10 years or so, which is unlikely. 9 a i 108 hectares ii 120 hectares iii 148 hectares b 15 h 10 a i 51.18 m ii 51.65 m iii 51.95 m iv 52.18 m b

d

0 2– 3

c 11 a b c 12 a b c

26 weeks A = P(1.05)n $25 000 90 °C i 76.3 °C 12 min 37 s 120 i 145

20

n

1 2 3 4

10

h 5 4 3 2 1

0

b $16 288.95 c 20 years b $14 427 c 10 years

f

n

1 2 3 4 5

2

N N = 120(1.1t) 120

t

x 0

1

2

3

4

log10(V) 4.5 4.4 4.3

0

g h i

1

2

3

4 t(years)

b The graph shows a slight upward curve

not starting from (0, 0); this suggests possible exponential growth. c 1.071 d 1.072, 1.082, 1.063. Estimated % annual growth is 7%. e

x(year)

4.6

700

iii 213

4

4.2

0

ii 176

3

V 4.653 4.550 4.447 4.352 4.255

800

ii 64.7 °C

2

down with a reducing slope — exponential decay. c 0.789, 0.789, 0.804, 0.800 d The ratio is about 0.795 or a 20.5% rate of depreciation. e

h = 5(0.7n), n ≥ 0, n ∈ J

1

b The curve starts from $45 000, curving

900

c 11 1 years

0

30

13 a V = 5000(1.1t) b $8857.81 c 8 years 14 a P(people)

t

0

⇒ P = 100.03t × 102.845 ⇒ P = 700 × 1.072t, i.e. 7.2% growth similar to d j P5 = 982 people; 2002 (t = 6), P6 = 1051 k During 2011 l During 2004

40

67 400 i 1005 ii 6340 iii 16 643 iv 17 146 17.48 months h = 5(0.7n) i 1.20 m ii 0.29 m

0 W = 3 log10(8t + 10)

t

4

g m ≈ 0.03; y-intercept = 2.845 h log10 (P) = 0.03 t + 2.845 i P = 10(0.03t + 2.845)

d = 50 + log10(15n)

W 3

3

15 a V($000)

52 51

3 The smaller a is, the steeper the graph. 4 a log3 (x) for x > 0 1 b 2 log2 (x) for x > 0 exercise 5h applications of exponential and logarithmic functions 1 a P = 10 000(2t) b i 80 000 ii 640 000 c 3.32 months 2 a 15 000 b i 15 528 ii 16 077 c 19 118 d 2022 3 a 3 kg b i 3.77 kg ii 5.1 kg iii 5.86 kg

2

50

x

c 5x

1

The points appear collinear.

ii No, the model suggests virtually no

1 0

−5– 4

0

t

0

y

c

W = 80(2−0.015t)

(10, 1) f(x) = log10(x)

1 0

d

2.9 2.8

y

e

7

3.0

y

1 0

4 5 6

3.1

ii 52.8 kg

W 80

d a a a b c a b

f log10(P)

t

0

P

2.845 2.875 2.905 2.940 2.966

1

2

3

4

j k l

1

2

3

4

x

The points are roughly collinear. m ≈ −0.099; y-intercept = 4.65 log10 (V ) = −0.099t + 4.65 log10 (V ) = −0.099t log10(10) + 4.65 log10(10) − = log10(10 0.099t) + log10(104.65) − ⇒ V = 10 0.099t × 104.65 − 4.65 = 10 × (10 0.099)t = 45 000(0.79)t The rate of depreciation is about 21%. In 2010 (x = 5), V = $13 847; in 2011, V = $10 939. During 2012

ChaPTeR 5 • Exponential and logarithmic functions

253

16 a

7 10 13 16 19

y y = 3x − 2+ 1

c

T 5000

—) (0,10 9

4000

(2, 2)

1

3000

x

0

1000 (0, 0)

1

2

3

4

5

R

The graph starts from (0, 0) and curves slightly up to the right like a positive exponential function. log10 R −

0.410

1.943

−0.140

2.352 2.563

0.183

2.837

0.716

3.637

log10 (T ) 4 3

d

y

i

ChaPTeR ReVieW ShoRT anSWeR

4 y2 3x 4 2 a x=5 3 a x=1

b x=5 b x = 0 or x = 1

4 a (0, 9 )

b y=1

1

10

254

1 −1 0

1

e The domain is R and the range is (−3, ∞). 6 a −3

b y=

(x + x

b 12

8 a x = 216

b x=5

14

y = 3x (1, 3)

1

y

x

y = 3x (1, 3)

3

y = log3(x)

(1, 1)

1

3

80 70 60 50 40 30 20 10

x

x=0

11 a 1500 b i 6000 ii 24 000 c 7.5 days

0

mUlTiPle ChoiCe

Maths Quest 11 Mathematical Methods CAS

2 C 5 E

3 A 6 C

−3

−1

1

x

iv Domain R, range ( −1, ∞) 2 a i 20 ii 25 b i L = 25, C = 28 ii L = 28, C = 30 c Lions by 1 year 1 month d 31 after 1 year 11 months 3 a y=4 b A = 0.2 c x = 0.651 d i 3.999 998 (≈ 4) ii −19 999 996 (≈ −2 × 107) e 0.831 f Domain R, range (−∞, 4) 4 a 80 °C b i 72.2 °C ii 30.6 °C c T

(3, 1) 01

y

c x= 3

0

1 B 4 B

left, vertical translation of 1 unit down

ii y = −1 iii y = 5−(x + 3) − 1

y = −1

3

y=x

v Domain R, range R+ b i Horizontal translation of 3 units to the

1)2

ey − 3 + 6 7 y

x

(−4, 4)

7 a log4 125 36

b

iii y = 0

x y = −3

y=0

1

E C B B

(−1, 5)

(1, 3)

0

−·4 −·3 −·2 −·1 0 ·1 ·2 ·3 ·4 ·5 ·6 ·7 ·8 log10 (R)

The points are very close to collinear. m ≈ 1.50 The y-intercept is 2.5614 or log10 (364.25) log10 (T ) = 1.50 log10 (R) + 2.5614 log10 (T ) = log10 (364.25 × R1.50) ⇒ T = 364.3 R1.50 T = 10 711.76 days. The difference may be due to rounding errors or ignoring the small effects of other planets and moons on Saturn’s orbit. k = 133 407.5625

9 12 15 18

y y = 5−x

y=0

10 a

1

h

D B D C

ii 5 , 5

f(x) = 3 × 2x − 3

3

9 x=

2

d e f g

1 a i 1 iv

d Domain = R, range = (1, ∞) 5 a y=0 b y = −3 c (1, 3) (for example)

log10 T

0

c

8 11 14 17

exTenDeD ReSPonSe

2000

b

A C A E C

d 5 a b c

T(x) = 60 (4−0.05t) + 20

(25, 30.6)

10 20 30 40 50 x

13.2 minutes e 20 °C 12 000 i 16 970 ii 40 363 11.3 days d 4 days e 4 more days

Chapter 6

Circular functions DiGital DoC doc-9757 10 Quick Questions

Chapter ContentS 6a 6B 6C 6D 6e 6F 6G 6h 6i

Trigonometric ratio revision The unit circle Radians Symmetry Identities Sine and cosine graphs Tangent graphs Solving trigonometric equations Applications

6a

trigonometric ratio revision

Recall that for a right-angled triangle: O = opposite A = adjacent H = hypotenuse. These ratios can be memorised using the term ‘SOH CAH TOA’.

sin (θ ) = H O θ

A

O H

cos (θ ) =

A H

tan (θ ) =

O A

WorkeD exaMple 1

Find the value of x in each of the following triangles. Express lengths correct to 2 decimal places and angles to 1 decimal place. a 9

b

64°

4.1

x x 27° c x

7.5

8.6

Chapter 6 • Circular functions

255

Write

think

a 1 Write the cosine ratio. 2

Replace A with x, H with 9 and θ with 64°.

3

Make x the subject of the equation.

4

Calculate x, correct to 2 decimal places.

cos (64 ) =

x 9

= 3.95 O b Use sin (θ ) = H

2

Replace O with 4.1, H with x and θ with 27°.

3

Make x the subject of the equation.

4

Calculate x, correct to 2 decimal places.

4.1 sin (27 ) = x x=

4.1 sin (27 )

= 9.03 c Use tan (θ ) =

c 1 Write the tangent ratio.

DiGital DoCS doc-9758 SkillSHEET 6.1 trigonometry review i

A H

x = 9 cos (64 °)

b 1 Write the sine ratio.

2

Replace O with 8.6, A with 7.5 and θ with x°.

3

Evaluate the right-hand side of the equation, keeping plenty of decimal places at this stage.

4

Make x the subject of the equation using inverse tan notation.

5

Calculate x, correct to 1 decimal place, using the inverse tan function of the calculator.

exercise 6a

a Use cos (θ ) =

tan ( x ) =

O A 8.6 7.5

= 1.146 67 x = tan 1 (1.14667) −

= 48.9°

trigonometric ratio revision

1 We1a Find the value of x in each of the following, correct to 2 decimal places. a b c 10 x 18°

doc-9759 SkillSHEET 6.2 trigonometry review ii

x

12 x

81°

2 We1b Find the value of y in each of the following. a b 8 54° y

36°

c

10.6 32° y

y 72°

6.8

2.4

3 We1c Find the value of a in each of the following. Give answers correct to the nearest tenth of a degree. b c a a 8 2.5 9 6.4 10.2 a a 3.7

256

Maths Quest 11 Mathematical Methods CAS

4 MC In the figure below: a

x

10 cm a x is nearest to: A 60 cm

21.98 cm 20°

y

B 22 cm

C 7.5 cm

D 8 cm

E 9 cm

B 64°

C 37°

D 26°

E 39°

B 6 cm

C 5 cm

D 10 cm

E 7 cm

b angle a is nearest to: A 53°

c y is nearest to: A 8 cm

5 MC If sin (a°) = 0.951, then tan a is equal to: A 0.000 29 D 1.32

B 0.0166 E 0.309

C 3.076

6 MC If 0° < a < 90° and cos a < 0.5, then which of the following is correct? A a < 30° D a < 45°

B a > 30° E a > 60°

C a < 60°

7 A tree 5 metres tall casts a shadow so that the angle of elevation from the end of the shadow to the top

of the tree is 35°. How long is the shadow?

5m 35° 8 A mathematically able tree removalist measures the angle of elevation of two points on a diseased

section of a large eucalypt at a distance of 40 m from the base of the tree. If the angles are 20° and 24° respectively, how tall is the diseased section of the tree? Diseased section

20° 24° 40 m 9 A 60-metre flying fox cable is set up to cross a river so that it will drop 10 metres vertically. What is the

angle of depression of the cable? 10 A soccer player has a shot for goal from the position shown in the figure below right. Find the ‘scoring

angle’, θ.

Goal 5m 7.3 m

12 m

θ

Chapter 6 • Circular functions 257

6B

The unit circle

So far in this chapter we have considered angles in a right-angled triangle. But trigonometric functions can be defined over a larger domain by considering their definition inside a unit circle. This is a circle with centre (0, 0) and a radius of 1 unit. A point P (x, y) is a point on the circle. Its location can also be defined as P (θ), where θ is the anticlockwise rotation from the positive x-axis. A negative θ value denotes a clockwise rotation. Using the SOH, CAH definitions in the diagram: x cosine (θ) = = x 1 y sine (θ) = = y 1 These may be abbreviated: x = cos (θ) y = sin (θ)

y 1 P(θ ) 1

y

θ 0 x

−1

1x

−1

The point P can also be denoted as (cos (θ), sin (θ)). Note: As θ increases, the x- and y-coordinates oscillate between −1 and +1, so −1 ≤ sin (θ) ≤ +1 and −1 ≤ cos (θ) ≤ +1 for all θ. The ‘tangent line’ on a unit circle is the line that is a tangent to the circle at the point (1, 0). Tangent (θ), or tan (θ) for short, is the height at which a line along an angle hits the tangent line. The diagrams below show tan θ for angles in the first and second quadrants. y

y

tan (θ ) θ

θ

x

x tan (θ)

The tangent line

sin (θ ) y . Note: Tangent (θ), or tan (θ), may also be defined as the ratio x . That is, tan (θ ) = cos (θ ) This formula and its derivation will be discussed in more detail later in this chapter.

Quadrants The coordinate axes divide the unit circle into four quadrants as shown in the diagram below left. The angle measurements, in degrees, between the quadrants are shown in the diagram below right. y

y 90°

2nd quadrant

1st quadrant θ°

0

3rd quadrant

x

180°

0

4th quadrant 270°

258 Maths Quest 11 Mathematical Methods CAS

x 0° or 360°

Therefore:

quadrant 1 is where 0° < θ < 90° quadrant 2 is where 90° < θ < 180° quadrant 3 is where 180° < θ < 270° quadrant 4 is where 270° < θ < 360°. y

Since y = sin (θ),

sine is positive in quadrants 1 and 2 and sine is negative in quadrants 3 and 4. Since x = cos (θ), cosine is positive in quadrants 1 and 4 and cosine is negative in quadrants 2 and 3. y Since tan (θ ) = , tangent is positive in quadrants 1 and 3 and x tangent is negative in quadrants 2 and 4. The diagram at right summarises which of the trigonometric functions is positive in each quadrant. The word ‘CAST’ will assist in recalling this. You could also use a short phrase, such as ‘All Stations To Croydon’.

Sine positive

All positive

Tangent positive

Cosine positive

x

WorkeD exaMple 2

If a is any angle, find the quadrant where: a sin (a°) > 0 and tan (a°) < 0 b sin (a°) < 0 and cos (a°) > 0. think

Write

a Sine is positive and tangent is negative in quadrant 2.

a Quadrant 2

b Sine is negative and cosine is positive in quadrant 4.

b Quadrant 4

exact values Calculated trigonometric values (sin, cos and tan) of most angles are rational approximations correct to several decimal places. However, for a few particular angles, exact trigonometric values can be determined. These include multiples of 90° and the ratios of isosceles and equilateral triangles. Exact values can be determined for 0° and any multiple of 90° using this diagram and the fact that y = sin (θ), x = cos (θ) and y tan (θ) . x The isosceles triangle with equal sides of one unit each gives the exact trigonometric values for 45°. The hypotenuse is calculated using Pythagoras’ theorem: hypotenuse =

12

0° or 360° x (1, 0)

0

(0, −1) 270°

45° 2 1 45°

SOH sin (45) =

180° (−1, 0)

(0, 1)

+ 12

= 2 So, using

y 90°

1 2

=

1 2 × 2 2

=

2 2

1

CAH

TOA

1 cos (45) = 2 =

2 2

tan (45) =

1 1

=1

Chapter 6 • Circular functions

259

The equilateral triangle with each side 2 units long, shown in the figure below, has been bisected.

Line of bisection

30° 2

60° 1

Using Pythagoras’ theorem: line of bisection = 22 − 12

= 3

So, using

SOH

CAH

TOA

1 3 1 3 × = 3 3 sin (30 ) =

1 2

3 2

cos (30) =

tan (30) =

3 3 3 tan (60 ) = 1

= sin (60 ) =

3 2

cos (60 ) =

1 2

= 3 These exact values are summarised in the table below. Angle (θ °) 0°

sin (θ) 0

cos (θ) 1

tan (θ) 0

30°

1 2

3 2

3 3

45°

2 2

2 2

1

60°

3 2

1 2

3

90°

1

0

Undefined

Worked Example 3

Without using a calculator, find: a sin (90 °) b cos (180 °) c tan (270 °). Think

Write

a 1 Sine corresponds to the y-coordinate on the unit circle. 2

The value of y at the point where θ = 90° is 1.

b 1 Cosine corresponds to the x-coordinate on the unit circle. 2

a

sin (90 °) = 1 b

The value of x at the point where θ = 180° is −1.

c 1 Tangent corresponds to the value of

y x on the unit circle.

2

The value of y is −1 and the value of x is 0 at the point where θ = 270°.

3

The value of

−1 0

is not defined.

260 Maths Quest 11 Mathematical Methods CAS

cos (180 °) = −1 c

tan (270 ) =

−1

0

tan (270 °) is undefined.

WorkeD exaMple 4

Without using a calculator, state the exact value of: a sin (45 °) b cos (60 °) c tan (30 °). think

Write

a Read the value from the table, or

determine using the special triangle, O sin (θ ) = . H

2

45° 1

45° 1

a sin (45) =

1 2

=

1 2

=

2 2

b cos (60) =

b Read the value from the table, or

determine using the special triangle, A cos (θ ) = . H

2

×

2 2

×

3

1 2

30° 3

60° 1 c Read the value from the table, or

determine using the special triangle, O tan (θ ) = . A

2

30°

1

2

3

1

=

1

=

3 3

3

60° 1

exercise 6B

c tan (30 ) =

3 3

3

the unit circle

a sin (40°) f sin (260°)

State whether the following values are positive (P ) or negative (N ). b sin (65°) c sin (110°) d sin (160°) g sin (215°) h sin (321°) i sin (425°) Compare your answers with those found using a calculator.

e sin (230°) j sin (−36°)

a cos (27°) f cos (295°)

State whether the following values are positive (P) or negative (N). b cos (68°) c cos (115°) d cos (200°) g cos (402°) h cos (−83°) i cos (−240°) Compare your answers with those found using a calculator.

e cos (250°) j cos (157°)

State whether the following values are positive (P ) or negative (N ). b tan (75°) c tan (118°) d tan (166°) g tan (308°) h tan (500°) i tan (−45°) Compare your answers with those found using a calculator.

e tan (199°) j tan (−137°)

a tan (12°) f tan (255°)

DiGital DoC doc-9760 the unit circle

4 We2 If a is any angle, find the quadrant where: a sin (a) < 0 and tan (a) > 0 b sin (a) > 0 and cos (a) > 0. 5 MC

a If 0° < a < 360°, sin (a) < 0 and cos (a) < 0, then which one of the following is true?

a 0° < a < 90° D 270° < a < 360°

B 90° < a < 180° e 0° < a < 180°

C 180° < a < 270°

a 0° < a < 90° D cos (a) > 0

B cos (a) < 0 e 270° < a < 360°

C 180° < a < 270°

a one solution D three solutions

B no solution e four solutions

C two solutions

a no solutions D one solution

B two solutions e four solutions

C three solutions

b If 0° < a < 360°, sin (a) > 0 and tan (a) < 0, then which one of the following is true?

c Given that 0° < a < 360°, then the equation cos (a) = −1 has: d If 0° < a < 360°, then the equation sin (a) = cos (a) has:

Chapter 6 • Circular functions

261

e If 0° ≤ a ≤ 180°, and sin (a) = cos (a), then a is equal to: a 60° D 0°

B 150° e 90°

6 We3 Find: a cos (180°) f cos (360°) 7 We4 DiGital DoC doc-9761 WorkSHEET 6.1

b sin (270°) g tan (270°)

c tan (360°) h tan (180°)

d sin (180°) i sin (630°)

e cos (270°) j cos (720°)

c tan (30°) h cos (0°)

d cos (60°) i sin (90°)

e sin (30°) j tan (90°)

State the exact value of:

a sin (60°) f tan (45°)

6C

C 45°

b cos (45°) g sin (45°)

radians

y

1

You are used to measuring angles in degrees (°), and will recall that there are 360° in a circle. An alternative unit for angle measurement is the radian (c). This is given on the calculator as (r). Radians are derived from the circumference of the unit circle, that is, a circle of radius 1 unit. In a unit circle, the circumference = 2πr = 2π(1) = 2π units. So, the angle swept in one revolution is equivalent to 2π c. Therefore 2π c = 360°, so: πc 180 1 = π c = 180° 1c = π 180

ius rad 1 radian

An arc length of one radian is 1c.

These formulas may be used to convert degrees to radians and vice versa. WorkeD exaMple 5

Convert the following angles into exact radians. a 120° b 270° using a CAS calculator think

a 1 Multiply 120 by

2

Write/DiSplaY

πc . 180

a 120 = 120 ×

=

2π c 3

=

3π 2

Simplify by dividing through by the highest common factor, 60.

b 1 Ensure your CAS calculator is in radians mode.

πc 180

b 270°

Enter the value of the angle in degrees. 2

Press ENTER to convert to radians.

3

Write the answer.

270 =

3π c 2

WorkeD exaMple 6

Convert the following angles to degrees. 3π c a using a CAS calculator 4 b 0.8π c think

a 1 Ensure your CAS calculator is in degrees mode.

Enter the value of the angle in radians. 262

Maths Quest 11 Mathematical Methods CAS

Write/DiSplaY

3π 4

a

r

x

2

Press ENTER to convert to degrees.

3

Write the answer.

b 1 Multiply 0.8π c by

= 135° 3π c = 135 4

180 . πc

b 0.8π c = 0.8π ×

180 π

2

Cancel out π c.

= 0.8 × 180°

3

Simplify.

= 144°

When working in degrees, all angles should be expressed to the nearest tenth of a degree unless otherwise stated.

Worked Example 7

Use a calculator to convert: a 57.2° to radians, correct to 3 decimal places b 2.75c to the nearest tenth of a degree. Think

Write

a 1 Multiply 57.2 by 2

a 57.2 = 57.2 ×

Evaluate and round off to 3 decimal places.

b 1 Multiply 2.75 by 2

πc . 180

180 . π

πc 180

= 0.998c b 2.75 = 2.75 ×

180 π

= 157.6°

Evaluate to 1 decimal place.

Exact values and radians Since π c = 180°,

πc πc πc πc = 90 = 30 = 45 and = 60 2 6 4 3 So the table of exact values can be written in terms of radian measurements: Angle (θ c)

sin (θ)

cos (θ)

tan (θ)

0

0

1

0

π 6

1 2

3 2

π 4

1 2

=

2 2

1 2

=

π 3

3 2

1 2

π 2

1

0

1 3

2 2

=

3 3

1 3 Undefined

Chapter 6 • Circular functions 263

Other important angles are shown on the circle below.

c 4–

π

45 °

5 —π c 6 15

35 °

0° 12

π

3 — c 4 1

60° –π c 3

c 2π— 3

90° –2

πc

y

c π– 6

0°

° 30

0° 0c

π c 180°

x 360° 2π c 330 1 1π c ° —

0° 21

6

7 —π c 4

c 5π— 3

πc 3— 270° 2

4π c — 3 24

0°

25 °

31 5°

0° 30

5 —π c 4 2

c 7π— 6

( π–6 sectors shaded) π – 6

Our special triangles can be updated to include radians as shown.

30° 2 2

Exercise 6C

π – 4

Radians

45°

1 π– 3

1

1 WE5 Convert the following angles into exact radians. a 30° b 45° d 20° e 50° g 270° h 360° j −225° k −420° 2 WE6 Convert the following angles to degrees. a

2π 3 −π f 6

π 5

b

e 3.6π

3

c f i l

2π 9 −π g 4 c

60°

1

60° 90° 150° 300° d

11π 6

h 0.375π

3 MC The smallest angle measurement listed below is: A 85°

B π

C 1.92c

D

2π 3

E

D

5π 4

E 3.84c

D

π 6

E 135°

4 MC The largest angle measurement listed below is: A −3π

4π 3

B

C 250°

5 MC If sin (a) = cos (a), then a could be equal to: A

π 3

B

π 4

C 50°

2π 5

6 WE7a Use a calculator to convert the following angles to radians, correct to 3 decimal places. a 49° b 78.2° c −125° d 191.4° e 234.6° f 327.5° g 170.25° h 216.8° 264 Maths Quest 11 Mathematical Methods CAS

7 We7b Use a calculator to convert the following radian measurements to degrees, to the nearest tenth

of a degree. a 0.35 e 15.6 8

b 1.47 f −2.71

c 3.142 g 4.06

Find the exact value of each of the following. π π a sin b cos 4 3 e tan

6D

π 4

f

π cos 6

d 0.958 h 8

c tan

π 3

d sin

π 6

g sin

π 2

h cos

π 4

Symmetry

The unit circle can be divided into symmetrical sections, as shown in the diagram below. Relationships between the circular functions — sine, cosine and tangent — can be established, based on these symmetrical properties. For simplicity, assume θ is an acute angle, although the following properties hold for any θ.

interaCtiVitY int-0265 Symmetry

c π – 2

P(π − θ )

1 y

P(θ ) = (cos (θ ), sin (θ )) = (x, y)

π c −x −1

θ

P( π + θ )

−y

0 or 2π c x 1 P(2π −θ )

−1 c 3— π 2

Quadrant 2 By symmetry: sin (π − θ ) = y = sin (θ ) cos (π − θ ) = − x = − cos (θ ) y tan (π − θ ) = − = − tan (θ ) x Quadrant 3 By symmetry: sin (π + θ ) = − y = − sin (θ ) cos (π + θ ) = − x = − cos (θ ) tan (π + θ ) =

− −

y = tan (θ ) x

Quadrant 1 As already seen: sin (θ ) = y cos (θ ) = x y tan (θ ) = x Quadrant 4 By symmetry: sin (2π − θ ) = − y = − sin (θ ) cos (2π − θ ) = x = cos (θ ) −y = − tan (θ ) tan (2π − θ ) = x

Note: An angle measurement is assumed to be in radians unless the degree symbolis given. Provided an angle is expressed as π ± θ or 2π ± θ, the trigonometric function remains the same, only the sign (+ or −) may change. Use this diagram to determine the sign.

S

A

T

C

Chapter 6 • Circular functions

265

WorkeD exaMple 8

a b c d

If sin (θ) = 0.93, find sin (180° + θ). If cos (θ) = 0.44, find cos (360° − θ). If tan (θ) = 1.72, find tan (π − θ). If cos (θ) = 0.83, find cos (π + θ).

tUtorial eles-1420 Worked example 8

think

a 1 Sketch the angle on a unit circle and relate it

Write/DraW

y

a

to the first quadrant. 180° sin (180° + θ ) + θ) (180°

2

By symmetry sin (180 ° + θ ) = −sin (θ ).

3

Replace sin (θ ) with 0.93.

b 1 Sketch the angle on a unit circle and relate

S θ

A θ

T

C

sin (θ ) x

sin (180 ° + θ) = −sin (θ) = −0.93 y

b

it to the first quadrant. S

A θ θ C

T

2

By symmetry cos (360 ° − θ ) = cos (θ ).

3

Replace cos (θ ) with 0.44.

c 1 Sketch the angle on a unit circle and relate

cos (θ ) x cos (360° − θ ) (36 0° −θ )

cos (360 ° − θ) = cos (θ) = 0.44 y

c

it to the first quadrant. π− π

θ

S

A

T

2

By symmetry tan (π − θ ) = −tan (θ ).

3

Replace tan (θ ) with 1.72.

d 1 Sketch the angle on a unit circle and

tan (θ )

θ

θ

x tan (π − θ )

C

tan (π − θ) = −tan (θ) = −1.72 y

d

relate it to the first quadrant. cos (π + θ ) π +θ

266

2

By symmetry cos (π + θ ) = −cos (θ ).

3

Replace cos (θ ) with 0.83.

Maths Quest 11 Mathematical Methods CAS

S

A θ

θ

T

cos (θ ) C

cos (π + θ) = −cos (θ) = −0.83

x

WorkeD exaMple 9

Find the exact value of each of the following. a tan (150 °) b sin (330 °)

tUtorial eles-1421 Worked example 9

7π d tan 6

3π 4

c cos

think

Write/DraW

a 1 Express tan (150 °) as tan (180 − 30 °). 2

a tan (150 °) = tan (180 − 30 °)

Sketch the angle on a unit circle and relate it to the first quadrant.

y

(18

0° −

30

°)

T

3

By symmetry tan (180° − 30 °) = −tan (30 °).

4

Replace tan (30 °) with its exact value,

1

or

3

3 . 3

=

− 3 3

A 30°

T

3

By symmetry sin (360° − 30 °) = −sin (30 °).

4

Replace sin (30 °) with its exact value, 2 .

C

sin (30°) x sin (360° − 30°)

(36

0°

−3

0°)

sin (330 °) = −sin (30 °)

1

=

π 3π c 1 Express cos . as cos π − 2

or

3

y

S

− 1

b sin (330 °) = sin (360 ° − 30 °)

Sketch the angle on a unit circle and relate it to the first quadrant.

4

C

tan (150 °) = −tan (30 °)

b 1 Express sin (330 °) as sin (360° − 30 °). 2

tan (30°) = tan (180° − 30°) x

A 30°

S

−1 2

π 3π = cos π − 4 4

c cos

4

y

π

Sketch the angle on a unit circle and relate it to the first quadrant.

π– 4

−

S

A π – 4

cos ( π–4 ) C

T

x

cos (π − π–4 )

3

π π By symmetry cos π − = − cos . 4 4

4

Replace cos π with its exact value of 4

3π π cos = − cos 4 4 1 2

or

2 . 2

=

− 1 2

or

− 2 2

Chapter 6 • Circular functions

267

7π as tan π + π . d 1 Express tan 6

2

7π = tan π + π 6 6

d tan

6

Sketch the angle on a unit circle and relate it to the first quadrant.

y

π+

3

π π By symmetry tan π + = tan . 6 6

4

π Replace tan with its exact value 6

π– 6

S

A

T

C

π– 6

tan (π–6 ) = tan (π + π–6 ) x

7π π tan = tan 6 6 1 3

=

3 . 3

or

1 3

3 3

or

WorkeD exaMple 10

3π If sin = 0.924 , evaluate each of the following. 8 11π . 8

5π 8

a sin

b sin

think

Write

3π 8π − 3π 5π a 1 Express sin = sin π − . as sin 8

8

8

8

2

3π 3π Using symmetry express sin π − as sin . 8 8

3π = sin 8

3

3π Replace sin with 0.924. 8

= 0.924

3π 11π as sin 8π + 3π = b 1 Express sin sin π + . 8

8

8

268

3

8

11π = sin 8

b sin

π + 3π 8

3π 3π By symmetry express sin π + as − sin . 8 8

3π = − sin 8

4

3π Replace sin with 0.924. 8

= −0.924

Symmetry

1 We 8 If sin (θ °) = 0.63, find: a sin (180 ° − θ ) b sin (180 ° + θ ) 2

2

exercise 6D

DiGital DoC doc-9760 the unit circle

3π 5π = sin π − 8 8

a sin

If cos (θ) = 0.25, find:

a cos (π − θ )

If tan (θ) = 2.1, find:

a tan (2π − θ )

Maths Quest 11 Mathematical Methods CAS

c sin (−θ )

d sin (360 ° − θ ).

b cos (π + θ )

c cos (2π − θ )

d cos (−θ ).

b tan (−θ)

c tan (π + θ )

d tan (π − θ ).

4

Given that sin (a) = 0.3, cos (b) = 0.7 and tan (c) = 0.9, write down the value of each of the following. a sin (180 ° + a) b cos (180 ° − b) c tan (360 ° − c) d sin (−a °) e sin (180 ° − a) f cos (−b °) g cos (360 ° − b) h tan (180 ° − c) i tan (180 ° + c)

5 We9a, b Find the exact value of each of the following. a sin (150 °) b cos (135 °) c tan (240 °) e sin (240 °) f cos (210 °) g tan (120 °) i cos (−60 °) j sin (−135 °) k tan (180 °) 6 We9c, d

5π a cos 4

Find the exact value of each of the following. 5π 7π b sin c tan 4 6

2π 3

d sin

7π 3

f

−π tan 6

g sin

h cos

11π tan 6

j

− 5π sin 4

k cos (π)

l

e cos i

d tan (330 °) h sin (300 °) l sin (270 °)

4π 3

7π 6

−π sin 2

π π π If sin = 0.383, cos = 0.924 and tan = 0.414, evaluate each of the following. 8 8 8 9π 17π 7π a sin b cos c tan 8 8 8

7 We 10

d cos

−π 8

e sin

7π 8

f

15π tan 8

Given that sin (75°) = 0.966, cos (75°) = 0.259 and tan (75°) = 3.732, find the value of each of the following. a sin (105 °) b cos (255 °) c tan (285 °) d sin (255 °) e cos (435 °) f tan (−75 °) 9 If sin (0.7) = 0.644, cos (0.7) = 0.765 and tan (0.7) = 0.842, find the value of each of the following. (Hint: π = 3.142, approximately.) a sin (2.442) b cos (3.842) c tan (5.584) d sin (−0.7) 8

6e

identities

y

An identity is a relationship that holds true for all legitimate values of a variable or variables. For example, a simple identity is x + x = 2x.

P(θ ) 1

the pythagorean identity Consider the right-angled triangle in the unit circle shown. Applying Pythagoras’ theorem to this triangle gives the identity: sin2 (θ ) + cos2 (θ ) = 1

sin (θ)

θ

0

cos (θ )

x

D

the tangent Consider the unit circle on the right. A tangent is drawn at A and extended to the point C, so that OC is an extension of OP. This tangent is called tangent (θ ), which is abbreviated to tan (θ ). Triangles ODP and OAC are similar, because they have their three corresponding angles equal. tan (θ ) sin (θ ) It follows that: = (corresponding sides) 1 cos (θ ) or tan (θ ) =

sin (θ ) cos (θ )

y 1

B P(θ ) C 1

0

sin (θ )

θ

cos (θ )

D 1A

tan (θ ) x

(as mentioned in an earlier section).

Chapter 6 • Circular functions

269

Another relationship between sine and cosine — complementary functions Consider the unit circle shown on the right. The triangles OAB and ODC are congruent because they have all corresponding angles equal and their hypotenuses equal (radius = 1). Therefore all corresponding sides are equal and it follows that: sin (90 ° − θ ) = cos (θ °) = x and cos (90 ° − θ ) = sin (θ °) = y or −1 π sin − θ = cos (θ ) 2

y 1 D x

y

θ

0

π and cos − θ = sin (θ ) 2

C(90° − θ ) 1

θ

1

x

B(θ ) y A1

x

−1

We say that sine and cosine are complementary functions. Though not required for this course, you may like to try to find the complementary function for tangent, that is, tan (90 ° − θ ) = ? Worked Example 11

If sin (θ) = 0.4 and 0° < θ < 90°, find, correct to 3 decimal places: a cos (θ) b tan (θ). Think

Write

a 1 Use the identity sin2 (θ ) + cos2 (θ ) = 1. 2

Substitute 0.4 for sin (θ ).

3

Solve the equation for cos (θ) correct to 3 decimal places.

4

Retain the positive answer only as cosine is positive in the first quadrant.

b 1 Use the identity tan (θ ) =

a sin2 (θ ) + cos2 (θ ) = 1

(0.4)2 + cos2 (θ ) = 1

sin (θ ) . cos (θ )

cos2 (θ ) = 1 − 0.16 = 0.84 cos (θ ) = ± 0.84 = 0.917 or −0.917 For 0° < θ < 90°, cos is positive so cos (θ ) = 0.917. b tan (θ ) =

2

Substitute 0.4 for sin (θ ) and 0.917 for cos (θ ).

3

Calculate the solution correct to 3 decimal places.

sin (θ ) cos (θ )

0.4 0.917 = 0.436 =

Worked Example 12

Find all possible values of sin (θ) if cos (θ) = 0.75. Think 1

Use the identity sin2 (θ ) + cos2 (θ ) = 1.

2

Substitute 0.75 for cos (θ ).

3

Solve the equation for sin (θ ) correct to 3 decimal places.

4

Retain both the positive and negative solutions, since the angle could be in either the first or fourth quadrants.

270 Maths Quest 11 Mathematical Methods CAS

Write

sin2 (θ ) + cos2 (θ ) = 1 sin2 (θ ) + (0.75)2 = 1 sin2 (θ ) = 1 − 0.5625 = 0.4375 sin (θ ) = ± 0.4375 = 0.661 or −0.661

WorkeD exaMple 13

Find a° if 0° < a° < 90° and

a sin (a°) = cos (42°)

think

Write

a sin (a °) = cos (42 °)

a 1 Write the equation. 2

Replace cos (42 °) with sin (90 ° − 42 °) (complementary functions).

sin (a °) = sin (90 ° − 42 °) sin (a °) = sin (48 °) a° = 48° b cos (a °) = sin (73 °)

b 1 Write the equation. 2

b cos (a°) = sin (73°).

Replace sin (73 °) with cos (90 ° − 73 °).

cos (a °) = cos (90 ° − 73 °) cos (a °) = cos (17 °) a° = 17°

WorkeD exaMple 14 2

If 0° < a° < 90° and cos (a°) = 3 , find the exact values of: a sin (a°) b tan (a°) c cos (90 − a°) d sin (180 + a)°. think

tUtorial eles-1422 Worked example 14

Write/DraW

1

Draw a right-angled triangle.

2

Mark in angle a°, its adjacent side (A = 2) and the hypotenuse (H = 3).

a° H=3

A=2

O= 5 3

Use Pythagoras’ theorem to calculate the opposite side (O) to a°.

a 1 Use the right-angled triangle to find

2

Substitute O = 5 and H = 3.

b 1 Use the right-angled triangle to find

2

O . H

O . A

Substitute O = 5 and A = 2.

c 1 Use the identity cos (90 ° − a °) = sin (a °). 2

Substitute sin (a ) =

5 . 3

d 1 Use the symmetry property

sin (180 ° + a °) =

2

−sin

Substitute sin (a) =

(a °).

5 . 3

O2 = 32 − 22 =5 O= 5 a sin (a) =

O H

=

5 3

b tan (a) =

O A

=

5 2

c cos (90 ° − a °) = sin (a °)

5 3

=

d sin (180 ° + a °) = −sin (a °)

=

−

5 3

(Note: The above results could have been obtained using the identities directly.)

Chapter 6 • Circular functions

271

exercise 6e

identities

1 Copy and complete the table below, correct to 3 decimal places:

θ°

30°

81°

129°

193°

260°

−47°

350°

sin2 (θ ) cos2 (θ ) sin2 (θ ) + cos2 (θ ) 2 We11a If sin (θ ) = 0.8 and 0° < θ < 90°, find, correct to 3 decimal places: a cos (θ ) b tan (θ ). 3 We11b If cos (θ ) = 0.3 and 0° < θ < 90°, find, correct to 3 decimal places: a sin (θ ) b tan (θ ). 4 We12 Find all possible values of the following, correct to 3 decimal places. a cos (x °) if sin (x °) = 0.4 b cos (x °) if sin (x °) = −0.7 c sin (x °) if cos (x °) = 0.24 d sin (x °) if cos (x °) = −0.9 5

6

7

Use the diagram at right to find the exact values of:

b sin (x °)

a c c cos (x °).

3

Use the diagram at right to find the exact values of: a b b cos (x) c tan (x).

x

Find the exact values of: 12 a cos (x) if sin (x) = 13 and 90° < x < 180° b sin (x) if cos (x) = c cos (x) if sin (x) =

−3

2 7 x

and x is in the third quadrant

5 −7 25

9 We13 Find a° if 0° ≤ a° ≤ 90° and: a sin (a °) = cos (20 °) b sin (a °) = cos (58 °) d cos (a °) = sin (82 °) e sin (8 °) = cos (a °) g sin (89 °) = cos (a °) h cos (17 °) = sin (a °).

a

36°

c 54° b

c cos (a °) = sin (39 °) f cos (44 °) = sin (a °)

Copy and complete the following table. sin (θ )

0.8

0.28

cos (θ )

0.6

0.96

tan (θ )

272

bc

8

and x is in the fourth quadrant 3π 3 < x < 2π . d sin (x) if cos (x) = 2 and 2 8 MC Examine the diagram at right and answer the following questions. a sin (54 °) is equal to: a cos (54 °) B cos (36 °) C tan (36 °) D sin (36 °) e tan (54 °) b cos (54 °) is equal to: a tan (36 °) B cos (36 °) C tan (54 °) D sin (36 °) e sin (54 °) c tan (36 °) is equal to: cos (36 ) sin (36 ) a B sin (36 °) cos (36 °) C sin (36 ) cos (36 ) D sin (54 °) cos (54 °) e sin (36 °) + cos (36 °) d tan (54 °) is equal to: cos (36 ) cos (54 ) a B sin (54°) − cos (54 °) C sin (36 ) sin (54 ) D sin (54 °) cos (54 °) e sin (36 °) cos (36 °)

10

5

Maths Quest 11 Mathematical Methods CAS

0.77 0.3 3.18

0.573 0.447

1.207

2

0.7

11 We14 a d g j

If 0° < a°, b°, c° < 90° and sin (a ) =

sin (b °) tan (a °) sin (90 ° − a °) sin (180 ° − a °)

b e h k

2 , 5

cos (b ) = 5 , tan (c ) =

tan (b °) sin (c °) cos (90 ° − b °) cos (180 ° + b °)

6F Sine and cosine the graph of y = sin (x)

11 , 5

3

c f i l

find:

cos (a °) cos (c °) sin (90 ° − c °) tan (180 ° + c °).

graphs

To get an idea of the shape of the y = sin (x) graph, we can construct a table of values, plot the points and join them with a smooth line. The values for the table can be generated from the unit circle. You may recall that continuing on the unit circle past 2π, the sine values begin repeating themselves. For example, 9π π π π sin = sin 2π + = sin . We have gone exactly once around the circle from . 4 4 4 4 x y = sin (x)

−π

0

4

−π 2

−π 4

0

π 4

π 2

3π 4

π

5π 4

3π 2

7π 4

2π

−0.71

−1

−0.71

0

0.71

1

0.71

0

−0.71

−1

−0.71

0

−3 π

interaCtiVitY int-0251 Sine and cosine graphs

y ( –2π , 1)

1 0.71 (−π, 0) −π

π ( –4 ,

(0, 0) 0

− π–2 − π–4

(− 3—4π , − 0.71) ( − π–2 , − 1)

–π 4

0.71) –π 2

y = sin (x)

π ( 3— , 0.71) 4

(2π, 0)

(π , 0) π

π ( 5— , −0.71) 4

−0.71

−1

2π

3π — 2

x

( 7—4π , −0.71) ( 3—2π , −1)

You can verify the shape of this curve using a CAS calculator. It can be observed that the curve repeats itself in cycles after an interval of 2π units. Due to this repetition it is called a periodic function; the period is the interval between repetitions. The y period of y = sin (x) is 2π radians (or 360°). When dealing with graphs of circular functions Period such as sin (x), unless otherwise stated, we assume that the units for x are radians. Amplitude The mean position of the graph of y = sin (x) is y = 0 and the maximum and minimum values are x 1 and −1 respectively. The distance from the mean Mean position to the maximum (or minimum) position position is called the amplitude of the periodic function. Period The amplitude of y = sin (x) is 1 unit.

the graph of y = cos (x) As for the graph of y = sin (x), the graph of y = cos (x) can be established by first completing a table of values. x

−π

−3π 4

−π 2

−π 4

0

π 4

π 2

3π 4

π

5π 4

3π 2

7π 4

2π

y = cos (x)

−1

−0.71

0

0.71

1

0.71

0

−0.71

−1

−0.71

0

0.71

0

Chapter 6 • Circular functions

273

When these points are plotted on a set of axes and joined with a smooth curve, the graph looks like the diagram at right. It can be seen that the basic shape of the curve y = cos (x) is the same as that of y = sin (x). It is also a periodic function with: period = 2π amplitude = 1.

y y = cos (x) (−π–4 , 0.71)

1 (0, 1) (π–4 , 0.71)

( π–2 , 0)

(− π–2 , 0) −π (− 3—4π, −0.71)

(−π , −1)

− π–2 − π–4

0

π –

π –

4

2

−0.71 3π −1 ( —4 , −0.71)

(2π , 1)

(7—4π , 0.71)

0.71)

( 3—2π , 0) 3π — 2

π

2π

x

( 5—4π , −0.71)

(π , −1)

π Note: The graph of y = cos (x) is exactly the same as that of y = sin (x) translated units or 90° to 2 the left. In general, the graphs of y = a sin (bx) + c and y = a cos (bx) + c (for a, b, c ∈R) have the following properties: 1. Amplitude = | a | 2π 360 2. Period = (or ) b b 3. Mean position at y = c 4. Maximum = c + a and minimum = c − a π 1 5. When c = 0, x-intercepts for cosine are at x = ± and every 2 period to the left and right of these; 2 b π 1 x-intercepts for sine are at x = ± and every 2 period to the left and right of these. b Note: | a | means the size or magnitude of a, expressed as a positive number, as the amplitude must always be positive. If a < 0, then the resulting graph is a reflection in the x-axis of the graph for which a > 0. For example, the graph of y = −2 sin (x) is the reflection of y = 2 sin (x) in the x-axis (think of the x-axis as a plane or flat mirror). Worked Example 15

State i the period and ii the amplitude of each of the following functions. a y b y = −1.5 sin (4 x) 4

0

3— π 2

3π

x

−4 Think

a 1 The cycle repeats after 3π units. 2

The distance from the mean position to the maximum position is 4 units.

b 1 Write the formula for the period of y = a sin (bx).

Write

a i Period = 3π ii Amplitude = 4

b i Period =

2π where b = 4 b

2

Substitute b = 4.

=

2π 4

3

Simplify.

=

π 2

4

By rule, the amplitude is | a | or | −1.5 |.

274 Maths Quest 11 Mathematical Methods CAS

ii Amplitude = | −1.5 |

= 1.5

WorkeD exaMple 16

Sketch the graphs of the following functions and state i the period and ii the amplitude of each. 2 x x ∈ [0, 12π] a y = 4 cos b y = sin (4 x) x ∈ [0, 2π] 3 3 think

tUtorial eles-1607 Worked example 16

Write/DraW

a 1 Write the formula for the period.

a i Period =

1

2π b 2π

2

Substitute b = 3 .

=

3

Simplify the value of the period.

= 6π

4

The amplitude is the value in front of cos, written as a positive value.

5

Draw a set of axes.

6

Because the amplitude is 4, mark or imagine horizontal guidelines at y = −4 and 4.

7

Sketch one cycle of the graph every period (every 6π) along the x-axis, for x ∈ [0, 12π], showing key x-values.

1 3

ii Amplitude = 4

y 4 0

3— π 2

9— π 2

3π

6π 15—2π 9π

π 21 — 2

12π x

5— π 4

3— π 2

−4

b 1 Write the formula for the period.

b i Period =

2π b

2

Substitute b = 4.

=

2π 4

3

Simplify the value of the period.

=

π 2

4

The amplitude is the value in front of sin, written as a positive value.

5

Draw a set of axes.

6

Because the amplitude is 3 , mark or imagine horizontal guidelines at −2 2 y = 3 and 3 .

2

ii Amplitude =

y 2– 3

0 − 2–3

7

Sketch one cycle of the graph every period (every π2 ) along the x-axis, for x ∈ [0, 2π].

8

Determine the x-intercepts and mark these on the graph.

2 3

–π 4

–π 2

3— π 4

π

7π — 4

2π x

Chapter 6 • Circular functions

275

WorkeD exaMple 17

Sketch the graph of y = cos (2x) − 3, x ∈ [0, 2π], and state: a the period b the amplitude c the maximum and minimum values. think

Write/DiSplaY

a 1 Write the formula for the period.

a Period =

2π b 2π 2

2

Substitute b = 2.

=

3

Simplify the value of the period.

=π

b 1 The amplitude is the value in front of cos,

written as a positive value. 2

3

4

Sketch the graph of y = cos (2x) first. Since the amplitude is 1, mark or imagine horizontal guidelines at y = +1 and y = −1. Sketch one cycle of the graph every period (every π) along the x-axis, for x ∈[0, 2π].

y 1 0

–π 2

−3.

y= The maximum and minimum are at c + a and c − a.

3— π 2

2π x

−1

Because the mean position is now at y = −3, translate y = cos (2x) vertically to this position.

c Maximum = −3 + 1

= −2 Minimum = −3 − 1 = −4 y 1 0 −1 −2 −3 −4 −5

π — 2

WorkeD exaMple 18

Sketch the graph of the following function. π f: [−2, 4] → R, f (x) = 2 cos x 2 think

276

π

Determine the x-intercepts and mark these on the graph.

c 1 The mean position of y = cos (2x) − 3 is at

2

b Amplitude = 1

Write/DraW

2π π , where b = . 2 b

1

The period of the function is

2

Simplify the value of the period.

3

The amplitude is 2.

4

Draw a set of axes.

5

Using an interval of 4 (the period), mark the x-axis from −2 to 4 (the specified domain).

Maths Quest 11 Mathematical Methods CAS

Period =

2π π 2

=4 Amplitude = 2

π

3— π 2

2π x

6

7

8

9

10

Show −2 and 2 on the y-axis, since the amplitude is 2.

πx y = 2 cos — 2 (4, 2)

( )

y 2

Visualise or check the general shape of the graph of y = a cos (bx).

−2

Starting from the point (0, 2), complete one cycle of the cosine function forward to (4, 2).

−1

0

1

2

3

x

4

−2

(−2, −2)

Complete half of a cycle back from (0, 2) to (−2, 2). The x-intercepts are halfway between the maximum and minimum points.

Sine and cosine graphs

exercise 6F

State i the period and ii the amplitude for each of the following functions.

1 We15a a y

b

1

2 0

2π

4π

x

x

3— π 2

−1.5

x

π

y 3 − –3π

0

−2 h

y 2.5 0

− –2π

π

−2.5

DiGital DoCS doc-9762 Sine graphs doc-9763 Cosine graphs

3π x

0

f

0

−π

−4 g

x

y 2

e

0

5— π 4

–π 4

−1

y 4

y 1.5

0

−2 d

c

y

π

x

π

2π

−3

y 0.5

y

i

8 6

x

0

–π 6

–π 2

x

4 2

−0.5 −2π

−π

0

x

2 We15b State i the period and ii the amplitude of each of the following functions. a y = sin (x) b y = 3 sin (x) c y = 2 sin (2x) d y = −4 sin (3x) g y = 0.4 cos

j

x +1 3

y = −sin (6x) + 4

e y=

1 2

x sin 3

h y = −3 cos (5x)

k y=

1 5

sin (π x )

f i

l

y = 2 cos (4x)

πx y = 2.5 cos 4 πx cos 2 y= 4

Sketch the graph of the following functions, showing one complete cycle, and state i the period and ii the amplitude of each. x a y = sin (2x) b y = 2 cos (x) c y = 3 sin 2

3 We16

1

d y = 4 cos (2x)

e y = 2 sin (3x)

x g y = 5 sin

h y = − 4 cos

3 j y = −3 sin (2x)

x 2

2 3

x cos 2

f

y=

i

y = 2 cos (4x)

Chapter 6 • Circular functions

277

4 MC Parts a to c refer to the graph shown below. y 3

0

x

–π 2

−3 a The amplitude of the function is:

π 2

a

B −3

C π

D 3

e 6

D 6

e 3

b The period of the function is:

π C π 2 c The equation of the function could be: a y = 3 sin (2x) B y = 3 sin (x) x D y = −3 sin (2x) e y = 3 sin 2 a 2π

B

C y = 3 cos (2x)

y

5 MC The equation of the curve at right could be:

x 2 x D y = cos 2

a y = 2 cos (x)

B y = 2 sin

C y = −2 cos (x) e

x y = − 2 cos 2

2

0

π

2π

−2

State the equation of each of the functions graphed below.

6

y

a

y

b

1.5

2 3π x

0 −1.5

0

−π–4

π – 4

x

3π — 4

−2 y

c

d

0

−2π

4π x

2π

0

y

−1

x

y

f

1 0

3π — 2

−4

−5 e

y 4

5

3 2π — 3

4π — 3

x − π– 6

0

−3

278

3π

Maths Quest 11 Mathematical Methods CAS

π– 6

x

x

7 We17 Sketch the graph of each of the following functions. b f (x) = 3 cos x for x ∈ [−4π, 4π]

a f (x) = cos (2x) for x ∈ [−2π, 2π] c f (x) = 2 sin (3x) for x ∈ [0, 2π]

d f (x) =

e f (x) = 1.5 sin x for x ∈ [−6π, 6π] 3

f

−3

2 cos (2x) for x ∈ [0, 2π]

f (x) = −4 sin (4x) for x ∈ [0, π]

8 We17 Sketch the graph of the following functions and state i the period, ii the amplitude, and iii the

maximum and minimum values of each. a y = cos (2x) + 1 x ∈ [0, 2π] x c y = 2 cos + 3 x ∈ [0, 2π] 2

b y = 2 sin (x) − 2

x ∈ [−2π, 2π]

x d y = sin – 1

x ∈ [0, 6π]

3

9 We18 Sketch the graphs of each of the following functions. a f : [−π, 2π] → R, f (x) =

3 2

2x sin 3

b f : [−6, 6] → R, f (x) = 1.8 cos

πx 3

3π x d f : [0, 8] → R, f (x) = −3 cos

c f : [0, 6] → R, f (x) = −1.4 sin π x

4

2

10 For each of the functions graphed below, state the rule using full function notation. a

b

y 3

y f(x)

5– 2

f(x) 0

8— π 6

0

x

−3

c

5π x

− 5– 2

d

y

y

2

1.8 f(x) 0

−1

f(x)

x

1

−1

−2

e

f

y

y

3

f(x)

f(x) x

− 1–3 0

y 0

−3

1

x

−2.4

−3

−2

x

2.4

0

−1

3

−1.8

3

g

0

y

h — 2

π

3— 2

2π

x

1 −π

−—2 −10 −2 −3 −4

— 2

π

2π x DiGital DoC doc-9764 WorkSHEET 6.2

Chapter 6 • Circular functions

279

6G Tangent graphs The graph of y = tan (x) To manually plot the graph of y = tan (x), the following table of values may be used. x y = tan (x)

−π

0

−3π

−π 2

−π 4

0

π 4

π 2

3π 4

π

5π 4

3π 2

7π 4

2π

Undefined

−1

0

1

Undefined

−1

0

1

Undefined

−1

0

4

1 π

Note: Multiples of 8 could be used to give more points to plot and a clearer indication of the shape of the graph. sin ( x ) Note the presence of some undefined y-values. This is because y = and cos (x) = 0 at these cos (x) values. These undefined values are shown as vertical asymptotes at the given value of x for which they occur. (An asymptote is a line that a graph approaches, but never quite reaches. In the case of y = tan (x), y approaches (but never actually reaches) −∞ and +∞ for particular x-values.) The graph of y = tan (x) is shown in the figure below. Vertical asymptotes y y = tan (x) 5π 3π (− — , 1) ( π–4 , 1) ( —4, 1) 4 1 (π , 0) (−π, 0) (0, 0) 0 π π π π −π π – – − –2 − –4 4 2 −1 ( 3—4π, −1)

(2 π , 0) 2π −1)

π 3— 2

x

( 7—4π,

The features of the graph of y = tan (x) are as follows: − π π 3π 1. It has vertical asymptotes through x = … , , ,… 2 2 2 2. It has no amplitude.

3. It has a period of π. 4. It has a range of R (the set of all real numbers). In general, the graph of y = a tan (bx) + c has the following properties. y

3π −— 2b

− π–b

π −— 2b

0

π — 2b

π– b

π 3— 2b

x

1. No amplitude π 2. Period = . b π 3. When c = 0, the graph has x-intercepts at x = ± and every period to the left and right of these. b −π π 4. It has asymptotes at x = and x = and every period to the left and right of these. 2b 2b (2n + 1)π 5. The following formula for asymptotes applies: xasymptote = ± , where n = 0, 1, 2, . . . 2b 6. Mean position at y = c. The graph is translated vertically to c. 280 Maths Quest 11 Mathematical Methods CAS

WorkeD exaMple 19

For each function below, state i the period and ii the equation of the two asymptotes closest to the y-axis. 1 x + 3 a y = 2 tan (4 x) b y = 3 tan 2 think

Write

a 1 Write the formula for the period.

a i Period =

π b

=

π 4

2

Substitute b = 4.

3

Write the formula to find the two closest asymptotes to the y-axis.

4

Substitute b = 4.

x=±

π 2×4

5

Simplify.

x=±

π 8

b 1 Write the formula for the period.

ii Asymptotes: x = ±

b i Period =

1 . 2

2

Substitute b =

=

3

Simplify.

4

Write the formula to find the two closest asymptotes to the y-axis.

5

Substitute b =

6

Simplify.

π 2b

π b π 1 2

= 2π

1 . 2

ii Asymptotes: x = ±

x=±

π 2b

π 1 2× 2

x=±π

WorkeD exaMple 20

π x over [−2, 4] without a Sketch the graph of the function y = 3 tan 2 CAS calculator. think

tUtorial eles-1423 Worked example 20

Write

π . 2

π π 2 =2

1

Find the period using b =

2

Find the two asymptotes closest to the y-axis π π by substituting b = into x = ± . 2 2b

Asymptotes: x = ±

State the other asymptotes by adding/subtracting the period, 2 units, to/from the first asymptotes, x = ± 1.

Other asymptotes are: x = −1 − 2, x = 1 + 2, x = 1 + 2 + 2 etc.

3

Period =

= ±1

π 2×

π 2

Chapter 6 • Circular functions

281

4

State all asymptotes in the domain [−2, 4].

5

Evaluate y when x = 2 and x = 2 to establish two definite points on the graph.

6

Draw a set of axes using x ∈ [−2, 4].

7

8

9

1

π 1 When x = 2 , y = 3 tan 4 =3×1 =3 −π −1 When x = 2 , y = 3 tan 4 = 3 × −1 = −3

−1

πx y = 3 tan — 2

( )

y

Mark in the vertical asymptotes at x = −1, x = 1, x = 3.

3

Using the asymptotes as a guide, sketch the standard tan curve.

0

1

−2 −1 − –2

1– 2

1

2

3

4

x

−3

Verify that this graph is correct by using a CAS calculator.

tangent graphs

exercise 6G DiGital DoC doc-9765 tangent graphs

All asymptotes in the domain [−2, 4] are: x = −1, x = 1, x = 3.

State i the period and ii the equation of the two asymptotes closest to the y-axis for each of the following. x x a y = tan (2x) b y = tan (3x) c y = tan d y = tan 3 2 x x e y = 3 tan (x) f y = 2 tan g y = − tan h y = −2 tan (2x) 4 2 i y = tan (x) − 1 j y = tan (2x) − 3

1 We19

2 Sketch the graph of each function in question 1, showing the first two cycles. 3 We20

The function y = 3 tan (4x) has a period equal to: π π B C π D 4 2

MC

a 4π

4 MC Use the graph to answer questions a and b. a The period of the function is equal to:

π 4 π C 2 e 3 b The equation of the function is: a

a y = 1.5 tan (x) C y = 1.5 tan (2x) e y = 3 tan (x)

B π D 2π

y

1.5 0

x 2 D y = 1.5 tan (3x)

e 2π

π – 2

π

2π

B y = 1.5 tan

5 We20 Sketch the graphs of each of the following functions over the given domain. a y = 3 tan DiGital DoC doc-9766 Investigation tangent graphs

282

x , [0, 2π] 2 2π x , [0, 5] 5

d y = 1.8 tan

Maths Quest 11 Mathematical Methods CAS

b y = tan

π x , [0, 3] 3

e y = −2 tan (π x), [−1, 1]

c y = 2 tan

x , [0, 2π] 3

3π

x

6H

Solving trigonometric equations

To find the solution to the equation sin ( x ) =

1

2

where x ∈ [0, 4π], we can consider the graph of y = sin (x).

y 1 1– 2

0

π – 2

π 3— 2

π

2π

π 5— 2

3π

π 7— 2

4π

x

y = 8 sin (θ )

−1 1

π

By drawing a horizontal line through y = , it can be seen that there are four solutions in the domain 2 [0, 4π]. π π The solution for 0 < x < , that is, in the first quadrant, is (from our knowledge of exact values). 2 4 Note: For inexact solutions in the first quadrant, use a calculator. The sine function is also positive in the second quadrant. y π 3π Using sine symmetry, the next solution is π − = 4 4 (different symmetry properties are used for cosine and sin ( π–4 ) sin (π − π–4 ) S A tangent). Since the graph is periodic, any further solutions are π – 4 found by adding (or subtracting) the period (in this case 2π) to x (or from) each of the first two solutions. T C For example, two further solutions are: 3π π + 2π and + 2π 4 4 9π 11 π = and 4 4 π 3π 9π 11π Therefore, four solutions in the specified domain are , , and . 4 4 4 4 However, if a domain is not specified, there are an infinite number of solutions, as multiples of 2π π 3π can be added (or subtracted) indefinitely to (or from) and . In this situation a general solution is 4 4 obtained where the solutions are in terms of a parameter, n, where n is an integer, i.e. n ∈ Z. π π The general solution for the first quadrant solution x = becomes x = 2nπ + where n ∈ Z. 4 4 3π 3π becomes x = 2nπ + The general solution for the second quadrant solution x = where n ∈ Z. 4 4 π π 3π Note that the general solution x = 2nπ + can be expressed as x = 2nπ + π − = (2n + 1)π − . 4 4 4 Substituting different integer values of n will give specific solutions as shown in the table below. π– 4

−

n

x = 2nπ +

π , n ∈Z 4

x = 2nπ +

3π , n ∈Z 4

−1

x = − 2π +

π − 7π = 4 4

x = − 2π +

3π − 5π = 4 4

0

x = 0+

1

x = 2π +

2

x = 4π +

π π = 4 4 π 9π = 4 4

π 17π = and so on 4 4

x = 0+

3π 3π = 4 4

x = 2π +

3 π 11π = 4 4

x = 4π +

3 π 19π = and so on 4 4

Chapter 6 • Circular functions 283

In general, if: − − • sin (x) = a, then x = 2nπ + sin 1 (a) and x = (2n + 1)π − sin 1 (a) where a ∈ [−1, 1] and n ∈ Z. −1 − • cos (x) = a, then x = 2nπ ± cos (a) where a ∈ [ 1, 1] and n ∈ Z − • tan (x) = a, then x = nπ + tan 1 (a) where a ∈ R and n ∈ Z. When solving trigonometric equations, the following need to be determined: 1. The first quadrant angle, irrespective of the sign 2. The two quadrants in which the given function is positive or negative 3. Two solutions between x = 0 and x = 2π (use the appropriate sine, cosine or tangent symmetry property). If more solutions are required: 1. Repeatedly add (or subtract) the period to the two solutions as many times as required, noting solutions after each addition or subtraction. 2. Stop when all solutions within the specified domain are found. 3. If no domain is given, a general solution is required. WorkeD exaMple 21

Find θ to the nearest tenth of a degree if cos (θ °) = −0.58, given that θ ∈ [0°, 360°]. think

Write/DraW

Write the equation.

cos (θ °) = −0.58

2

Find the first quadrant angle by solving cos (θ °) = 0.58 to the nearest tenth of a degree.

First quadrant angle = cos−1 (0.58) = 54.5°

3

Identify where the cosine function is negative.

Cosine is negative in quadrants 2 and 3.

4

Use cosine symmetry to find the solutions.

(1

1

80

y

− )°

.5

54 54.5° −0.58 + 0.58

180°

x

(1

80

+

54

.5

)°

(−0.58)

5

Simplify the solutions.

6

Since the period is 360°, no further solutions are required for θ ° ∈ [0°, 360°].

θ = (180 − 54.5)° or (180 + 54.5)° = 125.5° or 234.5°

WorkeD exaMple 22

Find the value of x in each of the following equations if x ∈ [0, 4π]. Give answers correct to 3 decimal places, unless exact answers may be found. a sin (x) = 0.3 −1 b cos ( x) = 2 think

a 1 Write the equation.

284

Write

a sin (x) = 0.3

x ∈ [0, 4π]

2

Use a CAS calculator to find the first quadrant angle (since no exact answer exists). Ensure your CAS calculator is set to radians.

First quadrant angle = sin−1 (0.3) = 0.305

3

Identify where sine is positive.

Sine is positive in quadrants 1 and 2.

Maths Quest 11 Mathematical Methods CAS

4

y

Use sine symmetry to find the two solutions for x ∈ [0, 2π]. 2.83

c

.305 0.3 x

7c

0.3

For x ∈ [0, 2π] x = 0.305 or (π − 0.305) = 0.305 or 2.837 5

Convert the specified domain to a decimal.

[0, 4π] = [0, 12.566]

6

The period of y = sin (x) is 2π, so add 2π (= 6.283) to each of the solutions above.

For x ∈ [0, 4π] x = 0.305, 2.837, (0.305 + 6.283), (2.837 + 6.283)

7

Adding 2π (= 6.283) to the last two solutions would give solutions beyond the specified domain, so stop here.

x = 0.305, 2.837, 6.588, 9.120

b 1 Write the equation.

b cos( x ) =

−1 2

x ∈ [0, 4π] −1 1 2

Recall a special triangle to find the first quadrant angle. Note: This is not a solution in this case.

First quadratic angle = cos

3

Identify where cosine is negative.

Cosine is negative in quadrants 2 and 3.

4

Use cosine symmetry to find the two solutions for x ∈ [0, 2π].

2

=

π 3

y –π 3

2— π 3 1 −— 2

1 — 2

x

4— π 3

For x ∈ [0, 2π] π π x = π − or π + 3 3 2π 4π = or 3 3 5

The period of y = cos (x) is 2π, so add 2π to each of the solutions above for x ∈ [0, 4π].

6

Adding 2π (=

6π ) to the last two solutions 3 would give solutions beyond the specified 12π domain (4π or ), so stop here. 3

For x ∈ [0, 4π] 2π 4π 2π 4π x= + 2π , + 2π , , 3 3 3 3 2π 4π 2π 6π 4π 6π = + + , , , 3 3 3 3 3 3 x=

2π 4π 8π 10π , , , 3 3 3 3

Chapter 6 • Circular functions 285

WorkeD exaMple 23

Find solutions to 2 sin (x) = 0.984 over the domain [0, 2π]. think

Write/DraW

x ∈ [0, 2π]

2 sin (x) = 0.984

1

Write the equation.

2

Divide both sides by 2 to get sin (x) by itself.

3

Determine the first quadrant angle in radians, correct to 3 decimal places.

First quadrant angle = sin−1 (0.492) = 0.514

4

Identify where the sine is positive.

Sine is positive in quadrants 1 and 2.

5

Use sine symmetry to find the solutions.

sin (x) = 0.492

y

(π −

0.51

4) c 0.492 π

0.492 x 0.514c

x = 0.514 or (π − 0.514) 6

Simplify the solutions.

7

Since the period is 2π, no further solutions are required over the domain [0, 2π].

= 0.514 or 2.628

Note: Any equation not in the form sin (A) = B (or cos or tan) should be transposed before the solutions are found.

WorkeD exaMple 24

Find the general solution of the following equations. 1 a cos ( x) = b 2 sin (x) − 1 = 0 c tan ( x) = 3 2 Hence, find all the solutions for −2π ≤ x ≤ 2π for each equation. think

a 1 Write the general solution for

cos (x) = a.

286

tUtorial eles-1424 Worked example 24

Write −1

a x = 2nπ ± cos

(a)

2

Substitute a =

1 into the general equation 2 − 1 , recognising that it and evaluate cos 1 2 is an exact angle.

− 1 x = 2nπ ± cos 1 2 π x = 2nπ ± 4

3

Write the two separate solutions and specify n ∈ Z.

x = 2nπ +

4

Substitute n = −1, n = 0 and n = 1 into each of the general solutions.

n = − 1: x =

Maths Quest 11 Mathematical Methods CAS

π π and x = 2nπ − , n ∈ Z 4 4 − 9π

and x =

−

7π 4

4 −π π n = 0: x = and x = 4 4 π 7π π 9π n = 1: x = 2π − = and x = 2π + = 4 4 4 4

5

Write the solutions for −2π ≤ x ≤ 2π.

b 1 Rearrange the equation and substitute

1 a = 2 into the general equations. Evaluate − 1 sin 1 ( 2 ), recognising that it is an exact angle.

x=

−

7π − π π 7π , , , 4 4 4 4 1 2

b sin ( x ) =

−1

x = 2nπ + sin x = 2nπ

Specify n ∈ Z.

(a) and x = (2n + 1)π − sin

− 1 + sin 1 ( 2 )

and x = (2n + 1)π

−1

(a)

− 1 − sin 1 ( 2 )

π π and x = (2n + 1)π − or 6 6 5π x = 2nπ + , where n ∈ Z 6 x = 2nπ +

2

Substitute n = −1 and n = 0 into each of the general solutions.

n = − 1: x = − 2π +

π − 11π = and 6 6

π − 7π = 6 6 π π 5π n = 0: x = and x = π − = 6 6 6 x = −π −

3

Write the solutions for −2π ≤ x ≤ 2π.

c 1 Write the general solution for tan x = a.

x=

−

11π − 7π π 5π , , , 6 6 6 6

c x = nπ + tan

−1

(a)

−

x = nπ + tan = ( 3) π x = nπ + , n ∈ Z 3 1

2

n = − 2 : x = − 2π +

Substitute n = −2, n = −1, n = 0 and n = 1 into each of the general solutions.

−5

π 3

=

n = − 1: x = − π + = n = 0: x =

−

π 3

3

Write the solutions for −2π ≤ x ≤ 2π.

− 5π

3

,

−

π 3

2π π 4π , , 3 3 3

Solving trigonometric equations

exercise 6h 1

x=

4π 3

π 3

2π 3

n = 1: x = π + =

π 3

Find the exact value of x in terms of π in each of the following equations, given that x is in the first quadrant. a cos ( x ) =

1 2

b sin ( x ) =

c tan ( x ) =

3

d cos ( x ) =

3 2

sin ( x ) =

3 2

e tan (x) = 1

f

1 2

DiGital DoC doc-9767 trigonometric equations

Chapter 6 • Circular functions

287

2 We21 Find the value of θ to the nearest tenth of a degree in each of the following equations, given

that θ ∈ [0°, 360°].

b cos (θ) = −0.25 e cos (θ) = 0.195 h cos (θ) = 0.757

a sin (θ) = 0.6 d sin (θ) = −0.85 g sin (θ) = −0.333

c tan (θ) = 5.72 f tan (θ) = −0.837

3 We22 Find the value of x in each of the following equations if x ∈ [0, 4π]. Give answers correct to

3 decimal places, unless exact answers are found. a sin (x) = 0.8 b cos (x) = −0.5 d tan ( x ) =

−1

e sin (x) =

3

c tan (x) = 1.5

−0.39

cos ( x ) =

f

− 2 2

2 2

and 0° < x < 360°, then x is equal to: a 150° or 210° B 135° or 225° C 225° D 135° or 315° e 120° π b If 3 tan (x) = 1 and 0 < x < , then x is equal to: 2 π π π π π a B C D e 6 2 3 4 5 5 We23 Find solutions, exact where possible, to the following over the domain [0, 2π]. a 2 sin (x) = 0.586 b 2 cos (x) = 1 c 2 tan ( x ) = 2 3 d 2 sin (x) + 0.893 = 0 3 tan ( x ) + 1 = 0 e 1.8 cos (x) + 1.236 = 0 f 4 MC a If sin ( x ) = cos ( x ) =

The solution to the equation 2 cos (x) + 1 = 0 over the domain [0, 2π] is: 5π 7π 2π 4 π π 2π π 5π a B C D , , , , 6 6 3 3 3 3 6 6

6 MC

e

4π 5π , 3 3

7 Solve each of the following, to the nearest tenth of a degree, over the domain [0°, 360°]. a 4 sin (x) = 1 b 3 cos (x) = −2 c 2 tan (x) − 7 = 0 d 4 + sin (x) = 3 e 1 + 2 cos (x) =

2

f

3 tan (x) + 9 = 0

8 Solve the following equations over [0°, 360°]. a sin (x + 30)° = 1 c tan (x + 45)° = 1

b cos (x − 60)° = 0 d 2 sin (x − 60)° = 1

2 cos (x + 90)° + 1 = 0 x =1 g 2 cos 2

f

e

i

3 tan (2 x ) = 1

k tan (2 ( x − 30) ) = 9 We24

3

sin (2x) = 0 x h 2 sin = 2 2 2 cos (3 x ) = − 1 j l

sin

(

1 2

)

( x − 40) = 0.6

Find the general solution of the following equations.

a 2 cos ( x ) − 3 = 0

1 3 2 sin ( x ) − 1 = 0 .

b tan ( x ) = c

Hence, find all solutions for −2π ≤ x ≤ 2π for each equation. 10 Find the general solution of the equation 2 sin (2x) − 1 = 0. Hence, find all solutions for −π ≤ x ≤ π. 11 Find the general solution of the equation 2 cos (3x) − 1 = 0. Hence, find all solutions for −π ≤ x ≤ π.

6i

applications

Many situations arise in science and nature where relationships between two variables exhibit periodic behaviour. Tide heights, sound waves, biorhythms and ovulation cycles are examples. In these situations trigonometric functions can be used to model the behaviour of the variables. The independent variable (x) is often a measurement such as time. When modelling with trigonometric functions you should work in radians unless otherwise instructed. 288

Maths Quest 11 Mathematical Methods CAS

WorkeD exaMple 25

E. coli is a type of bacterium. Its concentration, P parts per million (ppm), at a particular beach over a 12-hour period t hours after 6 am is described by the function πt P = 0.05 sin + 0.1. 12 a Find i the maximum and ii the minimum E. coli levels at this beach. b What is the level at 3 pm? c How long is the level above 0.125 ppm during the first 12 hours after 6 pm?

tUtorial eles-1425 Worked example 25

think

a Write the function.

Write/DraW

a P = 0.05sin

i 1 The maximum value of the sine function

π t + 0.1 12

i The maximum P occurs when sin

is 1. 2

3

πt Substitute sin = 1 into the equation 12 for P and evaluate.

Max. P = 0.05(1) + 0.1 = 0.15

State the solution.

The maximum E. coli level is 0.15 ppm.

ii 1 The minimum value of the sine function

ii The minimum P occurs when sin

is −1.

b

π t = 1. 12

π t = − 1. 12

2

πt Substitute sin = − 1 into the equation 12 for P and evaluate.

Min. P = 0.05(−1) + 0.1 = 0.05

3

State the solution.

The minimum E. coli level is 0.05 ppm.

1

At 3 pm it is 9 hours since 6 am.

2

Substitute t = 9 into the equation for P and evaluate.

b

At 3 pm, t = 9. 9π When t = 9, P = 0.05 sin + 0.1 12 3π = 0.05 sin + 0.1 4 1 + 0.1 2 = 0.035 + 0.1 = 0.135 = 0.05

c

3

State the solution.

The E. coli level at 3 pm is approximately 0.135 ppm.

1

A sketch graph will give a better understanding of this question.

2

State the amplitude.

Amplitude = 0.05

3

Calculate the period.

Period = 2π π 12 Period = 24

c

Chapter 6 • Circular functions

289

πt The basic graph is P = 0.05 sin . 12

4

Identify the basic graph.

5

State the translations needed.

6

Sketch the graph of P.

7

Draw a horizontal line through P = 0.125.

No horizontal translation is needed; the vertical translation is 0.1 units up. P (parts per million) 0.150 0.125 0.100 0.050 0

2

4

6

8

12 t (hours)

8

Identify where P > 0.125 from the graph.

The graph shows that P > 0.125 between the first two points where P = 0.125.

9

Solve the equation P = 0.125 to find the first two values of t.

When P = 0.125, πt 0.05 sin + 0.1 = 0.125 12 πt 0.05 sin = 0.025 12 πt sin = 0.5 12 πt π π = or π − 12 6 6 π 5π = or 6 6 t 1 5 = or 12 6 6 t = 2 or 10

10

Find the difference between the solutions t = 2 and t = 10.

P > 0.125, for 10 − 2 = 8 hours

11

State the solution.

The E. coli level is above 0.125 parts per million for 8 hours.

exercise 6i

applications

1 The weight of a rabbit over a period of time is modelled by the graph shown below. W (kg) 4 3 2 1 0

a State i the amplitude and ii the period. b Express W as a function of t. 290

10

Maths Quest 11 Mathematical Methods CAS

3

6

t (days)

2 The graph below shows the heart rate of an athlete during a particular hour of a workout. H (beats/min) 160 110 60 0

15

30

45

60

t (min)

a Find the initial heart rate. b State i the amplitude and ii the period. c Express H as a function of t. 3 WE25 The height above the ground, h metres, of a child on a swing at any time, t seconds, after being

released is:

πt h = 1 + 0.6 cos 2

Find: a the maximum height of the swing b the height after 4 i 3 seconds and ii seconds 3 c the length of time that the swing is below 1.5 metres, travelling from one side to the other.

4 The temperature, T (°C), inside a building on a given day is given by the function:

πt T = 8 sin + 18 12 where t is the number of hours after 8 am. a What is the maximum temperature in the building and the time at which it first occurs? b Find the temperature at i 8 pm, ii 6 pm and iii 12 am (midnight). 5 The displacement, x (in mm), of a harp string t seconds after it is initially plucked is modelled by the

function: x (t) = 12 sin (20π t) a b c d

What is i the amplitude and ii the period of this function? How many vibrations (that is, cycles) will it complete in one second? Find the displacement after 0.08 seconds. At what time will its displacement first be 6 mm? Chapter 6 • Circular functions 291

6 The height of a bungee jumper, h metres, above a pool of water at any time, t seconds, after jumping is

described by the function: h (t) = 20 cos (0.8t) + 20 a What is the initial height of the bungee jumper? b When, if at all, does the bungee jumper first touch the water? c Assuming the cord is perfectly elastic, how long is it until the bungee jumper returns to the lowest

position?

7 A cyclist rides one lap of a circular track at a constant speed so that her distance, d metres, from her

starting point at any time, t seconds, after starting is:

πt d = 50 − 50 cos 30 Find: a the time taken to complete one lap b the radius of the track c the maximum distance from the start d the length of the track e the distance from her starting point after i 15 seconds and ii 40 seconds f the times at which she is 93.3 metres from her starting points. 8 The depth of water, d metres, at a port entrance is given by the function:

πt d (t ) = 4.5 + 1.5 sin 12 where t is in hours. a Find i the maximum and ii the minimum depth at the port entrance. b A certain ship needs the depth at the port entrance to be more than 5 metres. The ship can be loaded and unloaded, and in and out of the port, in 9 hours. Assuming that the ship enters the port just as the depth at the entrance passes 5 metres, will the ship be able to exit 9 hours later? How long will it have to spare, or by how many minutes will it miss out? 9 The Australian dollar’s value (a) in US dollars was observed to follow the equation

a = 0.9 + 0.01t + 0.02 sin (0.5π t) over a period of 8 days (t represents the number of days). Using a CAS calculator, sketch a graph of a for 0 ≤ t ≤ 8. When will the Australian dollar first reach US$0.95? At what other times will it be worth US$0.95? Find all of the maximum turning points of the graph (to 3 decimal places). What is the highest value reached?

a b c d e

292 Maths Quest 11 Mathematical Methods CAS

Summary trigonometric ratio revision

• SOH CAH TOA O A cos (θ ) = • sin (θ ) = H H

tan (θ ) =

O A

H

θ

• Angles of elevation and depression:

O

A

n

tio leva of e

gle Horizontal

An

gle

An the unit circle

n

sio

res

ep of d

–π 2

• sin (θ) = y • cos (θ) = x

90° y 1st quadrant P(θ )

2nd quadrant S

A

T

θ cos (θ ) C

π , 180°

sin (θ ) x 0, 360°, 2π

3rd quadrant

4th quadrant 270° 3π — 2

• tan (θ) = the height of the point where the angle line meets the tangent line (at the RHS of the unit circle)

tan (θ )

y

θ

• tan (θ ) = radians

x

y x

• π c = 180° 180 • 1c = π c π • 1 = 180

Chapter 6 • Circular functions

293

π– 6

• For exact values use special triangles:

30° 2 2 π– 4

Angle (θ) 0 π or 30° 6 π or 45° 4 π or 60° 3 π or 90° 2

45°

cos (θ) 1

1 2

3 2

1 2

=

2 2

1 2

=

3 2

1 2

1

0

60°

π– 3

1

sin (θ) 0

3

1 1

tan (θ) 0 1 3

2 2

=

3 3

1 3 Undefined

Symmetry

• Relating a given angle back to the first quadrant of the unit circle is often helpful. Provided an angle is expressed as π ± θ or 2π ± θ, the trigonometric function (sin or cos) remains the same, 3π 4π 6π 8π only the sign (+ or −) may change. So π may be written as etc. , , , 3 4 6 8 • sin (π + θ) = −sin (θ) • sin (2π − θ) = −sin (θ) • sin (π − θ) = sin (θ) − − • cos (π − θ) = cos (θ) • cos (π + θ) = cos (θ) • cos (2π − θ) = cos (θ) • tan (π − θ) = −tan (θ) • tan (π + θ) = tan (θ) • tan (2π − θ) = −tan (θ)

identities

• sin2 (θ) + cos2 (θ) = 1 sin (θ ) • tan (θ ) = cos (θ ) π • sin (90 ° − θ °) = cos (θ °) or sin − θ = cos (θ °) 2

π • cos (90 ° − θ °) = sin (θ °) or cos − θ = sin (θ °) 2 Sine and cosine graphs

• Basic graph types: 1. y = a sin (bx) + c y = a cos (bx) + c 2π 2. Period = , amplitude = | a | b y a 0 −a

294

Maths Quest 11 Mathematical Methods CAS

y y = a sin (bx)

–π b

y = a cos (bx)

a

2— π b

x = period

0 −a

π — 2b

–π b

3— π 2b

2— π b

x

• When sketching: 1. Recall the basic graph type. 2. Find the period and amplitude. 3. Sketch in sections of one period. 4. Find x-intercepts between other known intercepts or minimum and maximum points. 5. Translate the graph vertically by c. tangent graphs

• y = a tan (bx) + c π • Period = , no amplitude b π • Asymptotes at x = ± and every period to the 2b left and right of these • The formula for asymptotes applies: (2n + 1)π xasymptote = ± , 2b where n = 0, 1, 2, . . .

y

3π −— 2b

− π–b

π −— 2b

0

π — 2b

π– b

x

π 3— 2b

Solving trigonometric equations

• When solving trigonometric equations, the following need to be determined: 1. The first quadrant angle, irrespective of the sign 2. The two quadrants in which the given function is positive or negative 3. Two solutions between x = 0 and x = 2π (use the appropriate sine, cosine or tangent symmetry property). • If more solutions are required: 1. Repeatedly add (or subtract) the period to the two solutions as many times as required, noting solutions after each addition or subtraction. 2. Stop when all solutions within the specified domain are found. 3. If a general solution is required, the answer will be in tems of n, n ∈ Z. In general, if: − − • sin (x) = a, then x = 2nπ + sin 1 (a) and x = (2n + 1)π − sin 1 (a) where a ∈ [−1, 1] and n ∈ Z. −1 • cos (x) = a, then x = 2nπ ± cos (a) where a ∈ [−1, 1] and n ∈ Z − • tan (x) = a, then x = nπ + tan 1 (a) where a ∈ R and n ∈ Z.

applications

• • • •

To find the maximum value of a function, replace sin x or cos x with +1. To find the minimum value of a function, replace sin x or cos x with −1. Initial values occur at t = 0. A sketch graph may provide greater understanding.

Chapter 6 • Circular functions

295

Chapter review S h ort anS Wer

1 The angle of elevation from an observer to an aircraft when it is 4.5 km away is 30°. How high is the

aircraft above the ground if the observer’s eye level is 1.75 m above the ground?

4.5 km

30° 1.75 m 2 Find the exact values of: a tan (120 °) 3 4

3 If cos (θ ) = a sin (θ)

5π . 6

b cos (150 °)

c sin

π

and 0 < θ < 2 , find the exact values of: b tan (θ).

4 Convert the following angles to exact radians. a 140° b 310°

c 1°

5 Convert the following radian measures to degrees. a

2π 5

b 1

c

6 If cos (θ) = 0.69, find: a sin (180 ° − θ °)

7π 2

b cos (2π − θ) c tan (π + θ). π 7 If cos (x) = and < x < π, evaluate: 2 a sin (x) b tan (x). 8 Sketch the graphs of the following functions. x a y = −4 sin over [−2π, 4π] b y = 1.5 cos (2x) over [−π, π] 2 9 Sketch the graph of: −1 x a y = 2 tan (4x) for x ∈ [−π, π] b y= tan for x ∈ [0, 12π]. 6 2 1 10 Find all of the solutions to the equation sin ( x ) = over the domain [0, 3π]. 2 11 Solve the equation 3 + 2 cos (x) = 0 over the domain [0°, 360°]. 3 12 Find the general solution of the equation sin ( x ) = . Hence find all solutions for −2π ≤ x ≤ 2π. 2 13 If cos (θ) = 0.40 and sin (θ) = 0.92, find: a tan (θ) b cos (180° − θ ) c sin (−θ) d cos (90° − θ ). 14 With the aid of a diagram if necessary, find exact values for: π 7π a sin b cos 6 6 −0.9

5π 3

c tan

d cos (π).

15 Solve the following.

3 2 1 c tan ( x ) = 3 e tan (x °) = −1 a sin ( x ) =

296

Maths Quest 11 Mathematical Methods CAS

b cos ( x ) =

−1

d cos (x °) = 1

2

16 The sound level of a siren follows the rule L (t) = 6 sin (π t) + 80, where L is the sound level measured

in decibels (dB) and t is the time in seconds. a What is the amplitude and period of L(t)? b What are the minimum and maximum sound levels of the siren? c How long does it take for the sound to reach its maximum level the first time? d Find the first time the sound level reaches 83 dB. e Sketch the graph of L(t) for 0 ≤ t ≤ 2. M U ltip l e C ho iC e

1 The relationship between a, x and y in the triangle shown is:

x a y a tan (a) =

y x

B

sin (a) =

x y

C tan (a) =

x y

D cos (a) =

y x

e sin (a) = xy

2 In the triangle shown, the value of x is closest to:

4.2

x 56°

a 2.35

B 2.83

C 5.07

D 7.51

e 3.48

3 The angle that the 3-metre ladder makes with the wall in this diagram is closest to:

3.0 m

1.2 m a 23.6°

B 21.8°

4 In which quadrants is tan (x) positive? a 1 and 2

B 1 and 4

C 66.4°

D 18°

e 76°

C 2 and 4

D 2 and 3

e 1 and 3

5 The value of cos (320°) is: a positive, as 320° is in the 1st quadrant C negative, as 320° is in the 3rd quadrant e positive, as 320° is in the 4th quadrant

B negative, as 320° is in the 2nd quadrant D negative, as 320° is in the 4th quadrant

6 If tan (a°) < 0, sin (a°) < 0 and 0° < a° < 360°, then which one of the following is correct? a 0° < a° < 90° D 0° < a° < 180°

B 180° < a° < 270° e 270° < a° < 360°

C 90° < a° < 180°

Chapter 6 • Circular functions

297

7 If tan (θ ) = a

3, then sin (θ) could be equal to:

1 2

3 2

B

C −1

D

5π 9

D

8 The angle 150° is equivalent to: a

2π 3

7π 6

B

9 The angle that is equivalent to a 405°

B 315° 4π is: 10 The value of cos 3 a

3 2

1

B 2

C

9π is: 4

C 540°

−

C

1

e 0

2

5π 6

D 270°

3 2

D

4π 3

e

e 300°

1

−1

e

2

π Use the following information to answer questions 11 to 13: sin (a) = 0.6 and 0 < a < . 2 11 sin (2π − a) is equal to: a 0.36 B 0.8 C −0.6 D −0.8 12 cos (a) is equal to: a 0.6 B 0.8 C −0.6 D −0.8 13 tan (π − a) is equal to: a 0.75 C 1.33 B −1.33 D −0.75 14 The one value that is equal to sin (53°) is: a tan (37 °) B cos (37 °) C cos (53 °) D sin (37 °) Questions 15 to 17 refer to the function f (x) = −2 sin (3x). 15 The amplitude of f (x) is equal to: a 3 D 2 B −2 C −3 16 The period of f (x) is equal to: 2π a 6π B 4π C D 2π 3 17 The range of f (x) is: a [0, 2] B [0, 4] C [−2, 0] D [−1, 1] 18 The rule for this graph is: y x a y = 3 cos 2 3

2

e 1 e 0 e 1.4 e cos (127 °)

e 1

e 3 e [−2, 2]

B y = 3 sin (2x)

x 3 D y = 3 cos (2x) e y = −3 cos (2x) C y = 2 cos

0

π

2π

3π

4π x

−3

19 The function y = 5 tan (4x) has a period and asymptote respectively of:

π 2 π π D 4π and x = 8π e and x = 4 2 20 If sin (3x) = 0.966, then x could be equal to: a 65° B 75° D 25° e 20° 21 If tan (2x) = 0.839 has one solution of x = 20°, then another solution could be: a 160° B 110° D 380° e 70° a 4π and x = 2π

298

Maths Quest 11 Mathematical Methods CAS

B π and x =

C

π π and x = 4 8

C 105°

C 200°

x 22 The maximum value of y = 3 sin − 1 is: a −1

B 0

2

C 2

23 The minimum value of h = cos (3π t) + 5 is: a 6

B 5

C 8

D 3

e −4

D 3

e 4

1 The temperature in an office is controlled by a thermostat. The preferred temperature, P, can be set to

values between 18 °C and 25 °C. The temperature, T (°C) in the office at time t hours after 9 am is given by the rule T = P + 2.4 sin (π t). If the preferred temperature on the thermostat has been set to 23 °C:

ex ten D eD r eS p o n S e

a find the maximum and minimum temperatures b find the temperature at i noon and ii 3.30 pm c sketch the graph of the function between 9 am and 5 pm.

Freddy feels thirsty if the temperature is above 24.2 °C. d Find the amount of time between 9 am and 5 pm that Freddy feels thirsty. 2 A ‘standing wave’ on a guitar string may be

y π x , 20 L where x cm and y cm are defined on the diagram x at right. a Find the period of the standing wave. b If the frets coincide with the mean positions of the wave, find the value of L. c If the frets were to be spaced at 16 cm, what would be the equation of a similar standing wave of amplitude 0.3 cm such that a fret is at each mean position?

approximated by the function y = 0.3 sin

3 The number of rabbits in a national park is observed for one year.

At any time t months after observation begins, the number is modelled by the function:

πt P = 2 − 0.8 sin 6 where P is in thousands. a Find: i the maximum number of rabbits ii the minimum number of rabbits iii the median number of rabbits. b Find i the period and ii the amplitude

of the function. Chapter 6 • Circular functions

299

c d e f

Sketch the graph of the function. Find the population after 5 months. How long is the population below 1600? How long is the population above 2100?

4 The height (in cm) that a clock’s pendulum swings above its base can be approximated by the function

7π H = 14 + 5.9 cos t at any time t seconds after being released. Give all answers correct to 4 3 decimal places. a Find i the maximum and ii the minimum heights that the pendulum reaches. b Find the height after i 1.5 seconds and ii 1 minute. c Sketch the graph of the function for the first 2 seconds. d On the same set of axes, sketch the median position. e Find the length of time that the pendulum is below 14 cm travelling from one side to the other. f Find the number of times the pendulum swings in 1 minute. The pendulum is found to be losing time and needs its swing adjusted to 75 swings per minute. g Find the new function H(t) that approximates the height of the pendulum.

5 The depth, h(t), of water in metres at a point on the coast at a time t hours after noon on a certain day is

2π (t + 2) given by h(t ) = 2.5 + 0.5 cos . 11

DiGital DoC doc-9768 Test Yourself Chapter 6

300

Use a CAS calculator to answer the following. a What is the depth of the water at noon (correct to 2 decimal places)? b What is the period of h(t)? c What is the depth of the water (and what time does each occur) at: i high tide? ii low tide? d Sketch the graph of h(t) for 0 ≤ t ≤ 12. e The local people wish to build a bonfire for New Year’s celebrations on a rock shelf near that point. They estimate that they can pass the point safely and not get splashed by waves if the depth of water is less than 2.25 m. Between what times can they work? f How long do they have?

Maths Quest 11 Mathematical Methods CAS

ICT activities Chapter opener DiGital DoC • 10 Quick Questions doc-9757: Warm up with ten quick questions on circular functions (page 255)

6a

trigonometric ratio revision

DiGital DoCS • SkillSHEET 6.1 doc-9758: Practise identifying the hypotenuse, adjacent and opposite sides of a right-angled triangle with respect to a given angle (page 256) • SkillSHEET 6.2 doc-9759: Practise calculating trigonometric ratios (page 256)

6B

the unit circle

DiGital DoCS • doc-9760: Investigate the unit circle (page 261) • WorkSHEET 6.1 doc-9761: Locating exact values in the unit circle, finding unknown angles and sides in a right-angled triangle and converting radians to degrees and degrees to radians (page 262)

6D

Symmetry

interaCtiVitY • Symmetry int-0265: Use the interactivity to consolidate your understanding of symmetry in the unit circle (page 265) tUtorialS • We 8 eles-1420: Using symmetry find the value of trigonometric functions of angles given in radians and degrees (page 266) • We 9 eles-1421: Find the value of trigonometric functions in exact values (page 267) DiGital DoC • doc-9760: Investigate the unit circle using a spreadsheet (page 268)

6e

identities

tUtorial • We 14 eles-1422: Watch how to find the value of trigonometric functions in exact values (page 271)

6F

Sine and cosine graphs

interaCtiVitY • Sine and cosine graphs int-0251: Use the interactivity to investigate the features of sine and cosine functions (page 273)

tUtorial • We 16 int-0298: Watch how to sketch the graphs of a sine and cosine function over a set domain, stating the amplitude and period of each (page 275) DiGital DoCS • doc-9762: Investigate sine graphs (page 277) • doc-9763: Investigate cosine graphs (page 277) • Worksheet 6.2 doc-9764: Sketching trigonometric graphs, solving trigonometric equations and recognising exact values (page 279)

6G

tangent graphs

tUtorial • We 20 eles-1423: Watch a tutorial on how to sketch the graph of a tangent function over a given domain (page 281) DiGital DoCS • doc-9765: Investigate tangent graphs (page 282) • Investigation doc-9766: Investigate tangent graphs (page 282)

6h

Solving trigonometric equations

tUtorial • We 24 eles-1424: Watch a tutorial on finding a general solution to three trigonometric equations using a CAS calculator (page 286) DiGital DoC • doc-9767: Investigate solutions to trigonometric equations (page 287)

6i

applications

tUtorial • We 25 eles-1425: Watch a tutorial on applying understanding of trigonometric graphs and equations to a real world model (page 289)

Chapter review DiGital DoC • Test Yourself doc-9768: take the end-of-chapter test to test your progress (page 300)

To access eBookPLUS activities, log on to www.jacplus.com.au

Chapter 6 • Circular functions

301

Answers CHAPTER 6 CirCUlar FUnCtionS exercise 6a

1 2 3 4 5 8

a a a a

3.09 8.41 18.2° D

C 3.25 m

trigonometric ratio revision b 1.88 c 4.00 b 2.97 c 12.50 b 24.3° c 51.1° bA c B 6E 7 7.14 m 9 9.6° 10 23.1°

4 a c e g i

−0.3

5 a

1 2 − 3 2

1 a e i 2 a e i 3 a e i 4 a b 5 a d 6 a d g j 7 a e

the unit circle P bP cP N f N gN P j N P bP cN N f P gP N j N P bP cN P f P gN N j P Quadrant 3 Quadrant 1 C bB B e C −1 b −1 0 e 0 Undefined h 0 1 3 2 1 2

b

i 1

e i

radians b f j

2 a b e f 3 E 4 C 5 B 6 a 0.855 d 3.341 g 2.971 7 a 20.1° d 54.9° g 232.6° 8 a d

=

2 2

1 2

g 1

302

π 4 π 2 − 5π 4 120° −30°

0.63 −0.63 −0.25 0.25 −2.1 2.1

c g k c g

1 3 2 2

1 2 −

i d P h N d N h P

e

d N h N

7 a d 8 a d 9 a c

c A c 0 f 1

=

d

i

d

j 2 2

b f

3 3

j

−0.383

h 2π

5π 3 d 330° h 67.5° l

c −2.182 f 5.716

b e h b e h

1.365 4.095 3.784 84.2° 893.8° 458.4°

b

1 2

c

e 1

f

c 180° f −155.3°

1 2 = 2 2

3 2

1 a i 4π

3 −

g

d

3

h

c −1 3 3

−

g

2 2

−0.63

i i 2π

2 a i 2π

3 2

h

−0.924

0.383 − 0.259 0.259 − 0.765 −0.644

c i π

− 3 2

l −1 c f c f

0.414 −0.414 − 3.732 −3.732

g i 6π

ii 0.4

i i 8

ii 2.5

k i 2

1 5

0

identities

sin2 θ + cos2 θ

30°

0.25

0.75

1

81°

0.976

0.024

1

129°

0.604

0.396

1

193°

0.051

0.949

1

260°

0.970

0.030

1

350°

0.030

0.970

1

−47°

0.535

0.465

1

2 a 0.6 3 a 0.954 4 a −0.917, 0.917 c ±0.971 5 a 2

b b

−5 13

b

8 a B 9 a 70° e 82°

b b b d

c

7 4

c c

5

bD b 32° f 46°

b

−2

c

24 25

0

d

3 7 7

d

c C c 51° g 1°

−1

d A d 8° h 73°

2

e

cos θ

0.6

0.96

d 2

0.25 −2.1 −2.1

g

1 5

j

2 5

Maths Quest 11 Mathematical Methods CAS

e

=

5 5

=

2 5 5

f

0.3 0.638 0.447 0.819

h k

c 11 6

4 5 −3 5

ii 2

i 4π

ii 3

i π

ii 4

f

5 6

i

5 6

l

=

3— π 4

i

π – 3

5 5

ii

1 2

i 4π ii

2 3

2—π x 3

y

0

− 2–3

g

π

2π

3π 4π x

i 6π

y 5 0

11 5

2π 3

2– 3

0.7

1 5

x

y

0

0.28 0.954 0.77 0.894 0.573

2

π

–π 2

–π 4

1– 2

0.8

4 3

i 2π

4π x

2π

y 4

−4

sin θ

b

ii 1

−3

10

4 5

i π

x

y

− 1–2

11 a

ii

3

0

=

3— π 2π 2

–π 2

−5

3π

6π

x

3 0.5 3 4

ii 1

x

y

0

2 3 3 7

π

–π 2

−1

4

ii 3

2

1.333 3.180 −0.714, 0.714 ±0.436

5 3

−4

y

1

ii 2

f i

1

cos2 θ

7 a

ii

ii

π 2 2π h i 5 π j i 3 l i 4

1 2

e i 6π

3 a

sin2 θ

−0.63 −0.25

g i π

3 2

d

k −1 b e b e b d

e i 2π

− 3 3 − 3 2

l −1

k 0

2

Sine and cosine graphs ii 2 ii b i π 3π ii 1.5 d i ii 2 4π ii 2 f i ii 3 2π ii 2.5 h i ii 3 ii 4 ii 1 b i 2π ii 2π ii 2 d i ii 3

c i 3π

tan θ 1.333 0.292 3.180 1.207

Symmetry b d b d b d

3

exercise 6F

0.3 0.7 −0.9

−

c

− 3 2 − 2 1 2 −

−0.7

θ

1 2

π 9

− 2 2

0.924 0.966 −0.966 0.644 −0.842

1

h 1

π 3 3π 2 − 7π 3 40° −45°

1 2 −

f

exercise 6e

i −1

3 3

6 a

b

6 a 6

h

exercise 6D

1 a c 2 a c 3 a c

g

j Undefined

π 6 5π 18 5π 6 36° 648°

1 2

c

f 1

exercise 6C

1 a

2 2

0.9 0.3 0.7 0.9

e

exercise 6B

b d f h

−

ii 5

1 4

i 4π

y

h

4 0

ii 4

f

y

π 2π 3π 4 π x π x

0

π i 2

y 2

π – 2

ii 2

–π 8

−2

j

–π 2 3— π 8

–π 4

−3

10 a f : 0,

−4

8 a i π 0

ii 1

iii 2, 0

y

x

1

i π

3

2π x

π

b i 2π

d f : [− 1, 3] → R, f ( x ) = 1.8cos

iii 0, −4

ii 2

4 a D 5E

bC

2x 3

d y = 4 cos

3x e y = − sin

f y = −3 cos ( 3x)

c i 4π

y

− 7—π − 5—π 4

4

−2π

− 3—π 2

− π–

−π

2

π– 4

0

π 3— 4

0 −1

π– 2

−1

π 7— 4

π 5— 4

π

π 3— 2

2π

x

d i 6π

2π

iii 0, −2

ii 1

y

y

π

2π

3π

4π

−1

3

−2 −4 π

0

4π

x

c

9 a

π 2— 3

0 − 2– 3

b

f i 4π g i 2π

ii ii

h i

ii

π 2 i i π

ii

π 2

ii

2π x

π

1 0 π π – –

3 3 − —— 4

8 4

π – 2

3— π 4

π – 3

π – 2

π

x

y 1.8

π

2π

0

−6

x

6

x

b

y

−1.8

c

y

y 1.4

1 0 π π — –

1.5 −6π

ii

2 a y

y

−π

2π x

π 4— 3

−3

e

e i π

2

y 3 0

x

ii

− 3–

−2

d

6π

ii

ii

j i

y 2 0

5π

x

π 4 π x=± 6 x = ±π 3π ± 2 π x=± 2 x = ±2π x = ±π π x=± 4 π x=± 2 π x=± 4

ii x = ±

d i 3π

−3

−3

tangent graphs

π 1 a i 2 π b i 3 c i 2π

π

0

b

exercise 6G

iii 5, 1

5 4 3 2 1

1 − 3—π − π– 4 4

ii 2

y

2x 3

2

g f : [0, 2π] → R, f (x) = −sin (x) − 1 h f : [−π, 2π] → R, f (x) = 2 sin (4x) – 1

−5

b y = 2 cos (2x)

x 2

−1 3π x − f f : [ 3 ,1] → R, f ( x ) = 2.4 cos 2

−3 −4

c y = 5 sin

7 a

−1 −2

c A

6 a y = 1.5 sin

2 x

0

−2

πx 2

3

x

−3

2x 5

2π x e f : [0, 3] → R, f ( x ) = − 3sin

y –π π 2

5 cos 2

c f : [−1, 1] → R, f (x) = 2 sin (π x)

0 −1

ii 3

3x 8π → R, f ( x ) = 3sin 2 6

b f : [0, 5π ] → R, f ( x ) =

2

y

0

8 x

0

−4

i

y 3

d

4

−1.5

12 6

0

6π x

0

2

4

6

π 2— 3

x

x

−1.4

Chapter 6 • Circular functions

303

c

y

y

j

x = π–

2 1 1 0 π π –

2π

2

d

0 −1 −2 −3 −4 −5 −6

3π 4π x

y

1 0 3π 3π 3π 9π — — — 4

e

2

2

y

π – 4

3— π 4

π – 2

b

0

π 3—π

2π

2

1 a

2 a b c d e f g h 3 a

3

4 2

π x

d

3

0 –π π–

equations

4

3 B 4 a D b B 5 a y

6π x

exercise 6h

x = 3—π

4

–π π 2

x

2π

c

x

d e

f

b

y

f

y

4 a 5 a

2

b

1

0 π 2π 4π 6π 8π x

0 3 –

3

3– 2

4

x

c d

g

e y

0

c

f

2

π – 2

3π 4π x

π 2π

0

−1

h

y

−2 3

y

d

3— π 3— π 2π 4 2

x

y

1.8 0 ππ – – 8 4

3— π 4

π – 2

π

x

0 5 5 – –

5– 2

8 4

15 — 4

5

x

−2

i

y

x = –π 2

304

1– 4 –π 2

π

3—π 2

2π x

−1 − 1– 0 2 −2

Maths Quest 11 Mathematical Methods CAS

π π π b c 4 6 3 π π π e f 6 4 3 36.9° and 143.1° 104.5° and 255.5° 80.1° and 260.1° 238.2° and 301.8° 78.8° and 281.2° 140.1° and 320.1° 199.5° and 340.5° 40.8° and 319.2° 0.927, 2.214, 7.210, 8.497 2π 4π 8π 10π , , , 3 3 3 3 0.983, 4.124, 7.266, 10.408 5π 11π 17π 23π , , , 6 6 6 6 3.542, 5.882, 9.826, 12.165 π 7π 9π 15π , , , 4 4 4 4 C b A 0.297, 2.844 π 5π , 3 3 π 4π , 3 3 5.820, 3.604 2.328, 3.955 5π 11π , 6 6

6 D 7 a 14.5°, 165.5° b 131.8°, 228.2° c 74.1°, 254.1° d 270° e 78°, 282° f 108.4°, 288.4° 8 a 60° b 150°, 330° c 0°, 180°, 360° d 90°, 210° e 45°, 135° f 0°, 90°, 180°, 270°, 360° g 120° h 90°, 270° i 15°, 105°, 195°, 285° j 45°, 75°, 165°, 195°, 285°, 315° k 60°, 150°, 240°, 330° l 113.74°, 326.26°

π , n ∈Z 6 − − 11π π π 11π For −2π ≤ x ≤ 2π : x = , , , 6 6 6 6 π b General solution: x = n π + , n ∈ Z 6 − 11π − 5π π 7π For −2π ≤ x ≤ 2π : x = , , , 6 6 6 6

9 a General solution: x = 2nπ −

y

2

3 2 1 0 −1 −2 −3

e

x = 3—π

Solving trigonometric

1– 2

1 x

c General solution: x = (2n + 1)π −

π or 4

3π π and 2n π + , n ∈ Z 4 4 − 7π − 5π π 3π For −2π ≤ x ≤ 2π : x = , , , 4 4 4 4 10 General solution: (12n + 5)π (12n + 1)π x= and , n ∈Z 12 12 − − 11π 7π π 5π For −π ≤ x ≤ π : x = , , , 12 12 12 12 (6n ± 1)π 11 General solution: x = , n ∈Z 9 −π ≤ x ≤ π : For − 7π − 5π − π π 5π 7π x= , , , , , 9 9 9 9 9 9 x = 2n π +

exercise 6i

applications

1 a i 1 kg ii 6 days b 2 a b c 3 a b c 4 a b

5 a

πt W = cos + 3 3 110 beats/min i 50 ii 60 min πt H = 50sin + 110 30 1.6 m i 1m ii 0.7 m 3.254 s 26 °C at 2 pm i 18 °C ii 22 °C iii Approx. 11.1 °C i 12 mm

c t = 6.582; t = 6.987 d (1.206, 0.931) and (5.266, 0.971) e $0.98 when t = 8

Chapter reVieW Short anSWer

1 2251.75 m 2 a 3 a 4 a 5 a b c 6 a b c 7 a b

−

b

3

7 4

b

7π 9 72° 57.3° 630° 0.724 0.690 1.049 0.436 − 0.484

b

−

3 2

c

1 2

c

π 180

7 3

31π 18

13 a

14 a c

c e 16 a b c

π 2π 3π 4π x

3 2 d −1

3

s

80 74

−π − –π 0 2

0

π x

–π 2

40 m 3.9 s 7.8 s 60 s 50 m 100 m 314.16 m i 50 m ii 75 m f 25 s and 35 s 8 a i 6 m ii 3 m b Yes, by approx. 24 minutes 6 a b c 7 a b c d e

1

t

2

MUltiple ChoiCe

1 6 11 16 21

−1.5

9 a

y 2

C E C C B

2 7 12 17 22

D B B E C

3 8 13 18 23

A D D A E

4 9 14 19

E A B C

5 10 15 20

E E D D

extenDeD reSponSe

1 S 120

1 a 25.4 °C, 20.6 °C b i 23 °C ii 25.4 °C c π − 7— 8

−π

π − 5— 8 π − 3— 4

π − 3— 8

− –2π

− π–8

0 − –4π

–π 8

3— π 8 –π 4

5—π 8 –π 2

7— π 8 π 3— 4

T (°C) π

x

27 26 25 24

25.4

23 −2

b y

22 21 20 19

20.6

18 (8, 0.98)

0

0.96 0.93 0

0.90 0

−

86

positive to the right, then the string is 11.41 mm to the left (or vice versa).

0.99

d 0.92

L (t)

1.5

1 S 10

9 a a ($A)

b

e

y

b

1 2 −

1 6

d

−4

b −0.40

π 2π 3π 5π b , , 3 3 4 4 π 7π d 0°, 360° , 6 6 135°, 315° Amplitude = 6, period = 2 Minimum = 74, maximum = 86 1 s 2

15 a

−2π −π 0

23 10 −

0.92

c

y 4

8 a

ii b 10 c −11.41 mm; if the displacement is

d

π 3π 9π 11π , , , 4 4 4 4 11 150° and 210° π 12 General solution: x = 2n π + and 3 π x = (2n + 1) π − , n ∈ Z 3 − 5π − 4π π 2π , , , For 2π ≤ x ≤ 2π : x = 3 3 3 3 10

1 2 3 4 5 6 7 8 t (days)

b t = 4.253 (3 d.p.), ∴ on the 5th day

3π 6π

9π 12π x

1

2

3

5 4 t (hours)

6

7

8 t

d 2 hours, 40 minutes 2 a 40 cm

π c 0.3sin x 16

b 80 cm

Chapter 6 • Circular functions

305

3 a i 2800 ii 1200 b i 12 months ii 0.8 c P (thousands)

iii 2000

306

5 a 2.71 m b 11 hours c i 3 m at 9 pm ii 2 m at 3:30 pm d h(t) (m)

2.0 0

ii 8.1 cm ii 8.1 cm

3.0 (0, 2.71) 2.5

8.1

3 6 9 12 t (months)

f Approx. 5 months 4 a i 19.9 cm b i 11.742 cm

g H = 14 + 5.9cos

14

d 1600 e 4 months 1 2

8π t 5

H 19.9

2.8 2 1.2 0

c

d See c. e 0.571 s f 52.5 times

Maths Quest 11 Mathematical Methods CAS

1

2

t

1.5 0

(9, 3) (12, 2.25)

(3.5, 2) 3

6

e 1:40 pm and 5:20 pm f 3 h 40 min

9

12 t (hours)

Exam practice 2 CHAPTERS 1–6 1 Determine the centre and radius of the circle described by x2 + 6x + y2 + 8y – 24 = 0.

4 marks

2 a Sketch the function f defined by:

20 minutes

2x + 2 f (x) = x + 2 − 2x + 8

−

3< x <0 0≤x<2 2<x ≤5

4 marks

b State the range of f.

1 mark

3 Solve for x in each of the following. a 23x = 5 b log2(x + 1) − log2(3x) = 2

2 marks 3 marks

M U ltip l e C ho iC e

1 For the function f (x) = (x − 2)2 + 3, what are the coordinates of the turning points for f (x − 1)? a (–3, 3)

B (–2, 3)

C (1, 3)

D (2, 3)

e (3, 3)

1] → R, where f (x) = 6 − 5x − is: C [6, ∞) D [–6, 1] 3 Which one of the following represents a many-to-one relationship? 2 The range for the function f: a [–5, 1]

[–5,

B

y

y

y

x

x

e

each question is worth one mark.

e [0, 12.25]

C

y

x

D

12 minutes

x2,

B [0, 6]

a

S ho rt a n S W er

y

x

x

4 62x × 3x × 2x + 1 simplifies to: 2 a 2 2 x + 2 x × 32 x

B 364x + 1

D 23x + 1 × 33x

2 e 62 x +3 x

3 2 C 62 x + 2 x

π x + 1 = 0 between the interval [−π, 2π] are: 3 −π 7π 5 7π 7 11 a B C and and and 6 6 2 2 2 2 − 1 7 11 11π D e and and 2 2 2 2 6 A possible equation for the rule of the function whose graph is shown below is: y a y = −2 sin (x) B y = −2 sin (3x) π x C y = − 2 sin D y = −2 cos (x) 3 x 5 The two x-values that satisfy the equation 2 sin

πx e y = − 2 cos

−3

0

3

3

Exam practice 2

307

7 Consider the function f (x) = 3 − −

2 . The equations of the asymptotes would be: (2 x + 1)

1 f (x) = 3 2 B x = −1 f (x) = −3 C x=0 f (x) = 0 1 D x= f (x) = 3 2 e x=2 f (x) = 1 8 For the function g(x) = 4 − 2 − x , the implied domain and range would be respectively: a (−∞, 2) and (4, ∞) B R /{2} and R + C [2, ∞) and [4, ∞) − − D ( ∞, 2] and ( ∞, 4] e R ∪ {0} and R+ a x=

e x t enDeD r e S ponS e 35 minutes

DiGital DoC doc-10162 Solutions exam practice 2

308

1 Rock pools in a tropical lagoon form when the depth of water in the lagoon falls below 6 metres. The

π (t − 3) depth of water in a lagoon can be modelled using d (t ) = 1.5 sin + 5.5, where d is the depth 6 in metres and t is the time in hours after 7 am in the morning. a Determine the maximum and minimum water depth. 2 marks b Determine the time, in hours, taken for the water depth to increase from the minimum depth to the maximum depth. 1 mark c Sketch the graph of d(t) in the time interval 0 ≤ t ≤ 24. 3 marks d At what time will the depth of water first be 6 metres deep? 1 mark e Billy and Tommy swim only in the safety of the rock pools. They arrive at the lagoon at 9:30 am and leave at 4:15 pm. i Determine how long they will be able to swim in the rock pools. Write your answer correct to 2 decimal places. 1 mark ii Determine the total time Billy and Tommy would be able to swim in the rock pools. Write your answer to the nearest whole minute. 2 marks kt 2 Over the summer months, the number of mosquito larvae in a dam was modelled using M = M0 e , where M0 is the initial population, t is the time in days and k is a positive constant. The number of mosquito larvae was recorded each morning at 8 am. The first day of recording was taken as t = 0. The recording lasted only until the end of summer (90 days). a On the first day of recording, there were 150 mosquito larvae. Determine the value of M0. 1 mark b At 8 am on the fifth day there were 235 mosquito larvae. i Write an equation that determines the number of larvae on the fifth day. 1 mark ii Show that the value of k is 0.11. 2 marks c If the mosquito larvae continue to increase according to the model, determine the expected number of larvae present in the dam at the end of summer (90 days). 2 marks d Frogs were introduced to the dam during the twentieth day. Mosquito larvae are part of the frogs’ diet. Since the frogs were introduced, the number of mosquito larvae has decreased. The decrease in mosquito larvae can be modelled using D = 200 e(20 – t) + 300, where D is the decrease in mosquito larvae after the twentieth day. The recording of the mosquito larvae has remains the same as before. i State the domain of the function D. 1 mark ii Determine the number of mosquito larvae expected in the dam over the long term. 1 mark iii By how much had the mosquito larvae decreased in the first twenty-four hours of recording after the frogs were introduced? 1 mark

Maths Quest 11 Mathematical Methods CAS

ChApTer 7

Matrices diGiTAL doC doc-9769 10 Quick Questions

ChApTer ConTenTS 7A 7B 7C 7d

Addition and subtraction of matrices Multiplying matrices Solving matrix equations Matrices and transformations

7A Addition introduction

and subtraction of matrices

In all walks of life we deal with the mathematical phenomenon called the matrix. A train timetable, a football scorecard and examination results are all examples of matrices. Anytime there is a rectangular grouping of numbers, there is a matrix. A summary of cricket scores after three matches is shown in the table below. Overs

Maidens

Wickets

Runs

Smith

10

3

2

35

Brown

14

2

3

29

Nguyen

9

2

3

39

Removing the headings from the table leaves only an array of numbers: 10 3 2 35 14 2 3 29 9 2 3 39 This is called a matrix. The plural of matrix is matrices. The matrix above is a (3 × 4) matrix as it has 3 rows and 4 columns. This number of rows and columns gives the order, or the dimensions, of the matrix.

6 3 5 −4

is a (4 × 1) matrix since it has 4 rows and 1 column.

A matrix with m rows and n columns is called an (m × n) matrix. ChApTer 7 • Matrices

309

Each individual entry in a matrix is called an element. In general, the elements of a matrix are identified by the row and column that pinpoint their position in the matrix. Thus if A is a (3 × 2) matrix, we write: a11 a12 A = a21 a22 a31 a32

The elements of the matrix A are referred to as aij, where i refers to the position of the element in the row and j refers to the position of the element in the column. So, a21 refers to the element that is in row 2, column 1.

Addition and subtraction of matrices Imagine, for the three players in the cricket team, that a fourth match gave the following results: Total

Overs

Maidens

Wickets

Runs

Smith

3

1

0

10

Brown

10

1

2

34

Nguyen

15

2

4

54

Clearly we can add these results to the previous results, giving the total figures for four matches: Total

Overs

Maidens

Wickets

Runs

Smith

13

4

2

45

Brown

24

3

5

63

Nguyen

24

4

7

93

In matrix form we write this as: 10 3 2 35 3 1 0 10 13 4 2 45 14 2 3 29 + 10 1 2 34 = 24 3 5 63 9 2 3 39 15 2 4 54 24 4 7 93 Matrices are added by adding corresponding elements. It follows that: 13 4 2 45 3 1 0 10 24 3 5 63 − 10 1 2 34 24 4 7 93 15 2 4 54

10 3 2 35 14 2 3 29 = 9 2 3 39

Matrices are subtracted by subtracting corresponding elements. The matrices we have added or subtracted have the same order, i.e. the same number of rows (three) and the same number of columns (four). Addition and subtraction of matrices can be performed only if the matrices are of the same order.

The commutative and associative laws for addition of matrices The Commutative Law Simply stated, if this law holds, matrices can be added in reverse position. Consequently for matrix addition, A + B = B + A, provided A and B are of the same order. And clearly with subtraction, A − B = A + (−B) = (−B) + A, if A and B are the same order. The Commutative Law holds for matrix addition for matrices of the same order. 310 Maths Quest 11 Mathematical Methods CAS

The Associative Law Simply stated, if this law holds, any number of matrices (of the same order) can be added together, adding any pair of matrices first. Thus: A + B + C + D = (A + B) + C + D = (B + C) + A + D = (B + D) + A + C provided the matrices are the same order. Thus for subtraction, A − B − C = [A + (−B) + (−C)] = [A + (−B)] + (−C) or A + [(−B) + (−C)] and so on. The Associative Law holds for matrix addition for matrices of the same order. Worked exAmpLe 1

Show that the Commutative Law holds for matrices A and B given below. 1 A= 0 2

−

2 1 0

1 0 4 , B = 5 3 0 0 3

−5

1 2

−4

Think 1

2

3

4

TUToriAL eles-1426 Worked example 1

WriTe

For the Commutative Law to apply, A + B = B + A. Both matrices have the same order so it is possible to add them. Find A + B by adding the corresponding elements in each matrix.

Find B + A by adding the corresponding elements in each matrix.

Write the answer.

Both matrices are (3 × 3) so it is possible for addition to occur. 1 A+ B = 0 2

2 −1 0

1 0 4 + 5 3 3 0 0

−5

2 2 = 5 2 2 0

−9

1 0 B+ A= 5 3 0 0

−4

2 2 = 5 2 2 0

−9

1 2

−4

5 5 1 1 + 0 2 2

2 −1

0

4 3

−5

5 5

The Commutative Law holds for matrices A and B as A + B = B + A.

Worked exAmpLe 2

1 A= 0 2

3 1 − 4

0 5 5 4 − , B = 1 0 and C = 7 0 2 3 3 1

4 2

−2

Find, if possible: A+B A−B A−C the answers to parts a, b and c using a calculator.

a b c d

ChApTer 7 • Matrices

311

Think

WriTe

1 3 0 5 a A and B have the same order, so it is a A+ B = 0 1 + −1 0 possible to add them. Add the corresponding 2 −4 2 3 elements of each matrix. 1 8 = −1 1 4 −1 1

b A and B have the same order, so it is possible b A − B = 0

to subtract them. Subtract the corresponding elements of each matrix.

c Subtraction cannot be performed since the

order of A is (3 × 2) and the order of C is (3 × 3).

d 1 Define the matrices using the CAS

calculator.

2

1 = 1 0

3 1 − 4

0 5 − − 1 0 2 3

1 −7 −2

c A – C cannot be calculated because A and C are of

different order. 1

d 0

2

3 1 −4

→a

0 5 − 1 0 →b 2 3 5 4 7 0 3 1

2

Use the CAS calculator to perform the matrix addition A + B.

3

Use the CAS calculator to perform the matrix subtraction A − B.

4

Use the CAS calculator to perform the matrix subtraction A − C.

4 →c 2

−2

1 a + b = −1 4 1 a−b= 1 0

8 1 −1

1 − 7 −2

ERROR (The two matrices are of different orders.)

Special types of matrices In life it is convenient to associate shapes with names. Thus we all know what a round shape looks like, what a pear-shaped person looks like and so on. A similar principle can be applied to describing the shapes of matrices. 312

Maths Quest 11 Mathematical Methods CAS

Type Column matrix

Shape Consists of a single column

Row matrix

There can be any number (1 × 2) of columns, but only one (1 × 10) row. (1 × 56) (1 × n) Consists of an Any number of rows. (1 × 1) equal number of Must have the same (2 × 2) rows and columns number of columns. (6 × 6) (20 × 20) A diagonal Must be a square matrix (1 × 1) matrix is a square (3 × 3) matrix with every (6 × 6) element, not on the (25 × 25) leading diagonal, equal to 0. A unit matrix Must be a square matrix (1 × 1) is a diagonal (2 × 2) matrix with all the (6 × 6) elements on the (20 × 20) leading diagonal equal to 1. Any A zero matrix can Any shape be any shape, but order all the elements are 0. Must Any shape If matrices have are equal, then both same their shape and all corresponding elements order must be identical.

Square matrix

Diagonal matrix

Unit matrix

Zero matrix

Equal matrices

Description Order There can be any number (2 × 1) of rows, but only one (5 × 1) column. (27 × 1) (m × 1)

Consists of a single row

Example

Order of example (4 × 1)

−1 3 2 0 [5

2 4

−1

3

0

(1 × 5)

4]

(2 × 2)

0

−5

2 0 0 0 0 0

0 0 −3

(3 × 3)

(3 × 3)

0 0 0 0 0 0

(2 × 3)

x y −1 3 2 p 3q = 4 − 9 1 4 1 4 x = –1, y = 3, p = 2, q = –3

Both (3 × 2)

1 0 0 0 1 0 0 0 1

Worked exAmpLe 3

2x + y Solve for x and y in the following equal matrices. 3 x − 2 y Think

2

Corresponding elements are equal, so write these as equations. Solve the simultaneous equations.

3

Write the answer.

1

4 = 13

WriTe

TUToriAL eles-1427 Worked example 3

2x + y = 4 3x – 2y = 13 2x + y = 4 [1] 3x – 2y = 13 [2] Multiply equation [1] by 2. 4x + 2y = 8 [3] Add equations [2] and [3]. 7x = 21 x=3 Substitute x = 3 into [1]. y = −2 x = 3 and y = −2

ChApTer 7 • Matrices

313

multiplication of a matrix by a scalar As in ordinary arithmetic, multiplication is simply a shorthand method of addition. Rather than 4 + 4 + 4 + 4 + 4 = 20, we say 5 × 4 = 20. Similarly, this method also applies to matrices. 2 1

2 + 0 1

−5

4 = 0 2

0

− 10

−5

is the same as: 2 2× 1

4 = 0 2

−5

0

− 10

Thus if A = [a b c], then 3A = [3a 3b 3c]. A scalar is a single (real) number. When a matrix is multiplied by a scalar, each element of the matrix is multiplied by the scalar.

Worked exAmpLe 4

If A = a 3A

1 and B = 7 b 4B c 3A + 4B

3

−2

4 5

−

0 2

, find : d 2(A + B)

Think

a Multiply each element of A by 3.

e 5(B − A) using a calculator.

WriTe

3

−2

a 3A = 3

4 5

9 12 = − 6 15 b Multiply each element of B by 4.

c Add the two matrices found in parts a and b.

1 7

0 −2

4 = 28

0 −8

b 4B = 4

9 12 4 + − 6 15 28

c 3A + 4B =

13 12 = 22 7 d 1 Find A + B by adding the

corresponding elements of A and B.

3 4 1 + − 2 5 7

b A+ B =

4 4 = 5 3 2

Multiply each element of the resulting matrix by 2.

4 4 2( A + B) = 2 5 3 8 8 = 10 6

314

Maths Quest 11 Mathematical Methods CAS

0 −2

0 −8

e 1 Define the matrices using your CAS

e

calculator.

3 −2

1 7 2

Perform the matrix arithmetic 5(b − a) and write the result.

4 5

→a

0 →b

−2

− 10 5(b − a) = 45

− 35

− 20

If A and B are matrices of the same order and a and b are real numbers, then: 1. aA + bA = (a + b)A 2. aA + aB = a(A + B) 3. (ab)A = a(bA). If aA = 0, then a = 0, or A is a zero matrix.

Addition and subtraction of matrices

exercise 7A

Show that the Commutative Law holds for matrices A and B given below.

1 We1

A=

3 1 6 and = B −4 2 2

1

−5

5 2 3 0 2 We2 , C = 1 − 1 2 1 4 find each of the following, if they exist. b B–A a A+B c B+D d A+C e C – D. 3 Find [2 −5 3 0 4] – [−3 4 1 6 −3]. 4 We3 Solve for x and y in the following equal matrices. 1 3 If A = 2 1

x + 2y 2 x − 3 y 5

=

5 −3

, B =

−

−

3 0 − 2

0 and D = 2 1

4 , −1

−3

5 − 11

State the order of each of the following matrices. 2 4 5 a 0 1 −2 b 4 −3 6 c [−2x + y]

d [1

6 Find the matrix P such that P – [−1

2

−5]

1 3 −4 0

4

5

3]

= [−3 4 2].

p 8 7 If 3 q = − 9 , find p, q and r. r 10 − 8 What does [ 5

2 4] − equal? 1 ChApTer 7 • Matrices

315

3[2 3 4] – 5[−1 A [11 −1 27] C [11 19 27] e [1 −1 27]

9 mC

−2

−3]

equals: B [0 d [1

19 19

−3]

27]

x

3 2 y = : − 1 2m 2 p 12

10 mC If 2

A x = 1, y = 3, p = −1, m = 3 C x = 1, y = 3, p = −1, m = 6 e x = 2, y = 6, p = −1, m = 12

11 We4

mC

B x = 1, y = 6, p = −1, m = 3 d x = 2, y = 3, p = −1, m = 3

If A = [0 4] and B = [−3 0], what is −4A − 3B?

−16]

A [9 C [9 −19] e [−13 −16]

B [5 d [5

−19] −16]

12 George, Linda and Mary sat two examinations in each of English, Mathematics and Science. In the first

examination they scored 40%, 50% and 60% respectively for English; 62%, 87% and 28% respectively for Mathematics; and 100%, 14% and 73% respectively for Science. a Write this information as a (3 × 3) matrix, labelling names on columns and subjects on rows. In the second examination their results were 50%, 65% and 80% respectively for English; 87%, 76% and 45% respectively for Mathematics; and 95%, 37% and 89% respectively for Science. b Write this information as a (3 × 3) matrix, labelling names on columns and subjects on rows. c Add the matrices together. d Hence calculate their average mark in each subject. e Write Mary’s average as a (1 × 3) matrix. 13 At the Queen Victoria market, 5 people each bought 3 apples and 2 oranges, 7 people each bought

1 apple and 6 oranges, and 8 people each bought 4 apples and 2 oranges. Write this information in matrix form, and use matrix addition to calculate the total number of apples and oranges bought by the group.

7B

multiplying matrices

Multiplying matrices is a procedure that takes place in two distinct stages: Stage 1: Identification of the correct conditions for multiplication to occur Stage 2: The multiplication itself. Stage 1: We need to establish whether multiplication is possible. This can best be thought of as two tiles, side by side, in the game of dominoes.

Here we can join the number three in each domino. A similar principle applies to the multiplication of matrices, if we think of each domino as row– column. Specifically, only those matrices that have the same number of columns in the first matrix as the number of rows in the second matrix can be multiplied. Further, if we can join the dominoes (matrices) as shown, that is by placing a three against a three, not only can we multiply the matrices, but the order of the resulting matrix is given by the two outside numbers, in this case a (2 × 4) matrix. In general we can multiply an (m × n) matrix by an (n × p) matrix, because the inside numbers are the same (both n). These inside numbers are called the inner product. The outer numbers (outer product) give the order of the resultant matrix (m × p). 316

Maths Quest 11 Mathematical Methods CAS

Stage 2: We know from our dominoes that we can multiply because the inner product is equal (3 × 3). The resulting answer matrix is of order (2 × 4). For example: 4 2 and B = − 1 7 3 0 2

2 5 A= − 1 2

−1

a A × B = AB = e

0 2 − 2

1 2 − 3

c d g h

b f

(2 × 3) × (3 × 4) = (2 × 4) In the answer matrix, the letter a marks the intersection of the first row and the first column. To calculate the value of a, we multiply the first row of the first matrix by the first column of the second matrix. (In matrix multiplication we always multiply rows of the first matrix by columns of the second matrix). −1

× –4 × −1 2 5 –1 × –0 a = 2 × 4 + 5 × −1 = 8 + −5 + 0 =3 2

5

−1

+

×0

To calculate the value of f, at the intersection of the second row and second column of the answer matrix, we multiply the second row of the first matrix by the second column of the second matrix. −1

2

3

f = −1 × 2 + f = −2 + 14 + 6 f = 18

2 × 7 2 2×7

+

3×2

This process is repeated for each element in the answer matrix, 3 37 so A × B = − 6 18 b11 b12 a11 a12 In general, if A = and B = b21 b22 a21 a22

12 −2

15 . −6

,

a11 × b11 + a12 × b21 a11 × b12 + a12 × b22 then A × B = a21 × b11 + a22 × b21 a21 × b12 + a22 × b22

.

Worked exAmpLe 5

4 1 −3 1 0 P= , Q = 2 5 7 3 2 a Write the order of the two matrices. c Does QP exist? Think

a 1 Matrix P has two rows and two columns. 2

Matrix Q has two rows and three columns.

b PQ is the product of a (2 × 2) and a

(2 × 3) matrix, so it exists.

TUToriAL eles-1428 Worked example 5

b Does PQ exist? d Calculate the product(s) that do exist. WriTe

a P is a (2 × 2) matrix.

Q is a (2 × 3) matrix. b PQ exists.

ChApTer 7 • Matrices

317

c QP is the product of a (2 × 3) and a

c QP does not exist.

d 1 Multiply the rows of matrix P

d PQ = 1

(2 × 2) matrix, so it does not exist.

0 4 1 3 2 2 5

with the columns of matrix Q.

7

−3

1× 4 + 0 × 2 1×1+ 0 × 5 1× −3 + 0 × 7 = − 3 × 4 + 2 × 2 3 × 1 + 2 × 5 3 × 3 + 2 × 7 2

4 1 = 16 13

Simplify PQ.

5

−3

Note: In worked example 5, PQ exists but QP does not exist, indicating that matrix multiplication is not commutative.

exercise 7B

multiplying matrices

1 3 A= 2 1

1 We5

, B =

5 −3

the following, if they exist. a AB b BA

diGiTAL doC doc-9770 WorkSHEET 7.1

Evaluate 2

3

1 3 ,Q = If P = 2 1

5 0 5 3 0 R= 1 4 − 1 4 1 0 −1 b R(P + S) c R(P + SQ).

a QR

−

0 5 and N = − − 0 3 4 a Show that MN ≠ NM. b Evaluate M2. 2 3 0 M+ c Find 2 N × 0 0 3 1 2

c BD

−

3 0 − 2

0 and D = 2 1

d AC

−1

0 and S = − 1

.

0 2

.

5 a Use multiplication to simplify the matrix equation below.

1 5 x 11 − = 3 4 y 5 b Write the answer to 5a in a form that does not contain matrices. c Solve for x and y. 6 A is a (2 × 2) square matrix. 2 −1 0 3 4 0 . 0 4

a Find A2 − 3 A − b Find A3. 318

4 . Find each of −1

−3

e CB

1 3 − × 5 0 3 − . 2 0

2

4 M=

5 2 3 0 , C = 1 − 1 2 1 4

−

Maths Quest 11 Mathematical Methods CAS

4 , evaluate:

−3

7 Write the simultaneous equations

2x + 5y = 11 + 2y = 12

−3x

as a matrix equation. −2 8 E = 2 3 4 and F = − 3 . −4 a Find EF. b Find FE. 1 9 If A = 3

c Does the Commutative Law hold?

1 0 4 2 and I = , find A + 5A − 3I. −2 0 1

3 4 3. , find P 1 2

10 a If P =

1

0 . 9 12

b Find the (2 × 2) matrix Q such that Q × P =

3 P= 1

11 mC

A

−5

3

−

3 9

2 P= 3

12 mC

1 9

−2

A

−3

0 and Q = −2

B

−3

3

−2

8 9

1 . PQ equals: 5

C

−5

3

3 − 11

d

d

−3

3

−

3 9

e

e

7 2 6 3

−3 −

5

−

8 9

. P2 equals: 0

−1

7 6

−

−1

B

3

−2 −

1 6

C

−2 −3

7 2 9 3

The order of the answer matrix when we multiply a matrix of order (2 × 3) by a matrix of order (1 × 2) is: A (1 × 2) B (1 × 3) C (2 × 1) d (3 × 2) e (3 × 1)

13 mC

14 The three matrices A, B and C have orders (a × b), (a × c) and (b × a) respectively. a For which pairs of matrices does the product exist? State the order of the answer(s). b Is it possible to multiply all three matrices together? If so, state the order of the answer. c For which pairs of matrices does the sum exist? State the order of the answer(s).

1 0 2 2 and I = , evaluate M and solve for a and b such that aM − bM − I = 0. 0 1 16 Fast-food chain McDonuts has outlets in the centre of the city. An overall view of sales is maintained to monitor demand. Hamburgers, drinks and chips were surveyed. In January, Store A sold 1200 hamburgers, 2367 drinks and 4219 serves of chips. a Write this as a (1 × 3) matrix. b The selling price of each product is $4.50, $1.95 and $2.45 respectively. Write this as a (3 × 1) matrix. c Multiply these matrices to calculate the total income from these three items. 17 Mayfleet, the international hotel chain, is building a new hotel in Arcadia Waters South. The guest accommodation comprises 80 standard rooms, 45 suites and 3 presidential suites. Each standard room has 1 king-size bed, 1 table, 2 chairs and 1 TV. Each suite has 1 king-size bed, 2 single beds, 2 tables, 5 chairs and 3 TVs. The presidential suites have 2 king-size beds, 6 single beds, 7 tables, 18 chairs and 6 TVs. a Set this information out in the form of two matrices. b Multiply these matrices to calculate the total number of pieces of furniture that Mayfleet must purchase to fit out the guest rooms.

15 If M =

3

−2

5 0

ChApTer 7 • Matrices

319

7C Solving matrix equations The multiplicative identity matrix The unit matrix, previously defined as a significant square matrix, does not alter any other matrix by which it is multiplied. It is denoted by I and is known as the multiplicative identity matrix. 1 0 0 I = 0 1 0 is an example of a multiplicative identity matrix. 0 0 1 2 3 If A = 0 2

−1 −4

4 3

1 0 and I = , then: 0 1 1 0 2 3 IA = 0 1 0 2

2 3 Notice that 0 2

−1 −

4 4 3

2 3 = 0 2

4 . 4 3

−1 −

1 4 1 0 is not possible. − 4 3 0 1 −

Thus, it follows that AI = IA = A, if A is a square matrix.

The inverse matrix When any square matrix is multiplied by its inverse we obtain the identity, I. The inverse of matrix A is written as A 1. By definition: −

A × A 1 = A 1 × A = I. −

−

1 0 Thus, if A is a (2 × 2) matrix, then A × A 1 = . 0 1 Similarly, if A is a (3 × 3) matrix, then: −

A× A

−1

1 0 0 = 0 1 0 0 0 1

.

Calculation of the inverse of a (2 × 2) matrix a b −1 Only square matrices have inverses. If A = , then its inverse A equals c d 1 d −b . The number (ad – bc) is called the determinant of the matrix A and is ad − bc − c a written as det (A) or | A |. The determinant may be positive, negative or zero. Notice that if det (A) = 0, the inverse does not exist, as 1 is undefined. In this case, matrix A is called a 0 singular matrix. Worked exAmpLe 6

If F =

1 −2

5 3

, find F 1 . −

Think

320

WriTe

1

Find det (F ).

det (F ) = (1 × 3) − (−2 × 5) = 13

2

Swap the elements in the main diagonal.

3 1

Maths Quest 11 Mathematical Methods CAS

3

Multiply the elements on the other diagonal by −1.

2

4

Write the inverse of F.

F

5

Alternatively, using your CAS calculator, define the matrix f.

1 5 − → f 2 3

6

Use the inverse feature of the CAS calculator to − determine f 1. Note that this is the same result as in step 4.

−1

− f 1

−5

=

1 3 13 2

=

3 13

−5 13

2 13

1 13

1

−5

Use of matrices to solve simultaneous linear equations Matrices may be used to solve simultaneous equations of the form: 2x + 5y = 11 −3x + 2y = 12 These simultaneous equations may be expressed as the matrix equation: 2 5 x 11 − = 3 2 y 12 2 5 x 11 If A = − , X = y and B = , 3 2 12 then AX = B, − − A 1AX = A 1B − IX = A 1B − X = A 1B det (A) = 2 × 2 − (−5 × 3) = 19 1 2 −5 − A 1= 19 3 2 x 1 2 y = 19 3 =

−

5 11 2 12

1 − 38 19 57

−2 = 3 x = –2, y = 3 Worked exAmpLe 7

Solve the simultaneous equations using matrix methods. 3x – y = −7 −2x + 5y = 9 Think 1

Write the simultaneous equations as a matrix equation.

WriTe

3 − 2

−

1 5

TUToriAL eles-1429 Worked example 7

x −7 y = 9 ChApTer 7 • Matrices

321

2

3

Calculate the inverse of the coefficient matrix.

Multiply both sides of the equation by this inverse matrix.

det

3 − 2

= 15 − 2 = 13 5

−1

The inverse is

1 5 1 . 13 2 3

1 5 1 13 2 3

3 − 2

1 x 1 5 1 = 5 y 13 2 3

−

−

7 9

x 1 − 26 I = y 13 13 4

Simplify.

5

Write the answers.

x −2 y = 1 x = −2 and y = 1.

Worked exAmpLe 8

Solve the simultaneous equations using matrix methods. Interpret your findings. 2x − 4y = 10 3x − 6y = 5 Think 1

WriTe

2 3

Write the simultaneous equations as a matrix equation.

2

Calculate the inverse of the coefficient matrix.

3

Interpret the result.

−4 −6

x 10 y = 5

2 −4 − det = 12 + 12 = 0 −6 3 The inverse does not exist as this is a singular matrix. There is no point of intersection between the lines. It follows that the lines are parallel.

Worked exAmpLe 9

Solve the following in terms of r.

rx – y = 1 x – ry = 2

Think

TUToriAL eles-1430 Worked example 9

WriTe

1

Write the simultaneous equations as a matrix equation.

r 1

x 1 y = −r 2

2

Calculate the inverse of the coefficient matrix.

r det 1

−1

−1 −r

= (r × − r ) − (1 × −1) = − r 2 + 1

The inverse is

322

Maths Quest 11 Mathematical Methods CAS

1 −r2

+ 1

−r −1

1 r

.

3

4

Multiply both sides of the equation by this inverse matrix.

x 1 y = − 2 r + 1

Write the answers.

Thus x= or x=

−r −1

1 1 r 2

−r

− +2 1 + 2r ,y= − 2 r +1 r +1

− 2

r−2 1 − 2r . ,y= 1− r2 1− r2

This method can be used to solve a range of problems efficiently. Worked exAmpLe 10

A cubic equation is of the form y = ax3 + bx2 + cx + d. The points (1, 1), (0, 3), (−1, 3), (2, −1) lie on the curve. Use matrices to find a, b, c and d. Think

WriTe

1

Substitute the coordinates given to form equations involving a, b, c and d.

y = ax3 + bx2 + cx + d 1 = a(1)3 + b(1)2 + c(1) + d 3 = a(0)3 + b(0)2 + c(0) + d 3 = a(−1)3 + b(−1)2 + c(−1) + d −1 = a(2)3 + b(2)2 + c(2) + d

2

Simplify.

1=a+b+c+d 3=0+0+0+d 3 = −a + b − c + d −1 = 8a + 4b + 2c + d

3

Set up a matrix equation, AX = B.

1 1 0 0 −1 1 8 4

1 0 −1 2

1 1 1 1

4

Using your CAS calculator, define matrix a.

1 1 0 0 −1 1 8 4

1 0 −1 2

1 1 1 1

→a

5

Using your CAS calculator, define matrix b.

6

Using the CAS calculator’s matrix multiplication − and matrix inverse features, calculate X = A 1B.

7

Write the answer.

1 3 3 −1

=

1 3 3 −1

→b

−1 A B= 1

a b c d

− 1 −4 3 3 1 3

a = 3 , b = − 1, c =

−4 3

,d =3

ChApTer 7 • Matrices

323

Solving matrix equations

exercise 7C

Find the determinant of each of the following.

1

a 1

2 3 4

diGiTAL doC doc-9771 WorkSHEET 7.2

3 − 6 5

c

x 2 3 y

f

b

d 0

0 2 4

−2

e

1 −4

3 −3

x2 x

−x −

6

x−2 9 − and det (A) = 3. Evaluate x. x + 1 3 2 −5 3 We6 a Find the inverse of matrix A = − 1 3 2 A=

. 1 1 b Show that A × A = A × A = I, where I is the multiplicative identity matrix. 4 If it exists, find the inverse of each of the following matrices. − −5 0 a 2 b 1 1 c − 1 − 2 2 0 15 − − 3 8 d 3 e 6 − − 1 2 1 3 −

−

x +1 2 If A = , for what values of x is A singular? x − 1 4 6 We7, 8 Solve each of the pairs of simultaneous equations by a matrix method. a 2x – 5y = 6 b −x – 3y = 7 c 2x + 4y = 1 −5x + y = − 7 x+y=2 2x + 4y = 0

−

3 9

5

Solve for x and y in terms of p.

7 We9

d x – y = −1

2x – y = 6

2px – y = 3 3x + py = 2

8 Solve for x, y and z, given 3x – 2y + z = 1, −x – y – 2z = 13 and x + 3y – 2z = 9. 9 mC A 30 d 16

10 mC

A

d

1 For what value of p does |D| = 0, where D = p B 20 e None of these The inverse of 1 3 is: −1 4

1 −1 7 1

−

−1 − 1

7 1

3 4 −

−

2 ? 3

B 4

1 3 1

3 4

1 4 7 1

e

C 13

C

1 4 7 1

1

−3 −

1

−3

x

11 mC If −x + y = 4 and 4x + 6y = 5, then equals: y

A

1 10 −

d

324

1 10

−

6 4 6 − 4

− −

1 4 1 5 −

1 − 1

4 5

Maths Quest 11 Mathematical Methods CAS

B

1 10 1 10

−

e

−

1

−4

1 −4

1 4 6 5

−

−

1 4 5

−6

C

−1

10

−

6 4

1 4 1 5

−

1 4

12 If A =

−2

−

12

1 1

12 , find A3 and A−1. − 2 1 0

13 P = and Q = 1 0 . Show that P and Q do not commute when multiplied. 0 1

cos (θ )

14 Find the multiplicative inverse of the matrix

sin (θ )

(θ ) . cos (θ )

− sin

0 1 1 0 and J = − , find the condition for pI + qJ to have a multiplicative inverse. 1 0 0 1

15 If I =

2 −1 1 0 1 1 and N = − . If MX = XN, show that X = k . 0 2 1 2 0 0 The point of intersection of the lines 2x + 3y = a and x – 2y = 5 is (7, 1). Use a matrix method to find a. We10 The points (1, 3), (2, 5), (−1, −1) and (3, 15) lie on the curve y = ax3 + bx2 + cx + d. Establish the equation of the curve. Solve for x, y and z in terms of r. rx + y + z = 1, 2x – ry – z = 3, 3x + 4y – rz = 5 Solve for a, b, c and d. a+b+c+d=0 a – 2b + 3c + d = 7 2a + b – 2c – d = −1 3a + 2b + c – 4d = 11 Show that the following equations have a unique solution only if b ≠ 1, −2. x+y+z=3 x + by + z = 3b −2x + 4y + bz = 2 Tercell, the clothing shop in Church Street, sells 15 suits, 45 shirts and 43 ties per week. It has another branch in High Street that sells 24 suits, 23 shirts and 37 ties per week. If the profit from each sale is $100 per suit, $18 per shirt and $14 per tie, use a matrix method to calculate the total weekly profit from the two shops. Slapstick, Brush and Company manufactures four main brands of paint, which they mix up using four basic ingredients — urg, blog, slug and nog. The four main brands of paint are Quix, Runns, Splash and Thix. Each litre of Quix requires 0.2 kg of urg, 0.5 kg of blog, 0.3 kg of slug and 1 kg of nog. Each litre of Runns requires 0.1 kg of urg, 0.4 kg of blog, 0.7 kg of slug and 0.8 kg of nog. Each litre of Splash requires 0.2 kg of urg, 0.7 kg of blog, 0.1 kg of slug and 1 kg of nog. Each litre of Thix requires 1.0 kg of urg, 0.5 kg of blog, 0.3 kg of slug and 0.0 kg of nog. Each week the company uses 2000 kg of urg, 1945 kg of blog, 1800 kg of slug and 3230 kg of nog. Use a matrix method to find out how many litres of each paint the company produces per week.

16 M = 17 18 19 20

21

22

23

7d

matrices and transformations

Matrices can be used to describe transformations (translation, reflection, rotation and dilation) of points or curves in the x–y plane. The point (x′, y′) is the image of the point (x, y) following a sequence of transformations.

Translation

inTerACTiViTY int-0266 matrices and transformations

A point P (x, y) is moved to the point P′ (x′, y′) by translating a units in the positive direction of the x-axis (right) and b units in the positive direction of the y-axis (up). A horizontal translation of a units and a vertical translation of b units can be represented by: x x' x a (x′, y′) = (x + a, y + b) or T = = y + y y' b

. ChApTer 7 • Matrices

325

Worked Example 11

− The point (2, −3) is translated by the matrix 4 . Find the new coordinates of the point. 2 Think

Write

1

x x' x a Use the formula T = = y + . y y' b

x' 2 y' = − + 3

2

Add the matrices to find the image point (x, y).

x' − 2 y' = − 1 (x′, y′) = (−2, −1)

2

−4

Other transformations by matrix multiplication It is useful to summarise the effects of various transformations as set out in the table below. Here we deal with (2 × 2) matrices that cause a change in the position of a point. These can be done on calculators or by hand, using the usual matrix multiplication techniques. Matrix

What it does

Calculation

Change

1 0 0 1

Nothing

x x' 1 0 x x T = = y = y y y' 0 1

0 1 1 0

Reflection in the line y=x

x x' 0 1 x y T = = y = y y' 1 0 x

0 − 1

−1

Reflection in the line y = −x

x x' 0 T = = − y y' 1

1 0

0 −1

Reflection in the x-axis

x x' 1 T = = y y' 0

−1 0 0 1

Reflection in the y-axis

x x' − 1 0 x T = y = = y y' 0 1

k 0 0 1

Dilation by factor k from the y-axis

x x' k 0 x kx T = = y = y y y' 0 1

(x′, y′) = (kx, y)

1 0 0 k

Dilation by factor k from the x-axis

x x' 1 0 x x T = = y = ky y y' 0 k

(x′, y′) = (x, ky)

Anticlockwise rotation of A degrees about the origin

x x' cos ( A) T = = y y' sin ( A)

0

cos ( A) sin ( A)

( A) cos ( A)

− sin

326 Maths Quest 11 Mathematical Methods CAS

x = 0 y

−1

0 x y = −1

(x′, y′) = (y, x)

− x

(x′, y′) = (−y, −x)

x

(x′, y′) = (x, −y)

y

(x′, y′) = (−x, y)

−y

−y

−x

( A) x cos ( A) y

− sin

(x′, y′) = (x, y)

x′ = x cos (A) − y sin (A) y′ = x sin (A) + y cos (A)

Worked exAmpLe 12

The point (−5, 1) is translated by the matrix y = x. Find its new coordinates.

and then reflected in the line 5

−3

Think 1

WriTe

Write the point (−5, 1) in matrix form.

TUToriAL eles-1431 Worked example 12

−5 (− 5, 1) = 1

x x' x a Use the formula T = = y + y y' b to translate the point.

x' − 5 − 3 + y' = 1 5

3

Add the matrices to find the image point (x′, y′).

x' − 8 y' = 6 (x′, y′) = (−8, 6)

4

0 1 The reflection matrix in y = x is from the 1 0 table.

x' 0 1 − 8 6 = y' = 1 0 6 − 8

2

x x' 0 1 x Use the formula T = = y y y' 1 0 to reflect the point. 5

The new point is (6, −8).

Write the answer.

Worked exAmpLe 13

Find the image of the point (3, 8) under 30° anticlockwise rotation about the origin. Think

WriTe

1

Write the point (3, 8) in matrix form.

2

Use the rotation formula x x' cos ( A) T = = y y' sin ( A) where A is 30°.

3

Evaluate using exact values.

4

Multiply the matrices.

5

Write the answer.

( A) x , cos ( A) y

− sin

3 (3, 8) = 8 x' cos (30 ) y' = sin (30 ) 3 2 = 1 2

−

1 2 3 2

3 3 −4 = 2 3 4 3+ 2

(30 ) 3 cos (30 ) 8

− sin

3 8

3 3 3 − 4, 4 3 + . As a The new point is 2 2 decimal approximation, the point is (−1.40, 8.43).

ChApTer 7 • Matrices

327

exercise 7d

matrices and transformations Find the new coordinates of the point P (3, −2) after the following transformations.

1 We11,12

− − 0 1 c 1 0 b 1 0 followed by 3 1 0 0 2 2 0 1 d Reflection in the y-axis e Translation of 6 units to the left and 4 units upwards, followed by reflection in the x-axis a

2

Find the new coordinates of Q (m, −n) after the following transformations.

0 1 followed by reflection in the line y = x 1 0 2n b Reflection in the line y = −x followed by a translation of − 3m 3 Two transformations as shown are applied to the point M (x, y). a Reflection by the matrix

3 0 0 1 x 0 1 1 0 y a Find the single (2 × 2) matrix that can be used to describe the combined transformations. b If M is the point (−1, 0), find its new coordinates after the transformations. 4 Write the matrices for the following rotations about O. a 90° clockwise b 180° clockwise d 360° clockwise e 90° anticlockwise g 120° clockwise h 150° anticlockwise j 120° anticlockwise

c 270° clockwise f 45° anticlockwise i 300° clockwise

5 mC Consider the point (−3, −5). a The image of this point after a reflection through the line y = x followed by a rotation of 180°

anticlockwise is: d (5, 3) A (−5, −3) B (−3, 5) C (−5, 3) b The image of the point (−3, −5) after a rotation of 270° clockwise followed by 4 a translation of − is: 2 A (1, −3)

B (9, −5)

c The image of the point

C (7, 1)

d (−1, 1)

e (3, 5)

e (6, 4)

(−3, −5)

after a reflection in the y-axis, followed by a reflection in the x-axis and then an anticlockwise rotation of 45° is: A (

−

2, − 2 2)

d ( −4 2,

2)

B (

−

2, −4 2)

− C (2 2, 2 2)

− 2, 4 2) e (

6 mC The point (x, y) is dilated by a factor of 3 from the x-axis, then reflected in the line y = x and then

translated 2 units to the left and 1 unit down. The coordinates of the final point are: B (3x + 2, y – 1) C (y – 2, 3x + 1) − − d ( 2 + 3y, 1 + x) e (y + 2, 3x + 1) A (3x – 2, y – 1)

7 a Combine a dilation of factor 2 from the x-axis and factor 3 from the y-axis to find the new

coordinates of the square (0, 0), (1, 0), (1, 1), (0, 1).

b Write down the single (2 × 2) matrix for the combined dilations. c Write down the single matrix that would represent dilations of factor n from both the x-axis and

the y-axis. 8 The point P (a, b) is subjected to a sequence of transformations as follows:

• a dilation of 2 away from the y-axis • a translation of 5 to the left and 3 downwards • a reflection in the x-axis • a reflection in the line y = x. Find the new coordinates of P after all of the above transformations have been completed in the given order, using matrix methods. 328

Maths Quest 11 Mathematical Methods CAS

9 WE13 Find the images of each of the following points under each of the anticlockwise rotations, i–v. a A (4, 0) i 30°

b B (5, 4) ii 45°

c C (−3, 6) iii 90°

d D (0, 9) iv 180°

v

−60°

10 Find the image of the point (7, −2) after a rotation of 270° anticlockwise followed by a translation

2 of . 4 11 The triangle ABC, with vertices A (2, −1), B (−4, 0) and C (5, 2), is rotated by a clockwise rotation of 45° (i.e. an anticlockwise rotation of −45°). a Find the new coordinates (A′, B′ and C′) of the rotated triangle. b Compare the area of triangle A′B′C′ to that of triangle ABC. 12 Consider the point A (a, b). a Is an anticlockwise rotation of 30° equivalent to a reflection in the line y = x followed by a clockwise rotation of 60°? Give reasons for your answer. b Is a reflection in the x-axis followed by an anticlockwise rotation of 45° equivalent to a transformation by the identity matrix? Give reasons for your answer. 13 The point P (3, 4) is transformed to become the point P′ ( − 2 3, 13). a Prove that this transformation represents an anticlockwise rotation of θ ° about the origin. b Calculate the value of θ in degrees.

Chapter 7 • Matrices 329

Summary Addition and subtraction of matrices

Matrix definitions: • A matrix is a rectangular array of numbers arranged in rows and columns. • An m × n matrix has m rows and n columns. • The individual entries in the matrix are called the elements of the matrix. Operations with matrices: • Addition and subtraction of matrices are performed by adding or subtracting corresponding elements. • Addition and subtraction can be performed only if the matrices are of the same order. • Scalar multiplication is performed by multiplying every element of the matrix by the scalar. x y ax ay a = p q ap aq • The Commutative Law holds: A + B = B + A • The Associative Law holds: A + B + C = (A + B) + C = A + (B + C )

multiplying matrices

• Matrices are multiplied in the following way. b11 b12 a11 a12 In general, if A = , and B = b21 b22 a21 a22 a11 × b11 + a12 × b21 a11 × b12 + a12 × b22 then A × B = . a21 × b11 + a22 × b21 a21 × b12 + a22 × b22 • The orders are (m × n) × (n × p) = (m × p). • Matrix multiplication is not usually commutative. That is, AB ≠ BA.

Solving matrix equations

• I is the multiplicative identity matrix, where AI = IA = A and A is a square matrix. • The multiplicative inverse of matrix A is A 1, and A × A 1 = A 1 × A = I. −

−

−

a b 1 d −b • If A = , then its inverse is A 1 = . ad − bc − c a c d The number (ad – bc) is called the determinant of the matrix A and is written as det (A) or |A|. If det (A) = 0, then A 1 does not exist and A is said to be singular. −

−

matrices and transformations

• Matrices can be used to represent transformations of points in the x–y plane. • A horizontal translation of a units and a vertical translation of b units can be represented by (x′, y′) = (x + a, y + b) x x' x a or T = = y + . y y' b • Pre-multiplying by a (2 × 2) matrix causes a change of the position of a point. For example, 0 1 pre-multiplying by causes a reflection in the line y = x. 1 0 x x' 0 1 x y T = = y = y y' 1 0 x • An anticlockwise rotation about the origin can be represented by x x' cos ( A) T = = y y' sin ( A)

330

Maths Quest 11 Mathematical Methods CAS

( A) x cos ( A) y

− sin

, where A is the angle of rotation.

Chapter review 1 State the order of:

a

1 0 8 4

b

0 − 1 2

2 Find the matrix D such that

1 2 4 0 4

2 5 0 −5 −1

S ho rT A n S W er

3 1 − D = 5 0 3

0 . 2

−2

−4

3 x = . Find x and y. − 1 y

3

4 A=

1 −2 − − and B = [1 0 4 1]. Evaluate: 0 3

2

5 P= , Q = [0 1

3] and R = [0 2

a PQ

−1].

a AB

Evaluate:

b QP

2 − 5

6 Find the inverse of

b BA

c PR

d P(Q + R)

. 6

−4

x y , show that AA−1 = I. 2 3 8 Solve for x, y, and z. 7 For A =

2x – 3y + z = 3 + y – 2z = 1 x–y–z=6

−3x

9 Find the new coordinates of the point R(−p, q) after reflection in the x-axis followed by a translation

of a . b 10 The point (3, 2) is dilated from the y-axis. Its new coordinates are (6, 2). Write down the (2 × 2) dilation matrix. 11 The point A (1, 2) is rotated clockwise about the origin through an angle of 60°. Find the new

coordinates of A in exact form. 2 1 0 1 − 12 M = . Find MN(NM) 1. , N = −1 4 3 5 1 4 0 13 P = − 2 0 , Q = − 2 −1 3

−

0 4 . Find (QP)−1. 5 1

1 3 1 3 0 , C = − . , B = −2 0 2 4 1 a Find |A|. b Find the inverse of C if it exists.

14 A =

c Calculate BC. ChApTer 7 • Matrices

331

3 2

0 , find P2 – 2P. Hence, express the inverse of P in terms of P.

15 If P =

−1

3 0

m U LT ip L e C h oiCe

5 c , N = . M and N commute under multiplication. Evaluate c. 0 3

5 −3

16 M =

−5 − 2 − 3 0 2

1

−1

equals:

x

−4

z A 2, 3, 0

3 2[−5 A

3

−6]

[−4

= 6 0

−y

2 If 3

– 3 [−2

−9]

2 − 1

4 If A =

d

5 −2 −6

e

−1 2

−4

9 , x, y and z are respectively equal to:

−4

B 2, −3, 0

−1]

equals:

B [−16

−15]

C 6, −9, −4

d 6, −9, 0

e 2, 3, −4

C [−4

d [−16

e [4

−15]

−9]

9]

and B = [3 0], AB equals:

6 0

0 0

B 4 − 3

A

4 C 5 −2

4 B −1 − 2

−6 A 5 2

C [6]

d [5]

e

6 0 − 3 0

1 2 2 , M – 2M equals: 4 0

5 If M =

−

7

A

− 12

−

d 3 − 4

8

−2

0

−6

8

B 11 − 12

−6

7 − 12

−6

e

C 7 4

8

−6

8

1

3 A (1 × 2)

B (2 × 3)

C (3 × 2)

d (2 × 1)

e (3 × 1)

B −7

C −5

d −6

e −4

C 4, −1

d −4, 1

e 1, −2

1 6 P = 2 , Q = [1 2] and R = 3

7

−1

−2

−3

1

A 7

2 4

. The order of PQR is:

equals:

x +1

2 is singular if x equals: −3 x − 4 B 1, 2 A −1, −2

8

332

Maths Quest 11 Mathematical Methods CAS

−1 2 is: 1 4

9 The inverse of

A

d

1 6

−4

−1

2

1

2 1

4 −1

−

B

2

e

−1

1 4 6 − 1 −1

6

−1

−2

−

4 −1

C

−1

1 2 − 1

−

2

−4

2

−1

. The coordinates of the new point are: 1 A (0, 0) C (0, 2) B (−4, 0) d (−4, −2) e (0, −2) −1 11 The point (a, b) is reflected in the line y = x and then translated by the matrix . Its new coordinates are: 3 A (a − 1, b + 3) B (a + 1, b − 3) C (b + 1, a − 3) d (b − 1, a + 3) e (−a + 1, −b + 3) 10 The point (−2, −1) is translated by the matrix

−2

12 Reflection in the line y = −x followed by a dilation of factor 2 from the y-axis is equivalent to pre-

multiplying by the single matrix: 1 2 0

A 0

B 0 − 2

0

−1

C

−1 −

0

0 2

d

1 0 0 2

e

0 − 1

0

−2

13 The point (2, −1) is reflected in the y-axis followed by an anticlockwise rotation of 90° about the origin.

The coordinates of the image point are: B (−1, −2)

A (1, −2)

C (−2, 1)

e (2, −1)

d (1, 2)

1 0

14 The coordinates of the image of (3, −4) under the transformation given by A (3, 4)

B (3, −4)

C (−3, 4)

0 are: −1

d (−3, −4)

e (4, 3) ex Ten d ed r eS p o n S e

1 A and B are both square matrices. a Show that A2 – B2 = (A – B)(A + B) only if A and B are commutative for multiplication.

1 9

b Use part a above to evaluate A2 – B2, where A =

7 B and = −3 6

−2

. −3

−2

2 A and B are both (2 × 2) matrices. a Show that (A + B)2 = A2 + 2AB + B2 only if A and B are commutative for multiplication. b Use the matrices in question 1b to prove the result in question 2a numerically. 3 Potatoes cost $p per kg, carrots cost $q per kg, onions cost $r per kg and eggs cost $s per dozen.

George buys 14 kg of potatoes, 10 kg of carrots, 8 kg of onions and 2 dozen eggs for a total cost of $60. Mary buys 10 kg of potatoes, 11 kg of carrots, 3 kg of onions and 5 dozen eggs for a total cost of $52. Cindy buys 4 kg of potatoes, 21 kg of carrots, 12 kg of onions and 1 dozen eggs for a total cost of $56. Amy buys 15 kg of potatoes, 5 kg of carrots, 11 kg of onions and 4 dozen eggs for a total cost of $69. Establish a matrix equation and hence calculate the values of p, q, r and s. a 4 a The point P (1, −3) is translated by the matrix . Find its new coordinates. b b The new point is reflected in the line y = −x. Find the new coordinates. c This point is now rotated counterclockwise through 90° before being reflected in the x-axis. Find

the new coordinates.

d Find the transformation(s) needed to return the point found in part b to the point (1, −3). ChApTer 7 • Matrices

333

4 0 . 0 1 a Find the coordinates of the new vertices, calling them O′, A′ and B′. b Has the area altered? If so, how? c The new triangle is now reflected in the line y = −x. Find the coordinates of the new vertices, calling them O″, A″ and B″. −3 d Finally, O″A″ B″ is translated by . Find the final coordinates of the vertices. 1 3i 0 , where i2 = −1. 6 A= −i 2 3 a Find A – 2A2. – b Find A 1. 1 0 2 3 4 c Show that A2 – A – 2I = (A – 2I)(A + I), where I = . Note: I = I = I = I etc. 0 1 7 Use a matrix method to solve the following simultaneous equations. mx − 3 y = 6 5 The triangle O (0, 0), A (2, −3), B (−5, −4) is transformed by the matrix

2x + m2 y = 3 8 Use a matrix method to solve the following simultaneous equations. a+b+c+d = 5 −a + b − c + d

= −5

8a + 4 b + 2c + d = 19 −

8a + 4 b − 2c + d = − 25

9 Tickets for a one-way trip on a Melbourne to Sydney passenger train can be purchased as either adult,

child (under 15 years old) or pensioner. The table below shows the number of passengers and the total takings for three trips. Number of adult passengers 145 130 142

Number of child passengers 103 110 115

Number of pensioner passengers 121 90 80

Total takings ($) 20 260 18 400 19 200

a Let x equal the cost of an adult’s ticket. Let y equal the cost of a child’s ticket. Let z equal the cost

of a pensioner’s ticket. Construct three equations in terms of x, y and z.

b Using matrices, express the equations in the form AX = B. − c Use your CAS calculator to find A 1. d Use your CAS calculator to determine the costs of a train ticket for an adult, a child and a

pensioner. 10 Describe the transformation given by the following matrices. It may be helpful to use a graph and consider the effect of the matrix on the points (a, b). b

0 1 1 0

f

e

diGiTAL doC doc-9772 Test Yourself Chapter 7

334

1 0 1

a 0

i

−3

0

−

0 3

Maths Quest 11 Mathematical Methods CAS

−1

0

0 − 1

0 1 − 1 0 0 1

j

0

−1

1 0

c

0 − 1

g

0 −1 0

−1

0 0 1

d

−1

3 0 0 3

h

ICT activities Chapter opener diGiTAL doC • 10 Quick Questions doc-9769: Warm up with ten quick questions on matrices (page 309)

7A

Addition and subtraction of matrices

TUToriALS • We1 eles-1426: Watch a tutorial on showing that the Commutative Law holds for two matrices (page 311) • We3 eles-1427: Watch a tutorial on solving for variables in two equivalent matrices (page 313)

7B

multiplying matrices

TUToriAL • We5 eles-1428: Watch a tutorial on recognising the order of matrices, and the product of two matrices if they exist (page 317) diGiTAL doC • WorkSHEET 7.1 doc-9770: Determine the order of a matrix, add, subtract and multiply by a scalar and practise matrix multiplication (page 318)

7C

Solving matrix equations

• We9 eles-1430: Watch a tutorial on using matrices to solve simultaneous literal equations (page 322) diGiTAL doC • WorkSHEET 7.2 doc-9771: Calculate determinants and inverses of matrices and solve matrix equations (page 324)

7d

matrices and transformations

inTerACTiViTY • Matrices and transformations int-0266: Use the interactivity to consolidate your understanding of matrices and transformations (page 325) TUToriAL • We12 eles-1431: Watch a tutorial on finding the new coordinates of a point after translating then reflecting it in the line y = x (page 327)

Chapter review diGiTAL doC • Test Yourself doc-9772: Take the end-of-chapter test to test your progress (page 334)

To access eBookPLUS activities, log on to www.jacplus.com.au

TUToriALS • We7 eles-1429: Watch a tutorial on using matrices to solve simultaneous equations (page 321)

ChApTer 7 • Matrices

335

Answers CHAPTER 7 mATriCeS exercise 7A

matrices

1 A+ B =

4 1 = B+ A − 2 3

− 5 2 a 1 6 − 2 1 3 c Does not exist. 5 −7 − 1 3 3 −1 [5 −9 2 −6 x = −1, y = 3 a 3×3 c 1×1 [−4 6 −3]

−

−

e 3 4 5 6

8

10 0 8 , NM = − 10 4 0 10

7] b 4×1 d 1×4

6 a

8 a [−29]

−4 − b 6 − 8 c No

11 12 13 14

57.5 70 81.5 36.5 97.5 25.5 81 45

42 + 32 16

−7

multiplying matrices b Does not exist.

9 − 2 3 − 6 8

2 [−13]

336

0 − 2 −3

27

− 19

−

6

−9 − 12

a AC, CA, CB c None

0

− 13

7 x=

1 c e 2

a

−2

1 3

b

0

−2

b Yes. ACB

1 0 1 2 1 45 3 1 2 2 5 3 2 6 7 18 6 Solving matrix equations b 8 d 0 f 5x2

9 xy − 6 −4, 3 −1 5 3 a 1 13 − 3 2

Maths Quest 11 Mathematical Methods CAS

−

1 4 0.5 1.5

16

−2

29

− 9

b x = 14, y = −7 d 7, 8

3p + 2 4p− 9 ,y= 2 2 p2 + 3 2p + 3

− 5 − 19 3

8 3, 3,

b [131 108 191 439 233] exercise 7C

0.5

− 0.5

d

c x = 22 , y = 22

8 − 12 − 16

0 1

b

3 3 6 5 x = ±3

−

−1

17 a 80

0 − 1 2

−2

1 e 1 3

4

59 92 23 36

6 a x= 7 ,y= 7

15 ; a = − 0.1, b = − 0.3 − 10 4.50 16 a [1200 2367 4219] b 1.95 2.45 c $20 352.20

1 a Does not exist.

28 d − 1

D C B

= 54 68

e

8 0

b

−1 4

c None

15 − 6

36.5 81

− 11

4

−2 −

10 a

d 74.5

5 2 −5 2 3 − 1 13 0 1 =I = 13 0 13 4 a

15 16 12 3

90 115 140 c 149 163 73 195 51 162

− 17

−5

−1 1 − 13 3

60 − 12

60

c − 108

9

50 65 80 b 87 76 45 95 37 89

6 5

0

40 50 60 12 a 62 87 28 100 14 73

c

1 2

− −

2 x + 5 y 11 7 − 3 x + 2 y = 12

9 C 10 B 11 A

exercise 7B

−6

2 4 ] − does not exist. 1

13 15 10 + 7

−

1 2

b

1 −1 5 − − 1 13 3 2 13 0 =I 0 13

2 3

1 = 13

4 a MN =

10

e 70

12 − 1 15

− 22

x + 5 y 11 = 5 a − 3 x + 4 y 5 b x + 5y = 11, −3x + 4y = 5 c x = 1, y = 2

7 p = 3 , q = −3, r = 3 8 [ −5

3 a

4

3 0 5 − 3 1 4 d Does not exist.

b

b

5 35 − b 5 1 −1 2 c R ( P + SQ ) does not exist.

Addition and subtraction of

9 E 10 E 11 D

12 I;

−1 2

3 2

− 3 2

−1 2

2 0 1 1 ≠ 1 0 1 1

13

sin(θ ) − sin(θ ) cos(θ ) p2 + q2 ≠ 0 Check with your teacher. a = 17 y = x3 – 2x2 + x + 3 − r 2 − 7r + 2) r 2 + 8r + 11 (3 x= 3 , ,y= r + 9r + 5 r 3 + 9r + 5 − 2 5r + 9r − 7 z= 3 r + 9r + 5 a = 1, b = −1, c = 2, d = −2 − 10 10 x= , y = 3, z = ; unique solution b+2 b+2 cos(θ )

14 15 16 17 18 19

20 21

if b ≠ 1 or −2. If b = −2, division by 0 If b = 1, equations 1 and 2 become identical. 22 $2912 + $3332 = $6244 23 2834, 120, 1120, 1037

iii (−9, 0) iv (0, −9)

exercise 7d

matrices and transformations b (−6, 0) c (3, −4) 1 a ( 2, 3) d (−3, −2) e (−3, −2) 2 a (m, −n) b (3n, −4m) −

0 3 1 0

3 a

1 0

−

3 −1

1 3

i

f

1 2

1 1

h

1 2

−

1

j

1 2

3 1 b B

−

1 2

1 1 − −

1 −1

−

3 3 5 a D c E 6D 7 a (0, 0), (3, 0), (3, 2), (0, 2) 3 0 b 0 2 n 0 0 n 8 (3 − b, 2a − 5) c

ii (2 2, 2 2) iii (0, 4)

b i

(

− 2,

1 2

9 2

ii

(

iii

(−4,

,

5 2

)

c

i ii

(

5 2

(

(

− 9 2

−5 3 2

− 3, 3

,

2

ChApTer reVieW

−3 2

)

)

d i ii

( (

(

−9 2

9 3 2

,

−9 2

,

9 2

)

)

0

8

0

0

0

− 12

1 2 0 −3

3 3 2

)

+3

1 5 9 13

5

d Impossible 6

y = − x 2 3x − 2 y 0 = 0 3x − 2 y

− 12

− 37

8 x = 11 , y = 11 , z = 11 9 (a – p, b – q)

2 0 10 0 1 11

b [−2]

6 4 5 2

(

1 2

+ 3,1 −

1 2

)

3

3

−

2 6 10 14

B C B A

3 A 7 B 11 D

4 E 8 B 12 E

1 a (A − B)(A + B) = A2 + AB − BA − B2 =

b [3]

C B E A

exTended reSponSe

3 4

−

8

17

+ 3 3,

2

4

1 0 0 1 =I

iv (3, −6) v

−

0

−1

1 3 x − 2 y

iii (−6, −3) −3

−3

y 1 × 2 3 3 x − 2 y

+3 3

)

mULTipLe ChoiCe

2

1 2 2 1 3 x = 3 and y = −1

7 x

+2

5

1 a 4×1 b 5×2

2 −6

− 1 27

16 c = 3

ShorT AnSWer

6

+ 2 3, 2

3 −1

)

5)

−3 3

1 3

+2 3

iv (−5, −4) v

) )

5 a

3)

5 3 2

,

0 6 0 3 0 4 −2 c − 0 2 1

iv (−4, 0) v (2,

2 −3 2

− b moves. Only if tan 1 = 22.5° would a this be true. 13 a OP = 5 = OP′. The transformation could be a rotation (in this case anticlockwise) about the origin. (It could also be a translation, but that is not the question.) b 80.72°

1 −2 4 a 0 3

9 a i (2 3, 2)

−2

2

7 2

b Areas are the same. 12 a No. Single matrix ≠ combined matrices. b No. Under the identity matrix nothing

−

3

)

−3)

( C' = (

d

−

1 2 −

0 − 1

9 3 9 , 2 2

− 3 , B' = (− 2 2, 2 2) 11 a A' = 1 ,

1 0 0 1

−1

0 e 1 g

−1 b 0

0

0 1

c

10 (0,

b (0, −1)

0 1 4 a − 1 0

(

v

1 5 12 62 − 19 13 64 − 257 67 7 4 14 a | A | = 6 b C is not a square matrix. No inverse. 3 c BC = 2 15 P(P − 2) = 3I P−2 1 1 0 = The inverse of P is . 3 3 2 − 3

12

A2 − B2 only if AB – BA, i.e., if A and B are commutative for multiplication. − 54 12 b − − 6 42 a (A + B)(A + B) = A2 + AB + BA + B2 = A2 + 2AB + B2 only if AB + BA = 2AB, i.e., if A and B are commutative for multiplication. b Not equal p = 2, q = 1, r = 2 and s = 3 a (a + 1, b − 3) b (3 − b, −a − 1) c (1 + a, −3 + b) d Translation by (−a, −b) a O′ (0, 0), A′ (8, −3) and B′ (−20, −4) b Yes, by a factor of four. The new triangle is now reflected in the line y = −x. c O″ (0, 0), A″ (3, −8) and B″ (4, 20). d The final coordinates of the vertices are (−3, 1), (0, −7) and (1, 21). − 27i + 18 0 a A3 − 2A2 = − − 14 − 8i 14 − 8i − i 0 − 1 b A 1= 3 − 2 3 i c A2 − A − 2I − 3i 0 2 0 = 9 0 − − − 2 −i 0 2 4i 1 =

0 − 3i 4i − 2 i − 3

− 11

(A − 2I )(A + I ) 3i − 2 = 2 =

−

0 3i + 1 i−2 2

−

0 i +1

0 − 3i 4i − 2 i − 3

− 11

ChApTer 7 • Matrices

337

x 1 6m 2 + 9 = 3 y m + 6 3m − 12

7

2 −1 = 3 1 9 a 145x + 103y + 121z = 20 260 130x + 110y + 90z = 18 400 142x + 115y + 80z = 19 200 145 103 121 x 20 260 b 130 110 90 y = 18 400 19 200 142 115 80 z 8

338

a b c d

−1

145 103 121 c 130 110 90 142 115 80 155 − 1135 101 6068 12 136 1517 − − = 119 2791 67 3034 30 340 1517 67 2049 − 64 6068 60 680 1517 d $70 for an adult’s ticket, $50 for a pensioner’s and $35 for a child’s. 10 a None b Rotation through 180° c Reflection in the x-axis

Maths Quest 11 Mathematical Methods CAS

Reflection in the y-axis Reflection in the line y = x Clockwise rotation of 90° Reflection in the line y = −x Dilation by a factor 3 from both the x-axis and the y-axis i Dilation by a factor 3 from both the x-axis and the y-axis and rotation through 180° j Anticlockwise rotation of 90° d e f g h

Exam practice 3 CHAPTERS 1–7 S ho rT A n S W er

3 1 1 Let A = . − 2 2 a Determine det (A) − b Hence, find A 1.

20 minutes

1 mark 2 marks

3 2 and A = 1 2

2 Let B = a B+A b AB c

1 2

. Determine the following.

−1

4 −2 0

2 marks 3 marks

A − 2B

3 marks

3 The point P (4, −2) has been dilated by a factor of k from the y-axis to form a new

coordinate P′. The new coordinate of P′ is (4, −4).

a Show that k = 1.

2 marks

b P′ is now translated by the matrix

2 −

3

to form a new coordinate R. The coordinates

of R are (2a, b − 1). Determine the value of a and b. 1 1 Let C = 3

−

a 1

A −3 d

2 marks

m U LTip L e C ho iC e

. If C is a singular matrix, then the value of a would be: − 1 B C 0 3

1 3

10 minutes

each question is worth one mark.

e 3

2 The point (1, 4) has been translated by the matrix

−

a . If the coordinates of the new point are b

(−1, 2), then the values of a and b would be: A a = − 2, b = 1 B a = − 2, b = − 2 2 − d a = − 1, b = 1

C a = − 2, b = 2

e a = − 1, b = − 2

2

3 A is a matrix of order m × n, B is a matrix of order p × n and C is a matrix of order n × p. Which one of

the following is correct? (AB)C would result in a matrix of order m × n. A + C would result in a matrix of order (m + n) × (n + p). BC would result in a matrix of order n × p. (AC)B would result in a matrix of order m × n. (BA)C would result in a matrix of order p × (nm) × p.

A B C d e

4 Which one of the following matrix equations would provide a solution to the simultaneous equations

x – 2y = 3, 3y – 2z = 1 and 3x + 4z = 2? x 3 1 A y = 1 = 3 z 2 3 x 1 d y = 3 z 3

2 − 2 4

−

−1

x 1 −2 B y = 0 4 z 3 −2

3 e 1 2

0 3 −2 0 4

−2

x 3 1 y = 1 0 z 2 3

−2

−1

3 1 2

0 − 3 2 0 4

x 1 C y = 3 z 3

−

2 − 2 4

−1

3 1 2

−1

Exam practice 3

339

e x T ended r e SponS e 35 minutes

1 A triangle ABC has been transformed by a factor of 3 from the x-axis and a reflection in the line y = −x.

The transformed triangle has vertices A′, B′ and C′. a i Write down the (2 × 2) matrix that can be used to describe a dilation by a factor of 3 from the x-axis. ii Write down the (2 × 2) matrix that can be used to describe a reflection in the line y = −x. iii Hence, write down the (2 × 2) matrix that can be used to describe the combined transformations. 1+1+1 = 3 marks b A′ has coordinates (−2, 3). Using your matrix from part a iii, determine the coordinates of A. 2 marks c The triangle A′B′C′ is rotated anticlockwise by 90° to form three new vertices, A″, B″ and C″. i Write down the (2 × 2) matrix to describe this transformation. ii Hence, determine the coordinates of B″ if B′ is (−5, −3). 1+1 = 2 marks 2 Jim owns a garden centre. He sells four varieties of fruit trees: apple, plum, nectarine and cherry. He

recorded the number of each tree he sold over a four-week period during spring. In the first week, he sold 4 apple trees, 2 plum trees, 3 nectarine trees and no cherry trees. In the second week, he sold 2 apple trees, 1 plum tree, 3 cherry trees and no nectarine trees. In the third week, he did not sell any plum or apple trees but he did sell 2 nectarine trees and 1 cherry tree. During the fourth week, Jim’s assistant did not record the number of plum trees sold. He did record that he sold 2 apple trees and 3 cherry trees but did not sell any nectarine trees. Let the unit cost, in dollars, of each type of tree sold be represented by the following variables: e = cost of an apple tree f = cost of a plum tree g = cost of a nectarine tree h = cost of a cherry tree. a Write a (1 × 4) matrix to record the number of apple, plum, nectarine and cherry trees sold in the first week. 1 mark b The (4 × 1) matrix C below shows the unit price, in dollars, of each type of fruit tree.

diGiTAL doC doc-10163 Solutions exam practice 3

340

e f C= g h 1 mark Using the matrix above, write the cost of 3 cherry trees. c The matrices W3 and C show the number of trees sold in the third week and the unit cost of each tree. e f W3 C = 0 0 2 1 g h i Write the order of the product of W3C. 1 mark ii In the context of this problem, what would the matrix W3C determine? 1 mark d The number of trees sold over weeks 2 to 4 can be represented in the following matrix, T. 2 1 0 3 T = 0 0 2 1 2 p 0 3 where p is the number of plum trees sold during the fourth week. By adding in your matrix from part a, write down the matrix that represents the total number of fruit trees sold over the four-week period. 1 mark The unit cost of a plum tree is $16.50. The amount of money Jim earns from selling the fruit trees for each of the first three weeks is $195.30, $175.35, $74.65 and ($158.85 + the money from the selling of p plum trees) respectively. e Write down a matrix, A, that shows the amount Jim earned over the four-week period. 1 mark f Set up a matrix equation that would determine the unknown values of e, g, h and p. 2 marks g Using an appropriate method, solve the matrix equation to find the unit cost of an apple tree, a nectarine tree and a cherry tree. 3 marks h The amount of money collected from selling the fruit trees during the fourth week was $208.35. Determine the number of plum trees that were sold during the fourth week. 1 mark

Maths Quest 11 Mathematical Methods CAS

ChapTer 8

Rates of change diGiTal doC doc-9773 10 Quick Questions

ChapTer ConTenTS 8a 8B 8C 8d 8e 8F 8G 8h 8i

Identifying rates Constant rates Variable rates Average rates of change Instantaneous rates Motion graphs (kinematics) Relating the gradient function to the original function Relating velocity–time graphs to position–time graphs Rates of change of polynomials

8a

identifying rates

In any subject dealing with things that move, grow or change it is often important to know the rate (‘speed’) with which quantities change. An example is the amount of water a city consumes per year. Knowing this rate is important so that measures can be taken to ensure there are adequate supplies of water stored in dams or reservoirs. Rates describe how much one quantity changes with respect to another. Example 1. The speed of a car is 70 km/h. This rate is read as 70 kilometres per hour, meaning that the car will travel 70 kilometres in an hour. Example 2. The council rates for a household are $560 per year. As with many rates, examples 1 and 2 involve time as the second variable. But this is not always the case. Example 3. The price of petrol is $1.129 per litre. This rate means that the petrol will cost 112.9 cents for every litre purchased. Rates usually have units associated with them but not always. 1 Example 4. The gradient of a road is 25 . This rate is read as 1 in 25 and means the road rises one unit vertically for every 25 units in length (horizontally). This is often expressed as a percentage, i.e. a 4% grade in this case. Road

1

25

In general, for two quantities A and B: Rate of change of A with respect to B = amount of A per B change in A . = change in B ChapTer 8 • Rates of change

341

Worked example 1

Which of the following represent a rate? a 20 m/s b 75 cents per packet

c $13

Think

WriTe

a 20 m/s is a rate as it compares distance and time. A

a 20 m/s is a rate.

distance of 20 metres is travelled each second. b 75 cents per packet is a rate as it compares cost and

b 75 cents per packet is a rate.

quantity of items. There is a cost of 75 cents for every packet. c $13 is not a rate as there is only one quantity.

e $13 is not a rate.

Worked example 2

A car travels 259 kilometres using 35 litres of petrol. Express this rate in km/L. Think

WriTe

distance travelled amount of petrol used

Rate =

259 km 35 L

Divide 259 by 35 to compare the number of kilometres travelled for each 1 litre of petrol.

=

7.4 km 1L

Express as a simplified rate.

= 7.4 km/L

1

Rate =

2

3

Note: Petrol consumption is usually expressed in L/100 km. 35 In this case, that would be × 100 = 13.5 L/100 km. 259

exercise 8a

identifying rates

1 We1 Which of the following represent a rate? a 100 km/h b $140 per week d 80 beats per minute e $2.75 g 25 kg h 10 000 litres j 18 degrees k 1.4% per quarter m 5 per person n A gradient of 0.8 p Cars per hour 2 a B C d e F G h i

3

342

c f i l o

64% $35.90 per m2 $11.00 per dozen 33 3.4

Match each of the rates you identified in question 1 to the most appropriate description below. The rental fee for a house The slope of a building’s roof The speed of a car The number of cards dealt in a game of poker The cost of oysters at a restaurant The pulse rate of a person walking The number of cars entering a car park The price of carpet The inflation rate one year ago

A horseriding school wishes to organise four-day riding treks for a group of twelve people. The following arrangements are made. a Each day the group will ride off at 9 am. b They will travel 20 km each day. c There will be four rest stops per day. d Each person must bring sleeping gear. e They will sleep two people per tent. f Meals will be supplied. g The cost will be $200 per person. h Three spare horses will be taken. Which of these arrangements represent a rate?

Maths Quest 11 Mathematical Methods CAS

4 Find the cost of the following items. a 5 dozen eggs at $2.12 per dozen c 27 hours of work at $11.25 per hour e 3.8 m3 of sand at $19.75 per m3

b 42.5 litres of petrol at 73.9 cents per litre d 4.9 kg of oranges at $1.14 per kg f 55 427 people at $11 per person

5 mC Francis delivers 440 pamphlets in 2 hours and earns $5 per 100 pamphlets delivered. a The rate of delivery of pamphlets per hour is: a 220 B 20 C 880 d 440 e 2.5 b How long will it take Francis to earn $55? a 4 hours B 4 hours, 20 mins C 5 hours d 6 hours e 3 hours 6 We2 Express the following as the rate indicated in the brackets. a Running 100 m in 12 seconds (m/s) b A rise of 21 °C in 6 hours (°C/h) c A meal costing $250 for 10 people ($/person) d 2.34 kg of washing powder added to 9 litres of water (g/L) e 32 m2 of lawn costing $149.50 ($/m2) f Losing 12.76 kg in 11 weeks (kg/week)

diGiTal doC doc-9774 SkillSHEET 8.1 introduction to rates of change

7 A 100-litre tank filled with water is leaking water at the rate of 2.5 litres per hour. a How long will it take for the tank to be half full?

Once it is half full, water is added to the tank at the rate of 3 litres per hour. b How much longer will it take until the tank is full again? c mC Which of the graphs below correctly shows the volume of water in the tank, v litres, as a function of time, t hours? a

v

100

100

50

50

50

120 t

v

20

120 t

20

120 t

20

120 t

v

e

100

100

50

50 100120 t

8B

v

C

100

20 d

v

B

Constant rates

When the rate of change of one quantity with respect to another does not alter, the rate is constant. For example, if petrol is $1.60 per litre, then every litre of petrol purchased at this rate always costs $1.60. This means 10 litres of petrol would cost $16.00 and 100 litres of petrol would cost $160.00. Calculating the gradient from the graph: rise run $(160 − 16) = (100 − 10) litres $144 = 90 litres

Gradient =

=

$1.60 1litre

Price ($)

160

16 10

100

= $1.60/litre Number of litres (L) = rate That is, the gradient of this graph represents a rate. A straight line graph shows that the rate of change between two quantities is constant.

ChapTer 8 • Rates of change

343

Worked Example 3

The graph shown represents the height, H, in metres (m) of a hot air balloon and time, T, in hours (h) after it is launched from a mountain top. Use the graph to find: a the gradient b the rate of change c the rule for height as a function of time.

Height (m)

1100

500

Time (h) Think

a Use the gradient of a straight line, m =

5

Write

rise . run

b Rate of change is the same as the gradient. Include

appropriate units. c As the graph is a straight line, express the rule in the

form y = mx + c. Remember that m is the gradient and c is the y-intercept.

1100 − 500 5− 0 600 = = 120 5

a Gradient =

b Rate of change = 120 m/h c The rule is of the form H = mT + c where

m = 20 and c = 500. H = 120T + 500

Worked Example 4

The table below shows the number of people, N, left inside an office building T minutes after the alarm goes during a fire drill. a Show that the relationship is linear. T 0 1 2 3 4 b Find the rate of change. N 140 105 70 35 0 c Express N as a function of T. Think

a 1 Plot the graph of the points given.

Write/draw

a

N 140 105 70 35 1

2

Check whether the points are collinear, that is, whether they lie on a straight line.

b 1 Calculate the gradient, m.

2

Use the gradient to state the rate of change. Include the appropriate units.

c Express the rule in the form y = mx + c where m = –35

and c = 140. Use N and T instead of y and x.

344 Maths Quest 11 Mathematical Methods CAS

2

3

4

T

A straight-line graph was produced, so the relationship is linear. 0 − 140 4−0 − 140 = = −35 4 Rate of change = −35 people/min The number of people left inside the building is decreasing at a constant rate.

bm=

c The rule is N = −35T + 140.

exercise 8B 1 a b c d e f g h i j k 2

a

Which of the following rates are constant? A person’s pulse rate when running 3 km The rate of growth of Australia’s population A person’s pulse rate when lying down The daily hire rate of a certain car The rate of growth of a baby The rate of temperature change during the day The commission rate of pay of a salesperson The rate at which the Earth spins on its axis The rate at which students arrive at school in the morning The rate at which water runs into a bath when the tap is left on The number of hours of daylight per day Which of the following graphs show a constant rate of change?

t

t

t

t

d

d

0 g

d

0 j

c

d

0 f

d

0 3

t

d

0 i

b

d

0 e

Constant rates

t

t

0 h

d

0

d

d

t

t

0

d

0

t

For the constant rates of change found in question 2, which of these rates are: i positive? ii negative? iii zero?

The distance travelled by a car is shown in this graph. The graph shows that the car is: a slowing down B speeding up C travelling uphill d travelling at constant speed e stationary 5 mC The volume of water, v litres, in a container is shown as a function of time, t minutes, in the graph. The rate of change of water in the container is: a 2 L/min B 8 L/min C −2 L/min d −8 L/min e 0 L/min

Distance (metres)

4 mC

0

Time (hours)

v (litres) 18 10

0

4

t (min)

ChapTer 8 • Rates of change

345

6 We3 For each of the functions graphed below, state: i the gradient ii the rate of change iii the rule. a d (m) b v (litres) 260

200

0 c

50 4

t (h)

0 d

w (kg) 100

30

t (min)

h (m) 75

60

0 e

40

0

t (weeks) f

w (g) 16

t (years)

10

w (g) 30

10

0

8

5 0

t (min)

120 v (litres)

20

7 mC Water is being pumped out of a swimming pool holding 30 000 litres at a constant rate. It takes

25 minutes to remove 1000 litres of water. a The rate of change of the volume of water in the pool is: a 400 L/min B 1200 L/min e 40 L/min d −40 L/min b The time taken to empty the pool is: a 12 hours, 5 minutes B 12 hours, 30 minutes d 2 days e 18 hours 8

C

−1200

C 6 hours, 30 minutes

The number of crates of fruit picked by a fruit-picker over the course of a day is shown in the graph.

Number of crates

13

7 6

5 pm

11 am 12 pm 1 pm

8 am

0 Time

If the fruit-picker is paid $12 per crate, answer the following. a What is the rate of pay per hour in the first 3 hours? b Explain what probably happened between 12 pm and 1 pm. c What is the rate of pay per hour in the last 4 hours? d Suggest two possible reasons why the line is not as steep in the afternoon. e How much is earned for the day? 346

L/min

Maths Quest 11 Mathematical Methods CAS

9 An athlete runs at a constant speed of 6 m/s for the first 420 m of a race. The next 400 m are run at a

constant speed of 5 m/s and the final 180 m at a constant speed of 6 m/s. a How far is the race? b How long does it take to run the first 420 m? c How long does it take to run the next 400 m? d How long does it take to run the final 180 m? e Sketch the graph of distance, D metres, versus time, T seconds. 10 We4 A spring is stretched a distance x cm beyond its natural length when a weight of W kg is attached to one end as shown in the following table. x

0

5

10

15

20

25

W

0

1

2

3

4

5

diGiTal doC doc-9775 plotting relations

a Show that there is a linear relationship between the variables. b What is the rate of change of W with respect to x? c Express W as a function of x.

A bucket contains 40 litres of water when it starts leaking through a hole in the bottom, such that the volume of water, V litres, in the bucket at any time, t hours, is given by the rule V = 40 − 6t. a At what constant rate, in litres/hour, is the water leaking? b How long will it take for the bucket to be empty? 12 Water is poured at a constant rate into the vessels a, b, c and d shown below. 11

a

b

c

d

Match the appropriate graph below, representing height of water, h, at any time, t, with each vessel above. i

ii

h

0

8C

t

iii h

h

0

t

0

Variable rates

If a rate is not constant (is changing), then it must be a variable rate. For example, the amount of electricity used per hour by a household is most likely to be a variable rate as the need for electricity will change throughout the day. Graphically, a variable rate will not be linear, because the gradient (or rate of change) is not constant. The distance–time graph at right shows a variable rate between t = 0 and t = 3, because the gradient is changing over this interval. However, it shows a constant rate between t = 0 and t = 1; t = 1 and t = 2; and t = 2 and t = 3, because the gradient is not changing over these intervals. Relationships that give curves when graphed show that the rate of change is continually changing; that is, a variable rate exists.

iv h

t

t

0

d

0

1

2

3

t

ChapTer 8 • Rates of change

347

Worked example 5

Consider the graph shown. For each interval, state whether: a the rate of change is constant or variable b the rate of change is positive, negative or zero.

B

C TUTorial eles-1432 Worked example 5

A Think

D

WriTe

a Consider whether the graph is linear for each

a Interval AB: variable rate of change

interval. The rate of change is constant if the graph is linear; otherwise it is variable.

Interval BC: constant rate of change Interval CD: constant rate of change

b Since the gradient of the graph gives a measure

of the rate of change, consider whether the gradient is positive, negative or zero in each interval.

exercise 8C 1 a b c d e f g h 2

a

b Interval AB: positive rate of change

Interval BC: zero rate of change Interval CD: negative rate of change

Variable rates

List which of the following rates are variable. The pulse rate of a person watching a horror movie The cost per kilogram of tomatoes at the local greengrocer today The speed of a car travelling through the city The cost per kilogram of tomatoes at the local greengrocer during the year The number of hours of sunlight per day The hourly rate of a receptionist The hourly rate of a private taxi driver The speed of a marble rolling down an incline Which of the following graphs show variable rates of change between points A and B? b

B

B A A

c

d

B

B A

A

f

e

A

B

B

A

348

Maths Quest 11 Mathematical Methods CAS

The graph at right shows how the distance of a runner from the starting line changes over the duration of a race. 100 For the graph shown, state whether: a the rate of change is constant or variable 50 b the rate of change is positive or negative. A c mC The rate of change of distance with respect to time is greatest at point: a A B B C C d D e E d The runner is travelling slowest (but still moving) at point: a A B B C C d D e E e At point E, the runner is: a moving very fast B speeding up d not moving e just beginning to move

3 We5

Distance (m)

4

E

D C B 6 Time (s)

12

C slowing down

The following table shows how weight, W (kg) varies with time, t (min), for a particular box as it is filled. t (min) W (kg) a b c d

5

0 0

1 3

2 6

3 9

4 12

Plot a graph of W versus t. What do you notice about the plotted points? Is the rate of change variable or constant? What is the rate of change in kg/min? The following table also shows weight, W (kg), varying with time, t (min). t (min) W (kg)

0 2

1 4

2 7

3 11

4 18

Plot these points on a graph. Do these points form a straight line? Is the rate of change variable or constant? What is the rate of change during: i the first minute? ii the second minute? iii the fourth minute? 6 The following tables of values show distance travelled, d km, at various times, t hours. Decide whether the rate of change of distance with respect to time appears constant or variable. a b c d

a

b

8d

t d

0 0

1 5

2 10

3 15

4 20

t d

0 0

1 15

2 20

3 45

4 80

diGiTal doC doc-9775 plotting relations

average rates of change

If a rate is variable, it is sometimes useful to know the average rate of change over a specified interval. For example, a tree grew from 6.2 m this time last year to 6.75 m now. The tree’s average rate of growth =

change in height change in time

(6.75 − 6.2) m 1 year = 0.55 m/year =

This means that the tree grew by 0.55 metres over the past year but not necessarily constantly at that rate during the year. ChapTer 8 • Rates of change

349

Worked example 6

Calculate the average rate of change between points P and Q marked on the graph. Q

Distance (m)

48 P

30

0

2

Think

Time (s)

7

WriTe

48 − 30 18 = = 3.6 5 7−2

1

Find the gradient of the chord PQ.

Gradient of PQ =

2

Use the gradient to state the average rate of change. Include the appropriate units.

Average rate of change = 3.6 m/s

Worked example 7

Find the average rate of change of height between t = 1 and t = 3 from the table below. t (min)

0

1

2

3

4

5

h (m)

20

60

90

130

140

145

Think

Calculate the average rate of change of height with respect to time by considering the change in each quantity. When the time changes from t = 1 min to t = 3 min, the height changes from 60 m to 130 m.

WriTe

Average rate of change of height change in height = change in time (130 − 60) m = (3 − 1) min 70 m = 2 min = 35 m/min

Worked example 8

Over a period of 6 hours, the temperature of a room is described by the function T(h) = h2 − 4h + 22, where T is the temperature in degrees Celsius after h hours. a What is the initial temperature of the room? b Sketch the graph of the function over the given time interval. c Draw a chord between the points where h = 1 and h = 5. d What is the gradient of this chord? e What is the average rate of change of temperature between h = 1 and h = 5? Think

a The initial temperature is the temperature at the

start of the time period. Substitute h = 0 into the function to find T(0).

350

Maths Quest 11 Mathematical Methods CAS

WriTe/draW

a When h = 0, T(0) = 0 − 0 + 22

= 22 The initial temperature is 22 °C.

TUTorial eles-1433 Worked example 8

b 1 To find the turning point, let h =

2

−

b . 2a

T (2) = 4 − 8 + 22 = 18 The turning point of the parabola is (2, 18).

Evaluate T (2).

c 1 Find the required points:

c T (°C) 27

T(1) = 12 − 4(1) + 22 = 19 T(5) = 52 − 4(5) + 22 = 27.

2

Indicate the points (1, 19) and (5, 27) on the graph and join with a straight line. rise and the points (1, 19) Use gradient = run and (5, 27).

d

Use the gradient to state the average rate of change. Include appropriate units.

e

exercise 8d

18

(2, 18) 1

2

3

5 H (hours)

4

27 − 19 5−1 8 = 4 =2

d Gradient =

e The average rate of change is 2 °C/h.

average rates of change b

y 8

y 10

P

0 −2

2

diGiTal doC doc-9776 SkillSHEET 8.2 Gradient of a straight line

Q

P

6

0 2

19

(5, 27)

22 (1, 19)

Calculate the average rate of change between the points P and Q marked on each graph.

1 We6 a

−

b 2a 4 = 2 =2

bh=

5

x

Q

5

x

Which of the following graphs have a straight line drawn that can aid in finding the average rate of change over a specified interval? a

y

b

x

0 c

y

0

0 d

x

y

x

y

0

x

ChapTer 8 • Rates of change

351

3

For each of the distance–time graphs below:

i draw a chord to the graph for the interval t = 1 to t = 3 ii find the gradient of this chord iii hence, find the average speed from t = 1 to t = 3. a d (km) b d (km) 70 70 50

10 0 c

1

0

t (h)

3

d

d (km) 70

1

3

t (h)

1

3

t (h)

1

3

t (h)

d (km) 70 60

40

0 e

1

0

t (h)

3

f

d (km) 70

d (km) 90 70

0 4

1

3

0

t (h)

The graph at right represents the weight of a non-uniform beam, W (kg), versus its length, L (m). a Find the gradient of the chord to the graph for the interval L = 5 and L = 20. b Find the gradient of the chord from L = 15 and L = 20. c What is the average rate of change between: i L = 5 and L = 20? ii L = 15 and L = 20? d Are these rates the same? Explain your answer.

W (kg) 32.5 30

10 0

Consider the graph at right. The chord with the highest gradient is the one from point 0 to point number: a 4 B 3 C 2 d 1 e 5

5 10 15 20 L (m)

5 mC

2

3

4

5

1

0 6

The total number of people at the zoo at various times of the day is shown in the table below. am T (time of day) N (number of people at the zoo)

10.00 0

pm 11.00 200

a Plot the graph of N versus T. b Draw chords to the graph for the interval: i 10.00 am to 1.00 pm ii 1.00 pm to 3.00 pm iii 3.00 pm to 5.00 pm. 352

Maths Quest 11 Mathematical Methods CAS

12.00 360

1.00 510

2.00 540

3.00 550

4.00 550

5.00 550

c Find the gradient of each of these chords. d What is the average rate of change from: i 10.00 am to 1.00 pm? ii 1.00 pm to 3.00 pm? iii 3.00 pm to 5.00 pm? e Briefly describe what these rates suggest about the number of people attending the zoo during the

course of the day. The height, h metres, reached by a balloon released from ground level after t minutes is shown in the table below.

7 We7

t (mins) h (m)

0 0

2 220

4 360

6 450

8 480

10 490

a Without drawing the graph, find the average rate of change of height with respect to time

between: i t = 0 and t = 2 ii t = 2 and t = 4 iii t = 4 and t = 6 iv t = 6 and t = 8 v t = 8 and t = 10. b Is the average rate of change for each 2-minute interval increasing or decreasing? 8 We8 The temperature of an iron rod placed in a furnace is described by the function T(t) = t2 + 20

between t = 0 and t = 10. T represents the temperature of the rod in degrees Celsius and t is the time in seconds. a What is the initial temperature of the iron rod? b Sketch the graph of the function over the given time interval. c Draw a chord between the points where t = 2 and t = 8. d What is the gradient of this chord? e What is the average rate of change of temperature between t = 2 and t = 8?

9 The weight of a rabbit in its first eight weeks is described by the function W(t) = t3 − 3t2 + 200, where

W represents the weight of the rabbit in grams and t is the time in weeks. a What is the weight of the rabbit at birth? b What is the average rate of change of weight of the rabbit between: i weeks 1 and 3? ii weeks 1 and 2? iii weeks 2 and 4? iv weeks 2 and 6?

8e

diGiTal doCS doc-9777 Gradient between two points doc-9778 WorkSHEET 8.1

instantaneous rates

If a rate is variable, it is often useful to know the rate of change at any given time or point, that is, the instantaneous rate of change. For example, a police radar gun is designed to give an instantaneous reading of a vehicle’s speed. This enables the police to make an immediate decision as to whether a car is breaking the speed limit or not. y

P 0

Tangent x

Instantaneous rates can be found from a curved graph by: 1. drawing a tangent to the curve at the point in question 2. calculating the gradient of the tangent over an appropriate interval (that is, between two points whose coordinates are easily identified). Note: The gradient of the curve at a point, P, is defined as the gradient of the tangent at that point. ChapTer 8 • Rates of change

353

Worked example 9

a Use the following graph to find the gradient of the tangent

W (kg)

at the point where L = 10. b Hence, find the instantaneous rate of change of weight, W, with respect to length, L, when L = 10.

10 Think

WriTe/draW

a 1 Draw in the required tangent. Do this ‘by eye’

a

(estimation). 2

W (kg) 12

To find the gradient of the tangent, choose a convenient interval (between the points where L = 5 and L = 15).

8 4 0

3

Use gradient =

appropriate units.

10 15 20 L (m)

5

4−8 15 − 5 −4 = 10 = − 0.4

rise . run

b Use the gradient to state the rate of change. Include

Gradient =

b Rate = −0.4 kg/m

The weight is decreasing with respect to length at a rate of 0.4 kg/m.

Worked example 10

a By plotting the graph of y = x2 − 2, find the instantaneous rate of change

where x = 1.

b Find the instantaneous rate of change of y = x2 − 1 at x = 2 using a calculator. Think

a 1 Plot the graph. 2

Draw a tangent to the curve at the point where x = 1. Do this by eye (estimation).

WriTe/draW

a

y 2 1 −2 −1 0 −1

1

2 x

−2 −3

3

354

L (m)

Find the gradient of the tangent. (Use points (1.5 O) and (0, −3) on the tangent line.)

Maths Quest 11 Mathematical Methods CAS

0 − (− 3) 1.5 − 0 3 = 1.5 =2

Gradient =

4

Use the gradient to state the rate of change. Note: There are no units to include as the required rate of change is the gradient.

b 1 Define the function using the CAS calculator.

Hence, the rate of change is 2.

b f1(x) = x2 − 1

2

Use the draw tangent feature of the CAS calculator to draw a tangent to f1(x) at (2, 3).

f1(2) = 22 − 1 =3

3

Use the coordinates and equations feature of the CAS calculator to determine the equation of the tangent line. Point the cursor to (2, 3) on the graph.

ytangent = 4x – 5

4

Answer the question.

The equation of the tangent is ytangent = 4x – 5. The gradient is 4. Hence, the instantaneous rate of change at (2, 3) is 4.

instantaneous rates

exercise 8e

Describe each of the rates below as either an instantaneous rate (I) or an average rate (A).

1

a Bill walks 12 kilometres in 2 hours. b An aircraft leaves the runway at 270 km/h. c A household uses 560 litres of water in one

day. d The pulse rate of a runner as he crosses the

finish line is 141 beats per minute. e A gas heater raises the temperature of a

room by 10 °C in half an hour. f A baby puts on 300 g in one week. g A road drops 20 m over a distance of 100 m. h Halfway along a flying fox, Jill is travelling

at 40 km/h. 2

a

Which of the following graphs have a tangent drawn at the point where x = 4? y

0 c

4

x

y

0 e

b

x

y

0

0 d

4

x

4

x

y

0 f

4

y

4

x

y

0

4

x

ChapTer 8 • Rates of change

355

g

y

h

0

x

4

y

0

x

4

In each of the following graphs, find an approximate value for the gradient of the tangent at the point indicated.

3 We9a a

y

b

x=1

3 2

y 20

x=2

10

1 0 c

1

x

2

0

y

d

5

2

y x=4

22 20

x = 1.7

x

4

2 0

1 1.7

4

x

0

4

x

8

Find the approximate value for the instantaneous rate of change of weight with respect to time at the time indicated on the graph.

4 We9b a

b

w (g)

t = 20

130

160

70

t = 10

80 0

c

w (g)

0

10 20 30 t (min)

w (g)

d

140

5 10 15 t (min)

w (g) 110 t=1

t=4 50

40 0

2

4

0 1

6 t (min)

5

t (min)

5 The weight of a person over a 40-week period is illustrated in the graph below. diGiTal doC doc-9779 Gradient at a point

Weight (kg)

90 80 70 60 50 40

0

10

20 30 Time (weeks)

40

a Estimate the maximum weight and the time at which it occurs. b Estimate the average rate of change of weight between week 10 and week 20. c Estimate the rate of change of weight at 30 weeks. 356

Maths Quest 11 Mathematical Methods CAS

6 The change of temperature in a restaurant kitchen over a 12-hour period is illustrated in the graph below.

Temperature (°C)

35 30 25 20 15 10 5 0

0 1 2 3 4 5 6 7 8 9 10111213

Time (hours) a When is the rate of temperature rise greatest? b Estimate the rate of temperature rise at 6 hours. 7 a Plot the graph of y = x2 from x = 0 to x = 4. b Draw tangents at the points where x = 1, x = 2 and x = 3. c mC The gradient of the tangent at x = 1 is closest to: a 0 C 2 B −1 d mC The gradient of the tangent at x = 2 is closest to: a 8 B 6 C 4 e mC The gradient of the tangent at x = 3 is closest to: a 6 B 9 C 3

d 3

e 4

d 2

e 0

d 7

e

−3

8 We10a a Plot the graph of y = 9 − x2 from x = −4 to x = 4. b Draw tangents at the points x = −2, x = 0 and x = 2. c Hence, estimate the instantaneous rate of change at the points: i x = −2 ii x = 0 iii x = 2. 9 We10b A tank is being filled at a variable rate. The depth of the water, H cm, at any time, t minutes, is

described by the rule H = t2 + 2t. At what rate is the depth of water changing after 2 minutes?

8F

motion graphs (kinematics)

In this section we will consider only objects moving in straight lines, either right and left, or up and down. Definitions 1. Position, x, describes the location of an object. 2. Distance, d, is how far an object has travelled. 3. Displacement, s, describes the change in an object’s position; that is, displacement = change in position = final position − initial position, or s = xfinal − xinitial. distance d 4. Speed = or speed = . time taken t 5. Velocity is the rate of change of position with respect to time, so change in position displacement s velocity = = , that is, v = . change in time t time taken Consider an object that begins at the –3-metre mark on a number line, moves 5 metres to the right, and then moves 5 metres to the left, taking 2 seconds to do so. For this example we have: x 1. Position (initially and finally) = −3 m −4−3−2−1 0 1 2 3 4 2. Distance = 10 m 3. Displacement = 0 m 10 m 4. Speed = = 5 m/s 2s 0m 5. Velocity = = 0 m/s 2s Distance and speed are (technically) always positive. Displacement and velocity can be either positive or negative, depending on the direction of motion.

inTeraCTiViTY int-0267 motion graphs (kinematics)

ChapTer 8 • Rates of change

357

Note: Some texts use displacement and position interchangeably (perhaps assuming displacement from a fixed origin). Worked example 11

Consider a lift starting from the ground floor, moving up to the top floor, stopping and then coming down to the fifth floor at the times shown on the diagram. Find: Top a the total distance travelled by the lift t = 14 s t = 20 s b the displacement of the lift after 25 s c the average speed of the lift d the average velocity of the lift.

TUTorial eles-1434 Worked example 11

Positive 100 m

150 m t = 25 s (5th floor)

t=0 Think

WriTe

a Add the distance travelled up (150 m) to the

distance travelled down (100 m). b At t = 0 s, the position of the lift is 0 m.

At t = 25 s, the position is +50 m.

c Average speed =

Ground floor

total distance travelled time taken

d Average velocity =

change in position change in time

a Total distance = 150 m + 100 m

= 250 m

b Displacement = change in position

= +50 – 0 = +50 m

250 m 25 s = 10 m/s

c Average speed =

(+ 50 − 0) m 25 s = +2 m/s

d Average velocity =

Worked example 12

Use the position–time graph at right, which shows the position of a particle, x, travelling in a horizontal straight line at any time, t, to find: a where the journey started b in which direction the particle moved initially c when and where the particle changed direction d when and where the particle finished its journey. Think

a Find x when t = 0.

x (m) 20 10 5 0

2

4 t (s)

WriTe

a When t = 0, x = 10 m. The journey started

10 metres from the origin or reference point. b Compare the position at t = 0 to the position a

short time after this. Notice that the graph has a negative gradient during this time.

358

Maths Quest 11 Mathematical Methods CAS

b As the position changes from 10 m at t = 0 s to

5 m at t = 2 s, the particle was moving towards the origin or reference point in the negative x-direction.

c Find x and t where the graph turns or changes

c After 2 s at a position of 5 m, the particle changed

direction. The particle changes direction at t = 2. When t = 2, x = 5.

direction.

d Find x and t where the curve stops.

d The particle finished its journey at a position of

20 m after 4 s. Worked example 13

Using the graph in worked example 12, consider the interval between t = 2 and t = 4 to decide whether: a the velocity of the particle is positive or negative b the particle is speeding up or slowing down. Think

WriTe

a 1 For this graph,

as

rise (or gradient) is the same run

a

change in position (or velocity). change in time So the gradient of a position–time graph gives the velocity. 2

Consider the gradient between t = 2 and t = 4.

From t = 2 to t = 4 the gradient is positive, so the velocity of the particle is positive.

b Consider how the gradient between t = 2 and

b The gradient is increasing, so the velocity is

t = 4 is changing.

increasing. This means the particle is speeding up.

Similarly, for the interval between t = 0 and t = 2 in worked example 12: x (m) 20

Positive gradient getting steeper Negative gradient getting less steep

10 5 0

2

4

t (s)

1. the velocity is negative because the gradient is negative 2. the magnitude of the velocity is decreasing because the magnitude of the gradient is decreasing. The negative gradient indicates the direction in which the particle is moving. 3. the speed is decreasing between t = 0 and t = 2 because the magnitude of the gradient is decreasing.

exercise 8F 1

motion graphs (kinematics)

Match the correct description (a, B, C or d) to each of the quantities (a, b, c or d) below. Quantity a Distance b Displacement c Speed d Velocity

a B C d

Description Rate of change of displacement with respect to time Change in position Length travelled Distance travelled with respect to time

2 State whether each of the following are true or false. a Speed can be positive or negative. b Displacement can be positive or negative. c Distance travelled is positive only. d Velocity is positive only. ChapTer 8 • Rates of change

359

A parachute ride takes people in a basket vertically up in the air from a platform 2 metres above the ground, then drops them back to the ground. Use the diagram below showing the position of the parachute basket at various times to find: a the total distance travelled by the parachute basket t = 55 s t = 60 s during a ride b the displacement of the parachute basket after 80 s c the average speed of the parachute basket during the ride d the average velocity of the parachute basket during the ride.

3 We11 diGiTal doC doc-9780 SkillSHEET 8.3 displacement and distance

19 m

t=0 2 m Platform 4

Consider the position and direction, at various times, of a particle travelling in a straight line as indicated below. t=5

a b c d e

t=2

t=0

−3 −2 −1

0

1

2

3

4

5

6

7

x

Where does the particle start? Where does the particle finish? In which direction does the particle move initially? When does the particle change direction? mC Which position–time graph below most accurately represents the particle’s motion? a

x (m) 6 5 4 3 2 1 0 −1 −2 −3

d x (m) 6 5 4 3 2 1 0 −1 −2 −3

360

t = 80 s Ground

B

2

5 t (s)

e

2

Maths Quest 11 Mathematical Methods CAS

5 t (s)

x (m) 6 5 4 3 2 1 0 −1 −2 −3

C

2

5 t (s)

2

5 t (s)

x (m) 6 5 4 3 2 1 0

x (m) 6 5 4 3 2 1 0 −1 −2 −3

2

5 t (s)

The total distance travelled in the first 5 seconds is: a 4m B 13 m d 14 m e 11 m g The displacement of the particle after 5 seconds is: B 14 m a −3 m e 0m d −4 m h The average speed in the first 2 seconds is: a 3 m/s B −2.5 m/s d 2.5 m/s e 1 m/s i The average velocity between t = 2 and t = 5 is: a 3 m/s B −2 m/s d 2 m/s e 1 m/s j The instantaneous speed when t = 2 is: a 2.5 m/s B 0 m/s d 2.8 m/s e 1 m/s f

C 9m

C 4m

C 6 m/s

C

−3

m/s

C 3 m/s

The following position–time graphs show the journey of a particle travelling in a straight line. For each graph, find: i where the journey started ii in which direction the particle moved initially iii when and where the particle changed direction iv when and where the particle finished its journey.

5 We12

a x (m) 8

0

b x (m) 12

c x (m) 12

10

10

4

3

1 2 3 4 5 t (s)

−3

0

d x (m) 18

e x (m)

0

1 2 3 4 5 6 t (s) f

5

1 2 3 4 5 6 7 8 t (s)

x (m) 5 4 2

0

1 2 3 4 5 t (s)

0

1 2 3 4 5 6 t (s)

−3 0

1 2 3 t (s)

−5

−5 6 For each position function of a particle given below (a–f ), sketch the position–time graph. In each case

explain: i where the particle started its journey ii in which direction it moved initially iii whether the particle changed its direction and, if so, when and where that happened iv where the particle finished its journey. a x(t) = 2t, t ∈ [0, 5] b x(t) = 3t − 2, t ∈ [0, 6] c x(t) = t2 − 2t, t ∈ [0, 5] d x(t) = 2t − t2, t ∈ [0, 4] e x(t) = t2 − 4t + 4, t ∈ [0, 5] f x(t) = t2 + t − 12, t ∈ [0, 5]

diGiTal doC doc-9781 SkillSHEET 8.4 interval notation

ChapTer 8 • Rates of change

361

7 a Plot the position–time graph for x(t) = 4t − t2. b Find the gradient at: i t=0 ii t = 1 iii t = 2 iv t = 3 v t=4 c Hence, give the instantaneous rate of change of position with respect to time (that is,

velocity) at: i t=0 ii t = 1 iv t = 3 v t=4 d Sketch the velocity–time graph from t = 0 to t = 5.

iii t = 2

8 We13 For each of the following position–time graphs, determine: i whether the velocity is positive or negative ii whether the particle is speeding up or slowing down. x b x c x a

0

d

e

x

0 9

t

0

t

t

x

0

t

Match the following position–time graphs (a–f ) with the corresponding velocity–time graphs (a–F ) on the opposite page. Position–time graphs a

c

x (m)

x (m)

3

3

3

2

2

2

1

1

1

0

d

b

x (m)

−1

1

2

3

0

4 t (s)

−1

1

2

3

0

4 t (s)

−1

−2

−2

−2

−3

−3

−3

e

x (m)

x (m)

f

3

3

2

2

2

1

1

1

−1

1

2

3

4 t (s)

0 −1

−2

−2

−3

−3

Maths Quest 11 Mathematical Methods CAS

1

2

3

4 t (s)

1

2

3

4 t (s)

1

2

3

4 t (s)

x (m) 4

3

0

362

0

t

0 −1 −2 −3 −4

Velocity–time graphs B v (m/s)

a v (m/s)

C v (m/s) 1

0

1

2

3

4 t (s)

1

2

3

1

2

3

4 t (s)

0

0

4 t (s)

e v (m/s)

d v (m/s)

0

0

1

2

3

4 t (s)

1

2

3

4 t (s)

F v (m/s)

1

2

3

0

4 t (s)

−1

10 A ball is projected vertically upwards from the top of a building 25 m high. Its position relative to the

11

The diagram at right shows the distance–time graph of two swimmers in a 200-metre race. a Who won the race? b Who was in front after 150 metres? c What was the approximate winning margin (in metres)? d What was the average speed of the winner? e For approximately how many seconds did Bill lead the race? f How many times did the lead change?

Distance (metres)

ground is given by the equation x = 25 + 20t − 5t2, where t is the time in seconds. Sketch a position–time graph for the ball and hence find: a the greatest height reached b when the ball reaches the ground c when the velocity of the ball is zero d an estimate for the velocity at which the ball is initially projected.

200 150

Alan

100 50 0

Bill 20 40 60 80 100 120 Time (seconds)

diGiTal doC doc-9782 WorkSHEET 8.2

ChapTer 8 • Rates of change

363

relating the gradient function to the original function 8G

Earlier we considered gradients calculated at a single point on a graph. However, if the gradients are calculated at every point within the domain of the function, the results can be graphed. This is the graph of the gradient function. The gradient of the function below has been calculated at several points as shown, and a table of gradient values has been constructed. Graph of original function y y = x3 − 2x2 + 2

6 4 Gradient = 7 2

0 1 −2 −1 −2

x Gradient

2

x

−1.0

−0.5

0

0.5

1.0

1.3

1.5

2.0

7

2.75

0

−1.25

−1

0

0.75

4

The table of values above may by be used to sketch a gradient function such as that at right. 4 Note: The gradient function is zero at x = 0 and x = 3 , which corresponds to the zero gradient (turning points) of f (x). The gradient function is negative in 4 the interval 0 < x < 3 , as is the gradient of f (x). The gradient function is positive elsewhere, which corresponds to the positive gradient of f (x).

Graph of gradient function y 8 6 4 2 −2 −1−2 0 1

2

relating the gradient function to the original function exercise 8G

1 Select several points (at least 6) on each of the graphs below and construct a table of gradient values. diGiTal doCS doc-9779 Gradient at a point doc-9775 plotting relations

Use each table to plot the gradient function in each case. You may use a CAS calculator to work through these problems. a

y y = x2 − 4 4

b

4

2

2

x −2−1 01 2 −2

e

y

2

x

−3−2−1 01 2 x −2 y = x3 + 3x2 + 2

2 1 y = ex

−3−2−1 0 1 2 3

f

y

4 2

4

−4

6

πx y = sin —– 4

0 1 2 3 4 5 6 7 8x

x

−1

y 6

y = 2x + 3

−3−2−1 0 1 2 3 −2

−4

d

c

y

y 40 32 24 16 8

y = x4

x −2−1 01 2

2 What do you notice about the gradient function of the quadratic function in question 1a? 3 Investigate other quadratic functions and their related gradient functions. What do you notice? 364

Maths Quest 11 Mathematical Methods CAS

x

4 Investigate the gradient functions of other cubics besides that shown in question 1c. What types of

functions are the gradient functions of cubics? 5 Which functions have a gradient function similar to the original function? 6 What do you notice about the original functions at x-values where the gradient function intersects the

x-axis? 7 When the gradient function is negative, what do you notice about the original function?

Relating velocity–time graphs to position–time graphs 8H

The gradient of a position–time graph gives the velocity, because velocity is the rate of change of position with respect to time. Therefore, by measuring the gradient of a position–time graph at various points, a velocity–time graph can be derived. Worked Example 14

The position–time graph for a particle moving in a straight line is shown below. x 3 2 Gradient = −2

1 Gradient = 2 0 −1

1

2

3

−2 Gradient = −1

4

5

t

Gradient = 1 Gradient = 0

The gradient of the curve at various times is indicated on the graph. Use this information to draw a velocity–time graph for the particle. Think 1

2

Set up a table of corresponding velocity and time values from the graph. Use the table of values to plot the velocity– time graph.

Write/draw

t v

0

1

−2

−1

2 0

3 1

4 2

v 3 2 1 −3 −2 −1 0 −1

1

2

3

4

5

t

−2

Similarly, a position–time graph can be derived from a velocity–time graph. The value of the velocity at a given time is equal to the gradient of the position–time graph at that time. By obtaining several of these values and knowing the initial position of a particle, the general shape of the position–time graph can be produced. The following example demonstrates the steps involved. Chapter 8 • Rates of change 365

Worked example 15

v 4 3 2 1

The velocity–time graph for a particle moving in a straight line and starting at the origin is shown in the diagram. Sketch the corresponding position–time graph.

0 −1 −2 −3 −4 Think 1

TUTorial eles-1435 Worked example 15

1 2 3 4 5

t

WriTe/draW

Set up a table of corresponding velocity and time values. (These velocity values represent the gradient of the position–time curve at the given times.)

t

0

v

1

4

2

2

3

4

0

−2

−4

x

2

Set up the axes for the position–time graph.

3

Draw in a curve starting at (0, 0) with a gradient of 4 decreasing to a gradient of 0 at t = 2 (the turning point). From t = 2 to t = 4, the gradient changes from 0 to −4. This means the curve will become steeper but with a negative slope.

0

1

2

3

4

5

t

relating velocity–time graphs to position–time graphs exercise 8h

For each position–time graph for a particle moving in a straight line, draw a velocity–time graph. The gradient of the curve at various times is indicated on each of the graphs.

1 We14

b

x

a

x Gradient = 0

Gradient = 3

Gradient = −3 0

1

2

3

4

5

Gradient = −2

t

6

Gradient = 2 Gradient = −2 Gradient = −1 2 mC

0

Gradient = 1 Gradient = 0

Gradients are shown at various points on a position–time graph. m=0

x

m = −6 m=6 m = −12

m = 12 0

366

Maths Quest 11 Mathematical Methods CAS

1

2

3

4

t

1 2 Gradient = 2

t

The velocity–time graph obtained from this position–time graph is: v 12

a

B

0

1

2

3

v 12

0

t

4

v 12

C

1

2

3

4

0

t

−12

2

3

t

4

−12

v 12

d

1

e

0

1

2

3

v 12

t

4

0

1

2

3

4

t

−12

Sketch the velocity–time graph for each of the following position–time graphs.

3 a

x 5

b

x 4

c

x 1

d

4

x 8 7

3

2

0

2 1

0

0 −1

1

2

3

t

1

2

3

4

t

1

6

t

2

5

−1

4 3

−2

2

−2

1 −4

−3

−1

−4 −5

0

1

2

3

4

t

−2

−6

−3 −4

−8

−5 −6 −7 −8

For each velocity–time graph shown below, sketch a position–time graph, given that the particle starts at the origin.

4 We15 a

b

v

c

v

−3

3

0

v

4

t

0

4

t

0 −1

1

4

t

−2

ChapTer 8 • Rates of change

367

v

d

v 8

e

v 3

f

3

0 −1

3

4

0

t

2

0

t

4

1

2

3

t

4

−3

−8

A particle starting at the origin moves with velocity v(t) = 2t − 4.

5

a Complete the table below.

t

0

1

2

3

4

v

−4

−2

0

2

4

Gradient of x–t graph b Sketch the position–time graph. 6

The velocities for a particle starting at the origin are given as a function of time. Sketch a position–time graph for each using t ∈ [0, 4]. a v=t+2 b v=2−t c v = 3t d v = −t

v 3

0

x

B

0

d

2

x

C

0

4 t

x

e

3

4

t

x

0

4 t

2

2

−3

A position–time graph is obtained from the velocity–time graph shown at right. For a particle starting at the origin, this position–time graph will look like:

7 mC

a

1

4 t

2

x

2 0

2

0

4 t

2

4 t

−1

8

Briefly describe the motion of a car during the first 30 seconds for each of the velocity–time graphs below. a v (m/s)

b v (m/s)

20

18

0

368

Maths Quest 11 Mathematical Methods CAS

30 t (s)

0

c v (m/s) 25

15

30 t (s)

0

10

25 30 t (s)

Sketch a velocity–time graph for each car described below for its first 20 seconds of motion.

9

a A car, initially at rest, increases its speed at a constant rate until it reaches a speed of 35 m/s after

10 s. It then maintains this speed. b A car is travelling at a constant speed 30 m/s. After 15 s the brakes are applied, decreasing its

speed at a constant rate until it stops after another 5 s. 10 A parachutist jumps from an aircraft and freefalls for 6 seconds. a If a parachutist falls y metres in t seconds where y = 5t2, find the average speed of the parachutist between: i t = 0 and t = 3 ii t = 3 and t = 6. b What is the speed of the parachutist after 6 seconds of freefall? c When the parachute is released (after 6 seconds of freefall), the speed of the parachutist is reduced by 2 m/s every second until a speed of 4 m/s is reached. How long after jumping from the aircraft does it take the parachutist to reach a speed of 4 m/s?

rates of change of polynomials

8i

We have seen that instantaneous rates of change can be found from a graph by finding the gradient of the tangent drawn through the point in question. The following method uses a series of approximations to find the gradient. Worked example 16

a Plot the graph of y = 2x2, x ∈ [0, 4]. b Draw a line through the following points on the curve and find the gradient of each line. i x = 2 and x = 3 ii x = 2 and x = 2.1 iii x = 2 and x = 2.01 iv x = 2 and x = 2.001 c Hence, predict the gradient of the tangent at the point on the curve where x = 2. Think

a

1

Use a table of values or a CAS calculator.

2

Plot the points using graph paper.

WriTe/draW

a

x y

0 0

1 2

2 8

3 18

4 32

y 32 i ii iii

18

iv

8 0 b i 1 Mark the points on the curve where x = 2 and

x = 3 and rule a line through them.

2

Select suitable values to calculate the rise . gradient using run When x = 2, y = 8; when x = 3, y = 18.

1

2

3

4

x

b i Line i

18 − 8 3− 2 10 = 1 = 10

Gradient =

ChapTer 8 • Rates of change

369

ii 1 Again, mark the points and rule a line

ii Line ii

through them. 2

When x = 2.1, y = 2(2.1)2 = 8.82

Select suitable points to calculate the gradient. Calculate the y-value for x = 2.1 using the rule y = 2x2.

8.82 − 8 2.1 − 2 0.82 = 0.1 = 8.2

Gradient =

iii 1 Again, mark the points and rule a line

iii Line iii

through them. 2

Select suitable points to calculate the gradient. This may be difficult, so select an x-value and calculate the y-value using the rule y = 2x2.

iv 1 Again, mark the points and rule a line

When x = 2.01, y = 2(2.01)2 = 8.0802 8.0802 − 8 Gradient = 2.01 − 2 0.0802 = 0.01 = 8.02 iv Line iv

through them. 2

When x = 2.001, y = 2(2.001)2 = 8.008 002 8.008 002 − 8 Gradient = 2.001 − 2 0.008 002 = 0.001 = 8.002

Select suitable points to calculate the gradient. Calculate the y-value for x = 2.001 using the rule y = 2x2.

c Continue to bring the second x-value closer and

c The gradient of the tangent at the point on

closer to 2. As the second x-value becomes closer to 2, the gradient of the line is becoming closer to 8. Notice that the slope of each line has progressively become closer to the slope of the tangent at x = 2.

the curve where x = 2 appears to be 8.

Worked example 17

Consider the function f (x) = 2 x2. a Using a CAS calculator, calculate the gradients of the chords connecting two points on the curve with the following coordinates: i x = 3 and x = 4 ii x = 3 and x = 3.1 iii x = 3 and x = 3.01 iv x = 3 and x = 3.001. b Hence, predict the gradient of the curve at x = 3. Think

a 1 Using the average rate of change

function of the CAS calculator, complete the function entry lines as: avgRC(2x2, x, 1) | x = 3 avgRC(2x2, x, 0.1) | x = 3 avgRC(2x2, x, 0.01) | x = 3 avgRC(2x2, x, 0.001) | x = 3 370

Maths Quest 11 Mathematical Methods CAS

WriTe

a i ii iii iv

avgRC(2x2, x, 1) | x = 3 avgRC(2x2, x, 0.1) | x = 3 avgRC(2x2, x, 0.01) | x = 3 avgRC(2x2, x, 0.001) | x = 3

2

Record the results.

b As the interval becomes smaller and

the two points come closer together, the chord approximates the tangent. The gradient of this tangent is the gradient of the curve at x = 3.

i Between x = 3 and x = 4, the gradient is 14. ii Between x = 3 and x = 3.1, the gradient is 12.2. iii Between x = 3 and x = 3.01, the gradient is 12.01. iv Between x = 3 and x = 3.001, the gradient is 12.002. b The gradient of f (x) = 2x2 approaches 12 as the interval

decreases. So, the gradient of f (x) = 2x2 at x = 3 is 12.

The previous example demonstrates that the gradient of a tangent to a function at a given point, x = a, can be approximated by finding the gradient of a line joining the point x = a and another point as close as possible to x = a.

Hybrid functions A hybrid function is one that has different rules over different sections of its domain. For example: x + 2, x ≤ 0 f (x) = 2 x>0 y x , The graph of f (x) is shown at right. This is a discontinuous function because there is a break (or jump) in 2 y = x2 the graph at x = 0. There is also no defined gradient at the point x = 0 on y=x+2 this graph, because at x = 0 two different values of the gradient can be obtained. x 0 −2 2 If the rate of change of y at x = 0 is calculated using points just to the left of x = 0, and then calculated using points just to the right of x = 0, different values are obtained: rise 2 = =1 Straight line: m = run 2 y Parabola: At the turning point (x = 0), gradient = 0. Not all hybrid functions have a break in them. For example: x , x ≤ 0 f (x) = 2 x , x > 0

Gradient of straight line is 1 at x = 0. 2

−2

0

y = x2

x

Gradient of parabola The graph of f (x) is shown at right. y=x is 0 at x = 0. However, even for this graph, there is no defined gradient at x = 0. The gradient of the tangent approaching x = 0 from the left is equal to the gradient of the straight line, that is, 1. The gradient of the tangent to the parabola approaching x = 0 from the right, is equal to zero. Therefore the gradient (or instantaneous rate of change) does not exist at x = 0. If the hybrid function is ‘smooth’ at the point where the combined functions meet, then the gradient does exist. A function is smooth at a given point if the left-hand gradient equals the right-hand gradient at that point. For example: y x ≤1 2 x , f (x) = 2 x + 1, x > 1 y = x2 + 1 The left-hand gradient of f (x) at x = 1 is the gradient of the line y = 2x, which equals 2. The right-hand gradient of f (x) at x = 1 is the gradient of the Both gradients 2 are 2 at x = 1. parabola y = x2 + 1, which also equals 2. So the left-hand gradient equals the right-hand gradient. Therefore, this hybrid function x 0 1 is smooth at x = 1, and the gradient (or the instantaneous rate of y = 2x change) at x = 1 is equal to 2. Chapter 8 • Rates of change 371

In summary, for a hybrid function to have a gradient (or instantaneous rate of change) at the point in the domain where the rule changes: 1. the function must be continuous at that point 2. the left-hand gradient must equal the right-hand gradient at that point. This makes the curve smooth. Worked example 18

Verify that the rate of change of the hybrid function 2 x, x≤2 f ( x) = 2 x>2 x , does not exist at x = 2, using a CAS calculator. Think

WriTe

1

Define a hybrid function using your CAS calculator.

2 x x ≤ 2 Define f ( x ) = 2 x x > 2

2

Set f 1(x) = f (x) for graphing.

f 1(x) = f (x)

3

Using the CAS calculator’s draw tangents feature, draw the tangent to y = 2x at x = 2.

Gradient = 2

4

Draw the tangent to y = x2 at x = 2.

Gradient = 4

5

Since the left-hand gradient and right-hand gradient are not equal, the rate of change does not exist at x = 2. Write the answer.

The left-hand gradient is 2 and the right-hand gradient is approximately 4 around the point x = 2. As the two results are not the same, the rate of change of the hybrid function 2 x , x≤2 f (x) = 2 x>2 x , does not exist at x = 2.

Worked example 19

4 − 2 t, t ∈ [0, 1] For the displacement function x( t ) = find: 2 3 − t , t ∈(1, 6] a the starting position b the finishing position c the displacement between t = 0 and t = 2 seconds d the instantaneous velocity at t = 1, if it exists. Think

372

WriTe

a Evaluate x(0).

a x(t) = 4 – 2t

b Evaluate x(6).

b x(t) = 3 – t2

c Evaluate x(2) − x(0). Note that x(0) = 4 from part a; x(2) = 3 − t2 at t = 2.

c x(2) – x(0) = (3 – 22) – [4 – 2(0)]

Maths Quest 11 Mathematical Methods CAS

x(0) = 4 – 2 × 0 =4 The starting position is 4 units. x(6) = 3 − 62 = 3 – 36 = −33 The finishing position is −33 units. =3–4–4 = −5 The displacement is −5 units.

d 1 Velocity is the rate of change of displacement,

d

therefore find the instantaneous rate of change of x at t = 1, if it exists. 2

To evaluate the left-hand gradient at t = 1, use the line x = 4 – 2t. Note that m = −2 in this rule.

Left-hand gradient = −2

3

To evaluate the right-hand gradient at t = 1, use the line x = 3 – t2. Calculate values for x when t = 1 and t = 1.001.

x(t) = 3 – t2 x(1) = 3 – 12 =2 x(t) = 3 – t2 x(1.001) = 3 – (1.001)2 = 1.998

4

Evaluate the right-hand gradient at t = 1.

5

The left-hand gradient equals the right-hand gradient, so the gradient exists and equals −2.

1.998 − 2 1.001 − 1 − 0.002 − Right-hand gradient ≈ = 2.001 0.001 Right-hand gradient ≈ −2 Right-hand gradient ≈

The instantaneous velocity at t = 1 is −2 units per second.

rates of change of polynomials

exercise 8i

1 We16 a Plot the graph of y = x2, x ∈ [0, 4]. b Draw a line through the following points on the curve and find the gradient of each line. i x = 2 and x = 4 ii x = 2 and x = 3 iii x = 2 and x = 2.5 iv x = 2 and x = 2.1 c Hence, predict the gradient of the tangent at the point on the curve where x = 2. d Use a CAS calculator to find the average rates of change for the intervals in part b. Does this

diGiTal doC doc-9779 Gradient at a point

change your prediction for the gradient? 2a Plot the graph of y = x3, x ∈ [0, 2]. b Draw a line joining the following points on the curve and find the gradient of each line. i x = 1 and x = 2 ii x = 1 and x = 1.5 iii x = 1 and x = 1.1 iv x = 1 and x = 1.01 c Hence, predict the gradient of the tangent to the curve where x = 1. d Use a CAS calculator to find the average rates of change for the intervals in part b.

Does this change your prediction for the gradient? 3 We17 a If y = x2 − x + 1, copy and complete the following table of values.

x

3

2

1.5

1.1

1.01

1

y b Hence, find the gradient of a line joining the points on the curve where x = 1 and: i x=3 ii x = 2 iii x = 1.5 iv x = 1.1 v x = 1.01. c Predict the gradient of the tangent at the point on the curve where x = 1.

The gradient of a tangent to a curve at the point where x = 3 is best approximated by the gradient of the line joining the points on the curve where: a x = 3 and x = 4 B x = 2 and x = 4 C x = 3 and x = 3.2 d x = 3 and x = 3.001 e x = 3 and x = 3.01

4 mC

ChapTer 8 • Rates of change

373

The rate of change at a point P on a curve is best approximated by finding the gradient of a line joining the point P and another point that is: a furthest from it B closest to it C anywhere on the curve d the origin e on the x-axis Use a CAS calculator to find the gradient of the following functions at the point indicated. a f (x) = x2 at x = 1 b f (x) = 3x2 at x = 2 c f (x) = x2 + 2x at x = 2 d f (x) = x3 + x at x = 1 e f (x) = 2 − 3x − x2 at x = 2 f f (x) = x3 − 2x + 3 at x = 1.5 g f (x) = x3 + x2 − 2x at x = 3 h f (x) = 12 x2 − 3x + 5 at x = 0.5 Hailstones fall x metres in t seconds where x = 4.9t2. Estimate the speed of a hailstone after: a 2 seconds b 3 seconds. h (m) The height, h metres, of a roller-coaster is given by − 1 3 3 2 50 h= d + d 80 8 where d is the horizontal distance from 0, in metres. Estimate the gradient of the curve describing the path of the rollercoaster, accurate to 2 decimal places, at the point where d is: 0 20 d (m) a 10 metres b 15 metres c 20 metres d 0 metres. We18 Verify that the rate of change of the function 2 x , x ≤1 f (x) = 2 x >1 2 x , does not exist at x = 1. We18 Verify that the rate of change of the function − 2 x , x<2 f (x) = − 2 x≥2 x , does not exist at x = 2. 5 − 4t , t ∈[0, 1] We19 For the displacement function x (t ) = , find: 2 t ∈(1, 6] 3 − 2t , a the starting position b the finishing position c the displacement between t = 0 and t = 2 seconds d the instantaneous velocity at t = 1, if it exists. 4t , t ∈[0, 2] For the displacement function x (t ) = , find: 2 t ∈(2, 6] 4+t , a the starting position b the finishing position c the displacement between t = 0 and t = 2 seconds d the instantaneous velocity at i t = 2 and ii t = 3, if it exists.

5 mC

6

7

8

9

10

11

12

374

Maths Quest 11 Mathematical Methods CAS

Summary identifying rates

• A rate describes how much one quantity changes with respect to another. • For two quantities A and B: change in A The rate of change of A with respect to B = amount of A per B = . change in B

Constant rates

• • • •

Variable rates

The gradient of a graph represents the rate of change. A positive gradient shows a positive rate of change or increasing function. A negative gradient shows a negative rate of change or decreasing function. A linear or straight line graph shows that the rate of change is constant.

v

t

• A variable rate exists when the rate of change is not constant. • On a graph, a variable rate of change can be seen when the points do not lie on a straight line. d

0

average rates of change

1

2

3 t

• Over a specified interval: change in A 1. Average rate of change of A with respect to B = . change in B 2. Average rate of change of a function f (t) over the time interval t = a to t = b is: change in f (t ) f (b) − f (a) = change in t b−a

y Rise Run

x

0

rise • When using a graph, find the gradient of the chord formed between the end points of the run interval. • When using a table of values or rules, find the change in each quantity over the specified interval. instantaneous rates

• The instantaneous rate of change is the rate of change at a particular instant rather than over an interval. • The instantaneous rate of change can be found by calculating the gradient of the tangent drawn to the curve at the required point.

y P Tangent 0

motion graphs (kinematics)

• Position, x, describes where an object is or was. • Distance, d, is how far an object has travelled. • Displacement, s, describes the change in an object’s position; that is, s = xfinal – xinitial. d distance • Speed = , or speed = . t time taken s • Velocity = displacement , or v = . t time taken

x

x

0

t

• The gradient of a position–time graph for an object gives the velocity of that object. • Distance and speed are always positive. Displacement and velocity can be either positive or negative depending on the direction of motion.

ChapTer 8 • Rates of change

375

relating velocity–time graphs (the gradient function) to position– time graphs (the original function)

• To draw a velocity–time graph (or gradient function) from a position–time graph (or original function): 1. Calculate the gradient of the graph at given time values. For an x–t graph this gives the velocity. 2. Construct a table of values showing the velocity (or gradient) at given time values (or x-values). 3. Use this table to plot the velocity–time graph (or gradient function). • To draw a position–time graph (or original function) from a velocity–time graph (or gradient function): 1. Set up a table of corresponding velocity (gradient) and time values (x-values) from the graph. 2. On a set of axes, plot a point for the initial position. Starting from this point, use the values in the table to estimate the appropriate relative slope at each given time (or x-value) to produce a curve.

rates of change of polynomials

• The gradient of a tangent to a function at a given point, x = a, can be approximated by finding the gradient of a line joining the point x = a and a point as close as possible to x = a. The closer the two points are, the more accurate the estimate of the gradient. • For a hybrid function to have a gradient (or instantaneous rate of change) at the point in the domain where the rule changes: 1. the function must be continuous at that point 2. the left-hand gradient must equal the right-hand gradient at that point.

376

Maths Quest 11 Mathematical Methods CAS

Chapter review 1 A car manufacturer claims a vehicle uses 12 litres of fuel per 100 km. a Express this rate in km/litre. b How far could the car travel using 35 litres? c How many litres are required to travel 765 km?

S ho rT a n S W er

2 a Determine the rate at which water is leaking from a tank, using the graph below. w (m3) 840 600

0

8

t (hours)

b How long does it take the tank to empty? 3 a Use the table below to plot the graph of M versus t.

t M

0

1

2

−9

−7

−1

3 9

4 23

b Is the rate of change of M constant or variable? Explain your answer. c Estimate the gradient when t = 2. 4 The amount of substance, A kg, in a container at any time, t hours, is A = t2 − 3t + 4, t ∈ [0, 5]. a Find the average rate of change during the first 4 hours. b At what rate is the amount changing after 4 hours? 5 Calculate the average rate of change of y = tan (x) between x =

π 4

π

and x = 3 . 6 A particle is moving in a straight line according to the position–time function x(t) = t2 − 2t + 5, where t ∈ [0, 5] and t is given in seconds. a Sketch the position–time graph. b Where did the journey start? c When and where did the particle change direction? x (m) d What was the distance travelled in the first 4 seconds? 7 From the position–time graph at right, find: 200 a the initial position b how long the object was not moving c the velocity during the first 5 seconds 30 d the velocity during the last 4 seconds e the total distance travelled 0 5 8 12 t (s) f the average speed for the journey. 8 For the velocity–time graph below, sketch a position–time graph, given that the object starts at the origin. v 2

0

3

6

t

−2 9 The height, h metres, of a projectile above the ground at any time, t seconds, is given by the function

h(t) = 5 + 12t − t2. a Find the rate of change of height when t equals: i 4 seconds ii 6 seconds iii 10 seconds. b Briefly describe the change of height of the projectile within the first 12 seconds. 2 x , x ≤ 2 10 Verify that the rate of change of the function f ( x ) = 2 does not exist at x = 2. x , x > 2

ChapTer 8 • Rates of change

377

2t ,

t ∈[0, 1]

11 For the displacement function x (t ) = 2 1 + t , a the starting position b the finishing position c the displacement between t = 0 and t = 2 d the instantaneous velocity at i t = 1 and ii t = 2, if it exists. m U lT ip l e C h oiCe

t ∈(1, 6]

, find:

1 Which one of the following is not a rate? a 35 km B 18 breaths per minute d 50 km/h e −8 °C/hour

C $2.50 per kg

2 Which one of the following is a constant rate? a The number of people entering the zoo per hour B The number of days it rains in Sydney per year C The hourly rate of pay of a tutor d The number of crates of fruit picked per hour by a fruit-picker e The number of patients visiting a doctor per day 3 The graph below shows the weight, w kg, of a tank of water t minutes after water is poured into it. w (kg) 70

20 0

t (min)

40

The rate of change of w in kg/min is: a 1.75 B 0.8 C 2.25 4 The graph below has a negative rate of change in the interval: a OP B PQ C QR

d 1.25

e 2

d RS

e OQ

d 6

e 2

y S

P R Q

x

O 5 The gradient of the chord PQ below is: a 3

B 12

C y

1 3

Q

21 P

9 0

5

9

x

6 From the table below, the average rate of change of H between t = 2 and t = 5 is: a 20 m/h B 38 m/h C 30 m/h d 190 m/h

t(h) H(m)

378

0 0

1 20

Maths Quest 11 Mathematical Methods CAS

2 40

3 70

4 120

5 190

6 280

e 50 m/h

7 The instantaneous rate of change of v at t = 4 on the graph below is: a 13.75 L/min B 12.5 L/min C 6.25 L/min d 55 L/min

e 8 L/min

v (L) 80 55 30 0

4

8 t (min)

8 Which one of the following describes an instantaneous rate? a The temperature dropped 15°C in 2 hours. B Harry was caught doing 85 km/h. C Jill ran 5 km in 28 minutes. d Jenny earned $80 in 7 12 hours. e A balloon rose 600 metres in 90 minutes. 9 The following diagram shows the position of a particle (x metres) travelling in a straight line at various

times (t seconds). t=5

t=2

t=0

−3 −2 −1

0

1

2

3

4

x

5

a The displacement after 5 seconds is: B 11 m C 7m a −2 m b The average speed in the first 5 seconds is: − a 11 m/s B 75 m/s C 11 m/s 5 5 c The average velocity in the first 5 seconds is: − a 11 m/s B 35 m/s C 35 m/s 5

d

−3

m

e 5m

d

−

7 5

m/s

e 1 m/s

d

−

11 5

m/s

e 1 m/s

10 The graph of a gradient function is shown below. y 2

Gradient function x

0

The graph of the original function could be: a

y 4

y

B

C

0 0 d

x

2

−2 e

y

4

y

x 0

x

y

2 1 0

1

x

0

1

x

ChapTer 8 • Rates of change

379

11 A position–time graph is shown below. x

0

t

2

Which one of the following diagrams represents its velocity–time graph? a

0

d

B

v

t

2

v

0

0

0

t

2

v

t

2

v

e

t

2

C

v

0

t

2

12 The velocity–time graph of a particle travelling in a straight line and starting at the origin is shown

below. v 3

0 −1

3

t

4

Which one of the following represents its position–time graph? a

x

0

d

B

3

4

0

e

3

4

Maths Quest 11 Mathematical Methods CAS

3

4

t

4

t

x

t 0

380

C

x

t

x

0

x

0

3

4

t

13 The rate of change of a polynomial f (x) when x = 3 is closest to: a

f (3.1) − f (3) 3.1 − 3

B

d

f (3) − f (0) 3− 0

e

f (3.001) − f (3) 3.001 − 3 f (3.00 001) − f (3) 3.00 001 − 3

14 The rate of change of y = x2 + x when x = 2 is: B 3 a 2 e 6 d 5

C

f (3.5) − f (3) 3.5 − 3

C 4

1 A soft-drink vending machine at a railway station dispenses cans of soft drink. It has a capacity of

600 cans when full. On a particular day: • the machine is half full at 6.00 am • in the next four hours drinks are dispensed at a constant rate of 15 per hour • at 10.00 am the machine is filled • in the next four hours the machine dispenses an average of 60 cans per hour at a steadily increasing rate • finally, from 2.00 pm to 6.00 pm an average of 30 cans per hour are dispensed at a steadily decreasing rate. a Sketch a graph showing the number of cans in the machine over the period from 6.00 am to 6.00 pm. Use the graph to: b find the number of cans in the machine at 6.00 pm c estimate at what time in the afternoon the machine is half full d estimate the rate at which cans are dispensed at 1.00 pm.

ex Ten d ed r eS p o n S e

2 The concentration (x mg/L) of a tranquilliser in a patient’s bloodstream at any time, t hours, after it is

administered is given by the following rule: 2t ,t≥0 t2 + 1 The tranquilliser is only effective if the concentration is at least 0.5 mg/L. a Determine the concentration at: i t=0 ii t = 1 iii t = 2 iv t = 4. b With the aid of a CAS calculator, sketch the graph for the concentration function. c Find the exact length of time that the tranquilliser is effective. d Determine the average rate at which the tranquilliser is absorbed into the bloodstream from t = 1 to t = 3 hours. Explain why the rate is negative. e Determine the instantaneous rate at which the tranquilliser is absorbed into the bloodstream at t = 0.5 and t = 2 hours. (Give answers to 2 decimal places.) f Verify that the rate at which the tranquilliser is absorbed into the bloodstream is 0 at t = 1. Hence, give the time and concentration when the concentration is at a maximum. x=

3 The position (x centimetres from the origin, 0) of an object travelling in a straight line at any time,

t seconds, is given by:

2t + 1, x= 2 t − 2t + 5,

t ∈[0, 2] t ∈(2, 5]

Determine the initial and final position of the object. Sketch the position–time graph. Verify that the velocity exists at t = 2 and determine this velocity. Find the displacement from t = 1 to t = 4. The average velocity from t = 1 to t = b is 2.9 cm/s. Find the value of b. Determine the time, t = c, for which the displacement from t = 0.5 to t = 2 is equal to the displacement from t = 2 to t = c. g Sketch the velocity–time graph for this object. a b c d e f

ChapTer 8 • Rates of change

381

4 A skateboarder starts on a 1 m long flat section before coming down on an inclined ramp. When

diGiTal doC doc-9783 Test Yourself Chapter 8

382

he reaches the top of the ramp a stopwatch is started. The distance he travels is known to follow an exponential curve d = at for the first 2 seconds of motion, where d is in metres and t in seconds. If his speed initially (at t = 0) is 1 m/s, find his equation of motion; that is, find a. Use the following steps to help. a Using a CAS calculator, find the gradient of d = 2t at t = 0. b Using a CAS calculator, find the gradient of d = 3t at t = 0. Clearly the value of a must be between 2 and 3. c Using trial and error, find the value of a correct to 2 decimal places. d Show that the accurate value is a = 2.718 281 8 (correct to 7 decimal places). e Find the value of 2.718 281 82 (correct to 5 decimal places). f Find the gradient of d = 2.718 281 8t at t = 2 (correct to 5 decimal places). g What do you notice? Note: 2.718 281 8 . . . is a very important transcendental number in mathematics and has the symbol e. You will find it on your calculator.

Maths Quest 11 Mathematical Methods CAS

ICT activities Chapter opener diGiTal doC • 10 Quick Questions doc-9773: Warm-up with ten quick questions on rates of change (page 341)

8a

identifying rates

diGiTal doC • SkillSHEET 8.1 doc-9774: Practise determining rates of change and using them in calculations (page 343)

8B

Constant rates

diGiTal doC • doc-9775: Investigate plotting relations using a spreadsheet (page 347)

8C

Variable rates

TUTorial • We 5 eles-1432: Watch a tutorial on identifying constant or variable rates of change (page 348) diGiTal doC • doc-9775: Investigate plotting relations using a spreadsheet (page 349)

8d

average rates of change

TUTorial • We 8 eles-1433: Watch a tutorial on calculating the average rate of change of the temperature of a room over a set period of time (page 350) diGiTal doCS • SkillSHEET 8.2 doc-9776: Practise finding the gradient of a straight line (page 351) • doc-9777: Investigate the gradient between two points using a spreadsheet (page 353) • WorkSHEET 8.1 doc-9778: Identify rates, form equations and make calculations involving rates (page 353)

8e

instantaneous rates

TUTorial • We 11 eles-1434: Watch a tutorial on calculating the average speed and velocity of a lift (page 358) diGiTal doCS • SkillSHEET 8.3 doc-9780: Practise calculating displacement and total distance travelled (page 360) • SkillSHEET 8.4 doc-9781: Practise describing intervals using interval notation (page 361) • WorkSHEET 8.2 doc-9782: Calculate displacement, instantaneous rates of change and starting and finishing positions, given position functions (page 363)

8G relating the gradient function to the original function diGiTal doCS • doc-9779: Investigate the gradient at a point using a spreadsheet (page 364) • doc-9775: Investigate plotting relations using a spreadsheet (page 364)

8h relating velocity–time graphs to position–time graphs TUTorial • We 15 eles-1435: Watch a tutorial on sketching a position–time graph of a particle moving in a straight line (page 366)

8i

rates of change of polynomials

diGiTal doC • doc-9779: Investigate the gradient at a point using a spreadsheet (page 373)

Chapter review diGiTal doC • Test Yourself doc-9783: Take the end-of-chapter test to test your progress (page 382)

To access eBookPLUS activities, log on to www.jacplus.com.au

diGiTal doC • doc-9779: Investigate the gradient at a point using a spreadsheet (page 356)

8F

motion graphs (kinematics)

inTeraCTiViTY • Motion graphs (kinematics) int-0267: Learn how distance-time, velocity-time and acceleration-time graphs are related graphically (page 357)

ChapTer 8 • Rates of change

383

Answers CHAPTER 8 raTeS oF ChanGe

exercise 8C

exercise 8a

1 2

3 4

5 6

7

identifying rates a, b, d, f, i, k, m, n, p a C bA d F f H i E k I mD n B p G b, c, e, g a $10.60 b $31.41 c $303.75 d $5.59 e $75.05 f $609 697 a A bC a 8.33 m/s b 3.5 °C/h c $25 per person d 260 g/L e $4.67/m2 f 1.16 kg/week a 20 h b 100 h c C

exercise 8B

e 9 a c e

b

throughout the time interval, except at the region E. E shows a constant rate of change. The rate of change is positive throughout the time interval, except at region E. E has a zero rate of change. D A D

c d e 4 a W (kg)

scarcer. $156 1000 m 80 s

b 70 s d 30 s

0

5 4 3 2 1 0

exercise 8d

average rates of change

2 5

70

Q

384

P 1

iii 0 km/h P Q

0

1

t (h)

3

iii −10 km/h b 0.5 ii 0.5 kg/m

70 50

N 600 500 400 300 200 100 0

0 10 am 11 am 12 pm 1 pm

b, c

T (°C) 120 (8, 84)

20 1

t (h)

3

c i d (km) 70

Q P

ii 15

Maths Quest 11 Mathematical Methods CAS

d e 9 a b

1

3

t (h)

iii 15 km/h

(2, 24) 2 4 6 8 10 t (s)

10 10 °C/s 200 g i 1 g/week iii 10 g/week

exercise 8e

b 6 h 0

T

c i 170 ii 20 iii 0 d i 170 people/h ii 20 people/h iii 0 people/h e Most people arrive in the morning, few

P

iii 10 km/h

40

2 pm 3 pm 4 pm 5 pm

Q

ii 10

b i d iii

t (h)

3

ii −10

iii 30 km/h

b i d (km)

2 3

t (h)

3

in the middle of the day and none later in the afternoon. 7 a i 110 m/min ii 70 m/min iii 45 m/min iv 15 m/min v 5 m/min b Decreasing 8 a 20 °C

b −4 2 a, c 3 a i d (km)

0

5 10 15 20 25 x (cm)

Q

1

0

b 0.2 kg/cm c W = 0.2x 11 a 6 L/h 12 a iv c ii

0

1 2 3 4 t (min)

b No c Variable d i 2 kg/min ii 3 kg/min iii 7 kg/min 6 a Constant b Variable

10 a w (kg)

P

70

change.

ii 30 150180 t (s)

t (h)

3

iii −5 km/h

e i d (km)

5 D 6 a, b

10 0

70

1

4 a 1.5 c i 1.5 kg/m d No, because it is a variable rate of

1000

0

0

1 2 3 4 t (min)

W (kg)

820

Q

ii −5

The points form a straight line. A constant 3 kg/min 20 16 12 8 4

1 a

P

70 60

90 70

0

b c d 5 a

d (metres)

420

d i d (km)

ii 0 f i d (km)

12 9 6 3

Constant rates

1 c, d, g, h, j 2 b, d, e, g, h, j 3 i b, e, h ii g, j iii d 4 D 5 C 6 a i 50 ii 50 m/h iii d = 50t b i 7 ii 7 L/min iii v = 7t + 50 c i −1 ii −1 kg/week iii w = 100 − t d i 0 ii 0 m/year iii h = 75 e i 0.75 ii 0.75 g/min iii w = 0.75t + 10 f i −0.25 ii −0.25 g/L iii w = −0.25v + 35 7 a D b B 8 a $24/h b Rest c $18/h d The picker is tiring a little or fruit is

Variable rates

1 a, c, d, e, g, h 2 b, d, e, f 3 a The rate of change is variable

1 a A e A 2 a, c, d, h

ii −2 g/week iv 28 g/week

instantaneous rates b I c A d I f A g A h I

3 a 1 c −1 4 a 2.5 g/min b 16 g/min c −25 g/min d 12 g/min 5 a 85 kg after 35 weeks b Approx. −1 kg/week c Approx. 2.0 kg/week 6 a Approx. 6 hours b Approx. 3.5 °C/hour y 7 a, b

b 2.5 d 0.25

x

6 a

4 0

0

i x=0 iii No b

ii Right iv x = 10

x 16

0 −2

4 1

2

3

x

4

d C

e A

c

−2

ii Right iv x = 16

x 15

5

−4 −3 −2 −1 0 −2

1

2

3

4

x

1 a c 2 a c 3 a b c d 4 a c e g i 5 a b c

d e f

ii 0

0 −1

1 2 3 4 5

t

i x=0 iii Yes, t = 1, x = −1

−7

exercise 8F

4 m/s m/s

−2

2 −4 2 m/s −4 m/s

iii 0 iii 0 m/s

x

0 −2 −4 −6

t

6

i x= iii No

y 9

9 6 cm/min

ii v ii v

−2

4 2

8

c i 4

4

b i iv c i iv d

12

c C 8 a, b

t

1 2 3 4 5

−5

t

1 2 3 4 5

16

0

x

7 a

10 8 6 4 2

iii −4

x

d

8 a i Positive ii Speeding up b i Negative ii Slowing down c i Negative ii Speeding up d i Positive ii Slowing down e i Positive ii Neither — velocity is constant. 9 a C b E c B d F e A f D 10 x (m) 45

1 0

motion graphs (kinematics) b B d A b True d False

C D False True 40 m −2 m (or 2 m below the platform) 0.5 m/s −0.025 m/s (or 0.025 m/s downwards) x=1 b x = −3 Right d t=2 C f D D h D C j B i x=0 ii Right iii t = 2, x = 8 iv t = 5, x = −3 i x=4 ii Right iii t = 4, x = 12 iv t = 6, x = 10 i x=0 ii Right iii t = 3, x = 12 and t = 6, x = 3 iv t = 8, x = 10 i x=0 ii Left iii t = 1, x = −5 iv t = 3, x = 18 i x = −3 ii Left 1 iii t = 1 , x = −6 iv t = 5, x = 5 2 i x=2 ii Left iii t = 3, x = −5 and t = 5, x = 5 iv t = 6, x = 4

ii Left iv x = 15

25

t

1 2 3 4 5

0

−8

i x=0 iii Yes, t = 1, x = 1 e

ii Right iv x = −8

x

9

a b c d 11 a c e

1

2

3

5 t (s)

4

45 m t=5 t=2 20 m/s Alan 12 metres 40 s

b Bill d 2.0 m/s f Twice

exercise 8G relating the gradient function to the original function

4

1 a

0

1 2 3 4 5

t

4

ii Left iv x = 9

−4 −2 0 2 −4

x 18

4

x

Gradient function

b 0

y 12 8

i x=4 iii Yes, t = 2, x = 0 f

t

1 2 3 4 5

1 2 3 4 5

y 12 8

t

4 −4 −2 0 2 −4

−12

i x = −12 iii No

Gradient function 4

x

ii Right iv x = 18

ChapTer 8 • Rates of change

385

c

v

3 a

y 12

5 a t

5

8

Gradient function

4

0 2

−4 −2 −4

x

4

0

1

2

t

3

0 −4

0 1

π – 4

− π–

c

x

2

1

2

t

4

3

0 2

4

6

x

8

d

0

2

4

2

3

t

4

6 a x

b

x

c

x

t

4

0 3 0

2

4 a

1 −3 −2 −1

−1 −2

1

2

t

1 2 3 4

3 x 0 0

b

−5

the gradient function of e is

t

7A 8 a It travels at a constant velocity of

c x

ex.

20 m/s.

b It starts at rest, increasing its velocity

at a constant rate for 15 seconds when it reaches 18 m/s. It then maintains this velocity. c It starts from rest, increasing its velocity at a constant rate for 10 seconds when it reaches 25 m/s. It maintains this velocity for a further 15 seconds, then decreases its velocity at a constant rate for 5 seconds, bringing it back to rest.

or inverted.

0

exercise 8h

relating velocity–time graphs to position–time graphs

1 2 3 4 5 t

d x

v 3

0

6 t

1

2

3

4 t

9 a v (m/s)

e x

35

−3 v 2

0

f 2

4

t

4

6 There are stationary (turning) points there. 7 The original function will also be negative

3

x

−8

4 Quadratic functions 5 The gradient function of sin (x) is cos (x); x

d

t

4

0

0

functions are always linear.

t

4

x

2 It is a straight line. 3 The gradient functions of quadratic

1

t

4

x

−3

0

2

12

−4

1

2

3

4

0

t

10

20 t (s)

b v (m/s)

x

t

30 0

2 C

386

−2

v

4 3

−2

4

12 y 5

0

1

0

2 t

1

−2

f

b

4

2

v

0

4

1 a

3

0

2

Gradient function

y

2

−8

−2 −1

e

−4

0

4

Gradient 4 (and original) function 2

Gradient of x–t graph

1 2

−

4

b x

8 y 8 6

−

v

b d

0

V

Maths Quest 11 Mathematical Methods CAS

1

2

3

4

t 0

15 20

t (s)

10 a i 15 m/s b 60 m/s c 34 seconds

ChapTer reVieW

ii 45 m/s

y

i (4, 16)

16

ii

0

(1, 1) 2

3

x

4

ii 5

7

ii

180

iii 3.31

iv 3.0301

ii 4.75

iii 3.31

iv 3.03, no

3

2

1.5

y

7

3

1.75 1.11 1.0101

1.1

ii 2 v 1.01

7 a 19.6 m/s b 29.4 m/s 8 a 3.75 b 2.81 c 0 d 0 9 Teacher to check. 10 Teacher to check. 11 a 5 c −10 12 a 0 b 40 c 8 d i 4

1.01

iii 1.5

1 1

1 0.5

0

1

2

12 4 4.75 −2.5

2

3

4

5

e 0.96 mg/L at t = 0.5, −0.24 mg/L at t = 2 f The maximum concentration is 1 mg/L

t

after 1 hour.

3 a Initial position is 1 cm, final position is

20 cm.

b

x (cm)

x = t2 − 2t + 5

20

(5, 20)

15 (4, 13) 10

3

x = 2t + 1

during first 6 seconds then decreases after that. 10 Teacher to check. 11 a 0 b 37 c 5 d i 2 ii 4 mUlTiple ChoiCe

ii 6

1 4 7 9 10 13

A B C

a D

D E

2 5 8 b 11 14

C A B C A D

3 D 6 E c B 12 C

(3, 8) (2, 5)

1 0

t

6

9 a i 4 m/s ii 0 m/s iii −8 m/s b The height of the projectile increases

b −69 d −4 m/s

t (hours)

concentration is decreasing.

1

5

b d f h

4

d −0.2 mg/L; it is negative because the

b x=5m c t = 1 s and x = 4 m d 10 m 7 a x = 30 m b 3s c 34 m/s d −50 m/s e 370 m f 30.83 m/s 8 x

0

3

c 2 3 hours x (t)

0

ii 4.75

4D 5B 6 a 2 c 6 e −7 g 31

b 240 c Approx. 3 pm d Approx. 84 cans/hour 2 a i 0 ii 1 iii 0.8 iv 0.471 mg/L b x (mg/L)

x 20

x

2

8.00 am 10.00 am12.00 pm 2.00 pm 4.00 pm 6.00 pm

Time

t

5

π

5 4

x

b i 3 iv 1.1 c 1

4

10

2 (1.1, 1.331) (1, 1) (1.01, 1.030301) 1

b i 7 c 3 d i 7 3 a

3

15

iii iv

1

2

line.

6 a

5

0

240

c Approx. 8.5 4 a 1 kg/h b 5 kg/h 12( 3 − 1) 5 Average gradient =

6

(1.5, 3.375)

1

b Variable, as the graph is not a straight

iv 4.1

i

(2, 8)

3

300

0

−9

iii 4.5

8

4

360

60

Tangent at t = 2

(2.1, 4.41)

(2, 4)

420

120

9

(2.5, 6.25)

b i 6 c 4 2 a y

480

M

8

1

540

iv

(3, 9)

0

600

23

iii

12

8.33 km/L 291.6 km 91.8 L 30 m3/h 28 hours

Number of cans

1 a b c 2 a b 3 a

rates of change of polynomials

4

1 a

ShorT anSWer

exercise 8i

1 a

exTended reSponSe

1

2

3

c 2 cm/s e b = 3.5 g v (cm/s)

4

5 t (s)

d 10 cm f c=3

8 7 6 5 4 3 2 1 1

2

3

4

5 t (s)

loge (2) b loge (3) a = 2.72, correct to 2 decimal places loge (2.7182818) = 1 e 7.389 06 7.389 06; 2.718 281 82 d = (2.718 281 8 x ) x = 2 dt g The gradient is the same as 2.71828182.

4 a c d f

ChapTer 8 • Rates of change

387

Chapter 9

Differentiation DiGitaL DoC doc-9784 10 Quick Questions

Chapter ContentS 9a 9B 9C 9D 9e 9F 9G

Introduction to limits Limits of discontinuous, rational and hybrid functions Differentiation using first principles Finding derivatives by rule Rates of change Sketching graphs containing stationary points Solving maximum and minimum problems

9a

introduction to limits

In mathematics it is important to understand the concept of a limit. This concept is especially important in the study of calculus. In everyday life we use the term limit to describe a restriction put on a quantity. For example, the legal blood alcohol concentration limit for a driver is normally 0.05 g/100 mL. As the number of standard alcoholic drinks consumed in 1 hour approaches 2, the average adult’s blood alcohol concentration approaches 0.05. Likewise, some time after a celebration, a person who has been drinking heavily at an earlier time may have a blood alcohol concentration that is approaching the legal limit of 0.05 from a higher level, as the number of drinks not yet metabolised by their body approaches 2. We could say that as the number of standard drinks remaining in the body approaches 2, the blood alcohol concentration approaches 0.05. In essence the blood alcohol concentration is a function, say f (x), of the number of drinks, x, remaining in the body.

Chapter 9 • Differentiation

389

WorkeD exampLe 1 1 1 Add the following series of numbers and state what value it is approaching. 12 + 14 + 18 + 16 + 32 +...

think

Write

1

Add the first 2 terms.

The sum of the first 2 terms is

3 4

(= 0.750).

2

Add the first 3 terms.

The sum of the first 3 terms is

7 8

(= 0.875).

3

Add the first 4 terms.

15 (≈ 0.938). The sum of the first 4 terms is 16

4

Add the first 5 terms.

The sum of the first 5 terms is

31 32

(≈ 0.969).

5

Add the first 6 terms.

The sum of the first 6 terms is

63 64

(≈ 0.984).

6

Give the upper limit.

The sum is approaching 1.

Expressing limits in mathematical language, we say that a limit can be used to describe the behaviour of a function, f (x), as the independent variable, x, approaches a certain value, say a. In some cases the function will not be defined at a. Using the correct notation for the example on blood alcohol concentration, we would write: lim f ( x ) = 0.05 x→2

This is read as ‘the limit of f (x) as x approaches 2 is equal to 0.05’. WorkeD exampLe 2

By investigating the behaviour of the function f (x) = x + 3 in the vicinity of x = 2, show that a by hand b using a CAS calculator. lim f ( x ) = 5: x→ 2

think

a 1 Create a table of values for x and f (x) in

the vicinity of x = 2.

2

Consider the values taken by f (x) as x approaches 2.

Write/DraW

a

x f(x)

1.95 1.99 1.995 2 2.005 2.01 2.05 4.95 4.99 4.995 5 5.005 5.01 5.05

As x approaches 2 from the left and the right, f (x) approaches a value of 5. So lim f ( x ) = 5.

y

3

x→2

−3 b 1 Using the Lists and Spreadsheets

390

f(x)

5

0

b 1.95

feature of the CAS calculator, enter the x-values into column A (a[ ]).

1.99 1.995 2 2.005 2.01 2.05

2

Enter a second column as a[ ] + 3.

3

Find the left- and right-hand limits as x approaches 2.

1.95 4.95 1.99 4.99 1.995 4.995 2 5 2.005 5.005 2.01 5.01 2.05 5.05 The left- and right-hand limits are equal, so when f (x) = x + 3, lim f ( x ) = 5 .

Maths Quest 11 Mathematical Methods CAS

x→2

2

x

Limits of continuous functions A continuous function has a graph that forms a continuous line; that is, it has no breaks. If a function is continuous at the point where a limit is being found, then the limit always exists and can easily be found by direct substitution. WorkeD exampLe 3

Find lim( x 2 + 2):

a by hand

x→ 2

b using a CAS calculator.

think

Write/DraW

a 1 To consider whether the function is continuous,

y

a

sketch the graph of y = x 2 + 2 in the vicinity of x = 2.

f(x)

6 2 0 2

2

As the graph is continuous, substitute x = 2 (from lim ) into x2 + 2 to evaluate the limit.

The function is continuous at x = 2. lim( x 2 + 2) = 22 + 2 x→2 =6

x→2

b lim( x 2 + 2)

b 1 Using the limit feature of the CAS calculator,

x→2

determine the required limit. 2

Record the result.

3

Write the answer.

x

6 lim( x 2 + 2) = 6

x→2

In general if a function, f (x), is continuous when x = a, then lim f ( x ) = f ( a). x→ a

theorems on limits theorem 1 For the constant function f where f (x) = c, then lim f ( x ) = c . x→a

If lim f ( x ) = A and lim g( x ) = B , then the following theorems apply. x→a

x→a

theorem 2 For example:

lim[ f ( x ) + g( x )] = lim f ( x ) + lim g( x ) = A + B

x→a

x→a

x→a

lim[ f ( x ) − g( x )] = lim f ( x ) − lim g( x ) = A − B

x→a

theorem 3 For example:

x→a

lim( x 2 x →1

+ 2x) =

For example:

x→a

+ lim 2 x = 1 + 2 = 3 x →1

lim[ f ( x ) × g( x )] = lim f ( x ) × lim g( x ) = A × B

x→a

theorem 4

lim x 2 x →1 x→a

x→a

lim [ x 2 (3 x )] = lim ( x 2 ) × lim (3 x ) = 1 × −3 = −3

x→− 1

x → −1

x → −1

f (x) A f ( x ) xlim lim = →a = if B ≠ 0 x → a g( x ) lim g( x ) B x→a

x 2 + 4) x + 4 xlim( →5 = lim x →5 x + 1 lim( x + 1) 2

x →5

29 = 6 Chapter 9 • Differentiation

391

These theorems on limits can be expressed in word form: 1. The limit of a sum equals the sum of the limits. 2. The limit of a difference equals the difference of the limits. 3. The limit of a product equals the product of the limits. 4. The limit of a quotient equals the quotient of the limits.

exercise 9a

introduction to limits

Add the following series of numbers and state what value it is approaching.

1 We1

4 + 2 + 1 + 12 + 14 + . . . 2 The diagram at right shows regular polygons with 3, 4 and 5 sides. As the number of sides gets very large (→ ∞), what shape emerges?

n=3

n=4

n=5

If n represents the number of sides of a regular polygon, then which of the following correctly describes the situation given in question 2 above? B lim C lim a n

3 mC

n→∞

D n→0

n→ 0

e n→5

1 as n gets infinitely large. n b Write this using limit notation. 1 + … and n represents the number of terms to be summed in the series, 5 a If S = 1 + 12 + 14 + 81 + 16 copy and complete the following table: a Find the value of

4

b

n

1

2

S

1

1 12

3

4

5

6

10

Which of the following is equal to lim S?

mC

n→∞

a 1.75

B 1.95

D 1

e 0

C 2

y

6 We2 By investigating the behaviour of the function f (x) = x + 5 in the vicinity of

3 2 1

x = 3, show that lim f ( x ) = 8. x →3

From the graph at right it can be seen that the lim f ( x ) is equal to:

7 mC DiGitaL DoC doc-9785 SkillSHEET 9.1 Substituting into a function

a 2

B −1

D 1

e 0

x→2

C 3

a

y 30

b

f(x)

4

c x

4

f(x)

y 2

f(x)

−3 0

d

x→4

y 0

14

0

x

4

x e

y 21

y

f

f(x)

4

y

f(x)

15 10

f(x)

5 0

392

0

−1

For each of the functions graphed below, find lim f ( x ).

8

4

Maths Quest 11 Mathematical Methods CAS

x

1 0

4

x

0

f(x)

4

x

2

x

9 We3a Evaluate the limits below. a lim(2 x + 5) b lim(10 − 2 k ) x→2

e lim(h 2 + 3h)

f

h→ 5

10 mC

c lim(49a + 1)

k →4

We3b

lim (3 p2 − 2 p + 3)

p→ − 2

x→0

B 3

a 4

g

lim(8 − 5 x 2 )

x→2

d lim( x 2 + 2) x→0

h lim( x 3 + 2 x ) x→0

If f (x) = 3x − 2, then the value of lim f ( x ) is equal to:

a 0

11 mC The

a→1

lim ( x 2 x→ − 3

C 2

D −2

e 1

C 9

D −4

e 0

− 5) is equal to: B −5

12 Find the values of the following limits.

x +3 x→2 x

a lim

2 x − 3x x→0 x +1

2

b lim

2

c

lim

x→ − 2

6x − 2 x

x + 3x − 2 x →3 x−2 2

d lim

DiGitaL DoC doc-9786 Investigation Sneaking up on a limit

Limits of discontinuous, rational and hybrid functions 9B

Limits of discontinuous functions

If a function is discontinuous at the point where the limit is being investigated, then the limit will exist only if the function is approaching the same value from the left as from the right. Consider the discontinuous functions graphed below. 1. From the left, lim f ( x ) = 1. (The symbol x → 1− indicates that we are x →1−

y

1

letting x approach 1 from the left side.) From the right, lim f ( x ) = 1.

0

x →1+

Left limit = right limit. Therefore, lim f ( x ) = 1.

x

1

x →1

y

2. From the left, lim f ( x ) = 2. x →1

−

f(x)

4

From the right, lim f ( x ) = 4. x →1+

2

Left limit ≠ right limit. Therefore, lim f ( x ) does not exist. x →1

0 1

Limits of rational functions

2

x

Finding the limit of a rational function involves simplifying the function before direct substitution can take place and recognising any values of x for which it is discontinuous.

WorkeD exampLe 4

x2 − x : x→1 x − 1 i by hand ii using a CAS calculator. x2 − x b Sketch the graph of f ( x ) = , stating the value of x for which it is discontinuous. x−1 a Find lim

think

a i 1 We cannot substitute x = 1 directly as we will get 00 ,

which is undefined, so we factorise the numerator.

Write/DraW

x2 − x x ( x − 1) = lim x →1 x − 1 x →1 x − 1

a i lim

Chapter 9 • Differentiation

393

2

Cancel (x – 1) from the numerator and denominator.

x, x ≠ 1 = lim x →1

3

Now substitute x = 1 (from lim) and evaluate.

=1

ii 1 Using the limit feature of the CAS calculator, enter

ii lim

x →1

x 2 − x =1 x→1 x − 1

the function and the value of x for the limit. 2

Record the result.

1

3

Write the answer.

x 2 − x lim =1 x→1 x − 1

x2 − x is the same as the graph x −1 of f (x) = x, except where the point (1, 1) does not exist.

b The graph of f ( x ) =

y

b

f(x) 1 0

1

x

The function is discontinuous at x = 1.

WorkeD exampLe 5

a By first factorising the numerator, simplify the rational function

x2 + 5 x + 6 , stating the value for which the function does not exist x+3 (that is, is discontinuous). b Find lim f ( x ), where a is the value at which f (x) is discontinuous. f ( x) =

tUtoriaL eles-1437 Worked example 5

x→ a

think

Write

x 2 + 5x + 6 x+3 ( x + 3)( x + 2) = ( x + 3)

a f (x) =

a 1 Factorise the numerator.

2

f ( x ) = x + 2, x ≠ − 3 so f (x) is discontinuous at x = –3

Cancel (x + 3) from the numerator and denominator.

b lim x + 2 = − 3 + 2

b Write an expression for lim f ( x ) and evaluate by

substituting x =

−3.

x→a

x→ − 3

= −1

Limits of hybrid functions Hybrid functions are functions that have different rules for different parts of the domain.

WorkeD exampLe 6

x 2 , x ∈( − ∞ , 2] . x − 1, x ∈(2, ∞ )

a Sketch the graph of the hybrid function f ( x ) = b Find

i lim f ( x ) x→ 2 −

ii lim f ( x ) x→ 2 +

iii lim f ( x ) if it exists. x→ 2

c Find solutions to parts a and b using a CAS calculator. 394

Maths Quest 11 Mathematical Methods CAS

think

Write/DraW

a 1 Sketch f (x) = x2 over the domain (−∞, 2].

y

a

4

Sketch (on the same axes) f (x) = x − 1 over the domain (2, ∞).

2

f(x)

1 0 i Substitute x = 2 into f (x) = x 2.

b

x

2

b i lim f ( x ) = lim x 2 x → 2−

x → 2−

x → 2+

x → 2+

ii Substitute x = 2 into f (x) = x − 1.

=4 ii lim f ( x ) = lim ( x − 1)

iii Are these limits equal?

=1 iii lim f ( x ) does not exist x→2

(as left limit ≠ right limit). c 1 Using the hybrid function feature of the CAS

calculator, define the functions and their domains. 2

2 − x ∞ < x ≤ 2 Define f ( x ) = x − 1 2 ≤ x < ∞

Plot the function using a suitable window on the CAS calculator.

y 4

f(x)

1 0 3

Use the limit feature of the CAS calculator to find the limit from the left.

x → 2−

4

Record the result.

4

5

Use the limit feature of the CAS calculator to find the limit from the right.

x → 2+

6

Record the result.

1

7

Write the answers.

x

2

lim ( x 2 )

lim ( x − 1)

c i lim ( f (x)) = 4 x → 2−

ii lim ( f (x)) = 1 x → 2+

iii lim( f (x)) does not exist as the left x→2

limit does not equal the right limit.

Limits of discontinuous, rational and hybrid functions exercise 9B

Which of the following graphs are discontinuous?

1 a

b

y

y

c

3

y

0 0

3

x

0

4

x

x

Chapter 9 • Differentiation

395

y

d

f

y

4

x

0

−1

y

e

0

2

0

x

x

3

For each discontinuous function above, state the value of x for which it is discontinuous.

2

x2 − 4x . x →4 x − 4

a Find lim

3 We4

x2 − 4x , stating the value of x for which it is discontinuous. x−4 x2 + x a Evaluate f (x) when x = 0 if f ( x ) = . Comment on this result. x For what value of x is f (x) discontinuous? Factorise the numerator of f (x). Now simplify f (x). Sketch the graph of f (x). Evaluate lim f ( x ) if it exists.

b Sketch the graph of f ( x ) = 4 b c d e f

x →0

By first factorising the numerator, simplify the following rational functions, stating the value for which the function does not exist (is discontinuous). 6 x − 18 x 2 − 5x x 2 + 5x + 4 x 2 + 3x b f (x) = c f (x) = d f (x) = a f (x) = x−3 x x+4 x

5 We5a

e f (x) =

x 2 − 7x + 6 x−6

f

f (x) =

x3 + 8 x+2

g f (x) =

x 2 + 3x − 4 x −1

h f (x) =

x 3 − 27 x−3

For each rational function in question 5, find lim f ( x ), where a is the value at which f (x) is x →a discontinuous. 7 We6a Sketch the graphs of the following hybrid functions. 6 We5b

4 − x , x ∈(−∞ , −1] − 3 x + 1, x ∈( 1, ∞ )

x + 3, x ∈(−∞ , 2) 4, x ∈[2, ∞ )

b g( x ) =

2 x , x ∈(−∞ , 0) 2 x , x ∈[0, ∞ )

d p( x ) =

a f (x) =

2 − x + 1, x ∈( ∞ , 1] x + 2, x ∈(1, ∞ )

c h( x ) = 8 We6b

For each of the corresponding functions in question 7, evaluate the following. ii lim f ( x ) iii lim f ( x )

a i lim f ( x ) x → 2−

x → 2+

x →2

b i lim g( x ) x→ 1

ii lim g( x ) x→ 1

iii lim− g( x )

c i lim h( x )

ii lim h( x )

iii lim h( x )

d i lim p( x )

ii lim p( x )

iii lim p( x )

− −

x →0− x →1−

x→ 1

− +

x →0+

x →0

x →1+

x →1

9 We6c Investigate whether the following limits exist. For those that do exist, state the limit.

x + 2, x ∈(− ∞ , 2) x → 2 3, x ∈[2, ∞ )

b lim

5 − x , x ∈(− ∞ , 3) x →3 2 x + 1, x ∈[3, ∞ )

d lim

2 − x , x ∈( ∞ , 2] x →1 x + 2, x ∈ (2, ∞ )

f

a lim

c lim

e lim

396

Maths Quest 11 Mathematical Methods CAS

x − 4, x ∈(− ∞ , 0] x → 2 x − 4, x ∈ (0, ∞ )

2 x − 3, x ∈(− ∞ , 1] x →1 2 x + 1, x ∈ (1, ∞ ) 2 − − 4 − x , x ∈( ∞ , 2) lim x → 2 x + 2, x ∈[ − 2, ∞ ) −

10 Evaluate the following. a lim( x 2 + 3 x + 2)

b lim

x →2

x→

3 x + 1, x ∈(− ∞ , 1) x →1 x 2 + 3, x ∈[1, ∞ )

− 1 2

14 x + 7 2x + 1

x 2 − 9 x − 10 1 x +1

c lim

d lim−

e lim(12 − 3 x − 2 x 2 )

f lim

x→

2 x − 2 x + 1, x ∈(− ∞, 0] x→ 0 x + 3, x ∈(0, ∞ )

x →2

x 2 + 3x + 2 x→ 1 x+2 ( x + 3)( x − 3) i lim x →3 x−3

h lim( x 3 + x 2 − 5 x )

g lim−

x →1

j

x 2 + 7x + 6 1 x +1

9C

( x + 2)(2 x − 3) x+2

x 2 − 6x + 8 x →4 x−4

k lim− x→

lim

x→ − 2

l lim

Differentiation using first principles

The gradient function is the rule for the instantaneous rate of change of a given function at any point. The gradient at any point (x, y) can be found by substitution into the gradient function. Consider the secant PQ drawn to the curve f (x) at right. The coordinates of P are [x, f (x)] and the coordinates of Q are [x + h, f (x + h)]. So the gradient of the secant PQ =

y

rise run

y = f(x) Q

f(x + h)

f ( x + h) − f ( x ) x+h−x f ( x + h) − f ( x ) = h

=

Tangent at P

P

f(x) 0

x+h x

x

As h approaches zero, that is, as Q draws as close as possible to P along the curve, PQ effectively becomes a tangent to the curve at P. We can therefore say that the gradient of the tangent at P is

lim

h→ 0

f ( x + h) − f ( x ) or h

f ′( x ) = lim

h→ 0

y Q

f ( x + h) − f ( x ) , h ≠ 0, h

where f ′(x) denotes the gradient of a tangent at any point, x, on the graph of f (x). That is, f ′(x) is the gradient function of f (x). The process of finding the gradient function lim

h→ 0

Q

Q moves closer to P as h approaches 0.

P Q 0

h

x

f ( x + h) − f ( x ) h

is called differentiation from first principles. Two different forms of notation are commonly used to represent a function and its derivative. 1. The European notation of Leibniz is: 2. The alternative notation is: (a) y for the function (a) f (x) for the function dy d (b) for the derivative. [ f (x)] or Dx ( f ) for the derivative. (b) f ′(x) or dx dx Chapter 9 • Differentiation 397

WorkeD exampLe 7

Find the derivative of x2 − 2 x from first principles: a by hand b using a CAS calculator. think

a 1 Define f (x).

3

The derivative is equal to: f ( x + h) − f ( x ) lim . h→ 0 h Find f (x) and f (x + h).

4

Simplify the numerator f (x + h) − f (x).

5

Factorise the numerator f (x + h) − f (x).

2

6

f ( x + h) − f ( x ) by h→ 0 h cancelling the common factor of h. Simplify lim

Write

a f (x) = x2 − 2x

f ′(x) = lim

h→ 0

f ( x + h) − f ( x ) h

f (x) = x2 – 2x f (x + h) = (x + h)2 – 2(x + h) f (x + h) − f (x) = (x + h)2 − 2(x + h) − (x2 − 2x) = x2 + 2xh + h2 − 2x − 2h − x2 + 2x = 2xh + h2 − 2h = h(2x + h − 2) lim

h→ 0

f ( x + h) − f ( x ) h h(2 x + h − 2) = lim h→ 0 h = lim(2 x + h − 2), h ≠ 0 h→ 0

Evaluate the limit by substituting h = 0.

= 2x − 2

b 1 Using the CAS calculator, define the function.

b Define f (x) = x2 − 2x

7

2

Define the formula used to evaluate the derivative.

f ( x + h) − f ( x ) h

3

Record the result.

2x + h − 2

4

Determine the limit as h → 0.

5

Record the result.

6

Write the answer.

lim(2 x + h − 2)

h→ 0

2x − 2 f ( x + h) − f ( x ) h f ′( x ) = 2 x − 2 f ′( x ) = lim h→ 0

WorkeD exampLe 8

If g (x) = 2 x2 + 5x − 2, find: a g ′(x) using first principles b the value(s) of x where the gradient equals 0. think

a 1 Let g (x) = 2x 2 + 5x − 2. 2

3

398

tUtoriaL eles-1438 Worked example 8

Write

a g(x) = 2x2 + 5x − 2

g( x + h) − g( x ) h

The derivative is equal to: g( x + h) − g( x ) . lim h→ 0 h

g ′(x) = lim

Find g(x) and g(x + h).

g(x) = 2x2 + 5x – 2 g(x + h) = 2(x + h)2 + 5(x + h) – 2

Maths Quest 11 Mathematical Methods CAS

h→ 0

4

Simplify the numerator g(x + h) − g(x).

5

Factorise.

6

g( x + h) − g( x ) by cancelling h→ 0 h the common factor of h.

g(x + h) − g(x) = 2(x + h)2 + 5(x + h) − 2 − (2x2 + 5x − 2) = 2(x2 + 2xh + h2) + 5x + 5h − 2 − 2x 2 − 5x + 2 = 2x2 + 4xh + 2h2 + 5x + 5h − 2 − 2x2 − 5x + 2 = 4xh + 2h2 + 5h = h(4x + 2h + 5) g( x + h) − g( x ) h→ 0 h h(4 x + 2h + 5) = lim h→ 0 h = lim(4 x + 2h + 5), h ≠ 0

Simplify lim

lim

h→ 0

7

Evaluate the limit by substituting h = 0.

b Solve g ′(x) = 0.

= 4x + 5 So g ′(x) = 4x + 5. b g ′(x) = 0

4x + 5 = 0 4x = −5 − x = 45 So the gradient equals 0 when x = − 45 .

Note: For any polynomial function, f (x), when the expression f (x + h) − f (x) is simplified, all of its terms have h as a factor.

exercise 9C

Differentiation using first principles

1 We7a Find the derivative of the following from first principles. a 5x − 7 b x2 + 10x c x2 − 8x

dy . dx b y = x2 − 3x + 1 e y = 6x − 2x2

d x3 + 2x

2 We7b Use first principles to find a y=x+3 d y = 9 − x2

c y = 4x2 f y = x3 + 5x − 4

DiGitaL DoCS doc-9787 SkillSHEET 9.2 Differentiating from first principles doc-9788 WorkSHEET 9.1

If g(x) = x2 − 6x, find: a g ′(x) using first principles. b the value(s) of x where the gradient equals 0.

3 We8

a If f (x) = x3 − 8, find f ′(x) using first principles.

4

b Hence, determine the value(s) of x where the gradient function is equal to 12. 5 By first deriving the gradient function f ′(x), evaluate f ′(3) when f (x) is equal to: a 7x + 5 b x2 + 4x c x2 − 3x + 2

d x3 − 5.

6 mC Which of the following do not denote the gradient at any point on a graph? (One or more answers

may be correct.) a f ′(x)

B lim

h→ 0

f ( x + h) − f ( x ) h

C lim

h→∞

f ( x + h) − f ( x ) h

f ( x + h) − f ( x ) h 7 mC The most accurate method for finding the gradient when x = 3 for the function f (x) = x2 + 2x is by: a sketching the graph and drawing a tangent at x = 3 to find the gradient B finding the gradient of the secant to the curve joining the points where x = 3 and x = 3.1 C finding f ′(x) using first principles and evaluating f ′(3) D guessing e finding the gradient of the line from the origin to the point (3, 15) 8 mC Given that f ′(x) = 4x if f (x) = 2x2 and g ′(x) = 3x2 + 1 if g(x) = x3 + x, then the derivative 3 + 2x 2 + x must be equal to: of x a 3x2 + 4x + 1 B 12x3 + 4x C 3x2 + 4x D 2x5 + 2x3 e 5x2 + 1 D

dy dx

e

Chapter 9 • Differentiation

399

9D

Finding derivatives by rule

Fortunately, the tedious process of finding derivatives from first principles need not be applied once rules are established. For polynomial functions, the following rules apply. Rule 1. If f (x) = x n, then f ′(x) = nx n − 1. Rule 2. If f (x) = ax n, then f ′(x) = nax n − 1. Rule 3. If f (x) = c, then f ′(x) = 0 (where c is constant). Rule 4. If f (x) = g(x) + h(x), then f ′(x) = g ′(x) + h′(x). WorkeD exampLe 9

Differentiate each of the following. a y = x8 b y = 3 x2 c y = 7x + 3

d y = 2 x5 +

think

3 5

x2 − 6 x

Write

a Write the expression for y. Apply rule 1

to find the derivative.

b Apply rule 2.

c Apply rules 2, 3 and 4. Remember that x0 = 1.

d Differentiate the 3 terms separately (that is,

apply rules 2 and 4).

a y = x8

dy = 8 x 8−1 dx = 8x 7 b y = 3x2 dy = 2(3 x 2−1 ) dx = 6x c y = 7x + 3 dy = 7 x1−1 + 0 dx = 7x 0 = 7 d y = 2x5 + 35 x2 – 6x

dy = 5(2 x 5−1 ) + 2( 35 x 2−1 ) − 1(6 x1−1 ) dx = 10 x 4 + 65 x − 6 x 0 = 10 x 4 + 65 x − 6

WorkeD exampLe 10

Find f ′(x) if f (x) = 3x(x − 2). think

Write

1

Write down f (x).

f (x) = 3x(x − 2)

2

Expand the brackets.

f (x) = 3x2 − 6x

3

Differentiate by rule.

f ′(x) = 6x − 6

WorkeD exampLe 11

If g( x ) =

4 x3 + 3 x2 , find g ′(x): a by hand x

think

a 1 Factorise the numerator, because at this

stage we can only differentiate a constant denominator.

400

Maths Quest 11 Mathematical Methods CAS

b using a CAS calculator. Write 3 2 a g( x ) = 4 x + 3 x

=

x

x x + 3) x

2 (4

2

Simplify g(x).

= x(4x + 3), x ≠ 0

3

Expand the brackets.

= 4x2 + 3x

4

Differentiate g(x) by rule.

b 1 Using the CAS calculator, define the function.

g ′(x) = 8x + 3 4 x 3 + 3x 2 x

b Define g( x ) =

2

Use the differentiation feature of the CAS calculator to find g′(x).

d ( g( x )) dx

3

Record the result.

8x + 3

4

Write the answer.

Given g(x) =

4 x 3 + 3x 2 , g ′(x) = 8x + 3 x

WorkeD exampLe 12

Differentiate each of the following. Express all answers with a positive index. 1 1 4 a f ( x) = x 3 b f ( x) = 7 c f ( x) = x 3 d f ( x) = . x x −

think

a 1 Write down f (x).

Write –3

a f ′(x) = x

−3 − 1

f ′(x) = −3x

−4

2

Differentiate by rule 1.

= −3x

3

Express the answer with a positive index.

=

b 1 Write down f (x). 2

Bring the x-term to the numerator using the index laws, as we can only differentiate a constant denominator.

3

Differentiate by rule 1.

4

Express the answer with a positive index.

c 1 Write down f (x). 2

Differentiate by rule 1.

b f (x) =

Express the answer with a positive index.

d 1 Write down f (x).

x4 1 x7 −7

= 1x

−7 − 1

f ′(x) = −7(1x − = −7x 8 =

−

)

7 x8 1

c f (x) = x 3 1 −1

f ′(x) = 13 (x 3 ) =

3

−3

=

x

−2 3

3 1 2

3x 3 4 d f (x) = x 4

2

Convert x to index form.

f (x) =

3

Bring the x-term to the numerator using the index laws.

f (x) = 4 x

1

x2 −1 2

Chapter 9 • Differentiation

401

4

f ′(x) =

Differentiate by rule 2.

−1 2

(4 x − 12 − 1)

= − 2x 5

6

=

Express with a positive index.

=

Express the power of x back in surd form.

−3 2

−2 3

x2 −

2

x3

To evaluate the gradient of a curve at a given point, substitute the given value of x into the gradient function or derivative. For the function f (x), the gradient at the point (a, f (a)) is found by evaluating f ′(a). This gives the gradient of the tangent at x = a, which equals the gradient of the curve only at that point.

WorkeD exampLe 13

For the function f (x) = 13 x3 − x2 − 5x + 3, evaluate f ′(−1), f ′(0) and f ′(4). think

Write 1

f (x) = 3 x3 – x2 – 5x + 3 f ′(x) = x2 − 2x − 5

1

Write the original function and then find the gradient function, f (x).

2

Substitute x = −1 into the gradient function.

f ′(−1) = (−1)2 − 2(−1) − 5 =1+2−5 = −2 − ∴ f ′( 1) = −2

3

Substitute x = 0 into the gradient function.

f ′(0) = (0)2 − 2(0) − 5 = −5 ∴ f ′(0) = −5

4

Using the CAS calculator, define the function.

5

Differentiate f (x) and evaluate the derivative at x = 4.

6

Record the result.

3

7

Write the answer.

f ′(4) = 3

1

Define f ( x ) = 3 x 3 − x 2 − 5 x + 3 d ( f ( x )) | x = 4 dx

tangents and normals As mentioned earlier in this chapter, the derivative f ′(x) is actually the gradient function. This means that the value of the gradient at any particular point on a curve is equal to the numerical value of the derivative at that point. Recall that if the gradient of a tangent to a curve at point P is m T, then the normal, mN, is a straight line perpendicular − 1 and passing (at right angles) to the tangent such that mN = mT through the point P as shown at right.

y Normal

Tangent

P 0

x

Also recall that the equation of a straight line is given by y − y1 = m(x − x1) where (x1, y1) is the point P, above, and m is the gradient. 402

Maths Quest 11 Mathematical Methods CAS

WorkeD exampLe 14

a Find the equation of the tangent to the curve f (x) = x2 + 6x − 8 at the point where the gradient

has a value of 8.

b Hence, find the equation of the normal at this point. c Find the equation of the tangent at this point using a CAS calculator. think

Write

a 1 Find the gradient function of the curve,

f ′(x).

2

Find x1, the value of x where f ′(x) = 8; that is, solve 2x + 6 = 8.

Alternatively, use the solve and differentiation features of a CAS calculator to find x1 when f ′(x1) = 8.

a f (x) = x2 + 6x – 8

f ′(x) = 2x + 6

For gradient = 8 2x + 6 = 8 2x = 2 x=1

So x1 = 1.

dy solve ( x 2 + 6 x − 8) = 8, dx

x

Record the result.

x=1

3

Find f (x1) to determine the value of y1.

y1 = f (x1) = f (1) = (1)2 + 6(1) − 8 = −1

4

Simplify the equation y − y1 = mT (x − x1) to find the equation of the tangent.

The equation of the tangent at the point (1, −1) is y − −1 = 8(x − 1) y + 1 = 8x − 8 y = 8x − 9

b 1 Find the gradient of the normal using

mN = 2

−1

mT

b mN =

−1 8

.

Simplify the equation y − y1 = mN (x − x1) to find the equation of the normal.

The equation of the normal at the point (1, −1) y − −1 =

is

y+1=

−1 (x − 1) 8 − ( x − 1)

8 8y + 8 = −x + 1 x + 8y + 7 = 0 c 1 The gradient is 8 at the point (1, −1).

c solve(y − −1 = 8(x − 1), y)

Substitute these values into the equation y − y1 = m(x − x1) and solve for y. 2

Record the result.

y = 8x − 9

3

Write the answer.

The equation of the tangent to the curve f (x) = x2 + 6x − 8, at the point (1, −1) with a gradient of 8 is: yT = 8x − 9.

Chapter 9 • Differentiation

403

WorkeD exampLe 15

For the function f (x) = x3 − x2 + 1, find the coordinates of the points where the gradient is perpendicular to the line y = −x + 3. Hence, find the equations of the tangents and normals at these points. think 1

Find the derivative.

2

Find the gradient of the tangent. As the tangent is perpendicular to the line y = −x + 3, the product of their gradients is −1, that is, m1m2 = −1. Let m1 equal the gradient of the tangent and let m2 equal the gradient of the line y = −x + 3.

3

Put f ′(x) = 1 and solve for x to find the point where the gradient of f(x) = 1.

Write

f (x) = x3 − x2 + 1 ∴ f ′(x) = 3x2 − 2x. − m1 = 1 where m2 = −1 m2

=

−1

(− 1) =1 ∴ mtangent = 1 3x2 − 2x = 1 − 2x − 1 = 0 (3x + 1)(x − 1) = 0 3x2

∴x= 4

Substitute x = 1 and x = − 13 into f (x) to find their y-coordinates, and state the points.

Find the equations of the tangent lines.

−1

or x = 1

3

f (1) = (1)3 − (1)2 + 1 = 1 f

−1 3

−1 3

=

∴ (1, 1) and 5

3

−1 3

–

−1 23 , 3 27

2

are the coordinates of the points.

23 27

(

= x

− − 1 3 1

y=x+3+ y=x+ Find the equations of the normal lines. −1 Use mnormal = to find the gradient mtangent of the normal.

23 27

+1=

The equation of the tangent at the point (1, 1) is y − 1 = 1(x − 1) y=x−1+1 ∴ y=x − The equation of the tangent at the point 1 , 23 is y–

6

)

23 27

32 27

−

23 27

−

(

= x

− − 1 3

y = −x – 13 + y = −x +

Maths Quest 11 Mathematical Methods CAS

3 27

1 1 = −1 The equation of the normal at the point (1, 1) is y − 1 = −1(x − 1) y = −x + 1 + 1 ∴ y = −x + 2 − The equation of the normal at the point 1 , 23 is mnormal =

y–

404

tUtoriaL eles-1440 Worked example 15

14 27

)

23 27

3 27

WorkeD exampLe 16

a Graph the function f (x) = x3 − 2x and f ′(x) on the same set of axes. b How can f (x) be used to predict the location and nature of the stationary points of f (x)? think

Write/DraW

a f1(x) = x3 − 2x

a 1 Using the CAS calculator, define f (x)

and f ′(x).

2

f2 ( x ) =

Plot f (x) and f ′(x) on the same axes.

d ( f1 ( x )) dx

f '(x)

y

f(x)

x

b The x-intercepts of f ′(x) predict the location

and nature of the stationary points of f (x).

exercise 9D

b At x = −0.816, f ′(x) goes from positive to negative

(left to right), indicating a maximum turning point for f (x) = x3 − 2x. At x = 0.816, f ′(x) goes from negative to positive (left to right), indicating a minimum turning point for f (x).

Finding derivatives by rule

Differentiate each of the following. d y = 4x4 + 23 x2 − 5x b y = 7x2 c y = 5x + 2 dy 2 Find if y is: dx a 3x4 b 8x7 c 5x 5 d −4x6 − 3 − e 2x f 7x. 3 Match the correct derivative from the set a to G below to each of the following. 1 We9

a y = x6

a x8 e

x4

+

2x3

−3

a 5x4 + 18x 2 − 4 e 6x2 − 4

b x4 + x 2 f x5 + 6x3 − 4x

c 2x3 − 4x + 7 g xp

d x 2 + 6x − 5

B 4x 3 + 2x F 2x + 6

C px p −1 G 4x3 + 6x 2

D 8x7

Differentiate the following.

4

a y = x6 + 3x 2 − 4 d y = 10x5 − 3x4 + 2x3 − 8x

b y = 5x4 − 7x3 + 6x e y=6

c y = x11 − 3x6 + 4x5 + 3x 2 f y = 3x4 + 5x4

Find f ′(x) if f (x) is:

5 a d

2 x4 3 3 x3 4

b

+ 12 x 2 − 3x

6 We10

e

a g( x ) =

+ 16 x3

c

6 x7 7

f

4x3 − 47 x 2 − 65 x + 8.

Find f ′(x) if f (x) is:

a f (x) = x(x + 3) d f (x) = 9(8 − 3x)2 7 We11

−5 2 x 8 5 2 x + 3 x4 5 4

b f (x) = 3x(2x − 5) e f (x) = (x + 2)3

c f (x) = (x + 4)2 f f (x) = (2x − 5)3.

Find g ′(x) by first simplifying g (x). x 3 + 5x x

b g( x ) =

8x3 − 6x 2x

c g( x ) =

3x 3 + 2 x 2 − 5x x

d g( x ) =

5x 4 + x 3 + 7 x 2 x2 Chapter 9 • Differentiation

405

Differentiate each of the following. Express all answers with a positive index.

8 We12 −4

−7

a x e

b x

−4x−6

i

5 x3 1

−4

−8

c 3x

d 5x

1 x4

f

−3x−5

g

j

10 x6

k 2x 2

1

2

m 4x 4

n 3x 5

o

q 4 x

r

3

s

x

x

h

1 x9

l

x3

p

2 x

2

1 x

3

Evaluate i f ′(1) ii f ′(−2) and iii f ′(0) for each of the following.

9 We13

a f (x) = 5x2 + 3x − 1 b f (x) = 13 x3 + 2x 2 − 4 c f (x) = 3x 2 − 2x + 6 d f (x) = x3 + 7x − 8 10 We13 a Find the x-intercepts of the parabola y = x 2 − 5x + 6. b Find the gradient of the parabola at the points where it crosses the x-axis. c Determine the value of x for which the gradient of the parabola is: i 0 ii 7 iii −3.

DiGitaL DoCS doc-9779 Gradient at a point doc-9789 tangent and normal

11 a Find the x-intercepts of the curve y = 2x 2 + 5x − 3. b Find the gradient of the curve at these points. c Find the coordinates of the point where the gradient is 0. 12

Find the coordinate(s) of the points on the curve x 3 − 3x2 where the tangent: a is parallel to the x-axis b is parallel to the line y = −3x + 2.

13 We14 Find the equations of the tangent and normal for each of the following curves. a f (x) = 5x2 + 3x − 1 at the point (1, 7) b f (x) = 13 x3 + 2x2 − 4 at x = −2 c f (x) = 3x2 − 2x + 6 at x = 2 d f (x) = x3 − 7x − 8 at the point (0, −8) e f (x) = x2 + 2 at x = a f f (x) = 3x2 − 4x at the point (2a, 12a2 − 8a) 14 We14 a Find the equation of the tangent at the point on the curve x 2 + 4x − 1 where the gradient is 6. b Hence, find the equation of the normal at this point. 15 Find the equation of the normal to the curve y = 2x 2 − 2x + 5 at the point where the curve crosses the

y-axis. 16 Find the equation of the normal to the curve y = −x2 + 4x at: a x=2 b x = 1.

Find the equation of the normal to the curve y = x3 + 2x2 − 3x + 1 at x = −2. 18 If the equation of the normal to the curve y = 2x2 + 4.5x + 1 is y = −2x + c, find: a the coordinates at the point of tangency b the value of c.

17

19 We15 Find the equations of the tangents to the curve f (x) = 2 x3 + 5 x2 − x + 2 that are parallel to the

line y − 2x = 4.

3

2

20

Find the coordinates of the points on the curve f (x) = 2x3 − x2 + 6 where the tangents are parallel to the line −5x + 2y = 12. Hence, find the equation of the tangents at these points.

21

Find the equation of the tangent to the curve f (x) = 2x2 − 4x + 4 that is perpendicular to the line y=

−1 4

x + 4.

Find the coordinates of the points where the tangents to the curve f (x) = x3 − 3x − 1 are perpendicular to the line x + 3y = 3. Hence, find the equations of the normals at these points.

22 We15

406

Maths Quest 11 Mathematical Methods CAS

The tangent to the curve y = ax2 + bx + c at the point (−2, −8) is parallel to the line −7x + y = 10. If the curve also passes through the point (−1, −3.5), find the values of a, b and c. Hence, find the equation of the tangent at the point (−2, −8). 24 a Find the equation of the tangent to the curve y = 2x2 − 2 at x = 2a. b Hence find the x-intercept of the tangent line in terms of a. c A straight line with equation y = 2x − 2 passes through the x-intercept of the tangent line. Find the value of a. d What is the equation of the tangent line? 25 We16 Sketch the graphs of: a f (x) = 4 x − 2 23

b

f (x) = x 2 − 4 x + 2

f ( x ) = x 3 + 3x 2 + 3 and the graph of f ′(x) for each on the same set of axes. How can f ′(x) be used to predict the location and nature of the stationary points of f (x) in each case?

c

9e

rates of change

The rate of change of a function refers to its gradient. For linear functions the gradient is constant; however, the gradient for other functions such as quadratic or cubic polynomials is continually changing.

The rate of change of position with respect to time is velocity.

The rate of change of velocity with respect to time is acceleration.

Differentiation provides us with a tool to describe the gradient of a function and hence determine its rate of change at any particular point. In essence, while average rates of change can be determined from the original function, differentiation of this function provides a new function that describes the instantaneous rate of change. y y = f(x2) Note: The term instantaneous rate of change is often referred to as rate of change. If P(x1, f (x1)) and Q(x2, f (x2)) are two points on the graph Q (x2, f(x2)) of the function with rule y = f (x), then the average rate of change of y with respect to x over the interval x ∈ [x1, x2] is equal to the gradient of the straight line PQ. change in f ( x ) change in x f ( x 2 ) − f ( x1 ) = x 2 − x1

Average rate of change =

P (x1, f(x1)) 0

x1

x2

x

The instantaneous rate of change finds the rate of change at a specific point. dy The instantaneous rate of change of y ( f (x)) with respect to x is given by the derivative ( f ′ (x)). dx Chapter 9 • Differentiation

407

WorkeD exampLe 17

If f (x) = x2 − 2x + 4, determine: a the average rate of change between x = 2 and x = 4 b a new function that describes the rate of change at any point x c the instantaneous rate of change when x = 4 d parts a, b and c using a calculator. think

Write

a f (x) = x2 − 2x + 4

a 1 Write the function. 2

Average rate of change =

change in f ( x ) change in x

f (4) − f (2) 4−2 12 − 4 = 2 =4

Average rate of change =

b Differentiate f (x).

b f ′(x) = 2x − 2

c Substitute x = 4 into f ′(x).

c f ′(4) = 2(4) − 2 = 6

d 1

Define the function on your CAS calculator.

So the rate of change when x = 4 is 6.

d Define f (x) = x2 − 2x + 4

2

Determine f (4) and f (2), and hence the expression f (4) − f (2) . 4−2

f (4) − f (2) 4−2

3

Record the result.

4

4 5

Use the differentiation feature of the CAS calculator to find the derivative of f (x). Record the result.

d ( f ( x )) dx 2x − 2

6

Evaluate the derivative at x = 4.

d ( f ( x )) x = 4 dx

7

Record the result.

6

8

Write the answers.

a The average rate of change between x = 2 and

x = 4 is 4. b f ′(x) = 2x − 2 c The instantaneous rate of change when x = 4 is 6.

WorkeD exampLe 18

A javelin is thrown so that its height, h metres, above the ground is given by the rule h(t) = 20t − 5t2 + 2, where t represents time in seconds.

a Find the rate of change of the height at any time, t. b Find the rate of change of the height when i t = 1 ii t = 2 iii t = 3. c Briefly explain why the rate of change is initially positive, then zero and then negative over the

first 3 seconds.

d Find the rate of change of the height when the javelin first reaches a height of 17 metres. 408

Maths Quest 11 Mathematical Methods CAS

think

a 1 Write the rule. 2

Differentiate h(t).

Write

a h(t) = 20t − 5t2 + 2

h′(t) = 20 − 10t

b 1 Evaluate h′(1).

b i h′(1) = 20 − 10(1)

Evaluate h′(2).

ii h′(2) = 20 − 10(2)

Evaluate h′(3).

iii h′(3) = 20 − 10(3)

2

3

c For rates of change:

Positive means increasing. Zero means neither increasing nor decreasing. Negative means decreasing.

d 1 Find the time at which the javelin is 17 m above

the ground by substituting h = 17 into h(t).

= 10 m/s = 0 m/s

= −10 m/s

c The javelin travels upwards during the first

2 seconds. When t = 2 seconds, the javelin has reached its maximum height. When t > 2 seconds, the javelin is travelling downwards. d 20t − 5t2 + 2 = 17 −5t2

2

Make RHS = 0.

+ 20t − 15 = 0

3

Divide both sides by −5.

4

Factorise, then solve for t. Note: The quadratic formula could also be used to solve for t.

5

The first time it reaches 17 m is the smaller value of t.

The javelin first reaches 17 m when t = 1 s.

6

Evaluate h′(1).

h′(1) = 20 − 10(1) = 10 m/s The rate of change of height is 10 m/s.

t2 − 4t + 3 = 0 (t − 1)(t − 3) = 0 t = 1 or 3

It is worth noting that there are two common ways of writing the derivative as a function. For example, the derivative of the function P(x) = x2 + 5x − 7 may be written as P′(x) = 2x + 5 dP = 2 x + 5. or as dx WorkeD exampLe 19

The shockwave from a nuclear blast spreads out at ground level in a circular manner. a Write down a relationship between the area of ground, A km2, over which the shockwave passes and its radius, r km. b Find the rate of change of A with respect to r. c Find the rate of change of A when the radius is 2 km. d What is the rate of change of A when the area covered is 314 km2?

Chapter 9 • Differentiation

409

Write

think

a State the formula for the area of a circle.

a A(r) = πr 2

b Differentiate A(r).

b A′(r) = 2πr

c Substitute r = 2 into A′(r).

c A′(2) = 2π(2)

d 1 Substitute A = 314 into the area function A(r)

d A(r) = πr2

Note: The units for the rate of change of A (km2) with respect to r (km) are km2 per km or km2/km.

= 12.57 The rate of change of A when the radius is 2 km is 12.57 km2/km. 314 = πr2 314 r2 = π = 99.95 r = 10 since r > 0

and solve for r.

2

A′(10) = 2π (10) = 62.8 The rate of change of A when area is 314 km2 is 62.83 km2/km.

Find the rate of change when r = 10.

exercise 9e

rates of change

If f (x) = x2 + 5x + 15, find: a the average rate of change between x = 3 and x = 5 b a new function that describes the rate of change c the instantaneous rate of change when x = 5.

1 We17

2 DiGitaL DoCS doc-9793 SkillSHEET 9.3 average rate of change doc-9777 Gradient between two points on a graph

A balloon is inflated so that its volume, V cm3, at any time, t seconds, is: V=

−8 3 t 5

+ 24t 2 , t ∈ [0, 10]

a What is the volume of the balloon when: i t = 0? ii t = 10? b Hence, find the average rate of change between t = 0 and t = 10. c Find the rate of change of volume when: i t=0 ii t = 5 iii t = 10.

The average rate of change between x = 1 and x = 3 for the function y = x2 + 3x + 5 is: B 9 C 5 D 3 e 7 3 2 4 mC The instantaneous rate of change of the function f (x) = x − 3x + 4x when x = −2 is: B −2 a 2 C 28 D 3 e 12 3 mC

a 1

5 mC If the rate of change of a function is described by

dy = 2 x 2 − 7 x , then the function dx

could be: a y = 6x3 − 14x

B y = 23 x 3 − 7 x

C y = 23 x 3 − 72 x 2 + 5

D y = x 3 − 72 x 2 + 2

e 2x2 − 7x + 5

In a baseball game the ball is hit so that its height above the ground, h metres, is h(t) = 1 + 18t − 3t2, t seconds after being struck. a Find the rate of change, h′(t). b Calculate the rate of change of height after: i 2 seconds ii 3 seconds iii 4 seconds. c What happens when t = 3 seconds? d Find the rate of change of height when the ball first reaches a height of 16 metres.

6 We18

DiGitaL DoC doc-9794 SkillSHEET 9.4 instantaneous rate of change

410

Maths Quest 11 Mathematical Methods CAS

7

The position, x metres, of a lift (above ground level) at any time, t seconds, is given by x(t) = −2t2 + 40t. a Find the rate of change of displacement (velocity) at any time, t. b Find the rate of change when: i t=5 ii t = 9 iii t = 11. c What happened between t = 9 and t = 11? d When and where is the rate of change zero?

8 The number of seats, N, occupied in a soccer stadium t hours after the gates are opened is given by:

N = 500t2 + 3500t, t ∈ [0, 5]

a Find N when: i t=1 ii t = 3. b What is the average rate of change between t = 1 and t = 3? c Find the instantaneous rate when: i t=0 ii t = 1 iii t = 3 d Why is the rate increasing in the first 4 hours?

iv t = 4.

9 The weight, W kg, of a foal at any time, t weeks, after birth is given by:

W = 80 + 12t − 103 t 2

where

0 ≤ t ≤ 20.

a What is the weight of the foal at birth? b Find an expression for the rate of change of weight at any time, t. c Find the rate of change after: i 5 weeks ii 10 weeks iii 15 weeks. d Is the rate of change of the foal’s weight increasing or decreasing? e When does the foal weigh 200 kg? 3

10 The weekly profit, P (hundreds of dollars), of a factory is given by P = 4.5n − n 2, where n is the number

of employees.

dP . dn b Hence, find the rate of change of profit, in dollars per employee, if the number of employees is: i 4 ii 16 iii 25. c Find n when the rate of change is zero. a Determine

11 Gas is escaping from a cylinder so that its volume, V cm3, t seconds after the leak starts, is described 1 2 by V = 2000 − 20t − 100 t . a Evaluate the rate of change after: i 10 seconds ii 50 seconds b Is the rate of change ever positive? Why?

iii 100 seconds.

12 We19 Assume an oil spill from an oil tanker is circular and remains that way. a Write down a relationship between the area of the spill, A m2, and the radius, r metres. b Find the rate of change of A with respect to the radius, r. c Find the rate of change of A when the radius is: i 10 m ii 50 m iii 100 m. d Is the area increasing more rapidly as the radius

increases? Why? 13 A spherical balloon is being inflated. a Express the volume of the balloon, V m3, as a function

of the radius, r metres. b Find the rate of change of V with respect to r. c Find the rate of change when the radius is: i 0.1 m ii 0.2 m iii 0.3 m. 14 A rectangular fish tank has a square base, with its height being

equal to half its base length. a Express the length and width of the base in terms of its height, h. b Hence, express the volume, V m3, in terms of the height, h, only. c Find the rate of change of V when: i h=1m ii h = 2 m iii h = 3 m. Chapter 9 • Differentiation

411

15 For the triangular package shown, find: a x in terms of h b the volume, V, as a function of h only c the rate of change of V when: i h = 0.5 m ii h = 1 m.

6 x 30° h

16 A new estate is to be established on the side of a hill. y

80

x

200

Regulations will not allow houses to be built on slopes where the gradient is greater than 0.45. If the equation of the cross-section of the hill is y = −0.000 02x3 + 0.006x2, find: dy a the gradient of the slope dx b the gradient of the slope when x equals: i 160 ii 100 iii 40 iv 20 c the values of x where the gradient is 0.45 d the range of heights for which houses cannot be built on the hill. 17 A bushfire burns out A hectares of land, t hours after it started, according to the rule A = 90t2 − 3t3. a At what rate, in hectares per hour, is the fire spreading at any time, t? b What is the rate when t equals: i 0? ii 4? iii 8? iv 10? v 12? vi 16? vii 20? c Briefly explain how the rate of burning changes during the first 20 hours. d Why isn’t there a negative rate of change in the first 20 hours? e What happens after 20 hours? f After how long is the rate of change equal to 756 hectares per hour?

Sketching graphs containing stationary points 9F

The derivative of a function gives its gradient function — that is, it gives the gradient of a tangent to the curve for any specified value of the independent variable. When the derivative equals zero, the tangent is horizontal. The point or points on the curve where this occurs are called stationary points. In other words, a function f (x) has stationary points when f ′(x) = 0. Stationary points can take the form of: 1. a local minimum turning point 2. a local maximum turning point 3. a stationary point of inflection. y

Local minimum turning point Just to the left of a, the gradient is negative; that is, if x < a, but close to a, then f ′(x) < 0. At the point where x = a, the gradient is zero; that is, at x = a, f ′(x) = 0. Just to the right of a the gradient is positive; that is, if x > a, but close to a, f ′(x) > 0. 412 Maths Quest 11 Mathematical Methods CAS

f '(x) < 0

f '(x) > 0 f '(a) = 0

0

a

x

In other words, for a stationary point at x = a, if the gradient changes from negative to positive as we move from left to right in the vicinity of a, it is a local minimum.

Local maximum turning point At x < a, but close to a, f ′(x) > 0. At x = a, f ′(x) = 0. At x > a, but close to a, f ′(x) < 0. In other words, for a stationary point at x = a, if the gradient changes from positive to negative as we move from left to right in the vicinity of a, it is a local maximum. The term local maximum or local minimum implies that the function has a maximum or minimum in the vicinity of x = a. This is important because some functions can have more than one stationary point.

y f '(a) = 0 f '(x) < 0

f '(x) > 0 0

x

a

Stationary point of inflection y

y f '(x) > 0 f'(x) < 0 f '(a) = 0

0

f '(x) > 0 a

f '(a) = 0 x

f '(x) < 0 a

0

x

At x < a, but close to a, f ′(x) > 0. At x < a, but close to a, f ′(x) < 0. At x = a, f ′(x) = 0. At x = a, f ′(x) = 0. At x > a, but close to a, f ′(x) > 0. At x > a, but close to a, f ′(x) < 0. In other words, for a stationary point at x = a, if the gradient remains positive or negative in the vicinity of a, it is a stationary point of inflection. Worked Example 20

If f (x) = x3 − 6x2 − 15x, find: a the value(s) of x where the gradient is zero b the stationary point(s). Think

a 1 Write the function.

Write

a f (x) = x3 − 6x2 − 15x

2

Differentiate f (x) to find the gradient function f ′(x).

f ′(x) = 3x2 − 12x − 15

3

Solve f ′(x) = 0 to find the x-values of each stationary point.

For stationary points: f ′(x) = 0 3x2 − 12x − 15 = 0 3(x2 − 4x − 5) = 0 x2 − 4x − 5 = 0 (x − 5)(x + 1) = 0 x = 5 or x = −1

b 1 Substitute each value of x into f (x) to find the

corresponding y-values.

2

Write the coordinates of each stationary point.

b f (5) = (5)3 − 6(5)2 − 15(5)

f (−1)

= −100 = (−1)3 − 6(−1)2 − 15(−1) =8

Stationary points occur at (5, −100) and (−1, 8).

Chapter 9 • Differentiation 413

Worked Example 21

Sketch the graph of the function f (x) = 5 + 4x − x2, labelling all intercepts and stationary points. Think

Write/Draw

1

Write the function.

f (x) = 5 + 4x − x2

2

Find the y-intercept by letting x equal 0.

y-intercept: x = 0, f (0) = 5 + 4(0) − (0)2 =5 so y-intercept is (0, 5).

3

Find the x-intercepts by letting f (x) = 0.

x-intercepts: if f (x) = 0, 5 + 4x − x2 = 0 x2 − 4x − 5 = 0 (x + 1)(x − 5) = 0 x = −1 or x = 5 so x-intercepts are (−1, 0) and (5, 0).

4

Differentiate f (x) to find the gradient function f ′(x).

f (x) = 5 + 4x – x2 f ′(x) = 4 − 2x

5

Solve f ′ (x) = 0 to find the x-value(s) of each stationary point.

For stationary points: f ′(x) = 0 4 − 2x = 0 −2x = −4 x=2

6

Substitute this value of x into f (x) = 5 + 4x − x2 to find the corresponding y-value.

f (2) = 5 + 4(2) − (2)2 =9 so there is a stationary point at (2, 9).

7

Determine the nature of the stationary point at x = 2 by evaluating f ′ (x) to the left and right, say at x = 1 and x = 3.

x < 2: f ′(1) = 4 − 2(1) =2 ∴ f ′(x) > 0 x > 2: f ′(3) = 4 − 2(3) = −2 ∴ f ′(x) < 0

8

Complete a gradient table. Since the gradient changes from positive to negative as we move from left to right in the vicinity of x = 2, the stationary point (2, 9) is a local maximum.

x

1

2

3

Sign of f ′(x)

+

0

−

Slope (2, 9) is a local maximum. 9

Sketch the graph.

y (2, 9) 9 f(x)

5

−1 0

414 Maths Quest 11 Mathematical Methods CAS

2

5 x

WorkeD exampLe 22

a Find the stationary points and determine their nature for the function

f(x) = x3 − x2 − 8x + 8. b Find the coordinates of all intercepts. c Sketch the graph of f (x) showing all stationary points and intercepts. think

Write/DraW

a 1 Write the rule for f (x).

a f (x) = x3 − x2 − 8x + 8

2

Differentiate f (x) to find f ′(x).

f ′(x) = 3x2 − 2x − 8

3

Solve f ′(x) = 0 to find the x-values of each stationary point.

For stationary points, solve f ′(x) = 0 for x. 3x2 − 2x − 8 = 0 (3x + 4)(x − 2) = 0 x = − 43 or x = 2

4

Substitute each value of x into f (x) to find the y-coordinates of the stationary points.

f ( − 43 ) = ( − 43 )3 − ( − 43 )2 − 8( − 43 ) + 8 =

392 27

= 14 14 27

∴ ( − 43 ,14 14 27 ) is one stationary point. f (2) = (2)3 − (2)2 − 8(2) + 8 = −4 ∴(2, −4) is another stationary point. Define f (x) = x3 − x2 − 8x + 8

5

Alternatively, all of this working can be done on a CAS calculator. First, define f (x).

6

Set f ′(x) = 0 and use the solve and differentiation features of the CAS calculator.

d solve ( f ( x )) = 0, x dx

7

Record the result.

x=

8

Evaluate f ( 43) and f (2).

f

9

Record the results.

392 − , 4 27

10

Determine the nature of the stationary point at x = − 43 by evaluating f ′(x) to the left and right. Choose x = −2 and x = −1.

x < 43 : f ′( − 2) = 3( − 2)2 − 2( − 2) − 8 = 8 ∴ f ′(x) > 0

−

11

Complete a gradient table and state the type of stationary point. Since the gradient changes from positive to negative as we move from left to right in the vicinity of x = − 43 , the stationary point ( − 43 , 14 14 27 ) is a local maximum.

12

Determine the nature of the stationary point at x = 2 by evaluating f ′(x) to the left and right. Choose x = 1 and x = 3.

13

Complete a gradient table and state the type of stationary point. Since the gradient changes from negative to positive as we move from left to right in the vicinity of x = 2, the stationary point (2, −4) is a local minimum.

−4 3

−4

, 3

or x = 2 f (2).

−

−

x > 43 : f ′( − 1) = 3( − 1)2 − 2( − 1) − 8 = − 3 ∴ f ′(x) < 0 x

−2

−4 3

−1

Sign of f ′(x)

+

0

−

Slope ∴ ( − 43 , 14 14 27 ) is a local maximum turning point. x < 2: f ′(1) = 3(1)2 − 2(1) − 8 = −7 ∴ f ′(x) < 0 x > 2: f ′(3) = 3(3)2 − 2(3) − 8 = 13 ∴ f ′(x) > 0 x

1

2

3

Sign of f ′(x)

−

0

+

Slope ∴ (2, −4) is a local minimum turning point.

Chapter 9 • Differentiation

415

b 1 To find the x-intercepts, factorise f (x) by long

division, or use another appropriate method.

2

b f (x) = x3 − x2 − 8x + 8

)= x2(x − 1) − 8(x − 1) )= (x − 1)(x2 − 8) = ( x − 1)( x − 2 2)( x + 2 2)

Solve f (x) = 0 for x-intercepts.

( x − 1) ( x − 2 2) ( x + 2 2) = 0 x = 1 or x = 2 2 or x = − 2 2 The coordinates of the x-intercepts are (1, 0), (2 2, 0) and ( − 2 2, 0).

3

f (0) = (0)3 − (0)2 − 8(0) + 8 = 8 ∴ (0, 8) is the intercept.

Evaluate f (0) to find the y-intercept.

c Sketch the graph of f (x) showing all stationary

c

—) (− —43 , 14 14 27

points and axes intercepts.

y

(0, 8)

(1, 0)

(−2 2, 0) 0

(2 2, 0) (2, −4)

x

WorkeD exampLe 23

The curve with equation y = ax2 + bx + 7 has a stationary point at (−2, 10). Find the values of a and b. think 1

Write the rule.

2

Differentiate y to find the gradient function.

3

Put

4

y = ax2 + bx + 7 dy = 2ax + b dx

dy = 0 and substitute x = −2 into dx

dy , as x = − 2 is a stationary point. dx Substitute x = −2 into y and put y = 10 to get another equation with a and b.

2a(−2) + b = 0 ∴ −4a + b = 0

a(−2)2 + b(−2) + 7 = 10 ∴ 4a − 2b = 3

5

Solve the simultaneous equations [1] and [2]. Add equations [1] and [2] to eliminate a and solve for b.

−2b

6

Substitute b = −3 into equation [1] to find a.

−4a

[1]

[2]

+b=3 −b = 3 ∴ b = −3 −3=0 =3 ∴ a = − 43 −4a

7

Write the values of a and b.

∴ a = − 43 and b = − 3

8

Alternatively, all of this working can be done on a CAS calculator. First, define f (x).

Define f (x) = ax2 + bx + 7

9

Use the solve and differentiation features of the CAS calculator to find a and b.

d solve f (− 2) = 10 and ( f ( x )) = 0 x = − 2, a dx

Record the result.

a=

10

416

Write

Maths Quest 11 Mathematical Methods CAS

−3 4

and b = −3

Sketching graphs containing stationary points exercise 9F

For each of the following functions, determine the value(s) of x where the gradient is zero. f (x) = + 2x b f (x) = x2 − 8x + 5 3 2 f (x) = x − 3x d f (x) = 2x3 + 6x2 − 18x + 1 y = (x + 6)(x − 2) f y = x2(x − 1) 2 1 3 2 y = 10 + 4x − x h y = 3 x − 3x + 5x − 2

1 We20a a c e g

x2

For each function in question 1, determine all of the stationary points.

2 We20b 3

If f (x) = x2 − 8x + 1: a show that there is a stationary point when x = 4 b evaluate f ′(3) and f ′(5) c state which type of stationary point it is.

4

For the function f ′(x) = 5 − x2: a find x when f ′(x) = 0 b state which type of stationary point it has.

5

If f (x) = x3 − 4 then: a show that there is a stationary point when x = 0 only b find f ′ (−1) and f ′(1) c state which type of stationary point it is.

6

If f ( x ) = 13 x 3 − x 2 − 3 x + 5: a show there are stationary points when x = −1 and x = 3 b evaluate f ′(−2), f ′(0) and f ′(4) c state which type of stationary points they are. When x = 1, the curve y = 2x2 − 3x + 1: a is decreasing C has a local minimum e does not exist

DiGitaL DoCS doc-9721 Quadratic graphs doc-9730 Cubic graphs

7 mC

B has a local maximum D is increasing

When x = −1 the function y = x3 − 2x2 − 7x: a is decreasing B has a local maximum C has a local minimum D is increasing e does not exist

8 mC

9 mC

if x >

−2

The graph below that best represents a function with f ′(−2) = 0, f ′(x) < 0 if x < −2, and f ′(x) > 0 is: y

a

y

B

y

C

f(x) f(x)

D

−2

0

x −2

f(x)

y

f(x)

e

y

0 −2

0

x

0

x

−2

0

x

f(x)

x

−2

Chapter 9 • Differentiation

417

f ′(1) = f ′(4) = 0 and f ′(x) < 0 if 1 < x < 4 and f ′(x) > 0 if x < 1 and x > 4. The graph that satisfies these conditions is:

10 mC a

y

B

f(x)

y f(x)

0 0

1

x

4

y

C

D

1

4

x

y

f(x) 0

1

x

4

0

1

4

x

f(x) e

y f(x)

0

4

1

x

11 We21 For each of the following, find the stationary points and determine their nature. a y = x2 + 6x + 2 b y = 8x − 2x2 c y = x3 − x2 1 2 1 3 1 2 f y = (x − 1)3 3 d y = x + x −3 e y = x − x − 2x 2

g y=

x3

+3

h y=

3

2

x3

− 27x + 5

12 We22 Use a CAS calculator to help sketch the graphs of the following functions, labelling all DiGitaL DoCS doc-9795 SkillSHEET 9.5 review of discriminant doc-9796 SkillSHEET 9.6 Solving cubic equations

intercepts and stationary points. f (x) = x2 − 2x − 3 b f (x) = x3 − 3x − 2 c f (x) = x3 − 2x2 + x 2 3 2 f (x) = x (3 − x) e f (x) = x + 4x + 4x f f (x) = x3 − 4x2 − 11x + 30 3 2 3 f (x) = (x + 2) h f (x) = 24 + 10x − 3x − x i f (x) = x3 − 2x2 − x + 2 3 f (x) = 8 − x 13 For the functions y = −2x2 − 5x + 8 and y = 2x3− x2 − 4x + 5: a use a CAS calculator to sketch the graph of each function, showing all coordinates of stationary points b hence, find the x-values for which: dy dy dy i ii iii =0 >0 < 0. dx dx dx a d g j

14 The curve with equation y = 2x2 + bx + 8 has a turning point at x = 3. Find the value of b. 15 We23 The curve with equation y = ax2 + bx has a stationary point at (3, 9). Find the values of a and b. 16 The curve with equation y = ax2 + bx + c passes through the point (2, −4) and has a stationary point at

(1, − 32 ). Find the values of a, b and c. 17 The curve with equation y = ax3 − x2 + bx + 2 has turning points at x = 1 and x = −2. Find the values of a and b.

Solving maximum and minimum problems 9G

There are many practical situations where it is necessary to determine the maximum or minimum value of a function. For quadratic functions, differentiation makes this a relatively simple task because, as we saw in the p

3RD EDIT ION

VCE M AT HEM AT I CS U N I T S 1 & 2

Third edition published 2013 by John Wiley & Sons Australia, Ltd 42 McDougall Street, Milton, Qld 4064 First edition published 2009 Second edition published 2010 Typeset in 10/12pt Times LT Std © John Wiley & Sons Australia, Ltd 2009, 2010, 2013 The moral rights of the authors have been asserted. National Library of Australia Cataloguing-in-Publication data Title: Edition: ISBN: Notes: Target audience: Subjects: Other authors/ contributors: Dewey number:

Maths quest 11 mathematical methods CAS/Robyn Williams . . . [et al.]. 3rd ed. 978 1 118 31058 8 (paperback) 978 1 118 31067 0 (flexisaver) 978 1 118 31060 1 (ebook) Includes index. For secondary school age. Mathematics — Textbooks. Williams, Robyn (Robyn Ellen) 1967– 510

Reproduction and communication for educational purposes The Australian Copyright Act 1968 (the Act) allows a maximum of one chapter or 10% of the pages of this work, whichever is the greater, to be reproduced and/or communicated by any educational institution for its educational purposes provided that the educational institution (or the body that administers it) has given a remuneration notice to Copyright Agency Limited (CAL). Reproduction and communication for other purposes Except as permitted under the Act (for example, a fair dealing for the purposes of study, research, criticism or review), no part of this book may be reproduced, stored in a retrieval system, communicated or transmitted in any form or by any means without prior written permission. All inquiries should be made to the publisher. Illustrated by Aptara and Wiley Composition Services Typeset in India by Aptara Printed in Singapore by Craft Print International Ltd 10 9 8 7 6 5 4 3 2 1

Contents 2K Using technology to solve quadratic equations

Introduction vi About eBookPLUS viii Acknowledgements ix

Exercise 2K 82 2L Simultaneous quadratic and linear equations Exercise 2L 87

Chapter 1

Linear functions

1

1A Solving linear equations and inequations 1B 1C 1D 1E 1F 1G

1H ■ ■ ■ ■

1 Exercise 1A 3 Rearrangement and substitution 4 Exercise 1B 6 Gradient of a straight line 8 Exercise 1C 10 Sketching linear functions 12 Exercise 1D 14 Simultaneous equations 15 Exercise 1E 17 Finding the equation of a straight line 18 Exercise 1F 20 Distance between two points and midpoint of a segment 21 Exercise 1G 22 Linear modelling 24 Exercise 1H 25

Summary 27 Chapter review 28 ICT activities 34 Answers 35

Chapter 2

Quadratic functions 2A Polynomials 2B 2C 2D 2E 2F

2G 2H 2I

2J

39

39 Exercise 2A 40 Expanding quadratic expressions 41 Exercise 2B 43 Factorising quadratic expressions 44 Exercise 2C 46 Factorising by completing the square 47 Exercise 2D 50 Solving quadratic equations — Null Factor Law 50 Exercise 2E 53 Solving quadratic equations — completing the square 55 Exercise 2F 57 The quadratic formula 58 Exercise 2G 61 The discriminant 62 Exercise 2H 66 Graphs of quadratic functions as power functions (turning point form) 66 Exercise 2I 68 Graphs of quadratic functions (intercepts method) 70 Exercise 2J 78

■ ■ ■ ■

80 83

Summary 90 Chapter review 92 ICT activities 95 Answers 96

Chapter 3

Cubic and quartic functions

105

Polynomials of degree 3 and 4 105 105 Exercise 3A 106 3B Long division of polynomials 107 Exercise 3B 109 3C Polynomial values 109 Exercise 3C 110 3D The remainder and factor theorems 111 Exercise 3D 113 3E Factorising polynomials 114 Exercise 3E 116 3F Sum and difference of two cubes 117 Exercise 3F 118 3G Solving polynomial equations 118 Exercise 3G 120 3H Cubic graphs — intercepts method 121 Exercise 3H 124 3I Quartic graphs — intercepts method 126 Exercise 3I 127 3J Graphs of cubic functions in power function form 128 Exercise 3J 130 3K Domain, range, maximums and minimums 132 Exercise 3K 134 3L Modelling using technology 136 Exercise 3L 137 3m Finite differences 138 Exercise 3M 142 3A Expanding

■ ■ ■ ■

Summary 144 Chapter review 146 ICT activities 150 Answers 151

ExAm prACtICE 1

Based on Chapters 1–3

159

Chapter 4

Relations, functions and transformations 4A Set notation

161 Exercise 4A 162 4B Relations and graphs Exercise 4B 165

163

161

4C Domain and range 167

Exercise 4C 170 4D Types of relations (including functions) 172 Exercise 4D 174 4E Power functions (hyperbola, truncus and square root function) 175 Exercise 4E 182 4F Function notation 183 Exercise 4F 186 4G Special types of function (including hybrid functions) 187 Exercise 4G 189 4H Inverse relations and functions 192 Exercise 4H 193 4I Circles 193 Exercise 4I 195 4J Functions and modelling 197 Exercise 4J 197 ■■ ■■ ■■ ■■

Summary 200 Chapter review 202 ICT activities 209 Answers 210

Chapter 5

Exponential and logarithmic functions 217 Introduction 217 5A Index laws 217 Exercise 5A 220 5B Negative and rational powers 221 Exercise 5B 224 5C Indicial equations 224 Exercise 5C 226 5D Graphs of exponential functions 227 Exercise 5D 230 5E Logarithms 231 Exercise 5E 233 5F Solving logarithmic equations 234 Exercise 5F 236 5G Logarithmic graphs 236 Exercise 5G 237 5H Applications of exponential and logarithmic functions 238 Exercise 5H 239 ■■ ■■ ■■ ■■

Summary 243 Chapter review 246 ICT activities 250 Answers 251

Chapter 6

Circular functions 255 6A Trigonometric ratio revision 255

Exercise 6A 256 6B The unit circle 258

Exercise 6B 261 6C Radians 262

Exercise 6C 264 6D Symmetry 265

Exercise 6D 268 iv Contents

6E Identities 269

Exercise 6E 272 6F Sine and cosine graphs 273

Exercise 6F 277 6G Tangent graphs 280

Exercise 6G 282 6H Solving trigonometric

equations 283 Exercise 6H 287 6I Applications 288 Exercise 6I 290 ■■ ■■ ■■ ■■

Summary 293 Chapter review 296 ICT activities 301 Answers 302

Exam practice 2

Based on Chapters 1– 6 307 Chapter 7

Matrices 309 7A Addition and subtraction of

matrices 309 Exercise 7A 315 7B Multiplying matrices 316 Exercise 7B 318 7C Solving matrix equations 320 Exercise 7C 324 7D Matrices and transformations 325 Exercise 7D 328 ■■ ■■ ■■ ■■

Summary 330 Chapter review 331 ICT activities 335 Answers 336

Exam practice 3

Based on Chapters 1– 7 339 Chapter 8

Rates of change 341 8A Identifying rates 341

Exercise 8A 342 8B Constant rates 343

Exercise 8B 345 8C Variable rates 347

Exercise 8C 348 8D Average rates of change 349

Exercise 8D 351 8E Instantaneous rates 353

Exercise 8E 355 8F Motion graphs (kinematics) 357

Exercise 8F 359 8G Relating the gradient function to the original

function 364 Exercise 8G 364 8H Relating velocity–time graphs to position–time graphs 365 Exercise 8H 366

8I Rates of change of polynomials 369

11B Calculating probabilities 478

Exercise 11B 481

Exercise 8I 373 ■■ ■■ ■■ ■■

Summary 375 Chapter review 377 ICT activities 383 Answers 384

Chapter 9

Differentiation 389 9A Introduction to limits 389

Exercise 9A 392 9B Limits of discontinuous, rational and hybrid functions 393 Exercise 9B 395 9C Differentiation using first principles 397 Exercise 9C 399 9D Finding derivatives by rule 400 Exercise 9D 405 9E Rates of change 407 Exercise 9E 410 9F Sketching graphs containing stationary points 412 Exercise 9F 417 9G Solving maximum and minimum problems 418 Exercise 9G 421 ■■ ■■ ■■ ■■

Summary 423 Chapter review 425 ICT activities 431 Answers 432

Chapter 10

Antidifferentiation (integration) 437 10A Antidifferentiation 437

Exercise 10A 439 10B Deriving the original function from the gradient

function 440 Exercise 10B 442 10C Approximating areas enclosed by functions 444 Exercise 10C 448 10D The fundamental theorem of integral calculus 452 Exercise 10D 456 10E Applications of antidifferentiation 457 Exercise 10E 459 ■■ ■■ ■■ ■■

Summary 463 Chapter review 465 ICT activities 468 Answers 469

Exam practice 4

Based on Chapters 1– 10 473 Chapter 11

Introductory probability 475 11A Introduction to experimental probability 475

Exercise 11A 477

11C Tree diagrams and lattice diagrams 483

Exercise 11C 485 11D The Addition Law of Probabilities 487

Exercise 11D 490 11E Karnaugh maps and probability tables 492

Exercise 11E 495 11F Conditional probability 497

Exercise 11F 500 11G Transition matrices and Markov chains 501

Exercise 11G 506 11H Independent events 507

Exercise 11H 511 11I Simulation 513

Exercise 11I 515 ■■ ■■ ■■ ■■

Summary 516 Chapter review 518 ICT activities 523 Answers 524

Chapter 12

Combinatorics 529 12A The addition and multiplication principles 529

Exercise 12A 531 12B Permutations 533

Exercise 12B 534 12C Factorials 536

Exercise 12C 537

12D Permutations using nPr 538

Exercise 12D 540

12E Permutations involving restrictions 541

Exercise 12E 544 12F Arrangements in a circle 545

Exercise 12F 546

12G Combinations using nCr 547

Exercise 12G 549

12H Applications to probability 551

Exercise 12H 554 ■■ ■■ ■■ ■■

Summary 557 Chapter review 558 ICT activities 561 Answers 562

Exam practice 5

Based on Chapters 1–12 565 ■■ Answers 567

Cumulative exam practice 569 Functions and graphs 569 Algebra 570 Rates of change and calculus 573 Probability 575 ■■ Answers 585

Index 587

Contents v

Introduction Maths Quest 11 Mathematical Methods CAS Third edition is specifically designed for the VCE Mathematical Methods CAS course and based on the award-winning Maths Quest series. This resource contains: • a student textbook with accompanying eBookPLUS • a TI-Nspire CAS calculator companion • a Casio ClassPad CAS calculator companion • a solutions manual • flexisaver versions of all print products • teacher support material available on eGuidePLUS.

Student textbook Full colour is used throughout to produce clearer graphs and headings, to provide bright, stimulating photos and to make navigation through the text easier. Cumulative exam practice questions assist students with exam preparation. Clear, concise theory sections contain worked examples and highlighted important text and remember boxes. Icons appear for the eBookPLUS to indicate that interactivities and eLessons are available online to help with the teaching and learning of particular concepts. Worked examples in a Think/Write format provide clear explanation of key steps and suggest presentation of solutions. Many worked examples have eBookPLUS icons to indicate that a Tutorial is available to elucidate the concepts being explained. Worked examples also have calculator icons that indicate support in the Calculator Companion books, which contain comprehensive step-by-step CAS calculator instructions, fully integrated into the examples, for the TI-Nspire CAS and Casio ClassPad calculators. Exercises contain many carefully graded skills and application problems, including multiple choice questions. Cross-references to relevant worked examples appear with the first ‘matching’ question throughout the exercises. A selection of questions are tagged as technology-free to indicate to students that they should avoid using their calculators or other technologies to assist them in finding a solution. Exam practice sections contain exam style questions, including time and mark allocations for each question. Fully worked solutions are available on the eBookPLUS for students. Each chapter concludes with a summary and chapter review exercise containing examination-style questions (multiple choice, short answer and extended response), which help consolidate students’ learning of new concepts. Also included are questions from past VCE exams along with relevant Exam tips.

Student website — eBookpLUS The accompanying eBookPLUS contains the entire student textbook in HTML plus additional exercises. Students may use the eBookPLUS on laptops, tablets and school or home computers, and cut and paste material for revision, assignments or the creation of notes for exams. Career profiles and History of mathematics place mathematical concepts in context. Investigations, often suggesting the use of technology, provide further discovery learning opportunities. WorkSHEET icons link to editable Word documents, and may be completed on screen or printed and completed by hand. SkillSHEET icons link to printable pages designed to help students revise required concepts, and contain additional examples and problems. Interactivity icons link to dynamic animations, which help students to understand difficult concepts. vi

Introduction

eLesson icons link to videos or animations designed to elucidate concepts in ways other than what the teacher can achieve in the classroom. Tutorial icons link to one-way engagement activities, which explain the worked examples in detail for students to view at home or in the classroom. Test yourself tests are also available and answers are provided for students to receive instant feedback.

Teacher website — eGuidePLUS The accompanying eGuidePLUS contains everything in the eBookPLUS and more. Two tests per chapter, fully worked solutions to WorkSHEETs, the work program and other curriculum advice in editable Word format are provided. Maths Quest is a rich collection of teaching and learning resources within one package. The eGuidePLUS provides NEW material from which teachers may set school-assessed coursework (SAC), such as application tasks, analysis tasks and semester exams, as well as additional investigations, worksheets and technology files.

Introduction vii

About eBookPLUS Next generation teaching and learning This book features eBookPLUS: an electronic version of the entire textbook and supporting multimedia resources. It is available for you online at the JacarandaPLUS website (www.jacplus.com.au).

Using the JacarandaPLUS website To access your eBookPLUS resources, simply log on to www.jacplus.com.au using your existing JacarandaPLUS login and enter the registration code. If you are new to JacarandaPLUS, follow the three easy steps below. Step 1. Create a user account The first time you use the JacarandaPLUS system, you will need to create a user account. Go to the JacarandaPLUS home page (www.jacplus.com.au), click on the button to create a new account and follow the instructions on screen. You can then use your nominated email address and password to log in to the JacarandaPLUS system. Step 2. Enter your registration code Once you have logged in, enter your unique registration code for this book, which is printed on the inside front cover of your textbook. The title of your textbook will appear in your bookshelf. Click on the link to open your eBookPLUS. Step 3. View or download eBookPLUS resources Your eBookPLUS and supporting resources are provided in a chapter-by-chapter format. Simply select the desired chapter from the drop-down list. Your eBookPLUS contains the entire textbook’s content in easy-to-use HTML. The student resources panel contains supporting multimedia resources for each chapter. Once you have created your account, you can use the same email address and password in the future to register any JacarandaPLUS titles you own.

viii About eBookPLUS

Using eBookPLUS references eBookPLUS logos are used throughout the printed books to inform you that a multimedia resource is available for the content you are studying. Searchlight IDs (e.g. INT-0001) give you instant access to multimedia resources. Once you are logged in, simply enter the searchlight ID for that resource and it will open immediately.

Minimum requirements • A modern internet browser such as Internet Explorer 7+, Mozilla Firefox 3+, Google Chrome 8+, Safari 3+ or Opera 9+ • Adobe Flash Player 10+ • Javascript must be enabled (most browsers are enabled by default).

Troubleshooting • Go to the JacarandaPLUS help page at www.jacplus.com.au/jsp/help.jsp. • Contact John Wiley & Sons Australia, Ltd. Email: [email protected] Phone: 1800 JAC PLUS (1800 522 7587)

Acknowledgements The authors and publisher would like to thank the following copyright holders, organisations and individuals for their assistance and for permission to reproduce copyright material in this book.

Images • iStockphoto: pp. 23 (upper)/© Danny Warren, 25/© MichaelSvoboda/, 26 (right)/© Aneese, 33/© Felix Alim, 41/© technotr, 135 (upper)/© Steve Maehl, 135 (lower)/© Scott Cramer, 136/© Keiichi Hiki, 161 (lower left)/© Adnan Ahmad, 163/© TIM MCCAIG, 240 (upper)/© Hal Bergman, 257/© Mac99, 429/© Grafissimo, 452/© Don Bayley, 460/© JamesReillyWilson/, 461/© Doug Sims, 515/uatp2, 522/ © Ivan Bajic, 544 (lower)/© Clerkenwell_Images • © Corbis Corporation: p. 240 (lower); p. 535/Jack Hollingsworth • © Digital Stock/Corbis Corporation: pp. 8 (left, right), 83, 198 • © Image Source: p. 241 • © ImageState: p. 421; p. 497/© John Foxx Images • John Wiley & Sons Australia: pp. 316, 541 (four images)/Photo by Renee Bryon; pp. 486, 544 (upper)/Malcolm Cross; p. 521 (lower)/Taken by Jo Patterson • © Newspix: p. 309/Robert McKell; p. 389/News Ltd/3rd Party Managed Reproduction & Supply Rights • © Photodisc, Inc.: pp. 40, 82, 161 (lower middle, lower right), 165, 196, 239, 242, 291, 292, 300, 360, 363, 425, 451, 482 (upper), 559 • photolibrary.com: p. 23 (lower); p. 217/Sydney/ SPL/CNRI; p. 467/Simon Fraser • Photolibrary Royalty Free: p. 11/© Huntstock/Photolibrary Inc. • Shutterstock: pp. 26 (left)/© Hannu Liivaar, 62/© Rudy Balasko, 88/© Marie Lumiere, 161 (upper left)/© C, 208/© Brendan Howard, 289/© Jeff Davies, 299 (lower)/© G Tipene, 299 (upper)/© Patricia Hofmeester, 341/© Colin Stitt, 355/© Kalim, 369/© Germanskydiver, 407 (left)/© 1971yes, 407 (right)/© Pete Niesen, 408/© Jamie Roach, 409/© James Thew, 411/© Gelpi, 478/© Andrejs Pidjass, 482 (lower)/© Kitch Bain, 487/© Roger Browning, 501/© Rich Carey, 512/© Claudio Gennari, 521 (upper)/© ronfromyork, 530/© arkasha1905, 540/© Lincoln Rogers, 545/© arkasha1905, 556/ © Lario Tus • © Stockbyte: p. 161 (upper right) • © Viewfinder Australia Photo Library: p. 199 • Visy Recycling: p. 197 Every effort has been made to trace the ownership of copyright material. Information that will enable the publisher to rectify any error or omission in subsequent editions will be welcome. In such cases, please contact the Permissions Section of John Wiley & Sons Australia, Ltd.

Acknowledgements

ix

Chapter 1

Linear functions diGital doC doc-9695 10 Quick Questions

Chapter ContentS 1a 1B 1C 1d 1e 1F 1G 1h

Solving linear equations and inequations Rearrangement and substitution Gradient of a straight line Sketching linear functions Simultaneous equations Finding the equation of a straight line Distance between two points and midpoint of a segment Linear modelling

Solving linear equations and inequations 1a

A linear equation is one that involves a variable raised to the power of 1 only. Recall that x1 is the same as x, so a linear equation involving x would contain only x’s, and not x , x2, x3, x −1, x −2 and so on. For 8 example, y = 7x − 3, ax + by = c and 3 x + 1 = 9 are all linear equations. 1 1 Note that y = is not a linear equation, as the is really x −1. x x To isolate a particular variable — known as ‘making x (or whatever the variable is) the subject’ — we focus on the variable by ‘undoing’ other terms and operations. Remember to do the same operation to both sides of an equation, in the reverse order to that originally used to make up the equation. Though the focus of this chapter is linear equations, some other types of equations will be included for skills practice. When there is only one variable involved in an equation, we may attempt to solve and find a numerical value by rearranging to make the variable the subject. 1 2

Worked example 1

Solve the following linear equations. 3x − 2 4x +5=1 a 7x − 4 = 17 b c 2 − 1 = 6 5 4 think

a 1 Write the equation. 2

Add 4 to both sides.

3

Divide both sides by 7.

b 1 Write the equation. 2

Subtract 5 from both sides.

Write

a 7 x − 4 = 17

7 x = 21 21 7 x=3 x=

b

3x − 2 +5=1 4 3x − 2 − = 4 4 Chapter 1 • Linear functions

1

3

Multiply both sides by 4.

4

Add 2 to both sides.

5

Divide both sides by 3.

3 x − 2 = − 16 3 x = − 14 x=

−14

3

4 x − 1 = 6 5 4x −1 = 3 5 4x =4 5

c 1 Write the equation.

c 2

2

Divide both sides by 2.

3

Add 1 to both sides.

4

Multiply both sides by 5.

5

Divide both sides by 4.

4 x = 20 20 4 x=5 x=

When an equation has variables on both sides, at some stage they must be gathered together on the same side of the equation. Worked example 2

Solve: a 4x − 3 = 3(6 − x)

b

9 x + 3 13 x + 7 = 2 3

c

3 − x 9( x + 7) = + 1. 4 10

think

a 1 Write the equation.

a 4x − 3 = 3(6 − x)

2

Expand the right-hand side (RHS).

4x − 3 = 18 − 3x

3

Collect x’s on one side, for example, the side that results in a positive x term, in this case, the left-hand side (LHS). (That is, add 3x to both sides.)

7x − 3 = 18

4

Add 3 to both sides.

5

Divide both sides by 7.

b 1 Write the equation.

7x = 21 21 7 x=3 x=

b

9 x + 3 13 x + 7 = 2 3 LCD = 6

2

Find the lowest common denominator for both terms.

3

Write all terms with the common denominator, adjusting numerators accordingly (so that numerator and denominator have been multiplied by the same amount).

3(9 x + 3) 2(13 x + 7) = 6 6

4

Now that all terms have the same denominator, the numerators must be equal. (Multiply each side by 6.)

3(9 x + 3) = 2(13 x + 7)

5

Expand all brackets.

6

Collect x’s on the LHS and numbers on the RHS.

7

Simplify and solve.

c 1 Write the equation. 2

2

Write

Find the lowest common denominator for all three terms.

Maths Quest 11 Mathematical Methods CAS

27 x + 9 = 26 x + 14 27 x − 26 x = 14 − 9 x=5 c

3 − x 9( x + 7) = +1 4 10 LCD = 20

3

Write all terms with the common denominator, adjusting numerators accordingly (so that numerator and denominator have been multiplied by the same amount).

5(3 − x ) 2 × 9( x + 7) 20 = + 20 20 20

4

Now that all terms have the same denominator, the numerators must be equal. (Multiply each term by 20.)

5(3 − x) = 18(x + 7) + 20

5

Expand all brackets.

15 − 5x = 18x + 126 + 20

6

Collect x’s on the RHS and numbers on the LHS.

15 − 126 − 20 = 18 x + 5 x − 131 =

7

−131

Simplify and solve.

23

23 x

=x −131

x=

23

There is only one solution to a linear equation. For example, the equation 3x + 2 = 8 has the solution x = 2. The solution to a linear equation can be represented as a single point on a number line. If the equals sign is replaced with an inequality sign, the solution is a portion of a number line. These expressions are referred to as linear inequations (sometimes called inequalities). Worked example 3

Solve the following linear inequations. a 6x − 7 ≥ 3x + 5 b −3m + 5 < −7 think

Write

a 6x − 7 ≥ 3x + 5

a 1 Write the inequations. 2

Combine the variable terms by subtracting 3x from both sides.

3

Add 7 to both sides.

4

Divide both sides by 3.

3x − 7 ≥ 5 3x ≥ 12 x≥4 b −3m + 5 < −7

b 1 Write the inequations.

−3m

2

Subtract 5 from both sides.

3

Divide both sides by −3. Reverse the inequality sign, as you are dividing by a negative number.

exercise 1a

m>4

Solving linear equations and inequations

1 We1 Solve the following linear equations. a 3x − 19 = −13 b 4x + 25 = −7

12 − 3 x =5 3 3 x − 1 = 10 h 2 5 2 We2 Solve the following linear equations. a 2x − 9 = 3(2x − 11) b 7x − 1 = 17(3x − 13) x+2 x−5 x + 11 2( x + 14) = = e f 6 3 3 9 6 x + 7 5x + 1 2 x + 29 x + 44 = +1 = +2 i j 5 4 3 8 3x − 1 =5 4 7x + 4 g − 8 = −9 3 d

< −12

diGital doC doc-9696 equation solvers

c 9x + 19 = −2

4x − 6 −7=3 3 x i 7 8− = 77 4

e

f

c x + 11 = 2(x + 12)

4 x + 66 13 − 3 x = 3 4 7 x − 9 21 − x = − 18 k 9 3 g

d 3x − 7 = 3(35 − 2x)

x + 10 2(7 − 3 x ) = 9 5 17 − x 49 + 2 x = +5 l 2 5

h

Chapter 1 • Linear functions

3

Solve the following linear inequations. 11 − 3 x b ≥6 a 15 − 6x ≤ 2 7

3 We3

d 16 − 4x < 7(1 − x)

1B

e 3x − 7 ≥ 2(35 − 2x)

19 − x + 3 > −1 − 4 4 − x 2( x + 1) ≤ f 3 6 c

rearrangement and substitution

When there is more than one variable involved in an equation, we may rearrange the equation to make a particular variable the subject using the same rules of equation solving described in the previous section. Worked example 4

Rearrange each of the following to make the variable in parentheses the subject. 1 m a 6x + 8y − 48 = 0 (y) b s = ut + at 2 k(u) c T = 2π k(k) 2 k think

a 1 Write the equation.

a 6x + 8y − 48 = 0

6x + 8y = 48

2

Add 48 to both sides.

3

Subtract 6x from both sides.

4

Divide both sides by 8.

y=

48 − 6 x 8

5

Cancel if possible. Here, divide the numerator and denominator by 2.

y=

24 − 3 x 4

6

Other ways of representing the answer are shown opposite.

b 1 Write the equation.

8y = 48 − 6x

24 3 − x 4 4 3 = 6− x 4 −3 = x+6 4

or y =

s = ut + 12 at 2

b

2

Subtract 12 at 2 from both sides.

s − 12 at 2 = ut

3

Multiply both sides by 2.

2s − at 2 = 2ut

4

Divide both sides by 2t.

5

Write the equation with the desired variable on the left.

c 1 Write the equation.

4

Write

2s − at 2 =u 2t 2s − at 2 u= 2t c

T = 2π T = 2π

m k

2

Divide both sides by 2π.

3

Square both sides.

T2 m = 4π 2 k

4

Form the reciprocal of both sides to make k the numerator.

4π 2 k = T2 m

Maths Quest 11 Mathematical Methods CAS

m k

5

Multiply both sides by m.

6

Write the equation with the desired variable on the left.

4π 2m =k T2 k=

4π 2m T2

Once a variable is isolated, we may substitute values of other variables to calculate various values of the isolated variable. The following worked example illustrates some practical applications. Worked Example 5

a The formula for converting temperature in degrees Fahrenheit (F) (which is the system used in

5( F − 32) the USA) to degrees Celsius (C ) is C = . 9 i Make F the subject. ii What is the temperature in degrees Fahrenheit when the temperature measured in degrees Celsius is 21° C? b The area (A) of a circle is given by A = π r 2, where r is the radius. Calculate the value of r correct to 2 decimal places when A = 20 cm2. Think

a i 1 Write the equation.

Write

C=

a i

5( F − 32) 9

2

Multiply both sides by 9.

9C = 5( F − 32)

3

Divide both sides by 5.

9C = F − 32 5

4

Add 32 to both sides.

5

Write the equation with F first. Sometimes it may be appropriate to use a common denominator.

9C + 32 = F 5 9C F= + 32 5 9C + 160 or F = 5

ii 1 Replace C with 21. Note that 9C means 9 × C. 2

9 × 21 + 32 5 189 + 32 F= 5 F = 37.8 + 32 F=

ii

Evaluate F.

F = 69.8 so 21°C = 69.8 °F b

1

Write the equation. Solve for r (make r the subject) as follows.

2

Divide both sides by π.

3

Take the square root of both sides, and write r first. As r is the radius, we take the positive root only.

4

Substitute A = 20 into the new formula.

b

A = πr2 A = r2 π r=

If A = 20, r=

5

Evaluate r.

A π

20 π

r = 2.52 cm, correct to 2 decimal places

Chapter 1 • Linear functions 5

Worked example 6

Note: In this example, m represents metres, m/s represents metres per second (velocity), and m/s2 represents metres per second per second (acceleration). The final velocity, v m/s, of an object that begins with velocity u m/s and accelerates at a m/s2 over a distance of s m is given by the equation v2 = u2 + 2 as. a Find the value of v when u = 16, a = 2 and s = 36. b Rearrange the given equation to make s the subject. c Find the distance travelled by an object that begins with a velocity of 10 m/s and has a final velocity of 4 m/s while accelerating at −1 m/s2. think

Write

a 1 Write the given equation and then make v the subject

a v 2 = u 2 + 2as

by finding the square root of both sides.

v = ± u 2 + 2as

2

Substitute u = 15, a = 2 and s = 36.

= ± 162 + 2 × 2 × 36

3

Simplify and evaluate v.

= ± 256 + 144 = ± 400 = ± 20 The final velocity is 20 m/s.

b 1 Begin with the given equation.

v 2 = u 2 + 2as

b

2

Subtract u2 from both sides.

v 2 − u 2 = 2as

3

Divide both sides by 2a.

v 2 − u2 =s 2a

4

Reverse so that s is given on the left.

c 1 Match the variables with the given information.

s=

v 2 − u2 2a

c u = 10

v=4 a = −1

3

Write the formula that has s as the subject (see part b above). Substitute the values given in step 1.

4

Simplify and evaluate.

5

Explain the answer in words.

2

exercise 1B

s= =

v 2 − u2 2a 4 2 − 10 2 2 × −1

16 − 100 − 2 − 84 = − 2 = 42 =

The object travels 42 m in its initial direction.

rearrangement and substitution

Each of the following is a real equation used in business, mathematics, physics or another area of science. Make the variable shown in parentheses the subject in each case. a A=L+P (P) b A = lw (l) d (t) c v= d C = 2πr (r) t

1 We4

6

Maths Quest 11 Mathematical Methods CAS

(β)

f

kQq r2

(r)

g Fd =

(v)

h v = γ rT

(γ)

i

(w)

j

e E = αθ + βθ 2

1 2 1 2 mv − mu 2 2 S = 2w(l + h) + 2lh

F=

S = 2πr2 + 2πrH

(H)

2 We5 Calculate the value of the subject (the first mentioned variable), given the values of the other

variables. k a I= 2 d b E = K + mgh c D = (n −

3

k = 60, d = 15 K = 250, m = 2, g = 10, h = 5

1 )λ 2

n = 3, λ = 2.8

d E = hf0 − W

h = 6.62, f0 = 5000, W = 20 000

e v = ω r 2 − y2

ω = 2, r = 1.6, y = 1

Make the variable in parentheses the subject and find its value using the given information. a A = l2

(l)

A = 60

4 πr3 3

(r)

V = 1000

c v = u + at

(a)

v = 25, u = 0, t = 6

(l)

T = 4, g = 9.8

(c)

K = 6.9, α = 0.05

b V=

d T = 2π e K=

l g

cα 2 1− α

4 We6 The perimeter, P, of a rectangle of length l and width w may be found using the equation

P = 2(l + w). a Find the perimeter of a rectangle of length 16 cm and width 5 cm. b Rearrange the equation to make w the subject. c Find the width of a rectangle that has perimeter 560 mm and length 240 mm.

5 The area of a trapezium (Figure A) is given by A =

a + b , where a and b are the lengths h 2

of the parallel sides and h is the height. a Find the area of the trapezium shown in Figure B. b Using Figure A, find an equation for determining side a in terms of the area A and side b. c Find a in Figure C. 9m a h

Area A

b Figure A

16 m

50 cm a

62 cm Area = 2000 cm2

21 m Figure B

Figure C

r 2 , 100 where A is the amount ($) in the account after two years, D is the initial deposit ($) and r is the interest rate (%). a Find the amount in such an account after two years if the initial deposit was $1000 and the interest rate was 6%. b Make r the subject of the equation. c Find the rate required for an initial deposit of $1000 to grow to $2000 after 2 years.

6 The size of a 2-year investment account with a particular bank is given by A = D 1 +

Chapter 1 • Linear functions

7

7 The object and image positions for a lens of focal length f are related by

1 1 1 + = , where u is the distance of the object from the lens u v f and v is the distance of the image from the lens. 4 2 3 a Make f the subject of the equation. b Make u the subject of the equation. c How far from the lens is the image when an object is 30 cm in front of a lens of focal length 25 cm? b 8 The length of a side of a right-angled triangle can be found using Pythagoras’ theorem: c2 = a2 + b2, where c is the length of the longest side, and a and b are the lengths of the two shorter sides. Find the value of b in the triangle above. 1 9 The volume of a cone is given by the rule V = 3 π r2h, where r is the radius of the widest r part of the cone and h is the vertical height of the cone. Given that the volume of a cone is 100 cm3 and its radius at the widest point is 12 cm, find the height of the h cone, expressing your answer in terms of π. the formula

diGital doC doc-9697 Career profile rick morris – vigneron

Gradient of a straight line

1C

cone

The gradient of a line describes its slope or steepness. y

y

y Negative gradient

Positive gradient

x

x

y (x2, y2) Rise (x1, y1) Run x

y

Zero gradient

Infinite gradient

x

x

The gradient may be calculated using the formula: y −y rise m = run or m = 2 1 x 2 − x1 These terms are illustrated at left. Here are two examples of where gradient can affect our everyday lives. Can you think of others?

A cliff face with a steeper gradient provides a greater challenge for climbers.

Scientists calculate the required gradient of solar panels so that the maximum amount of energy is absorbed.

8

Maths Quest 11 Mathematical Methods CAS

Worked example 7

Calculate the gradient of this linear graph using the intercepts shown. think 1

2

y

Write

Identify the rise and run. Rise = 14, run = 2. Calculate m = rise . run

14

14 2 =7

m=

−2

x

Worked example 8

Calculate the gradient of the line passing through the points (3, −6) and (−1, 8). think

Write

y2 − y1 . x 2 − x1

m=

1

Use the formula m =

2

Match up the terms in the formula with the values given. (x1, y1) (x2, y2) (3, −6) (−1, 8)

3

Substitute the given values.

4

Simplify.

5

Cancel if possible.

y2 − y1 x 2 − x1

8 − −6 − 1− 3 14 14 = = −− 44 − 7 = −7 = 2 2

m=

If the angle a line makes with the positive direction of the x-axis is known, the gradient may be found using trigonometry applied to the triangle shown below. y rise

θ run

x

tan (θ ) = opposite = rise = m adjacent run Worked example 9

y

a Calculate the gradient (accurate to 3 decimal places) of a line making

an angle of 40° to the positive x-axis.

b Calculate the gradient of the line shown at right. Express your answer

60°

to 2 decimal places.

think

a Since the angle the line makes with

the positive x-axis is given, the formula m = tan (θ) can be used.

Write

x

y

a m = tan (θ)

= tan (40°) = 0.839, correct to 3 decimal places

40°

x

Chapter 1 • Linear functions

9

b

1

2

b θ = 180° − 60°

The angle given is not the one between the graph and the positive direction of the x-axis. Calculate the required angle θ.

60° θ

Use m = tan (θ) to calculate m to 2 decimal places.

exercise 1C 1 We7

y

= 120°

m = tan (θ) = tan (120°) = −1.73

Gradient of a straight line

Calculate the gradient of each of the following linear graphs using the intercepts shown.

a

b

y

y

6

2

x

−3

6

x

Without drawing a graph, calculate the gradient of the line passing through: a (2, 4) and (10, 20) b (4, 4) and (6, 14) c (10, 4) and (3, 32) d (5, 31) and (− 7, 25). 3 We9a Calculate the gradient (accurate to 3 decimal places) of a line making the angle given with the positive x-axis. a 50° b 72° c 10° d − 30° e 150° f 0° g 45° h 89°

2 We8 diGital doC doc-9698 Gradient of a straight line

4 We9b

Calculate the gradient of each line below. Give answers to 2 decimal places. y

a

c

y

b

43°

x

y

69°

x

y

d

x 28°

15°

x

5 Which of these lines has: a a non-zero positive gradient? b a negative gradient? c a zero gradient? d an undefined gradient?

y

Maths Quest 11 Mathematical Methods CAS

B

5 4 3 2 1

A

10

x

−5−4−3−2−1 0 1 2 3 4 5 −1 −2 −3 −4 −5

D

C x

6 mC a Which of the following lines has a

b Which of the following lines has a

gradient of −2?

B

A

y 5 4 3 2 1

gradient of 3?

C

A

D

5 4 3 2 1

E

−5−4−3−2−1 0 1 2 3 4 5 −1 −2 −3 −4 −5

D

x

y

B C

−5−4−3−2−1 0 1 2 3 4 5 −1 −2 −3 −4 −5 E

x

7 Burghar plots the coordinates of a proposed driveway on a plan that is shown below. What is the

gradient of the proposed driveway?

Garage

way Drive 2 m

17 m 8 An assembly line is pictured below. What is the gradient of the sloping section? (Give your answer as a

fraction.) BOFFO Made in Austra lia

BOFFO Made in Austra lia

BOFFO Made in Austra lia

0.85 m 15 m

9 Determine the value of a in each case so the gradient joining the points is equal to the value given. a (3, 0) and (5, a), gradient 2 b (2, 1) and (8, a), gradient 5 c (0, 4) and (a, −11), gradient 3 d (a, 5) and (5, 1), gradient −2 10 For safety considerations, wheelchair

ramps are constructed under regulated specifications. One regulation requires that the maximum gradient of a ramp exceeding 1200 mm in length is 1 . 14 a Does a ramp 25 cm high with a horizontal length of 210 cm meet the requirements? 1 b Does a ramp with gradient meet 18 the specifications? c A 16 cm high ramp needs to be built. Find the horizontal length of the ramp required to meet the specifications.

diGital doC doc-9699 SkillSHEET 1.1 Using a gradient to find the value of a parameter

Chapter 1 • Linear functions

11

1d

Sketching linear functions

The general form for linear equations is y = mx + c, where m is the gradient of the line and c is the y-intercept. y ie

ad

Gr

y = mx + c Gradient

y

m nt

y-intercept x

y-intercept

x

x-intercept These lines have identical gradients (equal m values).

diGital doC doc-9700 WorkSHEET 1.1

To sketch a graph from a linear equation expressed in general form, follow these steps. Step 1 Plot the y-intercept on a set of axes. Step 2 Find and plot a second point on the line. Do this by substituting any value of x into the equation and finding the corresponding y-value. Step 3 Join the two points. Alternatively, you can use a CAS calculator or other graphing technology. To find the equation of a line given the gradient and y-intercept, simply substitute the values of m and c into y = mx + c.

Sketching linear graphs using intercepts

y

To draw a graph, only two points are needed. A line may then be drawn through the two points, and will include all other points that follow the given rule. The two points can be chosen at random; however, it is often easier to sketch a graph using the points where the graph crosses the axes. These points are called x- and y-intercepts. The x-intercept occurs when y = 0, and the y-intercept occurs when x = 0.

x

Worked example 10

Sketch the graph of y = −x + 6, showing x- and y-intercepts. think 1

2

Write/draW

Find the y-intercept (when x = 0). Substitute x = 0 into the equation.

If x = 0: y = −1 × 0 + 6 y=6

Find the x-intercept (when y = 0). Substitute y = 0 into the equation.

If y = 0: 0 = −x + 6

3

Mark the intercepts on a set of axes.

4

Join the intercepts with a straight line.

x=6

(6, 0)

y (0, 6)

(6, 0)

12

Maths Quest 11 Mathematical Methods CAS

(0, 6)

x

Worked example 11

Sketch the graph of 3x − 2y = 12. think

Write/draW

1

Find the y-intercept (when x = 0). Substitute x = 0 into the equation.

If x = 0: 3 × 0 − 2y = 12 −2y = 12 12 y= − 2 y = −6

2

Find the x-intercept (when y = 0). Substitute y = 0 into the equation.

If y = 0: 3x − 2 × 0 = 12 3x = 12 x=4

3

Mark the intercepts on a set of axes.

4

Join the intercepts with a straight line.

y

x

4

−6

The graphs of some equations do not have two intercepts, as they pass through the origin (0, 0). Such equations are of the form y = k x or a x + by = 0. To sketch graphs of such equations, we use (0, 0) and any other point, for example the point where x = 1. (We could choose any other non-zero value.) Worked example 12

Sketch the graph for the equation 4 x − 3y = 0. think 1

Try substituting x = 0 to find the y-intercept.

2

Note that the graph passes through (0, 0). There is no point substituting y = 0, as we know we’ll get x = 0.

3

Substitute another x-value. In this example we use x = 1.

4

Plot the points (0, 0) and (1, 43 ) on a set of axes. Note that 4 is 1 1 , which is a little less 3 3 than 1 1 . 2

Write/draW

4 × 0 − 3y = 0 −3y = 0 y=0

If x = 0:

tUtorial eles-1404 Worked example 12

(0, 0)

If x = 1:

4 × 1 − 3y = 0 4 − 3y = 0 4 = 3y 4 y= 3 4 (1, ) 3

y 2 (1, 4–3 )

1 (0, 0) 1

x

Chapter 1 • Linear functions

13

Sketching linear functions

exercise 1d

1 Use a CAS calculator or other method to sketch graphs of the following equations on the same set of axes. a y=x b y = 2x c y = 3x d y = −x e y = −2x 2 In question 1, what is the effect on the graph of the number in front of the x (the ‘x-coefficient’ or

‘gradient’)? 3 Use a CAS calculator or other method to sketch graphs of the following equations on the same set of axes. a y=x+1 b y=x+2 c y=x+3 d y=x−4 diGital doC doc-9701 linear graphs

4 Use a CAS calculator or other method to sketch graphs of the following equations on the same set of

axes. a y = 2x + 1

c y = −3x + 6

b y = 2x − 7

d y = 3x − 5

5 In questions 3 and 4, what is the effect on the graph of the number at the end of the equation (the

‘y-intercept’)? 6

Write the equation of a line having the following properties (where m = gradient and c = y-intercept). a m = 2, c = 7 b m = −3, c = 1 c m = 5, c = −2 −1 −3 2 1 f y-intercept 12, gradient −2 d m= ,c= e m = 4, c = 2 3 3 Rearrange the following equations and state the gradient and y-intercept for each.

7

a 2y = 8x + 10 e 21x + 3y = −27 i 15 − 6y + x = 0

b 3y = 12x − 24 f −10x + 5y = 25 j 2y + 7 + 5x = 0

−y

c g

= 3x − 1 − 2x = 66

d 16 − 4y = 8x h 8x + 3y − 2 = 0

−11y

8 State the equation for each of the following graphs. a

y − 1– 2

−1

y

b

6 5 4 3 2 1

3 2 1 −2

0 −1 −2

1

x

2

y

c

e

−1

0 −1 −2 −3 −4 −5

1

5 4 3 2 1

x

2

−1

0 −1 −2 −3 −4 −5

f

y

14

0 −1 −2 −3 −4 −5

1

2

3

Maths Quest 11 Mathematical Methods CAS

4

1

2

3

4

5

1

2

3x

y

9 8 7 6 5 4 3 2 1

5 4 3 2 1 −1

x

2

y

d

3 2 1 −2

0 −1 −2 −3 −4 −5

x −3

−2

−1 −10 −2 −3 −4 −5 −6 −7 −8 −9

6

x

9 We10 Sketch graphs of the following linear equations, showing x- and y-intercepts. a y = 6x + 18 b y = 3x − 21 c y = −2x − 3 d y = 10 − 5x e y = −9x + 30 f y = 2(x − 8) 10 We11 Sketch graphs for each of the following equations. a 2x + 3y = 6 b 4x + 5y = 20 c 6x − 3y = −18 d 7x − 5y = 35 11 Sketch the graph for each equation. a 6x + 7y + 42 = 0 c −3x + 4y − 16 = 0 12 We12 Sketch the graph for each equation. a x+y=0 c 2x + y = 0

b 5x − 2y + 20 = 0 d y − 3x + 6 = 0 b x−y=0

13 mC Which of the following is in the form ax + by = c? a 2x − 3y − 1 = 0 C 2x + 3y = x e y=

2 3

B 2x + 3y + 1 = 0 d 2x + 3y = 1

x−1

The x- and y-intercepts for the equation 2y = −3x + 12 are (respectively): a 2 and 3 B −3 and 12 − C 4 and 6 d −4 and −6 e 4 and 6

14 mC

15 mC Which of the following has a y-intercept of −3? a y = −3x − 3 C x + 3y = 9

B y = −3x + 3 d x − 3y + 9 = 0

e 3x + y + 9 = 0

1e

Simultaneous equations

Simultaneous equations are groups of equations containing two or more variables. In this section, we look at pairs of linear equations involving the variables x and y. Each equation, as we have learned in previous sections, may be represented by a linear graph that is true for many x- and y-values. If the graphs intersect (when wouldn’t they?), the values of x and y at the intersection are those that make both equations true.

y Both graphs have the same x- and x y-values here.

Graphical solution Finding the point of intersection of two straight lines can be done graphically; however, the accuracy of the graph determines the accuracy of the solution. Consequently, using a calculator to solve the equations graphically is more reliable than reading the solution from a hand-drawn graph.

Worked example 13

Use a CAS calculator to solve the following simultaneous equations graphically. y = −3x + 5 4x − 7y + 8 = 0 think 1

Using a CAS calculator, make y the subject of the second equation. Complete the entry line as: solve (4x − 7y + 8 = 0, y)

Write

solve (4x − 7y + 8 = 0, y) 4( x + 2) Result: y = 7

Chapter 1 • Linear functions

15

2

On a Graphs page, complete the function entry lines as: f 1(x) = −3x + 5 4 × ( x + 2) f 2( x ) = 7 Use the calculator to find the intersection point.

f 1(x) = −3x + 5 4( x + 2) f 2(x) = 7

3

Write the answer.

Solving y = −3x + 5 and 4x − 7y + 8 = 0 simultaneously produces the solution (1.08, 1.76).

algebraic solution It is possible to solve simultaneous equations without graphs, that is, algebraically. The methods of substitution and elimination taught in earlier years may be used. Worked example 14

Use the substitution method to solve the following simultaneous equations. y = 2x + 5 −x + 3y = 25 think 1

Write down and label the equations.

2

Substitute [1] into [2] and label the resulting equation [3].

3

Solve [3] for x and label the solution as [4].

4

Write

Use the solution to solve for y and label the solution as [5].

−x −x

+ 3(2x + 5) = 25

−x

+ 6x + 15 = 25 5x + 15 = 25 5x = 10 x=2

Substitute [4] into [1]. y = 2(2) + 5 y=4+5 y=9

5

State the complete answer.

Solution: (2, 9)

6

Optional check: substitute [4] and [5] into [2] to check that these values for x and y make [2] true.

Check in [2]. LHS = −x + 3y = −2 + 3(9) = 25 = RHS ✓

y = 2x + 5 + 3y = 25

[5]

Worked example 15

Use the elimination method to solve these simultaneous equations. 2 x + 9y = −5 5 x − 2y − 12 = 0 think

16

1

Write down and label the equations.

2

Rearrange [2] so it is in a similar form to [1]. Call this [3]. Write down [1] again.

Maths Quest 11 Mathematical Methods CAS

Write

2x + 9y = −5 5x − 2y − 12 = 0

[1] [2]

5x − 2y = 12 2x + 9y = −5

[3] [1]

[1] [2] [3]

[4]

3

Obtain 10x in both [1] and [3].

2 × [3] 10x − 4y = 24 5 × [1] 10x + 45y = −25

[4] [5]

4

Eliminate x as shown.

[5] − [4]:

5

Solve for y.

49y = −49 − 49 y= 49 y = −1

[6]

6

Substitute [6] into [1] to find x.

Substitute [6] into [1]. 2x + 9 (−1) = −5 2x − 9 = −5 2x = −5 + 9 2x = 4 x=2

[7]

−1)

7

State the solution.

Solution: (2,

8

Again, [6] and [7] may be checked in [2] if desired.

Check in [2]. LHS = 5x − 2y − 12 = 5(2) − 2(−1) − 12 = 10 + 2 − 12 =0 = RHS ✓

Worked example 16

Two shoppers buy the following at a fruit shop, paying the amounts given. What was the cost of each apple and each banana? Shopper 1: 4 apples and 3 bananas for $2.59 Shopper 2: 6 apples and 5 bananas for $4.11 think

Write

1

Decide on variable names for the unknown quantities.

Let a = cost of an apple (in cents). Let b = cost of a banana (in cents).

2

Write equations involving these variables. Work in terms of cents.

4a + 3b = 259 6a + 5b = 411

3

Choose a variable to eliminate, in this case b.

5 × [1]: 20a + 15b = 1295 3 × [2]: 18a + 15b = 1233

[3] [4]

4

Find [3] − [4] and solve for a.

[3] − [4]: 2a = 62 a = 31

[5]

Substitute [5] into [1]. 4 × 31 + 3b = 259 124 + 3b = 259 3b = 135 b = 45

[6]

5

6

Solve for b.

State the answer using [5] and [6] as a guide.

exercise 1e

[1] [2]

The cost of an apple is 31 cents, and the cost of a banana is 45 cents.

Simultaneous equations

1 We13 Use a CAS calculator to solve the following simultaneous equations. a y = −2x, y = −4x − 6 b y = 3x − 5, y = 20 c y = 3x + 5, y = 7x − 4 e y = 10x + 1, 2x + y = −6

+ y = −4, y = 6x + 5 f 9x + y = 17, x + y = 14 d

−3x

diGital doCS doc-9702 Simultaneous linear equations — graphical method

Chapter 1 • Linear functions

17

2 We14 Use the substitution method to solve the following simultaneously. a y = 3x + 1, y = 2x + 2 b y = 5x + 5, y = −x − 19 c y = x + 2, 3x − 4y = −1

− 3y = 2, y = −6x + 7 f y = 10 − x, 2x + 7y = 5 We15 Use the elimination method to solve these simultaneous equations. a 9x + 10y = 153, 3x − y = 12 b 7x − 11y = −13, x + y = 11 − c 6x − 2y = 10, 2x + 5y = 8 d −3x + y = 8, 4x + 2y = 21 e 7y − x = 11, x + y = 10 f x − 11y = −15, y + 6x = 9 We16 At the conclusion of a tour of Wonky Willy’s confectionery factory, Nutrina buys 10 choc balls and 8 fizz wizzers for $4.30, and her friend purchases 6 choc balls and 9 fizz wizzers for $4.05. Determine the cost of each type of lolly. The sum of two whole numbers, x and y, is 41. The difference between them is 3. Write two equations involving x and y and solve them to find the numbers. A farmer counts emus and sheep in a paddock, and notes there are 57 animals and 196 feet. Assuming no animal amputees, how many of each animal are there? A sports store supplies 24 basketballs and 16 cricket balls to one school for $275.60, and delivers 12 basketballs and 32 cricket balls to another school for $211. If delivery is free, how much did the supplier charge for each type of ball? e

3

4

5 6 7

d y = −2x + 3, −5x + 2y = 1

−4x

8 A businessperson hires a stretch limousine for 2 days and a sedan for 3 days while on an interstate trip.

If the total car hire cost of the trip was $675, and the limousine cost triple the price of the sedan, find the cost per day of the limousine. 9 mC A manufacturing plant produces square and circular metal panels in fixed sizes. If the mass of a square panel is 13 kg and that of a circular panel is 22 kg, how many of each panel are there in a truck loaded with 65 panels of total mass 1205 kg? The equations to solve are: a 13s + 22c = 1205, s + c = 65 B 22s + 13c = 1205, s + c = 65 C 13s + 22c = 65, s + c = 1205 d 22s + 13c = 65, s + c = 1205 e 13s + 22c = 1205, s + c = 35 10 mC Which of the following is a solution of 11x + 2y = −121 and 10x + 12y = −222? a (11, 2) C (10, 12) e (6, 10) B (−121, −222) d (−9, −11)

1F

Finding the equation of a straight line

Consider a general linear graph containing the particular points (x1, y1), (x2, y2) and the general point (x, y) (which could be any point). Using the first two of these points in the formula for gradient, we have y2 − y1 [1] m= x −x 2 1 Using the first point and the general point in the same formula yields y − y1 [2] m= x − x1 Putting [2] = [1] gives

y − y1 y2 − y1 , which may be rearranged to = x − x1 x 2 − x1 y −y y − y1 = 2 1 ( x − x1 ) [3] x 2 − x1

y

(x2, y2) (x, y) (x1, y1) x

y2 − y1 , equation [3] may be written x 2 − x1 y − y1 = m(x − x1) [4] This last formula may be used to find the equation of a straight line when two points are given or when the gradient and only one point are given. When two points are given, the gradient m may be found y2 − y1 first using m = x − x and substituted into the formula y − y1 = m(x − x1) along with one of the points. 2 1 Since m =

18

Maths Quest 11 Mathematical Methods CAS

Worked Example 17

Find the equation of the line having gradient 34 that passes through (7, 11). Express your answer in the forms i ax + by + c = 0 and ii y = mx + c. Think

Write

1

As one point and the gradient are known, use the formula y − y1 = m(x − x1).

y − y1 = m(x − x1)

2

List the given information.

m = 4 (x1, y1) (7, 11)

3

Substitute for all variables except x and y.

4

Simplify, expressing in the form ax + by + c = 0.

i 4y − 44 = 3(x − 7) 4y − 44 = 3x − 21 3x − 4y + 23 = 0

5

Express your answer in the form y = mx + c.

ii 3x + 23 = 4y

3

y − 11 =

3 4

(x − 7)

3

y= 4x+

23 4

Worked Example 18

Find the equation of the straight line containing the points (2, −5) and (−3, 1). Express your answer in the forms i ax + by + c = 0 and ii y = mx + c. Think

Write

1

Write down the points so they match the variables in the formula.

(x1, y1) (x2, y2) (2, −5) (−3, 1)

2

As two points are known, first use the y2 − y1 formula m = to find m. x 2 − x1

m=

y2 − y1 x 2 − x1

1− −5 − 3− 2 6 = − 5 =

= 3

Write the formula y − y1 = m(x − x1).

4

Substitute the calculated gradient m = 5 and the first point (x1, y1) = (2, −5). Leave x and y as they are.

5

Simplify and express in the two forms required.

−6 5

y − y1 = m(x − x1) −6

y − −5 =

−6 5

( x − 2)

6

y + 5 = −5 ( x − 2) −5y − 25 = 6(x − 2) −5y − 25 = 6x − 12 −5y = 6x + 13 i So 6x + 5y + 13 = 0, or ii

y=

−6 5

x−

13 5

Perpendicular lines Perpendicular lines are lines that meet at right angles. The gradients of two perpendicular lines, when multiplied together, equal −1. Stated mathematically: m1 × m2 = −1 Chapter 1 • Linear functions 19

−1 Another way to write this relationship is m1 = m . This type of relationship is known as a negative 2

reciprocal. For example, the negative reciprocal of gradients are negative reciprocals.

exercise 1F

2 3

is

−3 2

. So, two lines are perpendicular if their

Finding the equation of a straight line

1 Copy and complete the table below.

Gradient of equation 1 (m1)

Equation 1 f1(x)

diGital doC doc-9703 SkillSHEET 1.2 reciprocals and negative reciprocals

a

y = 2x + 1

y=

b

y = 3x − 4

y=

c

y = 4x + 6

d

y = 5x − 3

e

y=

−1 2 −1 3

x+2

y = −4x − 9

2

y=

7

−5 2

m1 × m2

x

1

−9

Gradient of equation 2 (m2)

Equation 2 f2(x)

x+4

7

y = 9x + 1

x

2 Sketch f1(x) and f2(x) for part a above using a calculator or using graph paper with identical scales on

each axis to produce accurate graphs. What do you notice about the graphs?

3 Repeat question 2 for graphs b to e. Do you notice anything special about each pair of graphs? 4 Find the gradient of a line perpendicular to another line that has a gradient of: a 4

b

−9

c

1 7

d

−8

9

e

7 2

f

1.

5 Find the gradient of a line that is perpendicular to the line with equation: a y = −5x + 2 d y=

−7 6

x−2

2 3

b y=x−1

c y=

x+1

e 2x + y = 5

f 3x − 4y = 7.

Find the equation of a straight line having the gradient given and passing through the point listed. Express your answer in the forms i ax + by + c = 0 and ii y = mx + c. a (1, 2) gradient 3 b (4, 1) gradient 5 c (3, −2) gradient −1 d (−3, 2) gradient 1

6 We17 diGital doC doc-9704 equation of a straight line

2

Find the equation of the line containing each pair of points. Express your answers in the forms i ax + by + c = 0 and ii y = mx + c. a (5, 2) (3, 1) b (1, 1) (5, 5) c (6, 3) (8, 2) d (2, −2) (0, 1) − Find the equation of the line passing through (3, 3) that makes an angle of 45° with the positive x-axis. Find the equation of the line containing (7, −2) that makes an angle of 71.565° with the positive x-axis. Find the equation of the line (in y = mx + c form) that: a is perpendicular to the line with equation y = 3x + 1, passing through (−3, 6) 2 b is parallel to the line with equation y = x − 9, passing through (4, −7) 5 c is parallel to the line with equation 3x + 6y = 8, passing through (2, 2) d is perpendicular to the line with equation −6x + 7y − 2 = 0, passing through (4, 0) e has gradient 2, passing through the intersection of the lines with equations y = 3x − 5 and y = −2x + 5 −3 f has gradient 4 , passing through the intersection of the lines with equations x + 4y = −14 and −5x + 2y = 4. Find the equation of the line that passes through the point of intersection of the lines whose equations are 7x − 3y − 19 = 0 and 3x + 2y + 5 = 0, given that the required line is parallel to the line with equation −5x − 2y = 3. Find the equation of a line containing the intersection of the lines with equations y = −3x + 4 and 5x − 3y + 40 = 0 that: 6 2 a has a gradient of 7 b is perpendicular to the line with gradient 3

7 We18

8 9 10

11

12

c passes through the point (−1, 9)

20

Maths Quest 11 Mathematical Methods CAS

d is parallel with the line joining (−8, 5) and (0, 4).

13 A line passes through the points (−8, −5), (4, −3) and (a, 12). Find the value of a. 14 The points (2, 7) and (6, 9) lie on the same straight line. Does the point (4, 8) also lie on this line? 15 The height of a particular young pine tree is found to increase in a linear manner each month in the first

year after planting. Find an equation connecting height with time in months after planting, using the information supplied in the diagram below.

diGital doC doc-9705 WorkSHEET 1.2

52 cm 34 cm

After 5 months

After 2 months

distance between two points and y (x , y ) midpoint of a segment y 1G

2

The distance, d, between any two points on the Cartesian plane may be found using Pythagoras’ theorem applied to a right-angled triangle as shown at right. Using Pythagoras’ theorem: c2 = a2 + b2 or c = a 2 + b 2 and replacing c with d, a with (x2 − x1) and b with (y2 − y1),

2

2

d

y1

(x1, y1)

(y2 − y1)

interaCtiVitY int-0260 distance between two points and midpoint of a segment

(x2 − x1)

we have d = ( x 2 − x1 )2 + ( y2 − y1 )2 . x1

x2 x

Worked example 19

Find the distance between the points (−3, 7) and (5, −2) correct to 3 decimal places. think

Write

1

Match up (−3, 7) and (5, −2) with (x1, y1) and (x2, y2).

(x1, y1) (−3, 7)

2

Substitute into the formula for d and simplify.

d = ( x 2 − x1 )2 + ( y2 − y1 )2

tUtorial eles-1405 Worked example 19

(x2, y2) (5, −2)

= (5 − − 3)2 + ( − 2 − 7)2 = (8)2 + ( − 9)2 = 64 + 81 = 145 = 12.042, correct to 3 decimal places

Chapter 1 • Linear functions

21

midpoint of a segment The middle or midpoint M (xm, ym) of a segment joining two general points A (x1, y1) and B (x2, y2) is shown on the Cartesian plane below. y

B (x2, y2) y2 − ym M (xm, ym) A (x1, y1)

ym − y1 xm − x1

x2 − xm

C

E D x

Consider the triangles ACM and MEB. ∠MAC = ∠BME (since AM and BM have the same slope) ∠CMA = ∠EMB (since ∠ACM and ∠MEB are both 90°) AM = BM (given, as M is the midpoint) Therefore ΔACM ≡ ΔMEB. Since the horizontal and vertical sides of the two triangles must be equal, we have: xm − x1 = x2 − xm and ym − y1 = y2 − ym Simplifying these, 2xm = x2 + x1 2ym = y2 + y1 xm =

x 2 − x1 2

ym =

y2 + y1 2

x +x y +y So the point M has coordinates 1 2 , 1 2 . 2 2 Worked example 20

Find the midpoint of the segment joining (5, 9) and (−3, 11). think 1

Match (5, 9) and (−3, 11) with (x1, y1) and (x2, y2).

2

Substitute values into the formula for M and simplify.

Write

(x1, y1) (5, 9) M=

(x2, y2) (−3, 11)

x1 + x 2 y1 + y2 , 2 2

5 + − 3 9 + 11 = , 2 2 2 20 = , 2 2 = (1,10)

distance between two points and midpoint of a segment exercise 1G

diGital doC doc-9706 distance between two points

1 We19 Find the distance between each of the following pairs of points. a (4, 5) and (1, 1) b (7, 14) and (15, 8) c (2, 4) and (2, 3)

d (12, 8) and (10, 8)

2 Calculate the distance between each of the pairs of points below, accurate to 3 decimal places. a (−14, 10) and (−8, 14) b (6, −7) and (13, 6) c (−11, 1) and (2, 2) 3 Find the distance between each of the following pairs of points in terms of the given variable(s). a (a, 1), (2, 3) b (5, b), (0, 6) c (c, 2), (4, c) d (d, 2d), (1, 5) 22

Maths Quest 11 Mathematical Methods CAS

4 Two hikers are about to hike from A to B (shown on the map below). How far is it from A to B ‘as the

crow flies’, that is, in a straight line?

N 50 m 100 m 200 m

100 m 200 m

B (E7, N4)

300 m Lake Phillios

E

W

A (W12, S5)

S

Grid spacing : 1 km

5 Using the coordinates shown

y (in metres)

on the aerial photo of the golf course, calculate (to the nearest metre): a the horizontal distance travelled by the golf ball for the shot down the fairway b the horizontal distance that needs to be covered in the next shot to reach the point labelled A in the bunker.

(225, 96)

A (320, 148) x (in metres)

(80, –64)

6 We20 Find the midpoint of the segment joining each of the following pairs of points. a (1, 3) and (3, 5)

b (6, 4) and (4, −2)

c (2, 3) and (12, 1)

7 Find the midpoint of the segment joining each of the following pairs of points. a (7, −2) and (−4, 13) b (0, 22) and (−6, −29) c (−15, 8) and (−4, 11)

d (6, 3) and (10, 15) d

(−3, 40) and (0, −27)

8 Find the coordinates of the midpoint of each of the following pairs of points, in terms of a variable or

variables where appropriate. a (2a, a) and (6a, 5a)

diGital doC doc-9707 midpoint of a segment

c (3f, 5) and (g, −1)

b (5, 3c) and (11, 3c)

9 Find the value of a in each question below so that the point M is the midpoint of the segment joining

points A and B. a A (−2, a), B (−6, 5), M (−4, 5)

3

b A (a, 0), B (7, 3), M (8, 2 ) Chapter 1 • Linear functions

23

y

10 A fun-run course is drawn (not to scale) at right.

If drink stations D1, D2 and D3 are to be placed at the middle of each straight section, give the map coordinates of each drink station. 11 Find the equation of a line that has a gradient of 5 and passes through the midpoint of the segment joining (−1, − 7) and (3, 3). 12 Find the equation of a line parallel to the line with equation 9x − 3y = 5 that passes through the midpoint of the segment connecting (0, −4) and (−2, 10).

1H

(1.5, 3.5)

(−4.5, 5)

D1

Official tent

D2

x

(1.5, −2) START/ FINISH

Coordinates are in kilometres.

(3, −7)

D3

(13, −8)

Linear modelling

Many real-life applications, such as fees charged for services, cost of manufacturing or running a business, patterns in nature, sporting records and so on, follow linear relationships. These relationships may take the form of a linear equation; for example, F = 50 + 30t may be used by a tradesperson to calculate her fee (in dollars) for t hours of work. Here, F is the fee in dollars, and t the time in hours. The 50 represents an initial fee for simply turning up, while the 30t is the amount charged for the time spent on the job. For example, if t = 2 hours, 30t = 60, so the total charge for the work would be $(50 + 60) = $110. Equations like F = 50 + 30t are sometimes referred to as ‘linear models’, a common form of which is: Total cost = Fixed cost + Cost per unit × Number of units. This is, of course, equivalent to y = mx + c. y = c + mx. Worked Example 21

A generator company charges a $200 delivery fee, and a rental fee of $1500 per day. a Find an expression relating total charge to the number of days for which the generator is hired. b Sketch a graph of the relationship. c What would be the charge for 4 weeks of rental? Think

a 1 Define convenient variables. 2

The fixed cost is $200, and the cost per unit is $1500. (c = 200, m = 1500)

b 1 Find the vertical intercept (when n = 0). 2

The total cost rises $1500 each day, so the graph must show this.

Write/draw

a Let T = total charge (in dollars) and

n = number of days the generator is hired. T = 200 + 1500n

b If n = 0, T = 200 T ($) 3200 1700 200 1

c 1 After 4 weeks, n = 28. Substitute this in the equation from part a.

2

Write the answer in words.

24 Maths Quest 11 Mathematical Methods CAS

2

n (Days)

c If n = 28

T = 200 + 1500 × 28 = 200 + 42 000 = 42 200 After 4 weeks, the total cost is $42 200.

Worked example 22

‘Rent-a-Chef’ provides food cooked and served by a qualified chef at parties. The company charges $120 as a booking fee, and an additional $30 per hour. Another company, ‘Greased Lightning’, provides fast food served by two students at a cost of $65 per hour, with no booking fee. Under what conditions would it be cheapest to hire ‘Greased Lightning’? think

tUtorial eles-1406 Worked example 22

Write

1

Define convenient variables.

Let C = cost (total) in dollars and t = time in hours.

2

Write an equation for the cost of hiring both organisations.

Rent-a-Chef C = 120 + 30t Greased Lightning C = 65t

3

Use simultaneous equations to find when the cost is the same with each group.

Put [1] = [2] 120 + 30t = 65t 120 = 35t 120 t= 35 = 3.4 hours

4

At 3.4 hours, the cost is the same. Since Greased Lightning has the higher per hour cost, after 3.4 hours, they would be more expensive.

It is cheaper to hire Greased Lightning for food preparation and service of less than 3.4 hours (3 hours and 26 minutes) duration.

[1] [2]

Notes 1. 0.4 hours = 0.4 × 60 minutes = 24 minutes. 2. An alternative approach would be to use a CAS calculator and find the point at which the two graphs crossed.

exercise 1h

linear modelling

The cost of hiring a floodlit tennis court consists of a booking fee and an hourly rate. a Use the photo to write an equation for the total hire cost in terms of the hourly rate. b Sketch a graph of the relationship. c What would be the charge for 3 hours?

1 We21

diGital doC doc-9708 Simultaneous linear equations

2 A singing telegram service charges a $60

appearance fee, and $8 per minute sung. a Write an equation for the total cost of a singing telegram in terms of the number of minutes sung. b Sketch a graph of the relationship. c What would be the charge for a 5-minute singing telegram? 3 Colleen delivers junk mail and is paid $32 to traverse a particular route, and a further 10 cents per

leaflet delivered. a Write an equation for the total payment she receives. b Sketch a graph of the relationship expressed in a. c What would be Colleen’s pay if she delivers 1650 leaflets along the route? 4 A pay-TV salesperson receives $300 per week plus $20 for every household he signs up to have pay-

TV connected. How much does the salesperson receive for a week in which he signs up 33 households? 5 We22 A computer firm, SuperComputers Inc., offers a back-up plan covering the ongoing service and

troubleshooting of its systems after sale. The cost of signing up for the service plan is $215, and there is an Chapter 1 • Linear functions

25

hourly rate of $65 for the serviceperson’s time. Purchasers not signing up for the plan are charged a flat rate of $150 per hour for service. Would it be advisable to sign up for the service plan if you expected to need 3 hours of service assistance during the life of a computer purchased from SuperComputers Inc? 6 Two amusement parks show the following information for school-age tourists in a promotional brochure. After how many rides does an excursion to Fun World become the cheaper option for the same number of rides?

A qua Wor ld $8.00 entry $2.50 per ride

$12 entry $1.50 per ride

7 A telephone company, Opus, offers calls to Biddelonia for a connection fee of $14, and thereafter

$1 per minute. Its rival, Elstra, offers calls for $2 per minute (no connection fee) to the same country. a Compare the cost of a 10-minute call to Biddelonia using each company. b At what point would it be cheaper to use Opus? 8 It costs you $6 to get into a taxi (the ‘flagfall’), and $1.50 per kilometre if you use PinkCabs, while NoTop taxis charge $8 flagfall, and $1.20 per kilometre. a How much would it cost with each company to travel 15 kilometres in one of its cabs? b When would it cost the same to use both companies? 9 Medirank, a health insurance company, charges $860 per year (for a single person) and requires customers to pay the first $100 of any hospital visit. HAB, on the other hand, charges an annual fee of $560 and requires its members to pay the first $150 of any hospital visit. Determine the number of hospital visits in a year for which the cost of health services is the same whichever company insures you. 10 Nifty is a car hire firm that charges insurance of $135 and $50 per day car hire. A competitor, Savus,

simply charges $65 per day and offers ‘free’ insurance. You are planning a holiday, and would prefer to use Savus. Under what conditions (days hired) could you justify this choice? 26 Maths Quest 11 Mathematical Methods CAS

Summary Solving linear equations and inequations

• Do the same to both sides and remember inverse operations + and −, × and ÷, √ and 2. • Aim to get a single variable by itself. • Solve inequations the same way as equations, keeping the original inequality sign at each step, unless multiplying or dividing by a negative number.

rearrangement and substitution

• ‘Make x the subject’ means manipulate into the form ‘x = …’. • ‘Substitute’ means to replace a variable with a value.

Gradient of a straight line

• m=

Sketching linear functions

• • • • •

Simultaneous equations

• Simultaneous equations can be solved with a calculator. • For equations of the form y = ax + b, y = cx + d, consider using substitution. • For equations of the form ax + by = c, dx + ey = f, consider using elimination.

Finding the equation of a straight line

• Formulas for finding the equation of a straight line: y −y y − y1 = 2 1 ( x − x1 ) y = (mx + c) y − y1 = m(x − x1) x 2 − x1 • For perpendicular lines, m1 × m2 = −1.

distance between two points and midpoint of a segment

• d = ( x 2 − x1 )2 + ( y2 − y1 )2

linear modelling

y2 − y1 m = tan (θ ) x 2 − x1 where θ is the angle the line makes with the positive direction of the x-axis.

The general equation for a straight line of gradient m and y-intercept c is y = mx + c. Lines with the same gradient (m) are parallel. To find the y-intercept, let x = 0 and find y. To find the x-intercept, let y = 0 and find x. If y = 0 when x = 0, substitute another x-value (for example x = 1) to find another point on the line. • Join two points and/or intercepts with a straight line.

x1 + x 2 , y1 + y2 • M= 2 2 • Total cost (y) = fixed cost (c) + cost per unit (m) × number of units (x) • y = c + mx

Chapter 1 • Linear functions

27

Chapter review S h ort anS Wer

3(5 x − 4) 6(4 x + 3) . = 7 5 7x + 8 = 4 x − 9. 2 Find the value of x where 3 10 1 Solve the equation

− 3( x

1− x . 4 2 4 The following formula may be used to study planetary motion. 3 Solve the inequality

+ 4)

≤

GmM m 4π 2 R = R2 T2 Make T the subject of the equation.

c

6

5 Using Pythagoras’ theorem, find the length of the hypotenuse of the

triangle shown. Express your answer: a in surd form b to 3 decimal places.

9

6 Calculate the gradient of each of the following lines. y a 6

c

1 grid square = 1 unit

6 5 4 3 2 1

−5 −4 −3 −2 −1 0 −1 −2 −3

x

−8

y

b

y

1 2 3 4 5 6x

d

y x (−12, −3)

x

(−4, −10) 7 Find the gradient of the line joining (−7, 15) and (2, −6). 8 Find the gradient of the line shown. y

77° x 9 State the gradient of the line below. y

5

28

Maths Quest 11 Mathematical Methods CAS

x

10 State the gradient and y-intercept (in that order) for each of the following. a y = 3x − 7

b 5x + 3y = 30

11 Find the equation for a linear graph having gradient 12 Sketch graphs of the following, showing intercepts. a y = −3x + 24

b −x + 8y = 40

c 2x − 4y − 8 = 0

and y-intercept −3.

2 5

c 9x − 7y − 63 = 0

d y + 6x = 0

−2x

13 Solve y = 3x + 10 and y = − 15 graphically. a Sketch the solution on a set of axes.

b State the solution (point of intersection).

14 Solve y = −3x, y = 6x − 15 using substitution. 15 Use the method of elimination to solve 4x − 7y = 21, −2x + y = 6. 16 A piggybank contains 67 coins. If there are only one- and two-dollar coins in the piggybank,

and there are 25 more one-dollar coins than two-dollar coins, how many of each type are there? 17 Find the gradient of a line perpendicular to 3x − 9y = 7. 18 Find the equation of the line containing (−4, 8) and (3, 1). −6

19 Find the equation of the line having gradient

7

that passes through (1, 4). 1 4

20 Find the equation of the line perpendicular to y = x − 5 that passes through (−8, 6). 21 The distance between (2, −7) and (a, −2) is

41 units. Find the value of a if it is positive.

22 Show that the triangle with vertices (3, 7), (3, 3) and (6, 3) is a right-angled triangle. 23 The midpoint of the line joining (k, 2h) and (9k, 6h + 2) is (20, −11). Find k and h. 24 The washing machine repair company ‘Washed out’ charges $75 to come to your house, as well as an

hourly charge of $65, calculated to the nearest half hour. a Write an equation that may be used to calculate the cost of any service call if the time taken by the repairer is known. b Sketch a graph of the relationship between repair cost and time taken to do a repair. 1 c How much would it cost to have a repair done that takes 3 2 hours? 7 x − 23 = 99 would be to: 3 a add 23 to both sides B divide both sides by 3 C divide both sides by 7 d multiply both sides by 3 e multiply both sides by 7

m U ltip l e C ho iC e

1 The first step in solving

2 x = −5 is a solution to the equation: a 3x + 7 = −8

B 2x − 7 = −5

d 2(x + 3) = 10

e

3 The solution to a x ≤ 10

− 5x

9

x + 14 ≥ 2( x + 2) is: 3 2 B x≥5

x + 25 =5 6

C

= 45

2

C x≤5

d x≤

−6

−2 5

e x≤ 5

4 When c2 = a2 + b2 is rearranged to make a the subject, the equation becomes: a c=

a2 + b2

B a2 = b2 + c2

d a=

c2 − b2

e a=b+c

C a2 =

c2 − b2

1

5 Which values, when substituted into K = 2 mv 2, give a value for K of 4? a m = 2, v = 4

B m = 4, v = 2

d m = 8, v = 1

e m = 1, v = 16

1

C m = 8, v = 2

6 Using the equation P = m1v1 + m2v2, if P = 10, m1 = 2, m2 = 6 and v1 = 4, v2 would equal: a

1 3

B

1 2

C 1

d 2

e 3

Chapter 1 • Linear functions

29

y

7 The line shown has a gradient of: a B C d e

−6

6

−3 −2

2 6 3

x

3

8 The gradient of the line shown at right is 3. The value of a must be: a −2 B −1 C 5 d 7 e 11

y (6, a)

9 The gradient of the line joining (−1, 0) and (4, −10) is: a

−4

B

d 4

−2

x

C 2

e 5

(2, −5)

10 Which of the graphs below has a gradient of y

a

6 ? 7

C y

y

B

7

6 6 7 x

y

d

x

−7

x

−6

y

e

x

−7

6

−6

−7

11 The gradient of the line with equation y = a −1

B

6 7

x − 1 is: 7

6 7

C 6

12 The y-intercept of the line with equation y = 12 x + a

x

2 3

B

3 2

2 3

d 6

e 7

d 3

e 12

is:

C 2

13 The gradient and y-intercept (in that order) of the line with equation 2x − 3y = 7 are: a 2 and −3 d

2 3

and

B 2 and 7

−7

e

3

−3

and

C 7 2

−2 3

and 7

14 Which of the following could be the graph of y = 2x + c? a

y

B

y

C

y

c x −c

30

Maths Quest 11 Mathematical Methods CAS

c x

x

y

d

−c

e y

c

x

x y

15 The equation of the line shown at right is: a 2x − 5y = 1 C 15x + 6y = −30 e

x y − =1 2 5

B 2x − y = 4 d 10y − x = −2

2

x

−5

16 To solve the equations 2x + y = 5 and 3x − 6y = 12 graphically on a calculator, you would enter the

equations in the function entry line as: f 1(x) = 2x + y and f 2(x) = 3x − 6y f 1(x) = 5 and f 2(x) = 12 f 1(x) = 5 − 2x and f 2(x) = 12 + 6y f 1(x) = 2x + 5 and f 2(x) = 3x + 12 x f 1(x) = −2x + 5 and f 2(x) = 2 − 2

a B C d e

17 Which of the following would be the most effective way to solve the following equations

simultaneously? y = 2x − 13 [1] y = 7x + 2 [2] a Multiply [1] by 2 and [2] by 13 and add the newly formed equations. B Multiply [1] by 7 and put it equal to [2]. C Multiply [2] by 2 and put it equal to [1]. d Multiply [1] by 2 and [2] by 7 and subtract the newly formed equations. e Put [1] equal to [2]. 18 The gradient of a line perpendicular to a line with a gradient of 7 is: a −1

B −7

d 7

e

7

7 1

19 The gradient of a line perpendicular to y = a

−1 7 5

d 16

20 The equation of the line containing (1,

C

+5 is: 16

1

C

−7

−2)

1

− 7x

B 7 e

−7

16 7

16

and (2, −3) could be expressed as:

a y−2=x−1

B y+2=1−x

d x − 2y = −3

e 3x − 5y = 1

21 A line with equation y − 7 = 5(x − 1) has: a gradient 5 and contains the point (7, 1) C gradient 5 and contains the point (1, 7) e gradient −1 and contains the point (5, 7)

C y+3=x−1

B gradient −7 and contains the point (−1, −7) d gradient −5 and contains the point (1, −7)

22 The distance between (4, 3) and (−2, 1) is equal to: a (−2 − 4)2 + (1 − 3)2

B

(4 − 3)2 + ( − 2 − 1)2

(4 + 2)2 + (3 − 1)2

e

(− 2 − 4)2 − (1 − 3)2

d

C

(4 2 + 32 ) − ( − 22 − 12 )

Chapter 1 • Linear functions

31

23 The midpoint of the segment joining (11, −3) and (−5, 17) is: 1 1 2 2 1 1 ( − 2 2 ,8 2 )

a ( − 5 , −1 )

B (3, 7)

d

e (4, 6)

C (6, 14)

24 Bote lives 5 kilometres from the nearest post office. At noon one day he begins cycling (from home) at

20 kilometres per hour in a constant direction away from the post office. At t hours after Bote begins cycling, the distance, D km, that he is from the post office is given by: a D = 5t B D = 20t C D = 5t + 20 d D = 20t + 5

e D = 20t − 5

25 The linear function f : D → R, f ( x ) = 6 − 2 x has range [−4, 12]. The domain D is: a [−3, 5] d e x t ended r e SponS e

[−14,

B [−5, 3]

18]

e

[−18,

C R

14]

1 The graph at right is a rough sketch of three points on a section of

y

sheet metal that are to be drilled by a programmed robotic drilling arm. Any deviation from a straight path, no matter how slight, means the arm must be programmed for more than one direction. 16 The coordinates marked are correct. Will the robotic arm be able to 12 move in one direction only to drill all three holes? 2 Points A, B and C have the coordinates (1, 6), (0, 0) and (−2, 2).

Find the coordinates for a point D so that the four points form a parallelogram. (−4, −2),

−1)

6

C B A

8

14

20

x

−7).

3 Consider the points (6, 2), (4, and (0, a Find the coordinates of the midpoints of each side of the quadrilateral formed by the points. b Show that the shape formed by the midpoints is a parallelogram. c Repeat parts a and b for a different set of starting points. d What can you conjecture based on your answers? 4 The cost of a parachuting course consists of a charge of $250, which covers equipment hire and tuition,

and a further expense of $55 per jump. a Express the total cost, C, as a function of j, the number of jumps. b How many jumps could a person doing the course afford if she was prepared to spend up to $1000? 5 A physics student conducts an experiment to find out how much a spring stretches when various

weights are hung from it. Her results are shown in the table below.

a b c d 32

Length of spring (cm)

Force applied (N)

4

0

7

10

12

20

16.5

30

20.5

40

25

50

What is the natural or ‘unstretched’ length of the spring? Plot a graph of the student’s results. Draw a straight line through the points that best describes the data. Select two points on the line and use them to fit a linear equation to the line.

Maths Quest 11 Mathematical Methods CAS

A second student conducts the same experiment on a similar spring. His results are shown below. Length of spring (cm)

Force applied (N)

5

0

10

10

16

20

21

30

24

40

28

50

e On the same set of axes you used in part b, plot the results of

the second experiment and join the points with a line of best fit. f Write an equation that describes the relationship between the

force applied and the length of the second spring. The gradients of graphs such as the ones you have drawn give an indication of the stiffness of a spring. The greater the gradient, the harder it is to stretch the spring. The lower the gradient, the easier it is to stretch the spring. g Comment on the stiffness of the two springs investigated by the students. h Is it likely that these two springs will ever be the same length at a given force; that is, is it likely that the intersection of the two graphs could ever really happen? Explain your answer.

diGital doC doc-9709 Test Yourself Chapter 1

Chapter 1 • Linear functions

33

ICT activities Chapter opener diGital doC • 10 Quick Questions doc-9695: Warm up with ten quick questions on linear functions (page 1)

1a

Solving linear equations and inequations

diGital doC • doc-9696: Use trial and error to balance an equation and hence solve for the unknown variable (page 3)

1B

rearrangement and substitution

diGital doC • Career profile doc-9697: Learn how a vigneron uses substitution and other areas of mathematics in his work (page 8)

1C

Gradient of a straight line

diGital doCS • doc-9698: Calculate the gradient between two given points (page 10) • SkillSHEET 1.1 doc-9699: Practise using the gradient to find the value of a parameter (page 11)

1d

Sketching linear functions

diGital doCS • WorkSHEET 1.1 doc-9700: Solve linear equations, use substitution and calculate gradients of straight line graphs (page 12) • doc-9701: Investigate the effect of changing the gradient and y-intercept for a linear graph (page 14) tUtorial • We 12 eles-1404: Watch a tutorial on how to sketch a linear graph by hand (page 13)

1e

Simultaneous equations

diGital doC • doc-9702: Use the graphical method to solve simultaneous linear equations (page 17)

34

Maths Quest 11 Mathematical Methods CAS

1F

Finding the equation of a straight line

diGital doCS • SkillSHEET 1.2 doc-9703: Practise writing positive and negative reciprocals of rational numbers (page 20) • doc-9704: Investigate the equations of straight lines (page 20) • WorkSHEET 1.2 doc-9705: Sketch linear graphs, determine equations of linear graphs and application problems (page 21)

1G distance between two points and midpoint of a segment interaCtiVitY • Distance between two points and midpoint of a segment int-0260: Consolidate your understanding of how to calculate the distance between two points and the midpoint of a segment (page 21) tUtorial • We 19 eles-1405: Watch how to calculate the distance between two points on a Cartesian plane (page 21) diGital doCS • doc-9706: Investigate the distance between two points (page 22) • doc-9707: Investigate the midpoint of a segment (page 23)

1h

linear modelling

diGital doC • doc-9708: Investigate simultaneous linear equations (page 25) tUtorial • We 22 eles-1406: Watch how to apply linear algebra skills to determine the conditions for which it would be cheapest to hire a catering company (page 25)

Chapter review diGital doC • Test Yourself doc-9709: Take the end-of-chapter test to test your progress (page 33)

To access eBookPLUS activities, log on to www.jacplus.com.au

Answers CHAPTER 1 linear FUnCtionS

3 a d g 4 a c 5 a c 6 a 2 7

exercise 1a

Solving linear equations and inequations b −8

c

7 −1 6

e −1 h 10 b 5

d 112

e 12

f i c f

−9

h

j −4

k

76 59 −9

1 a 2 d g 2 a

9

g

13

b x≤

3 a x≥ 6

d x < −3 exercise 1B

substitution

− 31

c x>3

e x ≥ 11

f x≥2

b 5 a b 6 a

f r=

j H= b 350 e 2.498

kQq F

S − 2π r 2 2π r

c 224 cm

x

7 −21 y

c

x

− 3–2

y=x+2

y

3 a–d

y=x+1

y

d

10

e

y=x+3

y

30 1 x

4 y=x− 4

1

c 18 cm

−1

−5

01

y = 2x − 7 x

−16

10 a

2

b

g 11 , 1 5 , 6 2

−6

y 4

y-axis (hence the name ‘y-intercept’). 6 a y = 2x + 7 b y = −3x + 1 2 1 c y = 5x − 2 d y= 3x+3

−2

x

3

5 The number is where the graph cuts the

1

y

y = −3x + 6

−7

e y= 4x− 2 7 a 4, 5 c −3, 1 e −7, −9

x

8

y = 2x + 1 y = 3x − 5

i

y

y

c 40 mm

d

x

10 — 3

−4

6

1 2

x

2

f

−1

c

x y

y = 2x y=x

4 a–d

Gradient of a straight line b5

x−5

y

b

graph. Positive values make the graph slope up when moving (or tracing) to the right; negative values make the graph slope down when moving to the right.

c 7

25 9 h= cm π

−4

2 −5

d y = 6x + 5 f y = −5x

−3

y = 3x

−5

18

2 The higher the number, the steeper the

−3

3

x−2

9 a

− 17 300

y

y = −2x

b y=

−3

8 b=2

b

4 3

Sketching linear functions

3V , 6.204 br=3 4π T 2 d l = g , 3.972 2π

A, 7.746

c y=

x

A A− D − 1 = 100 b r = 100 D D c 41.4% fv uv 7 a f = b u= v− f u+v c 150 cm

2 a 2

1 a–e

8 a y = 4x + 2

e y = 2x − 1

y = −x

v2 hγ = rT

v−u , 4.167 t (1 − α ) K c= , 2622 α2 42 cm P P − 2l w = − l or w = 2 2 240 m2 2A 2 A − bh a= − b or a = h h $1123.60

1 a 2

b 31 d 3 b Yes

c 0.176 f0

A w C d r= 2π

E − αθ θ2

exercise 1C

4 − 5 No

exercise 1d

3.078 − 0.577 57.290 2.61 − 3.73 C B E

b l=

c a=

4 a

3

b e h b d b d b 8

rearrangement and

2 a 0.267 d 13 100

e

9 a c 10 a

3

2 Fd + mu 2 g v= m S − 2lh i w= 2(l + h)

3 a l=

17

l −7

d v

e β=

3

9 −12 −13 −5

i 3

1 a P=A−L c t=

−7

1.192 − 0.577 1 0.93 − 0.53 D A B

f y = −2x + 12

4, −8 −2, 4 2, 5 −8 2 h , 3 3 b d f j

−5 − 7 , 2 2

5

c

x

y 6

−3

x

Chapter 1 • Linear functions

35

d

y

b

y

y = 20

x

5

exercise 1F Finding the equation of a straight line

11 a

e x

−5 −5

4 a

y

10

5 a 5

( 94 , 47 4)

d

x

y = 3x + 5

x

y

d

y 4 — − 16 3

5 y = 6x + 5 0

x

y = −4 + 3x x

y

f −1

7

e

c

1 2

f

ii y = 2 x + 2

7 a i x − 2y − 1 = 0 b i x−y=0

ii y = 2 x − 2 ii y = x

c i x + 2y − 12 = 0

ii y = 2 x + 6

d i 3x + 2y − 2 = 0

ii y = 2 x + 1 9 y = 3x − 23

−1

3

x

(1, −1)

−5

7 , − 29 (− 12 6 )

(1, 1)

f

y = 17 − 9x

y

x 14

17

( 38 ,

109 ) 8 y = 14 − x

y

2 a (1, 4)

(1, −2)

13 D

14 E

c (−7, −5)

15 A

Simultaneous equations y

1 a

e

23 − 20 (14 , 7 )

3 a (7, 9)

(−3, 6) y = −4x − 6

x

17 9

x

c e

1 −1−1 1

x y = −2x

4 6 7 8 9

(1,

−2)

59 21 (8, 8)

9 9

f (13,

−3)

b (6, 5) d

1 19 ( 2, 2 )

f

84 99 ( 67 , 67 )

15 cents and 35 cents 5 22 and 19 16 emus, 41 sheep Basketballs $9.45, cricket balls $3.05 Limousine $225 (sedan $75) A 10 D

Maths Quest 11 Mathematical Methods CAS

1

2

43

−7

14

−3

9

−3 2

x +7

b y= 5x− 5

x+3

d y= 6x+ 3

−1 2

f y= 4x − 2

3 82

12 a y = 7 x + 7

b y=

−x

39

d y= 8 + 4 14 Yes

exercise 1G distance between two points and midpoint of a segment 1 a 5 b 10 c 1 d 2 2 a 7.211 b 14.765 c 13.038

3 a b (−4, −15) 5 17 d( , )

1

x+5

13 94 15 H = 22 + 6t

y

7

−3

c y = −x + 8

y = −6 − 2x

1

−1

11 y = 2 x − 2

5

3

d i x − 2y + 7 = 0

6

x

2 −4

ii y = 3x − 1 ii y = 5x − 19 ii y = −x + 1

−5

y = 10x + 1

−3

6 a i 3x − y − 1 = 0 b i 5x − y − 19 = 0 c i x+y−1=0

e y = 2x − 3 5

36

6 7

c y=

y

e

y

exercise 1e

−2

b −1

10 a y =

(−3, −13)

−6

c

c −7

8 y=x−6

x

2

b

e

1

−4

12 a

1

b 9

4

d 8

y = 7x − 4

d

−1 9

−6

c

−9 7 , , −1 7 9

y

b

d 5 , 2 , −1

2 They are perpendicular. 3 They are perpendicular.

x

5

c

2 −5

1

y −7

b 3, 3, −1

c 4 , −4, −1

y = 3x − 5

−7

−1

−1

1 a 2, 2 , −1

, 20 ) ( 25 3

20

c

a2 − 4a + 8

b

b 2 − 12b + 61

2(c 2 − 6c + 10)

d

5d 2 − 22d + 26

4 21.024 km 5 a 216 6 a (2, 4) c (7, 2) 1 2

1 2

7 a (1 , 5 ) 1

1

− c ( 9 2 , 9 2)

8 a (4a, 3a)

c 3 f + g , 2

b 108 b (5, 1) d (8, 9) − − 1 b ( 3, 3 2 ) 1

1

− d ( 1 2 , 6 2)

b (8, 3c)

2 9 a 5 b 9 10 D1 (−1.5, 4.25), D2 (−1.5, 1.5), D3 (8, −7.5) 11 y = 5x − 7 12 y = 3x + 6

linear modelling

9 Undefined

1 a C = 25 + 5t

−

10 a 3, 7

b Cost ($)

1 2

c

35 30 25

b

−5

3,

22 Teacher to check. 23 k = 4, h = −3 24 a C = 75 + 65t b C ($)

10

, −2 2

11 y = 5 x − 3 12 a

1 2 Time (h)

140

y

75

24

c $40 2 a C = 60 + 8m

1

76 68 60

1 5 9 13 17 21 25

x

8 y

b 1 2 Time (min)

5

c $100 3 a P = 32 + 0.1n

x

−40

b Payment ($) 34 33 32

c $197

$960 Yes ($410 compared to $450) After 4 rides a Opus $24, Elstra $20 b After 14 minutes 8 a PinkCabs $28.50, NoTop $26 2 b After 6.7 km (6 3 km) 9 6 visits 10 Savus would be cheaper for up to 9 days hire.

13 a

5 a 6 a

3 4 5

c 11 7

−7 3

b (−5, −5)

y f(x) = 3x + 10 5

Chapter reVieW

x

−5 (−5, −5) −5

f(x) = −2x − 15

4π 2 R3

or 2π R

R GM b 3.873 −7

b 11 d

−7

− 63 − 33

5 14 ( , − 5)

15 ( 10 , 5 )

3

16 17 18

21 two-dollar and 46 one-dollar coins −3 y = −x + 4 −6

C C B E C B

4 8 12 16 20 24

D D A E B D

a parallelogram.

(1, −6)

1 26 3 x ≥ −14

3 7 11 15 19 23

4 a C = 250 + 55j b 13 jumps 5 a 4 cm b, c, e

x

−2

A A B B A D

d The midpoints of any quadrilateral form

y

4 5 6 7

Short anSWer

2 6 10 14 18 22

be shown by calculating gradients or equations for lines joining different pairs of points. 2 (−1, 8) −9 1 3 a Midpoints: (1, 0), (5, 2 ), (2, −4), (−2, 2 )

x

7

d

D D B D E C A

1 No, the points are not collinear. This may

−9

15

t (h)

extended reSponSe

y

c

10 20 Number of leaflets

GM

2

mUltiple ChoiCe

b Cost ($)

4 T=

c $302.50

205

8

19 y = 7 x + 7

8 4.331

20 y = −4x − 26

34

21 6

Force applied to spring (N)

exercise 1h

60 50 40 30 20

Student 1 Student 2

10 0

5

10 15 20 25 Length of spring (cm)

30

d Answers will vary. One possible answer 30

95

is y = 13 x − 13 . f Answers will vary. One possible answer is y = 2x − 10. g The first spring is stiffer than the second, as the gradient of its graph is greater than that of the second spring. h The graphs intersect at the point (−8.75, −27.5). It is not possible for the springs to have a negative length, so this point is not achievable.

Chapter 1 • Linear functions

37

Chapter 2

Quadratic functions diGital doC doc-9710 10 Quick Questions

Chapter ContentS 2a 2B 2C 2d 2e 2F 2G 2h 2i 2J 2k 2l

Polynomials Expanding quadratic expressions Factorising quadratic expressions Factorising by completing the square Solving quadratic equations — Null Factor Law Solving quadratic equations — completing the square The quadratic formula The discriminant Graphs of quadratic functions as power functions (turning point form) Graphs of quadratic functions (intercepts method) Using technology to solve quadratic equations Simultaneous quadratic and linear equations

2a

polynomials

A polynomial in x, sometimes denoted by P(x), is an expression containing only non-negative whole number powers of x. The degree of the polynomial is given by the highest power of the variable x. Examples of some polynomials are: 10 3x + 1 x2 + 4x + 7 −5x3 + 0.5x2 2x4 − x3 + 2x2 + x + 5

is a polynomial of degree 0 as 10 can be written as 10x0 is a polynomial of degree 1, or a linear polynomial is a polynomial of degree 2, or a quadratic polynomial is a polynomial of degree 3, or a cubic polynomial is a polynomial of degree 4, or a quartic polynomial.

2 − Expressions such as 3x2 − 4x + , −5x4 + x3 − 2 x and x 2 + sin (x) + 1 are not polynomials. x The general form of a degree n polynomial is given by P(x) = an x n + an − 1xn − 1 + . . . + a2x2 + a1x + a0 where n is a positive whole number and an, an − 1, . . . a2, a1, a0 are called coefficients (or constants) provided the leading coefficient an ≠ 0. This chapter will deal with polynomials of degree 2, or quadratics. The general form of a quadratic polynomial is P(x) = a2x2 + a1x + a0. It is also commonly written as y = ax2 + bx + c when graphing applications are involved. In the polynomial equation P(x) = 13x2 − x + 1: x is the variable. 13 is the coefficient of x2 and 13x2 is the quadratic term. −1 is the coefficient of x and −x is the linear term. The constant term is 1. The degree of the polynomial is 2. The leading term is 13x2 as it is the term with the highest power of x. Chapter 2 • Quadratic functions

39

An example of where polynomials are useful is shown below.

The surface area, S, of a plant hothouse of length L and height x can be approximated by the quadratic polynomial S ( x ) = π x 2 + Lπ x − 4.

Value of a polynomial diGital doC doc-9711 History of mathematics Évariste Galois

A polynomial may be evaluated by substitution of a number for the variable. For example, for the polynomial P(x) = 13x2 − x + 1, the value of P(x) when x = −3 is given symbolically as P(−3) and is found by substitution: P(− 3) = 13(− 3)2 − (− 3) + 1 = 13 × 9 + 3 + 1 = 121 This process can be completed on your CAS calculator. A CAS calculator is particularly useful for performing multiple substitutions simultaneously. 1. Using a CAS calculator, define the polynomial 1. Define t(x) = 3x2 − 2x + 16. t(x) = 3x2 − 2x + 16. 2. To evaluate t(x) for x-values of −3, 2 and 5, 2. t({− 3, 2, 5}) complete the entry line as: The result is (49, 24, 26.5279). t({− 3,2, 5})

exercise 2a 1

polynomials

State the degree of each of the following polynomials. b 65 + 2t u4 1 d x6 − 3x5 + 2x4 + 6x + 1 e 2u − − 6 + u5 3 2 a x3 − 9x2 + 19x + 7

c 2x − 8 + 3x2

2 State the variable for each polynomial in question 1. 3 Which polynomial in question 1 is: a linear? b quadratic?

c cubic?

4 For each of the following expressions:

• if it is a polynomial, write P • if it is not a polynomial, write N and circle the term(s) that made it so. x2 5 +x a 7x + 6x2 + b c 3x 4 − 2 x 3 − 3 x − 4 9 x d k 40

−

2

+ k − 3k 3 + 7

Maths Quest 11 Mathematical Methods CAS

e 2x − 8x + 1

5 Consider the polynomial P( x) = −2x3 − 3x + 4x2 + 5. a What is the degree of the polynomial? b What is the variable? c What is the coefficient of the linear term? d What is the value of the constant term? e Describe the role of the ‘4’ using the terminology of polynomials. 6 For the polynomial T(m) = 5m − 2m2 − 1, evaluate the following. a T(2) b T(−2)

c T(x + 1)

7 A sports scientist determines the following equation for the action of a breaststroke swimmer during

one complete stroke: v(t) = 63.876t 6 − 247.65t 5 + 360.39t 4 − 219.41t 3 + 53.816t 2 + 0.4746t where t = time (sec) and v(t) = velocity (m/s). a b c d

What is the degree of the polynomial? What is the variable? Evaluate v(0.2) and interpret the meaning of your answer. How much faster is she swimming at 0.7 seconds than she is at 0.5 seconds?

2B

Expanding quadratic expressions

A quadratic expression is a polynomial of degree 2. It must contain a quadratic term; any others (a linear term and a constant term) are optional. Examples of quadratic expressions include: x2 − 3x − 1

−5x

− 10x2

3t 2 + 1

Quadratic expressions can be produced when two linear expressions are multiplied. Consider the expansion (4x + 9)(2x − 3). When expanding brackets, ‘multiply everything by everything else’ as shown on the diagram at right. That is, first term × everything in the second brackets, then second term × everything in the second brackets. The results of each part of the expansion are: 1 4x × 2x = 8x2 2 4x × −3 = −12x 3 9 × 2x = 18x Parts 2 and 3 may be combined to give 6x, so we have:

4

2

1

(4x + 9)(2x − 3) 3

4

9 × −3 = −27

(4x + 9)(2x − 3) = 8x2 + 6x − 27 This method can be easily extended to deal with brackets containing more than two terms. The above method can be used on all types of binomial expansions, though a couple of shortcuts for special cases are shown in worked examples 1, 2 and 3. Chapter 2 • Quadratic functions 41

Worked Example 1

Expand the following. a (3x + 5)(6x − 7) b (2x − 9)2 c (5x − 3)(5x + 3) d −2(x − 9)(4 − x) Think

a 1 Write the expression and mark the required

multiplications.

Write

a (3x + 5)(6x − 7)

2

First term × everything in the second brackets gives 18x2 − 21x.

3

Second term × everything in the second brackets gives 30x − 35.

= 18x2 − 21x + 30x − 35

4

Combine the middle (x) terms.

= 18x2 + 9x − 35

b 1 Write the expression. Recognise it as a ‘perfect square’. 2

Remember the shortcut: Square the first term, double the product of the two terms and square the last term.

3

Square the first term to get 4x2.

4

Double the product of the two terms to get 2 × (2x) × (−9) = −36x.

5

Square the last term to get 81.

c 1 Write the expression. Recognise that the only difference

between the two sets of brackets is the sign in the middle, so the answer is a ‘difference of two squares’. 2

The shortcut for the expansion of this type is (first term)2 − (second term)2.

d 1 Write the expression.

b (2x − 9)2

= 4x2 − 36x + 81 c (5x − 3)(5x + 3)

= (5x)2 − 32 = 25x2 − 9 d −2(x − 9)(4 − x)

2

Rewrite the question so x is the first term in both brackets. This is not essential as long as all combinations of terms are multiplied in the next step.

= −2(x − 9)(−x + 4)

3

Expand the brackets first.

= −2(−x2 + 13x − 36)

4

Multiply the brackets’ contents by −2.

= 2x2 − 26x + 72

Worked Example 2

Expand (x − 7)(x + 2) − (2x − 1)(x + 4) and simplify. Think

Write

1

Write the expression.

(x − 7)(x + 2) − (2x − 1)(x + 4)

2

Expand the first pair of brackets.

(x − 7)(x + 2) = x2 − 5x − 14

3

Expand the second pair of brackets.

(2x − 1)(x + 4) = 2x2 + 7x − 4

4

Subtract the two expanded groups in the order given. Use new brackets for clarity as shown.

(x − 7)(x + 2) − (2x − 1)(x + 4) = (x2 − 5x − 14) − (2x2 + 7x − 4)

5

Apply the negative sign to the contents of the second brackets.

= x2 − 5x − 14 − 2x2 − 7x + 4

6

Collect like terms and simplify.

= −x2 − 12x − 10

42 Maths Quest 11 Mathematical Methods CAS

Worked example 3

If P(x) = (2x − 3)2 − (2x + 1)(3x + 9): a evaluate P(1) b expand and simplify P(x) c use your result in b to confirm your answer in a. think

Write

a 1 Substitute x = 1 into the rule for P(x).

a P(1) = (2 × 1 − 3)2 − (2 × 1 + 1)(3 × 1 + 9)

2

Simplify each bracketed term.

P(1) = (−1)2 − (3)(12)

3

Simplify P(1).

P(1) = 1 − 36 = −35

b 1 Expand the first bracketed term.

b (2x − 3)2 = 4x2 − 12x + 9

2

Expand the second and third bracketed terms.

(2x + 1)(3x + 9) = 6x2 + 21x + 9

3

Subtract the second result from the first result to obtain P(x).

P(x) = 4x2 − 12x + 9 − (6x2 + 21x + 9)

4

Simplify.

P(x) = −2x2 − 33x

c 1 Evaluate P(1) using the formula from part b.

2

c P(1) = −2(1)2 − 33 × 1

= −2 − 33 = −35

The answers are the same.

Compare with the result from part a.

exercise 2B

expanding quadratic expressions

Simplify your answers to questions in this exercise as fully as possible. 1 We1a Expand the following. a (2x + 6)(x + 5) c (5x − 7)(5x + 2) e (x + 4)(7x − 9) g (6x − 13)(2 − 3x) i (x − 9)(9x − 1)

b d f h j

(3x + 1)(4x − 3) (8x − 3)(6x − 1) (x + 6)(x + 15) (5 − x)(6 − x) (4x + 21)(x − 3)

2 We1b Expand using the shortcut for perfect squares. a (2x + 3)2 b (3x − 5)2 2 c (6x + 1) d (7x − 6)2 2 e (x − 8) f (x + 13)2 − 2 g ( 2x + 9) h (2x + 9)2 2 i (4 − 3x) j (6 − x)2 3 We1c Expand (remember the shortcut for difference of squares). a (2x − 6)(2x + 6) b (3x + 5)(3x − 5) c (6x + 1)(6x − 1) d (2x − 9)(2x + 9) e (11x + 3)(11x − 3) f (x − 12)(x + 12) g (x + 6)(x − 6) h (7 − 2x)(7 + 2x) i (1 − x)(1 + x) j (5x + 1)(5x − 1) 4 We1d Expand. a 2(x + 6)(3x + 5) c −4(x + 2)(2x − 5) e 8(3x − 1)(4x − 1) g 4(4 − x)(7 − x) i a(2x + 9)(x − 6)

diGital doCS doc-9712 SkillSHEET 2.1 expanding perfect squares doc-9713 SkillSHEET 2.2 expanding differences of squares

b 3(x − 4)(2x + 7) d 3(4x − 9)(2x − 1) f −7(2x + 3)(5x − 10) h −5(7x − 4)(2 − x) j −b(8 − 2x)(x + 4) Chapter 2 • Quadratic functions

43

5

Expand the following. a (6x − 13)(6x + 13) c −3(7 + 2x)(x − 8) e (2x − 14)(2x + 14) g (x + 16)(x − 9) i −4(5x − 3)2

b d f h j

6 We2 Expand and simplify. a (x − 9)(x + 2) + (x + 4)(x − 4) c 2(x + 5)2 + 5(2x + 7)(x − 3) e 4(x − 3)(3x + 5) − 2(2x + 1)2 g ( x + 3)( x − 3) i 12 (2x + 1)(4x + 7)

(5x + 2)(12x − 5) (3x + 11)2 (6x − 5)2 −(x − 4)(x + 4) 5(2x + 7)(2x − 7)

b (3x + 7)(2x − 1) + (4x − 3)(3x − 4) d (x − 12)(x + 12) − 3(x + 1)(x + 5) f 6(2x − 5)(2x + 5) − (x − 6)(x + 6) h (3 x + 2)(4 x − 2) j (5x + 13)(3x − 5)

7 We3 Expand. Use a calculator to verify the answers. a (2a + 4)(2a + b) b (x + 2y)(3x − 5y) c (6 − 7c)(2 − 7c) d (u + 4v)2 e (6r − s)(2r + 5s) f (3u − 2t)(3u + 2t) g 2(h − 8k)(h + 8k) h −3(m − 6n)2

Factorising quadratic expressions

2C

Factorising is the reverse process to expanding. It involves writing an expression as a product of two or more factors. Four methods of factorising will be considered.

highest common factor (hCF) Always look for a common factor first. Take out the highest common factor from every term in the expression and place it in front of the expression. This makes the factorisation process simpler if further factorisation is required. For example, 4x2 + 8x − 12 can be factorised to 4(x2 + 2x − 3). We would then use inspection (see below) on the bracketed quadratic.

Factorisation by inspection A quadratic polynomial of the form ax2 + bx + c is called a quadratic trinomial. It should first be checked to see if it is a perfect square: a2x2 − 2abx + b2 = (ax − b)2 or a2x2 + 2abx + b2 = (ax + b)2 If not, factorisation is carried out by inspection or ‘trial and error’. This method involves finding factors of a × c that add up to b, i.e. the coefficient of the x term.

perfect squares a2x2 + 2abx + b2 = (ax + b)2 A test for a perfect square is as follows: 1. Arrange the expression in order of decreasing powers of x. 2. Does first term coefficient × last term × 2 = middle term coefficient? (Coefficients are numbers or variables in front of x2 and x terms.) 3. If yes, you have a perfect square.

For example: 4x 2 + 20x + 25 Double 2

10 Multiply (2 × 5)

difference of two squares (dotS) ax2 − b =

(

ax + b

For example, 44

49x2

)(

)

a x − b where a and b may simplify to a rational number.

− 9 = (7x + 3)(7x − 3).

Maths Quest 11 Mathematical Methods CAS

5

Worked example 4

Factorise the following. a −6x2y + 15xy b x2 + 7x + 12 2 d 27x − 75 e 9x2 − 30x + 25

c 6x2 + 28x − 48

think

Write

a −6x2y + 15xy

a 1 Write the expression. 2

Take out a common factor of −3xy. Make the common factor negative so the leading term inside the brackets will be positive.

= −3xy(2x − 5)

b x2 + 7x + 12

b 1 Write the expression. 2

Look for a common factor. (There isn’t one.)

3

Write (x )(x ) and look for factors of +12 that add to give +7. +4 and +3 achieve this.

= (x + 4)(x + 3) c 6x2 + 28x − 48

c 1 Write the expression. 2

Is there a common factor? Yes (2).

3

Attempt to factorise by inspection. Write 2(3x and try factors of −24. A few possibilities are: (3x + 12)(x − 2) = 3x2 + 6x − 24 NO (3x + 1)(x − 24) = 3x2 − 71x − 24 NO (3x + 4)(x − 6) = 3x2 − 14x − 24 NO (Nearly, just a wrong sign on the x term.) (3x − 4)(x + 6) = 3x2 + 14x − 24 YES

= 2(3x2 + 14x − 24) )(x

)

= 2(3x − 4)(x + 6) d 27x2 − 75

d 1 Write the expression. 2

Look for a common factor. There is one (3).

= 3(9x2 − 25)

3

Recognise the difference of squares in the brackets.

= 3(3x + 5)(3x − 5) e 9x2 − 30x + 25

e 1 Write the expression. 2

There are no common factors.

3

There seem to be a few square numbers in the expression, which looks suspiciously like a perfect square. The square root of the first term is 3x, and the square root of the last term is 5 or −5. Since we need a negative middle term, take −5. Double the product of these is 2 × 3 × −5 = −30, which is the middle term, so we have a perfect square.

= (3x − 5)2

Worked example 5

Factorise: a (x + 3)2 − 24

b (x − 6)2 + 5(x − 6) + 6.

think

Write

a 1 Write the expression, and recognise a difference of

squares. 2

( (

) second term )

a (x + 3)2 − 24

first term + second term ×

= ( x + 3) + 24 ( x + 3) − 24

first term −

= x +3+2 6 x +3−2 6

(

)(

)

Chapter 2 • Quadratic functions

45

b (x − 6)2 + 5(x − 6) + 6

b 1 Write the expression, and notice the linear expression

(x − 6), which is in the usual position of the variable.

2

Let X = (x − 6) and rewrite the expression. Note: This is often called a substitution method.

Let X = (x − 6) X2 + 5X + 6

3

Factorise the new version of the expression.

= [X + 2][X + 3]

4

Replace X with x − 6.

= [(x − 6) + 2][(x − 6) + 3]

5

Simplify.

= (x − 4)(x − 3)

Worked example 6

Factorise the following using an appropriate method. a −6x2y + 15xy b 50 − 2(x − 4)2 think

Write

a −6x2y + 15xy

a 1 Remove the common factor of 3xy. 2

= 3xy(−2x + 5)

= 3xy(5 − 2x) or −3xy(2x − 5)

Simplify.

b 50 − 2(x − 4)2 = 2(25 − (x − 4)2)

b 1 Remove the common factor of 2. 2

Factorise the expression within the brackets using the difference of two squares method.

= 2(52 − (x − 4)2) = 2(5 − (x − 4))(5 + (x − 4))

3

Simplify.

= 2(5 − x + 4)(5 + x − 4) = 2(9 − x)(1 + x)

exercise 2C

Factorising quadratic expressions

1 We4a Factorise the following. a 5x2y3 + 20xy2 b −8ax2 + 14ax 4 3 2 d 22r s + 11r s e 5x2 + 10x + 75 2 We4b, c Factorise the following. a x2 + 16x + 63 b x2 − 17x + 66 − 2 d 3x − 6x + 24 e −x2 − 4x + 96 2 g 6x − 37x + 45 h −40x2 + 46x + 14

c 2x2 + 22x + 48 f 6x2 − 11x − 35

3 We4d Factorise the following. 2 a 4x − 9 b 81x2 − 25 d 18x2 − 162 e (x + 1)2 − 4 g 98 − 2(x + 3)2 h −3(5x − 8)2 + 27

c 2x2 − 32 f (3x − 4)2 − 1

4 We4e Factorise the following. a x2 + 14x + 49 b 25x2 − 10x + 1

− 12x + 12 2 g 4x − 4 5 x + 5 d

3x2

−72x2

+ 24x − 2 2 h 4x − 12 2x + 18 e

5 We5a Factorise the following. a (x + 3)2 − 9 b (x − 3)2 − 16 2 d (3x − 2) − 81 e 2(x + 1)2 − 8 2 g 50(3x + 2) − 98 h 1 − (6 − x)2 46

c 7p2q − 21p + 7 f −3x2 − 18x + 6

Maths Quest 11 Mathematical Methods CAS

c 2x2 + 12x + 18 f

x2 + 2 3 x + 3

c 36 − (2x + 7)2 f −3(x − 4)2 + 48 i 64 − (2 − 3x)2

6 We5b Factorise the following. a (x + 7)2 + 9(x + 7) + 20 c (2x + 7)2 − 9(2x + 7) + 8

b 6(x − 3)2 + 13(x − 3) + 5 d 12(3x − 11)2 − 19(3x − 11) − 18

7 We6 Factorise the following, using an appropriate method. a 3x2 − 24x − 27 b 25x2 + 10x + 1 2 c (x − 13) − 2(x − 13) + 1 d x2 − 28x + 196 2 e 18 x − 6 x f 60x2 + 40x + 5 2 g 60x − 5x h 9 − 9x2y2

4(3x − 1)2 − (x + 2)2 j −12x2 + 70x + 98 8 mC a Which of the factors below could be multiplied by (x + 5) to get x2 + 14x + 45? a (x − 19) B (x + 5) C (x + 9) d (x + 14) e (x + 31) i

b The factorised form of 6x2 − 67x − 60 is: a 2(3x + 5)(x + 6) d (6x − 5)(x + 12)

B 2(3x − 5)(x − 6) e (6x + 5)(x − 12)

c A factor of 6x2 − 54 is: a (x − 3)

B (x − 6)

C (3x + 6)(2x + 10)

C (x − 9)

d (6x − 9)

e 6x

9 Quadratic polynomials are said to be identical if they have the same factorised forms and the same

expanded forms; the relevant coefficients must be equal. Suppose that P(x) = −5(x + 3)2 + 35(x + 3), Q(x) = 60 − 5x2 + kx and R(x) = −5(x + 3)(x + m). Use written algebra to find the values of k and m, if it is known that the three polynomials are identical. Use a mental arithmetic substitution technique to verify your answers.

2d

Factorising by completing the square

Factorising by inspection works best when whole numbers are involved — for example (2x + 3)(x − 7). Sometimes it is clear that it is impossible to find whole numbers to complete a factorisation. For example, with x2 + 6x − 1, there is no pair of factors of 6 that add to −1. In such cases, completing the square may be used to factorise a quadratic. The method of completing the square involves ‘manufacturing’ a perfect square so that an expression may be factorised as a difference of squares. Because this method often produces surds, factorising this way is sometimes referred to as ‘factorising over R’, where R is the set of real numbers that includes surds. Worked example 7

Use the method of completing the square to factorise the following over R. a x2 + 6x − 1 b x2 − 7x + 8 think

a 1 Write the expression.

Write

a x2 + 6x − 1

x 2 + 6 x + ( 62 )2 − ( 62 )2 − 1

2

Halve and square the x-coefficient (6), and then add and subtract it from the equation. Since the same value has been added and subtracted, the expression is equivalent to that in the question.

3

Evaluate the brackets.

= x2 + 6x + 9 − 9 − 1

4

Combine the first three terms as a perfect square, as x2 + 6x + 9 = (x + 3)2.

= (x + 3)2 − 10

5

Recognise a difference of two squares.

= ( x + 3) + 10 ( x + 3) − 10

6

Simplify.

= x + 3 + 10 x + 3 − 10

(

)(

)

Chapter 2 • Quadratic functions

47

b x2 − 7x + 8

b 1 Write the expression. 2

Halve and square the x-coefficient (−7), and then add and subtract it from the equation.

3

Evaluate the brackets.

4

Combine the first three terms as a perfect square, 7 2. since x 2 − 7 x + 49 4 = (x − 2 )

5

Combine the last two terms.

x 2 − 7x + (

) − ( −27 )2 + 8

−7 2 2

49 = x 2 − 7 x + 49 4 − 4 +8 8 = ( x − 72 ) − 49 4 +1 2

32 = ( x − 72 ) − 49 4 + 4 2

= ( x − 72 ) − 17 4 2

6

Recognise a difference of squares.

= ( x − 72 ) +

7

Simplify the surd part if possible.

= x − 72 +

8

Use a common denominator for the last two terms if desired.

2

(

17 2

(

17 4

( x − 7 )2 − 2

)( x −

7 2

)(

−

17 2

or x − 7 −2 17 x − 7 +2 17

17 4

)

)

Worked example 8

Consider the quadratic polynomials represented by P(x) = x2 + bx + 4, where b can be any real number. Answer the following questions. a Factorise P(x). b Find the linear factors of P(x), if they exist for b = 3, 4, 5 and 6. Discuss the differences in the formats of your answers. think

a 1 Write the expression.

Write

a x2 + bx + 4

b2 b2 = x 2 + bx + − + 4 2 2

2

Complete the square: halve and square the x-coefficient (b), and then add and subtract it in the expression.

3

Recognise the first three terms as a perfect square.

b 2 b2 = x + − + 4 2 4

4

Simplify.

b 2 b 2 − 16 = x + − 2 4

5

Recognise a difference of two squares.

b b 2 − 16 b b 2 − 16 =x+ + x+ − 2 4 2 4

6

Simplify the surds.

b b 2 − 16 b b 2 − 16 =x+ + x+ − 2 2 2 2

7

Put all the terms over a common denominator (2).

8

State the factors.

2 x + b + b 2 − 16 2 x + b − b 2 − 16 = 2 2 The factors of P(x) are x+

48

Maths Quest 11 Mathematical Methods CAS

± b 2 − 16 + b . 2

2 x + b ± b 2 − 16 or 2

b 1 Evaluate the factors for b = 3.

b x+

± 32 − 16 + 3 2

32 − 16 =

−7

2

Evaluate the surd.

3

Since the surd is the square root of a negative number, there are no possible linear factors.

There are no linear factors for b = 3.

4

Evaluate the factors for b = 4.

x+

5

Simplify.

6

State the factors.

When b = 4, x + 2 is a repeated factor, so P(x) is a perfect square.

7

Evaluate the factors for b = 5.

x+

8

Simplify.

± 4 2 − 16 + 4 2

± 0 +4 2 =x+2 = x+

± 52 − 16 + 5 2

= x+

± 25 − 16 + 5 2

± 9 +5 2 ±3 + 5 = x+ 2 8 2 = x + or x + 2 2 = x + 4 or x + 1 = x+

State the factors.

The factors are x + 4 and x + 1.

10

Evaluate the factors for b = 6.

x+

11

Simplify.

9

± 62 − 16 + 6 2

= x+

± 36 − 16 + 6 2

= x+

± 20 + 6 2

= x+

±2 5 + 6 2

= x ± 5+3 12

State the factors.

The factors are x + 5 + 3 and x − 5 + 3.

13

Note the differences in the number and format of the linear factors; when b = 3, there were no linear factors. Write your observations.

b = 3: There are no linear factors. b = 4: x + 2 is a repeated linear factor so P(x) is a perfect square. b = 5: The linear factors are x + 1 and x + 4. b = 6: The linear factors are x + 5 + 3 and x − 5 + 3; they are found by completing the square.

Chapter 2 • Quadratic functions 49

Exercise 2D

Factorising by completing the square

1 WE7a Use the method of completing the square to factorise the following over R. a x2 + 4x − 3 b x2 + 10x + 20 2 c x + 6x + 7 d x2 + 2x − 7 e x2 + 8x + 13 f x2 − 4x − 1 g x2 − 12x + 19 h x2 − 2x − 5 2 i x − 8x + 10 j x2 − 6x − 4 2 WE7b Use the method of completing the square to factorise the following over R. a x2 + 3x + 1 b x2 + 5x − 3 2 c x − 7x + 2 d x2 − x − 1 2 e x + 9x + 4 f x2 + 11x − 6 2 g x − 3x + 5 h x2 + 5x + 2 2 i x − 13x − 1 j x2 + x − 3 3 MC Which of the following could be added to and subtracted from x2 + 36x + 1 to assist factorising

using completing the square? A 6 b 18

c 72

d 144

4 MC Which of the following could be added to and subtracted from

using completing the square? A 4 b 17 4 5 MC The expression

x2

c 16

+ 6x − 2 factorises to:

( x − 3 + 2 )( x − 3 − 2 ) c ( x + 3 + 7 )( x + 3 − 7 ) e ( x + 3 + 6 )( x + 3 − 6 ) A

x2

e 324

− 17x − 8 to assist factorising

d 17

e 289

2

(x + 3 + d ( x − 3 + b

)( 10 )( x − 3 −

4

) 10 )

11 x + 3 − 11

6 MC The expression x2 + 4x + 12: A b c d e

can be factorised using whole numbers can be factorised using the method of completing the square cannot be factorised using the methods covered in this topic so far is a linear expression, and so cannot be factorised is already factorised

7 Challenge: Factorise the following. a 3x2 − 18x − 3 c −5x2 − 10x + 15

b 2x2 + 10x + 4 d −12x2 + 4x − 8

Solving quadratic equations — Null Factor Law

2E

As a quadratic equation is a degree 2 polynomial (highest x power of 2), it will have at most two solutions. To solve the quadratic equation x2 − 4x + 3 = 0, factorise by inspection. x2 − 4x + 3 = 0 (x − 1)(x − 3) = 0 or A × B = 0 where A = (x − 1) and B = (x − 3). For A × B to equal zero, either A or B or both must be zero. This is known as the ‘Null Factor Law’. That is, A = 0 or B=0 So (x − 1) = 0 or (x − 3) = 0 Solving these two equations gives: x − 1 = 0 or x−3=0 x = 1 or x=3 50 Maths Quest 11 Mathematical Methods CAS

Worked example 9

Solve the following. a (x + 5)(7x − 11) = 0 c 4x2 + 25 = 20x

b 12x2 − 11x − 15 = 0 d 6x2 = 54

think

a 1 Write the equation. 2

Note that the equation is already factorised.

3

Apply the Null Factor Law and solve two equations.

b 1 Write the equation. 2

Factorise by inspection. This may take several attempts.

3

Solve two equations.

c 1 Write the equation. 2

Rearrange (transpose) so all terms are on the side on which the x2 term is positive. Write terms in order of decreasing powers of x.

3

Factorise (in this case by recognising a perfect square).

4

Solve (one equation here, or two identical equations if you like).

d 1 Write the equation.

Write

a (x + 5)(7x − 11) = 0

x+5=0 x = −5

or or

7x − 11 = 0 7x = 11 x = 11 7

or or or

3x − 5 = 0 3x = 5 x = 53

or or

x−3=0 x=3

b 12x2 − 11x − 15 = 0

(4x + 3)(3x − 5) = 0 4x + 3 = 0 4x = −3 − x = 43 4x2 + 25 = 20x

c

4x2 − 20x + 25 = 0

(2x − 5)2 = 0 2x − 5 = 0 2x = 5 x = 52 6x2 = 54

d

2

Rearrange.

6x2 − 54 = 0

3

Take out the common factor, 6.

6(x2 − 9) = 0

4

Notice a difference of squares.

6(x + 3)(x − 3) = 0

5

Solve.

x+3=0 x = −3

Worked example 10

The height of a triangle is 5 cm more than its base length. If the area of the triangle is 18 cm2, find the base length and height. think 1

Define the length and height in terms of a variable. Here, x is used.

2

Draw a diagram.

Write

tUtorial eles-1407 Worked example 10

Let base length = x Then height = x + 5 x+5 x

Chapter 2 • Quadratic functions

51

3

Use the given information about area to form an equation involving the variable.

Area = 12 bh 18 = 12 (x)(x + 5) 36 = x(x + 5) 36 = x2 + 5x x2 + 5x − 36 = 0

4

Rearrange and form a quadratic equation.

5

Factorise.

(x + 9)(x − 4) = 0

6

Solve.

x = −9 or x = 4

7

Discard the negative answer, as length in this context must be positive.

x=4

8

Calculate the height (x + 5).

h=x+5 =9

9

Write the answer in words.

Base length = 4 cm, height = 9 cm.

(as x > 0)

Worked example 11

Consider the quadratic equations below. Equation 1: (2x + 1)(x − 3) = −4x Equation 2: (2x + 1)(x − 3) = −4x2 Use a calculator to determine which equation(s) has rational solutions. think

Write

Equation 1 (2x + 1)(x − 3) = −4x 2x2 − 5x − 3 = −4x

1

Expand the brackets.

2

Put all the terms on the left-hand side (add −4x to both sides).

3

Factorise.

(2x − 3)(x + 1) = 0

4

Apply the Null Factor Law.

2x − 3 = 0 3 x=2

5

Evaluate the solutions.

Both solutions are rational.

2x2 − x − 3 = 0

x+1=0 x = −1

or or

Equation 2 1

Expand the brackets.

2

Put all the terms on the left-hand side (add −4x2 to both sides).

3

Divide all the terms by 6.

4

Complete the square.

(2x + 1)(x − 3) = −4x2 2x2 − 5x − 3 = −4x2 6x2 − 5x − 3 = 0 x2 − 56 x − 12 = 0 x2 − 5 x + 6

25 144

25 − 144 − 12 = 0

+ 72 =0 ( x − ) − 25144 (x − ) − ( ) = 0 5 2 12

5 2 12

52

5

97 12

97 12

2

5

Apply the Null Factor Law.

x − 12 +

6

Evaluate the solutions.

Both solutions are irrational. Only equation 1 has rational solutions.

Maths Quest 11 Mathematical Methods CAS

= 0 or x − 12 −

97 12

5

=0

Fixed point iteration Fixed point (or simple) iteration is a way of solving equations numerically rather than algebraically. To use this method, the equation to be solved must be manipulated into the form x = g(x). For example, if the equation to be solved is f (x) = x2 − 7x + 1 = 0, a possible manipulation is x2 − 7x + 1 = 0 x2 + 1 = 7x x2 + 1 =x 7 In this case, we have x = g(x), where g( x ) = Fixed point iteration works as follows: 1. Make an initial guess, say x = 1.

x2 + 1 . 7

2. Substitute the guess into g(x).

g( x ) =

x 2 + 1 12 + 1 2 = = 7 = 0.2857 7 7

3. Substitute the improved guess, 0.2857.

g( x ) =

x 2 + 1 0.28572 + 1 = = 0.1545 7 7

4. Substitute the improved guess, 0.1545.

g( x ) =

x 2 + 1 0.15452 + 1 = = 0.1463 7 7

5. Substitute the improved guess, 0.1463.

g(x) =

x 2 + 1 0.14632 + 1 = = 0.1459 7 7

6. Repeat the process until successive values for g(x) are equal (within a tolerance of, say, 0.0001).

g( x ) =

x 2 + 1 0.14592 + 1 = = 0.14589 7 7

7. In this example, the equal successive values of g(x) are 0.1459 and 0.14589, within the tolerance of 0.0001.

x = 0.1459 is a solution to x2 − 7x + 1 = 0.

Solving quadratic equations — null Factor law exercise 2e

Solve the following.

1 We9a

a (x + 6)(2x + 3) = 0

b (x + 1)(6x − 1) = 0

c (x − 5)(x + 2) = 0

d (7x + 12)(9x + 2) = 0

e (16x + 8)(2x − 6) = 0

f

g (6 − 7x)(x + 6) = 0

h

−3x(x

2 We9b Factorise and solve the following. a x2 + 6x − 72 = 0 b 2x2 − 7x − 4 = 0 d 2x2 + 12x − 110 = 0

e 3x2 − 27 = 0

g 12 −

h 15x −

75x2

3 We9c, d

=0

d

=1

g 32x2 = 162 j

f

32x2 − 16x + 2 = 0

=0

b 6x2 + 7x = 49 e

−32x

=

6x2

+ 10

h 5x2 = 8x

+5 =x 13

6x 2

c 1 + 8x = −16x2 f i

27 + 12x2 = 36x 10 − x x2 = 2

The solutions to the equation (x − 3)(x + 7) = 0 are: x = and x = −7 x = −3 and x = 7 x = 3 and x = 7 x = 3 and x = −7 x = 4 and x = −21

4 mC a B C d e

3x2

c 4x2 + 13x + 10 = 0

diGital doC doc-9714 Quadratic expressions

Rearrange and solve the following.

a x2 + 45 = 14x

36x2

(7 + x)(23 − x) = 0

− 1) = 0

−3

Chapter 2 • Quadratic functions

53

5 mC a

−2 13

Which of the following is a solution of 2x2 − 11x = 13? B

− 13 2

2 C 13

d 1

e 13 2

The solutions to a quadratic equation are x = −9 and x = 25 . The equation could be: a (x − 9)(x − 25 ) = 0 B 5(x − 9)(x − 2) = 0 C 2(x + 9)(x − 5) = 0 d (x + 9)(5x − 2) = 0 e (x + 9)(2x − 5) = 0 x 7 We10 The width of a rectangle is 3 cm less than its 2 length. If the area of the rectangle is 40 cm , find the length and width. 8 We11 A rectangular piece of metal with length 6 cm and width 4 cm is to be enlarged by increasing both of these 4 cm dimensions by x cm as shown at right. a If the value of x is 2, describe the change in area that occurs. b Find the exact dimensions of the enlarged rectangle if its 6 cm area is (20 2 + 32) cm2. 6 mC

x

9 The temperature, T (°C), in an office follows the equation T = 85 t 2 − 5t + 30, where t is the time (in

10

11

12 diGital doC doc-9715 WorkSHEET 2.1

13

14

15

hours) the office’s climate control system has been operating. How long does it take the temperature to reach 20 °C? A number pattern follows the rule n2 + 3n + 2; that is, the number pattern is 6, 12, 20, . . . (found by substituting n = 1, n = 2, n = 3 into the rule). a Find the next value by substituting n = 4 into the rule. b What positive n value gives the number 210 when substituted into the rule? The diagram at right demonstrates the idea of ‘rectangular’ numbers. The formula N = w(w + 1) gives the value of a rectangular number based on a shape with a width of w dots. w=1 w=2 w=3 a Which rectangular number has a N=1 N=6 N = 12 width of 6? b What is the width of the rectangular number 272? The amount of bending, B mm, of a particular wooden beam under a load is given by B = 0.2m2 + 0.5m + 2.5, where m kg is the mass (or load) on the end of the beam. What mass will produce a bend of 8.8 mm? A window washer drops a squeegee from a scaffold 100 m off the ground. The relationship between the height of the squeegee (h), in metres, and the length of time it has been falling (t), in seconds, is given by h = 100 − 5t 2. a When does the squeegee pass a window 30 m off the ground? b Exactly how long does it take for the squeegee to hit the ground? The following equations are already in the form x = g(x). Find a solution using fixed point iteration with an initial guess of x = 1. x2 + 6 x2 − 2 a x= b x= 5 3 Find a solution to each of the following (correct to 3 decimal places) using fixed point iteration. First manipulate each equation into the form x = g(x) in a similar way to that shown on page 53. a x2 − 9x + 2 = 0 b x2 + 5x − 7 = 0 2 c x − 4x + 1 = 0 d 2x2 − 11x + 8 = 0

16 Show that an alternative manipulation of x2 − 7x + 1 = 0 into the form x = g(x) is x =

7 x − 1. Does this form converge using fixed point iteration? If so, state the solution found. 17 Find a manipulation of 4x2 + x − 3 = 0 that converges using fixed point iteration. State the solution. 18 Investigate when equations of the form x2 − bx + 1 are likely to converge to a solution using x2 + 1 fixed point iteration with a manipulation of the form x = . b

54

Maths Quest 11 Mathematical Methods CAS

Solving quadratic equations — completing the square 2F

Recall that when you cannot factorise quadratics by the method of sensibly guessing whole numbers, the method of completing the square may be used. Completing the square may also be used to solve quadratic equations that don’t appear to easily factorise. Worked example 12

Solve the following, giving answers in exact (surd) form. a x2 + 10x + 2 = 0 b x2 − 7x + 1 = 0 c x2 − 8 = 0 think

a 1 Write the equation. Notice that there are no

factors of 2 that add to 10, so this equation is a candidate for completing the square.

tUtorial eles-1408 Worked example 12

Write

a

x2 + 10x + 2 = 0

x 2 + 10 x + ( 102 )2 − ( 102 )2 + 2 = 0

2

Halve and square the x-coefficient and then add and subtract it.

3

Evaluate the terms in brackets.

4

Partially simplify (the last two values) as shown.

(x2 + 10x + 25) − 23 = 0

5

Form a perfect square with the first three terms.

(x + 5)2 − 23 = 0

6

Take the constant term (in this case the −23) to the other side of the equation, remembering to change the sign.

7

Take the square root of both sides.

8

Solve for x.

b 1 Write the equation.

x2 + 10x + 25 − 25 + 2 = 0

(x + 5)2 = 23

x + 5 = ± 23 x = −5 ± 23 b x2 − 7x + 1 = 0

2

Decide that completing the square is appropriate.

3

Halve and square the x-coefficient. Add and subtract this term.

4

Evaluate the terms in the brackets.

49 x 2 − 7 x + 49 4 − 4 +1 = 0

5

Partially simplify.

49 4 x 2 − 7 x + 49 4 − 4 +4 =0

x 2 − 7 x + ( 72 )2 − ( 72 )2 + 1 = 0

45 ( x 2 − 7 x + 49 4 )− 4 = 0

6

Form a perfect square.

7

Solve for x. Note that the surd part has been simplified where possible.

( x − 72 )2 − 45 4 =0 ( x − 72 )2 =

45 4

x − 72 = ± =±

45 4 9×5 4

= ± 3 25 x = 72 ± 3 2 5 =

7±3 5 2

Chapter 2 • Quadratic functions

55

c 1 The equation may be solved for x immediately,

without completing the square. 2

Rearrange the equation.

3

Take the square root of both sides and simplify.

c x2 − 8 = 0

x2 = 8 x=± 8 = ± 4×2 = ±2 2

Worked example 13

Use the method of completing the square to solve the following equations, or to explain why there are no solutions. a 2(2x − 3)2 − 2(2x − 3) − 2 = 0 b 3x2 + 6x + 4 = 0. think

a 1 Write the equation. Notice the repeated appearance

of the linear expression (2x − 3), suggesting a substitution technique. Let a = 2x − 3.

Write

a 2(2x − 3)2 − 2(2x − 3) − 2 = 0

Let a = 2x − 3 ∴ 2a2 − 2a − 2 = 0

2

Divide all terms (i.e. both sides) of the equation by 2, the coefficient of a2.

a2 − a − 1 = 0

3

Halve and square the coefficient of a and then add and subtract it.

a2 − a +

4

Solve for a by completing the square.

(a

2

( ) − ( ) −1 = 0 1 2

2

1 2

1

2

)

1

− a + 4 − 4 −1 = 0

(a − )

1 2 2

5

−4=0

(a − )

1 2 2

=

5 4

1

a− 2 = ± 1

a= 2± 5

Replace a with 2x − 3, and solve for x.

1

2x − 3 = 2 ±

1

5 2

1

5 4

x = 32 + 4 ±

6

Write the answer.

Maths Quest 11 Mathematical Methods CAS

5 4

The solution to 2(2x − 3)2 − 2(2x − 3) − 2 = 0 is x=

56

5 2

5 2

2x = 3 + 2 ±

x = 47 ±

5 4

7± 5 . 4

b 1 Write the equation. 2

Divide all terms of the equation by 3, the coefficient of x2.

3

Apply the process of completing the square. Notice here that the next step, finding the square root of both sides of the equation, is not possible as negative does not exist. Note: You will soon learn a way to predict when this will happen.

4

3x2 + 6x + 4 = 0

b

Write the answer.

x 2 + 2x + 4 = 0 3

x 2 + 2 x + ( 22 )2 − ( 22 )2 + 43 = 0 x 2 + 2 x + 1 − 1 + 43 = 0 ( x + 1)2 + 13 = 0

−1

( x + 1)2 =

3

There are no solutions to 3x2 + 6x + 4 = 0.

Solving quadratic equations — completing the square exercise 2F

diGital doCS doc-9716 SkillSHEET 2.3 Solving equations in the complete square form doc-9717 SkillSHEET 2.4 Simplifying surds

1 We12a Solve, giving the solution in exact (surd) form. a x2 + 8x + 1 = 0 b x2 + 12x + 3 = 0 2 c x + 4x − 2 = 0 d x2 − 6x + 4 = 0 e x2 − 10x + 18 = 0 f x2 − 6x + 6 = 0 2 We12b Solve, giving the solution in surd form. a x2 + 3x − 1 = 0 b x2 + 5x + 2 = 0 2 c x − 7x + 5 = 0 d x2 − 9x − 2 = 0 2 e x + 11x + 4 = 0 f x2 − x − 6 = 0

Rearrange and solve, giving the solution in exact form. b x2 + 2 = 6x d 4 − x2 = 7x f x2 − 3(5x − 2) = 0 x 2 + 3x − g 14x − x2 = −1 h = 2 4 4 We12c Solve. 2 a x − 14 = 0 b 6 − x2 = 0 c 3x2 = 36 d −2x2 + 18 = 0 3

a x2 = 4x + 1 c 9x − 2 = x2 e 2(3x + 5) = x2

When completing the square to solve x2 + 16x + 1 = 0, the perfect square part of the expression will be: a 8 B 16 C 64 d (x + 8) e (x + 8)2

5 mC

When completing the square to solve x2 − 20 x − 4 = 0, the value to be simultaneously added and subtracted is: a 4 B 10 C 5 d 100 e 20

6 mC

7 mC

An equation that is being solved using completing the square is at the stage shown below. (x − 9)2 − 15 = 0

The solution is: a (x − 9)2 = 15

B x = ± 24

d x = 9 ± 15

e

C x = − 9 ± 15

x = 3 ± 15

8 We13 For each of the following, find exact solutions (if they exist) or prove there are no solutions. a 5x2 − 2x + 3 = 0 b 3(2x − 5)2 + 6(2x − 5) − 4 = 0 c 4x2 − 2x = 6 d 5(4x + 3) = (4x + 3)2 + 9 Chapter 2 • Quadratic functions

57

2G

The quadratic formula

An alternative to the methods of factorising by inspection or completing the square is to use the quadratic formula. The derivation of the formula follows, and is based on the method of completing the square, but all you have to remember is the formula in the last step. ax2 + bx + c = 0 (where a ≠ 0) Consider the general quadratic equation. b c Divide every term by a. x2 + x + = 0 a a Use the completing-the-square method. 2 2 b b2 b2 c b x2 + x + 2 − 2 + = 0 Halve and square to get b = b a 4a 4a a a 2a 4a2 and add and subtract it to the equation. b b2 b2 c x2 + x + 2 = 2 − Rearrange. a 4a 4a a The left side is now a perfect square. Get 2 2− x + b = b 4 ac a common denominator on the right side. 2a 4a2 Take the square root of both sides. Remember b b 2 − 4 ac x+ =± to take both positive and negative square roots. 2a 4a2 x+

Simplify the RHS denominator.

b ± b 2 − 4 ac = 2a 2a

b from both sides. 2a

x=

Write using the one denominator (2a).

x=

Subtract

−b

2a −

±

b 2 − 4 ac 2a

b ± b 2 − 4 ac 2a

The following examples show how the quadratic formula may be used. Worked Example 14

Use the quadratic formula to solve the following without a calculator. a 3x2 − 8x − 9 = 0 b x2 + 5x + 6 = 0 Give answers in exact (surd) form and comment on the usefulness of this formula for each example. Think

a 1 Write the equation and match up a, b and c.

Write

c a a b

3x2 − 8x − 9 = 0

2

Write the quadratic formula.

x=

3

Substitute the matched values for a, b and c.

x=

4

Simplify.

x=

5

At this stage, it is tempting to cancel 2 out of the 8 and 6, but both terms of the numerator must possess this factor, and initially they don’t. However, in simplifying 172 into 4 × 43 , the factor of 2 emerges; then the cancelling is possible in step 6.

x=

6

Cancel down a common factor of 2 from the numerator and denominator.

x=

58 Maths Quest 11 Mathematical Methods CAS

−b ±

b 2 − 4 ac 2a

− ( − 8) ± +8 ±

(− 8)2 − (4 × 3 × − 9) 2×3

64 + 108 6

8 ± 172 6 8 ± 4 × 43 x= 6 8 ± 2 43 x= 6 4 ± 43 3

4 + 43 4 − 43 or x = 3 3

7

Write the two solutions separately.

x=

8

In the original equation, the coefficient of x2 was 3; also, there are surds in the answer. These facts suggest that neither completing the square nor standard factorisation would have been straightforward methods.

Using the quadratic formula was the most appropriate method for solving this equation.

b 1 Write the equation and match up a, b and c.

ba

b c 1x2 + 5x + 6 = 0

2

Write the quadratic formula.

x=

3

Substitute the matched values for a, b and c.

x=

4

Simplify.

x=

−b ±

b 2 − 4 ac 2a

−5 ±

52 − (4 × 1 × 6) 2 ×1

x= x= x= x =

−

5 ± 25 − 24 2

−5±

1

2 −5±1

2 −5+1

2 −

or x =

−5−1

2

−

4 6 or x = 2 2

x = − 2 or x = − 3 5

In the original equation, the coefficient of x2 was 1; also, there are no surds in the answer. This suggests that an easier factorisation method (i.e. inspection) would be more suitable. Note: This applies only if the question does not specify a particular method.

Using the quadratic formula is not the most suitable method for solving this equation. Factorisation of the left side in the format (x + a)(x + b) would be more efficient.

Worked example 15

Use the quadratic formula to solve 2x2 − 2kx + 4k = 0, and use your solution to solve 2x2 + 6x − 12 = 0. think

Write

1

Recognise a common factor of 2 (divide each term by 2).

2x2 − 2kx + 4k = 0 x2 − kx + 2k = 0

2

Identify a, b and c for the quadratic formula.

a = 1, b = −k, c = 2k

3

Substitute the values for a, b and c into the quadratic formula.

x=

k ± ( − k )2 − 4 × 1 × 2 k 2 ×1

4

Simplify.

x=

k ± k 2 − 8k 2 Chapter 2 • Quadratic functions

59

5

Compare 2x2 − 2kx + 4k = 0 to 2x2 + 6x − 12 = 0.

6

Substitute k = −3 into the solution found in step 4 and simplify.

−2k

= 7 and 4k = −12, so k = −3.

( − 3)2 − 8 × − 3 2 −3 ± 9 + 24 x= 2 −3 ± 33 x= 2 x=

−3 ±

Worked example 16

Consider the set of quadratic equations in the form 4x2 + bx + 9 = 0, where b is 11, 12 or 13. Use the quadratic formula to explain the solutions that occur with each b-value. think

Write

1

Write the equation and match up a, b and c.

a b c 4x2 + bx + 9 = 0

2

Substitute the values for a, b and c into the quadratic formula.

x=

3

Simplify.

x=

4

Substitute b = 11 and simplify.

x= x= x=

−b ±

b2 − 4 × 4 × 9 2×4

−b ±

b 2 − 144 8

− 11 ±

112 − 144 8

− 11 ±

121 − 144 8

− 11

− 23

± 8

5

Evaluate the result.

There is no solution, as we cannot evaluate

6

Substitute b = 12 and simplify.

x=

− 12 ±

122 − 144 8

− 12 ±

144 − 144 8 − 12 ± 0 x= 8 −3 x= 2 x=

7

Evaluate the result.

There is one solution, x =

8

Substitute b = 13 and simplify.

x= x= x= x=

60

Maths Quest 11 Mathematical Methods CAS

− 13 ±

132 − 144 8

− 13 ±

169 − 144 8

− 13

± 25 8

− 13 ± 5

8

−3 2

.

− 23

.

9 10

and x = −1.

There are two solutions, x =

The b2 values have had a clear impact on the solutions; recall that ± 0 = 0 and ± 25 = ±5.

The structure of the quadratic formula means the changing value of the expression under the square root sign, b2 − 144, is critical. b = 11: There was no solution (as negative cannot be resolved). b = 12: There was one solution (as ± 0 = 0). b = 13: There are 2 solutions (as ± 25 = ±5).

exercise 2G 1

−9

Evaluate the result.

4

the quadratic formula

Identify a, b and c (the coefficients of x2, x and the constant respectively) in each of the following quadratic equations. a x2 + 4x − 3 = 0 b x2 − 7x + 9 = 0 2 c 9 − 4x + x = 0 d 7 − 3x − 6x2 = 0 2 1 e 2 x − 7x + 5 = 0 f (x + 1)2 = 0 g (2x − 3)2 = 0 h 5 − 2(x2 + 2) = 0 − 6x 2 + 4 i 3x2 − 10x + 4 + 4x2 − 11x = 0 j =0 2 Use the quadratic formula to solve the following (even though the equations may be factorised by inspection). a x2 + 9x + 20 = 0 b x2 − 10x + 16 = 0 c 2x2 − 13x − 24 = 0 d −4x2 + 13x − 3 = 0

2 We14

3 We15 Find exact (surd) solutions to each of the following. a x2 + 5x + 3 = 0 b x2 + 8x + 5 = 0 2 c x + 3x + 1 = 0 d x2 + 10x + 12 = 0 e x2 − 6x + 2 = 0 f x2 − 7x + 6 = 0 g x2 − 4x − 2 = 0 h x2 − 9x − 8 = 0 − 2 i 2x + 3x + 1 = 0 j −4x2 + 12x − 1 = 0 − 2 k 2kx + 4x + 6k = 0 l 2kx2 − (k + 1)x + 4 = 0 4 a Use a CAS calculator and your answer for question 3k above to solve x2 + 4x − 3 = 0, and write the

value of k you used to do this.

b Use a CAS calculator and your answer for question 3l above to solve 62x2 − 32x + 4 = 0, and write

the value of k you used to do this. 5 Use a CAS calculator to find approximate solutions (3 decimal places) to each of the following. a x2 − 6x − 2 = 0 b x2 + 3x − 9 = 0 c −2x2 + 7x + 1 = 0 d −9x2 − 2x − 2 = 0 e −x2 − 8x + 1 = 0 f 3x2 + x + 9 = 0 6 Use a CAS calculator to find approximate solutions to the following. a x2 + 6x = 11 b 2x2 = 7 − 4x − 2 c 10x + 2 = 5x d x2 = 8x − 6 2 e 5 = 9x − 2x f x2 − 2 = 7x + 4 2 g 5x + 6x + 2 = 0 h −x2 + 4x = 8 7 The population of a colony of rare African ants is given by the equation N = x2 + 2x + 300, where N is

the number of ants, and x is the height of the anthill in centimetres. How high could the anthill be when there are 850 ants in the colony? 8 The profit, $P, made when a particular make of car is manufactured in t hours is given by the equation

P = 40t2 − 7t − 5000. In order to just ‘break even’, how long must the manufacturing process take?

9 The position of an item x metres from the start of an assembly line is given by x = −4t2 + 20t + 5 where t

is the time in minutes elapsed since the item started on the line. When is the item at position x = 0 metres on the assembly line? Chapter 2 • Quadratic functions

61

10 A golf ball is hit from the bottom of a bunker as shown

at right. The height, h metres, of the ball above the ground is given by h = 5.2t − 1.5t2 − 1, where t seconds is the time the ball has been in flight. a How deep is the bunker? b When is the ball first level with the top of the bunker? c At which times is the ball at a height of 3 metres? 11 Debbi plans to pave sections of her backyard, as shown at right. She has a total of 12 square metres of pavers. a Find the value of x if she wishes to use her entire supply of 5 pavers. She then finds that the value of x (found above) will need to be rounded either up or down to a multiple of 0.5 m. Calculate the effect this will have on her existing supply of pavers if she rounds: b up c down. 12 The surface area (in m2) of cement transport containers made by a certain company is given by 4π r2 + 24π r, where r is the radius of the container. If the surface area of a particular container is 60 m2, determine its radius. 13 The Gateway Arch in St Louis, Missouri, was designed by Eero Saarinen in 1948 but not completed until 1964. The line of the arch can be represented by the quadratic function 2 (x − 95)2 + 190 where y is the height y = − 95 of the arch at a horizontal distance x along the base (ground level) from one side of the arch. Both x and y are in metres. a Re-express this function in an expanded form. b How wide is the arch at its base? c This is the tallest arch in the world. How high is the Gateway Arch?

2H

h

xm

4m

xm

Sloppo Cement Co.

r

The discriminant

You may have found on occasions that no solutions (or ‘roots’) can be found for a quadratic equation. If you were using the quadratic formula at the time, you would have found the trouble started when you tried to evaluate the square root part of the formula. The expression under the square root sign is called ‘the discriminant’. The discriminant is used to determine how many roots of an equation exist and is denoted by the upper case Greek letter ‘delta’ (Δ). −

b ± b 2 − 4 ac Δ = b2 − 4ac 2a The quadratic formula The discriminant If Δ < 0 (that is, negative), then no real solutions exist, as we cannot yet evaluate the square root of a negative value. The word ‘real’ is used to describe numbers we can deal with at present. The set of real numbers includes positives, negatives, fractions, decimals, surds, rationals (numbers that may be expressed as a ratio — for example 49 ) and irrationals. (In further studies of maths, you will learn about a way of dealing with square roots of negative numbers using what are known as ‘imaginary numbers’.) If Δ > 0, then the discriminant can be evaluated, and its square root can be added and subtracted in the quadratic formula (see above) to produce two real solutions. If Δ = 0, then there is nothing to add or subtract in the quadratic formula, and so there is only one real solution to evaluate.

62 Maths Quest 11 Mathematical Methods CAS

x=

Worked example 17

Find the value of the discriminant for the equation 3x2 − 5x + 2 = 0. think

Write

1

Write the expression. Note the value of a, b and c.

3x2 − 5x + 2 = 0 a = 3, b = −5, c = 2

2

Write the formula for Δ.

Δ = b2 − 4ac

3

Substitute for a, b and c.

Δ = (−5)2 − 4(3)(2)

4

Evaluate.

Δ = 25 − 24 Δ=1

Worked example 18

How many real solutions are there to the equation −7x2 + 3x − 1 = 0? think

Write

1

Write the expression. Note the value of a, b and c.

−7x2

2

Write the formula for Δ.

Δ = b2 − 4ac

3

Substitute for a, b and c.

Δ = (3)2 − 4(−7)(−1)

4

Simplify.

Δ = 9 − 28 = −19

5

Comment on the number of solutions.

Since Δ < 0, there are no real solutions.

+ 3x − 1 = 0 a = −7, b = 3, c = −1

Worked example 19

For what values of k does −x2 + 2kx − 9 = 0 have: i two distinct solutions? ii one solution? iii no solutions? think

tUtorial eles-1409 Worked example 19

Write

1

Write the expression. Note the value of a, b and c.

−x2

2

Write the formula for Δ.

Δ = b2 − 4ac

3

Substitute for a, b and c.

Δ = (2k)2 − 4(−1)(−9)

4

Simplify.

Δ = 4k2 − 36

5

Consider case i, two distinct solutions. Note: |k| means the ‘size’ or ‘absolute value’ of k without regard to + or −. Since k2 is always positive, we require k > 3 (values such as 3.1, 4, 5 etc.) or k < −3 (values such as −3.1, −4, −5 etc.) if k2 is to be > 9.

iii Two distinct solutions, require Δ > 0.

+ 2kx − 9 = 0 a = −1, b = 2k, c = −9

So

4k2 − 36 > 0 4k2 > 36 k2 > 9 |k| > 3 k > 3 or k < −3

Chapter 2 • Quadratic functions

63

6

7

Consider case ii, one solution. The ‘size’ of k must be = 3. (That is, 3 or −3 are our only choices.)

iii One solution, require Δ = 0.

Consider case iii, no solutions. The ‘size’ of k must be < 3 (values such as −2.9, −2, −1, 0, 1, 2, 2.9 etc.). An alternative method is shown in the next example.

iii No solutions, require Δ < 0.

4k2 − 36 = 0 4k2 = 36 k2 = 9 |k| = 3 k = 3 or k = −3 k = ±3

So

4k2 − 36 < 0 4k2 < 36 k2 < 9 |k| < 3 −3 < k < 3

So

Worked example 20

For which values of k does 2x2 + 2kx + (k + 4) = 0 have: i two solutions? ii one solution? iii no solutions? think

Write/draW

1

Write the expression. Note the value of a, b and c.

2x2 + 2kx + (k + 4) = 0 a = 2, b = 2k, c = k + 4

2

Write the formula for Δ.

Δ = b2 − 4ac

3

Substitute for a, b and c.

Δ = (2k)2 − 4(2)(k + 4)

4

Simplify and factorise.

Δ = 4k2 − 8(k + 4) = 4k2 − 8k − 32 = 4(k2 − 2k − 8) = 4(k + 2)(k − 4)

5

Since Δ is a more complicated expression than those in the previous example, a graph of Δ versus k (Δ on the vertical axis, k on the horizontal axis) is useful. (Recall how you sketched quadratic graphs in previous work, or see the next section.)

Δ>0

Δ

−2

4

−32 6

7

8

64

Δ>0 Δ=0

Δ<0

For case i, we require Δ > 0 (quadrants 1 and 2) From the graph, Δ > 0 when k < −2 or k > 4.

iii Two solutions, Δ > 0.

For case ii, we require Δ = 0 (k-intercepts) From the graph, Δ > 0 when k = −2 or k = 4.

iii One solution, Δ = 0.

For case iii, we require Δ < 0 (quadrants 3 and 4) From the graph, Δ < 0 when −2 < k < 4. Note: This method involving sketching a graph of Δ may be used as an alternative to the method shown in the previous example.

iii No solutions, Δ < 0.

Maths Quest 11 Mathematical Methods CAS

So k < −2 or k > 4.

So k = −2 or k = 4. So −2 < k < 4.

k

Worked example 21

Find an expression for the discriminant of the equation x2 + (2k +1)x + 4k = 0. Use this result to: a find exact and approximate values of k for which the equation has no solution b explain the result obtained in part a through substitution of appropriate k-values into the equation. think

Write

1

Identify the coefficients required.

x2 + (2k + 1)x + 4k = 0 a = 1, b = 2k + 1, c = 4k

2

Write the formula for Δ.

Δ = b2 − 4ac

3

Substitute for a, b and c.

Δ = (2k + 1)2 − 4 × 1 × 4k

4

Simplify.

Δ = 4k2 + 4k + 1 − 16k = 4k2 − 12k + 1

a 1 For the original equation to have no solution,

we require Δ < 0.

a 4k2 − 12k + 1 < 0

12 ± ( − 12)2 − 4 × 4 × 1 2×4 12 ± 144 − 16 = 8 12 ± 128 = 8

2

Solve the left-hand side using the quadratic formula.

k=

3

Simplify.

k=

12 ± 8 2 8

k = 32 ± 2 4

Interpret the result for k.

If 32 − 2 < k < 32 + 2 , then Δ < 0 and the equation has no solutions.

5

Evaluate approximate values for k.

If 0.086 < k < 2.914, then Δ < 0 and the equation has no solutions.

b 1 A selection of values for k will produce Δ < 0,

Δ = 0 and Δ > 0. Find a k-value that gives Δ > 0.

b Δ = 4k2 − 12k + 1

Let k = −2 Δ = 4(−2)2 − 12 × −2 + 1 Δ = 16 + 24 + 1 Δ = 41 Δ > 0, so there are two solutions.

2

Find a k-value that gives Δ < 0. (Pick a value in the range from step 4 of part a.)

Let k = 1 Δ = 4(1)2 − 12 × 1 + 1 Δ = 16 + 24 + 1 Δ = −7 Δ < 0, so there are no solutions.

3

From step 3 of part a, we know the k-values for which Δ = 0.

3 If k = 2 ± 2 , Δ = 0, so there is one solution.

4

Evaluate the result.

If 3 − 2 < k < 3 + 2 , there are no solutions 2 2 to the original equation.

Chapter 2 • Quadratic functions

65

exercise 2h

the discriminant

1 We17 Find the value of the discriminant in each case. a x2 + 9x + 2 = 0 b x2 − 4x − 1 = 0 2 d 2x − 3x + 10 = 0 e −3x2 + x + 3 = 0 2 g x + 15x = 1 h 9 − 7x = 4x2

c 5x2 + 6x − 7 = 0 f −x2 − 2x − 6 = 0 i −3x2 = 5

2 We18 How many real solutions are there to the following equations?

Do not actually work out any solutions. diGital doC doc-9718 Calculating the discriminant

b −x2 + 4x + 4 = 0 e −2x2 − 8x − 8 = 0 h 6 − 6x = x2

a 5x2 + x + 2 = 0 d 3x2 + 6x + 3 = 0 g 5x2 = 2 − x

c 3x2 − 3x + 1 = 0 f 9 − x2 + x = 0 i 12x = 9x2 + 4

3 We19 Find an expression for the discriminant of each of the following. a x2 + ax + 1 = 0 b ax2 + 2x + 3 = 0 c x2 + 6x + a = 0 d ax2 + bx + 1 = 0 e mx2 + 2mx + 1 = 0 f x2 + (m + 1)x + 3 = 0 2 2 g x − mx − (m + 4) = 0 h (k − 1)x − kx + 2 = 0 4 mC Which values of m below both result in the expression x2 − mx + 5 = 0 having two

distinct solutions? a −20 and 3

B 0 and 20

C 4 and 5

5 We19, 20 For what values of k does each equation have: i two distinct solutions? ii one solution? a x2 + kx + 4 = 0 b x2 − 4x + k = 0 2 d kx − 18x + 20 = 0 e x2 − 4x + (k + 1) = 0 2 g 4kx + 12kx + 9k = 0 h (k + 4)x2 + 10x + 5 = 0

d −6 and −5

iii c f i

e 0 and 5

no solutions? x2 + 4kx + 4 = 0 6x2 + 4kx + (k + 3) = 0 (k − 1)x2 − (k + 1)x + 2 = 0

6 We21 Assume k = −1 in each of questions 5 a–i. Use substitution and your previous answers to question 5 to complete the following table. (There is no need to actually solve the equations.)

Question

Equation after substitution k = −1

Number of solutions

Reasoning

a b c d e f g h i

Graphs of quadratic functions as power functions (turning point form) 2i

Quadratic functions are also power functions. Power functions are functions of the form f (x) = xn, n ∈ R. The value of the power, n, determines the type of function. When n = 1, f (x) = x, the function is linear. When n = 2, f (x) = x2, the function is quadratic. Other power functions will be discussed later. Under a sequence of transformations of f (x) = xn, n ∈ R, the general form of a power function is f (x) = a(x − b)n + c (where a, b, c, and n ∈R). All linear and quadratic polynomials are also linear and quadratic power functions, because all linear and quadratic functions are transformations of f (x) = x and f (x) = x2 respectively. When a quadratic function is written in turning point form, it is written in power form. For example, the quadratic function y = x2 + 4x + 6 can also be represented as the power function y = (x + 2)2 + 2. 66

Maths Quest 11 Mathematical Methods CAS

In previous years’ work, you will have discovered the following connections between a quadratic function in turning point (TP) form, and its graph. Summary of the sequence of transformations y = a(x − b)2 + c a is the dilation factor from the x-axis. If a > 1, the graph gets thinner or narrower. If 0 < a < 1, the graph gets wider. If a < 0, the graph is turned upside down (or, reflected in the x-axis).

x-coordinate of turning point (horizontal shift)

y

y-coordinate of turning point (vertical shift) Minimum value of y for ∪ shape graphs Maximum value of y for ∩ shape graphs

a positive, increasing

y x Turning point (b, c) x

a negative, increasingly negative

Worked Example 22

For the graph of y = −3(x + 2)2 − 1: i state the turning point coordinates ii describe the width of the graph as ‘same’, ‘thinner’ or ‘wider’ compared to the width of y = x2 iii state whether the graph is a minimum (∪) or maximum (∩) type, and state the maximum or minimum value of y iv find the y-intercept v sketch the graph (x-intercepts are not required). Think

Write/draw

y = −3(x + 2)2 − 1

1

Write the equation. Compare y = −3(x + 2)2 − 1 with y = a(x − b)2 + c a = −3, b = −2, c = −1.

2

TP x-coordinate: b = −2. TP y-coordinate: c = −1

i TP (−2, −1)

3

y dilation factor = −3 (magnitude greater than 1 means it is thinner than y = x2)

ii Thinner

4

a is negative, so the graph is ‘upside down’ or a maximum shape.

iii The graph is a maximum (∩) type.

5

Maximum y-value is c, which equals −1.

The maximum value of y is −1.

6

The y-intercept can be found by substituting x = 0 into the original expression and simplifying.

iv y-intercept: y = −3(0 + 2)2 − 1

Use the information above to sketch the graph.

v

7

= −3(2)2 − 1 = −13

y x

(−2, −1)

−13

Chapter 2 • Quadratic functions 67

If a quadratic function is not in power form or turning point form, it must be manipulated in order to answer questions like those posed in the previous example. To do this, we use the method of completing the square as demonstrated in the following example.

Worked example 23

Convert the function y = x2 − 6x + 1 to power form. State the coordinates of the turning point and the maximum or minimum value of y. think

Write

1

Write the equation. Ensure the x2 coefficient is 1. It is. (If not, divide the equation by whatever will change the x2 coefficient to 1.)

y = x2 − 6x + 1

2

Halve the x-coefficient and square it. Add and subtract this value after the x term.

y = x2 − 6x + 9 − 9 + 1

3

Group terms to make a perfect square.

y = (x2 − 6x + 9) − 8

4

Factorise the perfect square part.

y = (x − 3)2 − 8

5

Compare with y = a(x − b)2 + c.

a = 1, b = 3, c = −8

6

The turning point is (b, c).

The turning point is (3, −8).

7

a is positive, so there is a minimum value of y = −8 when x = 3.

The minimum value of y is −8.

Worked example 24

Identify the coordinates of the turning point of the graph of y = 2x2 + x − 4 using algebraic methods. think

Write

y = 2x2 + x − 4 1 = 2[ x 2 + 2 x − 2]

1

Write the rule for the function. Begin the process of completing the square by first taking out 2 as the common factor.

2

Half of shown.

3

Continue on with completing the square, within the square [ ] brackets.

32 1 = 2[( x + 14 )2 − 16 − 16 ]

4

Multiply the 2 through the square brackets, leaving it as a factor of the curved brackets.

= 2( x + 14 )2 − 33 8

5

Read the coordinates of the turning point from this form. Write your answer.

1 2

1 , so add and subtract this as is 14 , and ( 14 )2 = 16

1 1 = 2[( x 2 + 12 x + 16 ) − 2 − 16 ]

33 = 2[( x + 14 )2 − 16 ]

The turning point has coordinates − − ( 14 , 33 8 ).

Graphs of quadratic functions as power functions (turning point form) exercise 2i

1 We22 State the turning point coordinates for each of the following. a y = (x − 5)2 b y = (x + 7)2 + 3 2 c y = (x − 2) − 7 d y = (x − 1)2 + 8 e y = 2(x + 3)2 − 4 f y = −6(x − 2)2 + 2 68

Maths Quest 11 Mathematical Methods CAS

2 We22 For each of the following: i state the turning point coordinates ii describe the width of the graph as same width, thinner or wider than y = x2 iii state whether the graph is a minimum (∪) or maximum (∩) type, and state the maximum or

minimum value of y iv find the y-intercept v sketch the graph (x-intercepts are not required). a y = (x − 4)2 + 2 c y = (x + 5)2 − 8

b y = (x − 2)2 − 3 d y = (x − 1)2 − 1

e y = 2(x + 5)2 − 9

f

g y = 65 (x + 1)2 − 12

h y = −7(x − 3)2 + 3

i

y = −(x − 8)2 − 9

j

y = 13 (x − 4)2 + 4 y = (1 − x)2 + 20

3 Sketch graphs having the following properties, but do not show intercepts. a Turning point (2, 5), dilation factor of 3 from the x-axis b Turning point (−1, 3), dilation factor of 1 from the x-axis c Turning point (0, −4), dilation factor of 2 from the x-axis d Turning point (6, 0), dilation factor of −1 from the x-axis e Turning point (7, −7), dilation factor of −4 from the x-axis f Turning point (0, 2), dilation factor of 12 from the x-axis 4 Determine the function rules of the following graphs, given that they are all of the power function form

y = (x − b)2 + c (that is, dilation factors all equal to 1). a

y

b

(−1, 5)

y

c

y

4

8 (2, 4)

x

x

−1

x

(2, −5) d (−3, 1)

e

y

y

y 7

x −8

f

(6, 6) x

x

−30

(−3, −2)

Find the y-intercepts of graphs with the following properties.

5

a Turning point (1, 4), dilation factor of 2 from the x-axis b Turning point (−2, −3), dilation factor of −1 from the x-axis 6 We23 Convert each of the following to power form, and state the coordinates of the turning point and

the maximum or minimum value of y. y = x2 − 4x + 9 y = x2 − 12x + 37 y = x2 − 4x − 5 y = x2 + 18x + 0 y = 3x2 + 12x + 15

a c e g i

b d f h

y = x2 − 6x + 17 y = x2 + 8x + 13 y = x2 + 7 y = 2x2 − 12x + 22

7 We24 For each of the following quadratic functions: i use by-hand algebraic methods to find the exact coordinates of the turning point ii use a CAS calculator to verify this algebra and find decimal equivalents for these coordinates. a y = 3x2 − 2x + 1 b y = −2x2 − 5x − 4 c y = 6x2 − 3x − 4 Chapter 2 • Quadratic functions

69

8 A section of perspex for a parabolic window is drawn on a

coordinate grid as shown at right. What equation should be programmed into the automatic glass cutting machine, using the grid system on the diagram?

8 6 4 2 0

2

4

6

8

What is the effect on the graph of increasing k in the equation y = k(x − 2)2 + 1? The graph is: a raised B lowered C thinned d widened e translated left F translated right 10 mC What is the effect on the graph of increasing k in the equation y = (x + k)2? The graph is: a raised B lowered C thinned d widened e translated left F translated right 11 mC What is the effect on the graph of increasing k in the equation y = (x + 1)2 − k? The graph is: a raised B lowered C thinned d widened e translated left F translated right 9 mC

diGital doC doc-9719 WorkSHEET 2.2

12 For the graph y = M(x + h)2 + n, state: a the x-coordinate of the turning point b the y-coordinate of the turning point c the dilation factor from the x-axis d the coefficient of x2 in the expanded version e the coefficient of x in the expanded version f the constant term in the expanded version.

Graphs of quadratic functions (intercepts method) 2J

In this section we will consider graphs of quadratic functions of the form y = ax2 + bx + c. When we talk about sketching a graph, we mean drawing a diagram showing the main features — not a true scale graph showing every point plotted accurately using a computer package or other means. To sketch a quadratic graph, the following features should generally be apparent or labelled.

1 the y-intercept

y

The y-intercept is found by substituting x = 0 into the rule and solving the equation for the corresponding y-value. That is, it is the c in the general form y = ax2 + bx + c. All quadratics have a y-intercept.

y-intercept

c x

2 any x-intercepts if they exist Any x-intercepts are found by substituting y = 0 into the rule and solving the equation for the corresponding x-value. Let’s say the rule factorises to y = (x − d)(x − e). Then substituting y = 0 into the rule gives the equation 0 = (x − d)(x − e). For this to be true, we must have x = d or x = e.

70

Maths Quest 11 Mathematical Methods CAS

y x-intercepts c d

e

x

3 The turning point coordinates Consider the general quadratic function, and complete the square as follows: y = ax 2 + bx + c

y

Axis of symmetry x-intercepts

y-intercept

b c = a x 2 + x + a a

x

b b 2 b 2 c = a x 2 + x + − + 2a 2a a a

Turning point b b2 — — (− 2a , c − 4a )

b 2 b2 c = a x + − 2 + a 2a 4a b 2 b2 +c = a x + − 2a 4a b 2 b2 = a x + + c − 2a 4a You may recognise this form as turning point form. In this case, the coordinates of the turning point are b2 −b 2a , c − 4 a Using xt and yt for the coordinates of the turning point, we have −

b b2 and yt = c − 2a 4a If a quadratic function has two x-intercepts, the x-coordinate of the turning point may be found by averaging them. xt =

y

d

d+e —— 2

e

x

If there is only one x-intercept, then it is the x‑coordinate of the turning point. y

d

x

The y-coordinate of the turning point may be found using the general turning point form above, by completing the square from scratch or by substituting the x-coordinate into the original equation. Recall also the two main types of parabolas: y

y

x

Minimum parabola y = ax2 . . . (a > 0)

x

Maximum parabola y = ax2 . . . (a < 0) Chapter 2 • Quadratic functions 71

Worked Example 25

Sketch the graphs of the following, showing all intercepts and the turning point in each case. a y = x2 − 4x − 32 b y = x2 + 10x + 25 c y = −2x2 + 11x − 15 Think

a 1 Write the rule.

Write/draw

a y = x2 − 4x − 32

2

Find the y-intercept (when x = 0).

If x = 0, y = −32

3

Factorise before finding x-intercepts.

y = (x + 4)(x − 8)

4

Find the x-intercepts (when y = 0).

If y = 0, 0 = (x + 4)(x − 8) x = −4 or x = 8

5

Find the x-coordinate of the turning point (TP).

, b = −4, a = 1 2a − ( − 4) = 2(1) = 2

6

Substitute into the original rule to find the y-coordinate of the TP. Alternatively, use b2 yt = c − . 4a Write the turning point coordinates.

7 8

TP xt =

−b

yt = 22 − 4(2) − 32 = 4 − 8 − 32 = −36 TP (2, −36)

Combine the above information and sketch the graph.

y

−4

8

x

(2, −36) b 1 Write the rule.

b y = x2 + 10x + 25

2

Find the y-intercept (when x = 0).

If x = 0, y = 25

3

Factorise before finding x-intercepts.

y = (x + 5)(x + 5) = (x + 5)2

4

Find the x-intercept (when y = 0).

If y = 0, 0 = (x + 5)2 x = −5

5

Find the x-coordinate of the turning point. Alternatively, since there is only one x‑intercept (−5), it must be the turning point x-coordinate.

TP

6

Substitute into the original rule to find the y-coordinate of the turning point.

7

Write the turning point coordinates.

8

Combine the information and sketch the graph.

xt =

, b = 10, a = 1 2a − 10 = 2(1) = −5

yt = (−5)2 + 10(−5) + 25 = 25 − 50 + 25 =0 TP (−5, 0) y 25 (−5, 0)

72 Maths Quest 11 Mathematical Methods CAS

−b

x

c 1 Write the rule.

c y = −2x2 + 11x − 15

2

Find the y-intercept (when x = 0).

If x = 0, y = −15

3

Factorise before finding x-intercepts.

y = −(2x2 − 11x + 15) = −(2x − 5)(x − 3)

4

Find the x-intercepts (when y = 0).

If y = 0, 0 = −(2x − 5)(x − 3) 2x − 5 = 0 or x − 3 = 0 2x = 5 or x = 3 5 x = 2 or x = 3

5

Find the x-coordinate of the turning point.

TP xt =

−b

2a

, b = 11, a = −2 − 11

= = 6

Substitute into the original rule to find the y-coordinate of the turning point.

7

Write the turning point coordinates.

8

Sketch the graph, showing all important features.

yt = − 2

11 4

(or 2.75)

( ) + 11( ) − 15 2

11 4

− 242 + 16 − 242 = 16 + 2 = 16 1 = 8 (or

=

11

TP

2(− 2)

11 4

121 − 15 4 484 240 − 16 16

0.125)

1

( 4 , 8 ) y 5– 2

—, 1– ) (11 4 8

3

x

−15

Don’t be put off if asked to sketch a quadratic graph whose equation doesn’t have 3 terms. Such cases are easier to sketch, as the following example shows. Worked example 26

Sketch the graphs of the following equations. a y = x2 − 16 b y = −2x2 + 14x think

a 1 Write the rule.

Write/draW

a y = x2 − 16

2

Find the y-intercept (when x = 0).

If x = 0, y = 02 − 16 = −16

3

Factorise before finding x-intercepts. In this case, recognise a difference of squares.

y = (x + 4)(x − 4)

4

Find the x-intercepts (when y = 0).

If y = 0, So

0 = (x + 4)(x − 4) x = −4 or x = 4 Chapter 2 • Quadratic functions

73

xt =

−b

5

Find the x-coordinate of the turning point.

, b = 0, a = 1 2a 0 xt = 2(1) =0

6

Substitute into the original rule to find the y-coordinate of the turning point.

yt = 02 − 16 = −16

7

Write the turning point coordinates.

8

Sketch the graph, showing all important features.

TP (0, −16) y

−4

x

4

(0, −16) b 1 Write the rule.

b y = −2x2 + 14x

2

Find the y-intercept (when x = 0).

If x = 0, y = −2(0)2 + 14(0) =0

3

Factorise before finding x-intercepts. In this case, use a common factor of −2x.

y = −2x(x − 7)

4

Find the x-intercepts (when y = 0).

If y = 0, So

5

Find the x-coordinate of the turning point.

xt =

= Substitute into the original rule to find the y-coordinate of the turning point. b2 We could also use yt = c − . 4a

Write the turning point coordinates.

8

Sketch the graph.

14 4 7 2

(or 3.5)

( ) + 14 ( ) 2 ( ) + 49

yt = − 2 = = =

7

−

b , b = 14, a = − 2 2a − 14 xt = − 2( 2) =

6

0 = −2x(x − 7) x = 0 or x = 7

−

7 2

49 4

− 49 2 49 2

(

TP

2

7 2

+

98 2

(or 24.5)

7 49 , 2 2

)

y 7– 49 ,— 2 2

0

74

Maths Quest 11 Mathematical Methods CAS

7

x

Worked example 27

Sketch the graphs of the following equations. a y = 3x2 + 6x − 1 b y = −2x2 + x − 7 think

a 1 Write the rule.

Write/draW

a y = 3x2 + 6x − 1

2

Find the y-intercept (when x = 0).

If x = 0, y = −1

3

Factorising before finding x-intercepts (when y = 0) is not easily done, so use the quadratic formula.

If y = 0, 0 = 3x2 + 6x − 1 −b ±

b 2 − 4 ac 2a a = 3, b = 6, c = −1 x=

x= = = = =

−6 ±

62 − 4(3)(− 1) 2(3)

−6 ±

36 + 12 6

−

6 ± 48 6

−

6 ± 16 × 3 6

−

6±4 3 6

−

3±2 3 3 x = 0.155 or =

4

x = −2.155

xt =

−b

, b = 6, a = 3 2a − 6 xt = 2(3)

Find the x-coordinate of the turning point.

= −1 5

Substitute into the original rule to find the y-coordinate of the turning point.

6

Sketch the graph, showing all important features.

yt = 3(−1)2 + 6(−1) − 1 =3−6−1 = −4 TP (−1, −4) y −2.155

0.155 x −1

(−1, −4) b 1 Write the rule. 2

Find the y-intercept (when x = 0).

b y = −2x2 + x − 7

If x = 0, y = −2(0) + (0) − 7 = −7

Chapter 2 • Quadratic functions

75

3

Factorising before finding x-intercepts (when y = 0) is not easily done, so try the quadratic formula.

If y = 0, 0 = −2x2 + x − 7 b 2 − 4 ac 2a a = −2, b = 1, c = −7 −1 ±

x=

−1 ±

−1 ±

cannot be evaluated so there are no x-intercepts.

No x-intercepts

5

Find the x-coordinate of the turning point.

xt =

−

b 2a −

=

1 2( 2)

=

−1 −4

=

1 4

−

(or 0.25)

yt = − 2 ( 14 ) + 14 − 7 2

1 = − 2 ( 16 ) + 14 − 7 −1 8+ −1 = 8 + − 55 = 8

=

7

− 55

−4

4

Substitute into the original rule to find the y-coordinate of the turning point.

12 − 56 −4

=

6

12 − 4(− 2)(− 7) 2(− 2)

=

− 55

−b ±

x=

Sketch the graph, showing all important features.

1 4 2 8

−7 − 56 8

(or − 6.875)

y —) ( 1–4 , − 55 8

x

−7

Worked example 28

Consider the graph of y = x2 − 5x + 9. Use written algebra to: a show that the parabola has no x-intercepts b find how many units the parabola needs to be translated down, or ‘lowered’, so that it has exactly one x-intercept c find how many units the original parabola needs to be lowered so that its x-intercepts are 13 units apart. think

a 1 Write the rule for the function. Any potential

x-intercepts depend on the solutions of the equation 0 = x2 − 5x + 9. 76

Maths Quest 11 Mathematical Methods CAS

Write

a y = x2 − 5x + 9

2

Calculate the discriminant, given a = 1, b = −5 and c = 9.

a = 1, b = −5 and c = 9 Δ = (−5)2 − 4 × 1 × 9 = 25 − 36 = −11

3

The discriminant is negative. Interpret the meaning of this.

The discriminant is negative, so there are no solutions to the quadratic equation 0 = x2 − 5x + 9; that is, the parabola has no x‑intercepts.

b 1 Replace the original equation with a ‘lowered’

one. Change 9 to a variable, c.

b y = x2 − 5x + c

2

For this equation to have one x-intercept, the discriminant, Δ, must equal 0. Calculate Δ.

a = 1, b = −5 Δ = (−5)2 − 4 × 1 × c = 25 − 4c

3

Set Δ = 0 and solve for c.

25 − 4c = 0 c = 6.25

4

Write the new equation.

y = x2 − 5x + 6.25

5

Compare the new equation with the original equation.

The graph is lowered by 9 − 6.25 = 2.75 units.

c 1 Calculate the axis of symmetry of

y = x2 − 5x + 9.

c Axis of symmetry =

b = −5, a = 1 Axis of symmetry =

−b

2a − ( − 5)

2 ×1 5 = 2 = 2.5 The turning point is at x = 2.5.

2

The x-coordinate of the turning point (here, 2.5) is always halfway between any x-intercepts; the required distance between the intercepts is 13 units.

13 ÷ 2 = 6.5 2.5 + 6.5 = 9 and 2.5 − 6.5 = −4 The x-intercepts must be 9 and −4.

3

The x-intercepts of a function y = (x − d)(x − e) are d and e given a dilation factor parallel to the x-axis of 1.

The rule of the function is y = (x + 4)(x − 9).

4

Expand to find the y-intercept.

y = (x + 4)(x − 9) y = x2 − 5x − 36 The y-intercept is −36; the answer is correct as the first two terms are identical to those in the previous functions.

5

The amount of vertical translation will be the difference between the original and final y-intercepts. Answer the question.

9 − (−36) = 45 The original parabola must be lowered by 45 units.

Using the discriminant Since finding x-intercepts for a quadratic graph involves solving a quadratic equation, we can use the discriminant to decide the number of x-intercepts such a graph has. Chapter 2 • Quadratic functions 77

y

y

y x

x x

Δ>0 2 x-intercepts

Δ=0 1 x-intercept

Δ<0 No x-intercepts

Graphs of quadratic functions (intercepts method) exercise 2J

diGital doCS doc-9720 Quadratic graphs — general form doc-9721 Quadratic graphs — turning point form

1

Find the y-intercepts for each of the following. b y = −3x2 + 6x − 4 e y = 5x + 2 − 4x2

a y = x2 + 9x + 2 d y = 6 − x2 a d g j

State the x-intercepts for each of the following. y = (x − 1)(x − 6) b y = (x − 3)(x + 2) y = (x − 4)(x − 5) e y = x(x − 2) y = (x + 5)(5 − x) h y = −x(x + 8) − y = 3(2 − x)(x + 10)

a c e g

State the x-intercepts for each of the following. y = (2x − 5)(x + 1) y = (x − 9)(4x − 9) y = (2x + 3)(4x + 1) y = (Ax + a)(Bx + b)

2

3

4

c y = 4x2 + 2x f y = 1 − x − x2 c y = (x + 5)(x + 1) f y = (3 − x)(4 − x) i y = −(x + 9)2

b y = (−3x + 1)(x + 2) d y = −(x + 1)(6 − 5x) f y = 3x(x + 4)

Use the quadratic formula to find exact values (if possible) for the x-intercepts of: b y = −x2 + 7x − 7 d y = 10 − 3x + 3x2.

a y = 4x2 − 2x + 3 c y = −2x2 − 9x − 1

5 We25a, b Sketch graphs of the following, showing all intercepts and the turning point in each case.

(Hint : Factorise first.) y = x2 − 4x + 3 y = x2 + 6x + 8 y = x2 − 8x + 12 y = x2 + 3x + 2 y = x2 − 11x − 12 y = x2 − 16x + 64

a c e g i k

b d f h j l

y = x2 + 2x + 1 y = x2 + 12x + 35 y = x2 + 2x − 63 y = x2 − 5x + 6 y = x2 + 14x + 49 y = x2 + 8x − 153

6 Sketch the following graphs, showing all intercepts and the turning point in each case. a y = 3x2 + 2x − 8 b y = 5x2 + 18x − 8 c y = 3x2 − 4x − 15 d y = 4x2 − 8x + 3 e y = 8x2 − 10x + 3 f y = 7x2 + 18x − 9 2 2 g y = 15x + 48x + 9 h y = 9x − 2x − 7 i y = 2x2 + x − 28 2 2 j y = 3x + 5x + 2 k y = 2x − 3x − 9 7 We25c Sketch the graphs of the following. a y = −x2 − 8x + 33 b y = −x2 + 2x + 3 − 2 d y = x + 18x − 81 e y = −4x2 + 12x − 5

78

c y = −x2 − 18x − 45 f y = −8x2 − 6x + 5

8 We26 Sketch the graphs of the following. a y = x2 − 25 b y = x2 − 121 2 d y=3−x e y = 2x2 − 18 2 g y = x + 5x h y = x2 − 8x j y = −4x2 − 24x k y = 21x − 3x2

c y = −x2 + 1 f y = −3x2 + 12 i y = 5x2 − 10x

9 We27 Sketch the graphs of the following. a y = x2 + 2x − 7 b y = x2 + 4x + 1 d y = −4x2 + 2x + 3 e y = −x2 − 18x − 1 2 g y = x − 7x − 2 h y = 2x2 + 7x + 4 − 2 j y = 3x + 17 k y = 4x2 − 5x

c y = x2 + 8x + 3 f y = x2 − 3x + 1 i y = 3x2 − 9x − 5

Maths Quest 11 Mathematical Methods CAS

If a and b are positive numbers, which of the following graphs could be that of y = (ax + 1)(x + b)?

10 mC a

B

y

C

y

y

x x x y

d

e

y

x x

If k and p are positive numbers, which of the following could be the graph of y = x2 − 2kx + p?

11 mC

y

a

B

y

x

x

y

d

e

x

12

y

C

x

y x

Without sketching, determine how many x-intercepts each of the following graphs have. b y = −3x2 − 4x + 8 d y = 4x2 + x + 9 f y = 4x2 + 72x + 324

a y = x2 + 37x + 208 c y = −9x2 + 78x − 169 e y = −x2 − 12x − 35

13 Consider the value of the discriminant in deciding which graph matches which equation below. a y = 5x2 + x + 1 b y = 6x2 − 2x − 1 c y = −7x2 − x + 2 d y = −4x2 + 8x − 4 a

y

B

y x

x

C

d

y

y x

x 14 We28b Use a CAS calculator and written algebra to find how many units and in which direction the

graph of y = 2x2 − 3x − 2 should be translated vertically so that it has exactly one x-intercept. Also, state the function rule for this new parabola in both expanded and turning point forms.

15 We28c Use a CAS calculator and written algebra to find how many units and in which direction the

graph of y = −3 (x + 2)2 − 4 should be translated vertically so that its x-intercepts are exactly 4 units apart. Also, state the function rule for this new parabola in both expanded and turning point forms.

Chapter 2 • Quadratic functions

79

Using technology to solve quadratic equations 2k

Quadratic expressions, equations and functions are linked closely, as you have previously seen. Study the following table carefully. Classification

Characteristics

Examples

Quadratic expression

Degree 2 polynomial, no equals sign Can be written in expanded and/or factorised form

4x2 − 3.4x + 2 (x − 7)(x + 3) x2 + 5x − 6

Quadratic equation

One variable, one equals sign 0, 1 or 2 solutions exist

4x2 − 3.4x + 2 = 19 (x − 7)(x + 3) = 0 x2 + 5x − 6 = −4

Quadratic function

Two variables, one equals sign in the rule Can be sketched, as it is a set of ordered pairs Sketch is a parabola, with 0, 1 or 2 x-intercepts

y = 4x2 − 3.4x + 2 y = (x − 7)(x + 3) y = x2 + 5x − 6

Note: The solutions (also known as the roots) of a quadratic equation (say, 3x2 − 4.3x − 1.68 = 0) are identical to the x-intercepts of its related parabola (here, the sketch of y = 3x2 − 4.3x − 1.68). They are also known as the zeros of the related expression (here, 3x2 − 4.3x − 1.68). Can you see why the solutions of x2 + 5x − 6 = −4 become the x-intercepts of the parabola given by y = x2 + 5x − 2? Worked example 29

Consider the expressions x2 − 8x + 12, x2 − 10x + 21, and x2 − 12x + 32. Use written algebra and/or a CAS calculator to: a find the zeros and factors of the expressions b find the turning points of their related parabolas c use the patterns seen in these answers to predict the next three and previous two quadratic expressions, along with the features of their related parabolas. Now consider a general quadratic function (variable x) whose graph is an upright parabola with a dilation factor from the x-axis of 1. Its x-intercepts are j and k. d Find the rule for this quadratic function. e Find the function rule for a second parabola that has been translated 3 units to the left of the original parabola. f Verify your results for d and e (algebraically and graphically) by letting j = −5 and k = 2. think

a 1 Factorise x2 − 8x + 12.

a x2 − 8x + 12 = (x − 2)(x − 6)

Set the factorised expression equal to 0 and solve.

(x − 2)(x − 6) = 0 x−2=0 or x=2 or

3

Factorise x2 − 10x + 21.

x2 − 10x + 21 = (x − 3)(x − 6)

4

Set the factorised expression equal to 0 and solve.

(x − 3)(x − 7) = 0 x−3=0 or x=3 or

5

Factorise x2 − 12x + 32.

x2 − 12x + 32 = (x − 4)(x − 8)

6

Set the factorised expression equal to 0 and solve.

(x − 4)(x − 8) = 0 x−4=0 or x=4 or

2

80

Write/draW

Maths Quest 11 Mathematical Methods CAS

x−6=0 x=6

x−7=0 x=7

x−8=0 x=8

b 1 The x-value of the turning point is halfway

between the roots (zeros) of the associated parabola.

2+6 =4 2 For x2 − 10x + 21, 3+ 7 TP x-value = =5 2 For x2 − 12x + 32, 4+8 TP x-value = =6 2

TP x-value =

2

Find the y-values of each turning point by substituting the x-values in the corresponding functions.

For x2 − 8x + 12, 42 − 8 × 4 + 12 = −4 For x2 − 10x + 21, 52 − 10 × 5 + 21 = −4 For x2 − 12x + 32, 62 − 12 × 6 + 32 = −4

3

State the turning points of each parabola.

The turning points of these three parabolas are (4, −4), (5, −4) and (6, −4).

c 1 Predict the next three expressions based upon the patterns from parts a and b. 2

Predict the previous two expressions.

3

The intercepts and turning points are moving to the right (horizontally) by 1 unit for each step in the progression. This means that only the x-coordinates change. Also, the coefficient of the x term in the expanded expression is the negative sum of the zeros, and the constant term is the product of the zeros.

d 1 Assign variables to the x-values of the zeros

c (x − 5)(x − 9)

(x − 6)(x − 10) (x − 7)(x − 11) (x − 1)(x − 5) (x − 0)(x − 4) Expression

Zeros

TP

Factors

x2 − 4x

0, 4

(2, −4)

(x)(x − 4)

x2 − 6x + 5

1, 5

(3, −4)

(x − 1)(x − 5)

x − 8x + 12

2, 6

(4,

−4)

(x − 2)(x − 6)

x2 − 10x + 21

3, 7

(5, −4)

(x − 3)(x − 7)

x2 − 12x + 32

4, 8

(6, −4)

(x − 4)(x − 8)

x2 − 14x + 45

5, 9

(7, −4)

(x − 5)(x − 9)

x2 − 16x + 60

6, 10

(8, −4) (x − 6)(x − 10)

x2 − 18x + 77

7, 11

(9, −4) (x − 7)(x − 11)

2

d Let j and k be the x-values of the zeros

of each quadratic function.

(x-intercepts) of the quadratic function.

2

State the general function.

y = (x − j)(x − k)

3

Expand.

y = x2 − ( j + k)x + jk

e 1 When the parabola moves 3 units to the left,

3 is subtracted from both zeros (j and k).

f

b For x2 − 8x + 12,

e The zeros are now j − 3 and k − 3.

2

State the new function.

y = (x − ( j − 3))(x − (k − 3))

3

Expand.

y = x2 − ( j + k − 6)x + ( j − 3)(k − 3)

1

Given j = −5 and k = 2, state the quadratic function.

2

Translate this function 3 units to the left by subtracting 3 from j and k.

f y = (x + 5)(x − 2)

y = x2 + 3x − 10

j − 3 = −5 − 3 = −8 k − 3 = 2 − 3 = −1 Chapter 2 • Quadratic functions

81

3

State the new function.

4

Sketch the graphs of y = x2 + 3x − 10 and y = x2 + 9x + 8 on the same set of axes.

y = (x + 8)(x + 1) y = x2 + 9x + 8 y

−8

−5

(TP)

5

Observe the features of the two parabolas.

−1

y = x2 + 9x + 8 y = x2 + 3x − 10

2

x

(TP)

The turning points and zeros have been translated 3 units to the left. The second parabola has been translated 3 units to the left from the first parabola. The answers are verified.

Using technology to solve quadratic equations exercise 2k

1 Use a CAS calculator to find all solutions of each of the following. a x2 + 4x + 1 = 0 b x2 = 11x + 2 − 2 c x + 4x = 2 d x2 = 8x − 8 2 e x + 12x + 9 = 0 f −3x2 + 5x − 1 = 0

diGital doC doc-9722 Simultaneous, quadratic and linear equations

2 Find all roots of the quadratics below using a CAS calculator. a 2.3x2 + 0.7x − 0.59 = 0 b −0.811x2 − 5.2x − 3.1 = 0 c −3.97x2 + 17x + 8.05 = 0 d 5.18x2 = 2.66x + 9 e 0.006x2 + 0.923x + 0.361 = 0 f x2 + 500x = 47 3 The distance, d, of a comet from one of the moons of Jupiter is given by the equation

d = 47.9t2 + 0.03t − 908.7, where t is the number of hours since the comet was first discovered on 28 June 2001. At what value of t will the comet reach this moon?

4 The number of marine organisms, N, in a marine research organisation’s testing tank is found to follow

the equation (or model) N = −0.0751h2 + 0.69h + 200, where h is the number of hours since the tank was supplied with nutrient and stocked with 200 organisms. How long after being fed could the colony survive without further food before none were left?

82

Maths Quest 11 Mathematical Methods CAS

5 A diver follows a parabolic path from the diving board to the water, given by the function below. What

is the horizontal distance travelled by the diver from leaving the diving board to entering the water? (h represents the height of the diver above the water for a distance, d, from the diving board in the equation shown. Both h and d are in metres.)

h = −0.5d2 + 2d + 5

6 We29 As part of a Year 11 Maths extension activity, Harry and Christine were each asked by their

teacher to come up with a quadratic expression whose zeros had a sum of 10.25 (or as close to that as possible). The coefficient of the linear term had to be −5. When they returned to class the next day, Harry announced his expression was 0.48x2 − 5x − 2.3, and Christine said hers was 0.49x2 − 5x + 1.9. Who was closest, and by how much? 7 One of the solutions of the equation 10x2 + 11x = k (where k is a constant) is −1.6. a Find the value of k. b Find the other solution to the original quadratic equation.

Simultaneous quadratic and linear equations 2l

In previous studies you have dealt with pairs of simultaneous linear equations and solved these using algebra. The solution could also be represented graphically. The same is true when we have one linear and one quadratic equation as a pair of simultaneous equations. Consider the following pair of simultaneous equations: y = x2 + x − 2 and y = 3x + 1 If x = 3 is substituted into the first equation, y = 32 + 3 − 2 = 10 is obtained. If x = 3 is substituted into the second equation, y = 3 × 3 + 1 = 10 is obtained. That is, the coordinate pair (3, 10) ‘fits’ both equations, so it is a solution. If x = −1 is substituted into the first equation, y = (−1)2 + (−1) − 2 = −2 is obtained. If x = −1 is substituted into the second equation, y = 3(−1) + 1 = −2 is obtained. That is, the coordinate pair (−1, −2) is also a solution. To illustrate the situation graphically, the related linear and quadratic functions must be sketched on the same set of axes. As shown below, there are three possible relationships: • no points in common (hence, no solutions) • one point in common (hence, one solution) • two points in common (hence, two solutions).

interaCtiVitY int-0261 Simultaneous quadratic and linear equations

Chapter 2 • Quadratic functions

83

y

y

x

y

One solution

No solution

x x

Two solutions

When a linear equation and a quadratic equation are solved simultaneously, a new quadratic equation is formed, as you will see in the following examples. The number of solutions relates to the discriminant (positive = 2 solutions; negative = no solutions; zero = one solution) of the ‘new’ quadratic. Worked Example 30

a Solve the system of equations y = x2 + x − 2 and y = 3x + 1. b Illustrate the solution using a sketch graph. The turning point of the quadratic graph is not

required.

Think

a 1 Write the original equations and label them.

Write/draw

a y = x2 + x − 2

[1] [2]

y = 3x + 1

2

Equate [1] and [2]. Put RHS [1] = RHS [2].

3

Collect terms on the side that makes the x2 term positive.

4

Combine like terms.

x2 − 2x − 3 = 0

5

Factorise if possible.

(x − 3)(x + 1) = 0

6

Solve for x.

7

Substitute x-values into the linear equation [2] to find the corresponding y-values.

If x = 3, y = 3(3) + 1 = 10 If x = −1, y = 3(−1) + 1 = −2

8

Write the solution coordinates.

Solutions: (3, 10) and (−1, −2)

b 1 Find intercepts for a sketch of the linear graph.

2

Find intercepts for a sketch of the quadratic graph. (The turning point is not required here.)

3

Sketch the linear and quadratic graphs on the same axes, and include the points of intersection found in part a.

x2 + x − 2 = 3x + 1 x2 + x − 2 − 3x − 1 = 0

x = 3 or x = −1

b For y = 3x + 1,

if x = 0, y = 1 if y = 0, 0 = 3x + 1 −1 = 3x −1 x = 3

For y = x2 + x − 2, if x = 0, y = −2 if y = 0, 0 = x2 + x − 2 so 0 = (x + 2)(x − 1) and x = −2 or x = 1 y (3, 10)

− 1–3 −2 (−1, −2)

84 Maths Quest 11 Mathematical Methods CAS

1 1 −2

x

Worked example 31

Solve the equations y = −0.5x2 − 4x + 2 and 11x + 2y = 6. think 1

Write the two equations and label them.

2

Solve [2] for y.

Write

y = −0.5x2 − 4x + 2 11x + 2y = 6

[1] [2]

2y = −11x + 6 y = −5.5x + 3 −0.5x2

[3]

− 4x + 2 = −5.5x + 3

3

Set [1] = [3].

4

Simplify.

5

Multiply by −2.

6

Factorise and state the solutions.

7

Find the corresponding y-values and substitute in [3].

When x = 2, y = −5.5(2) + 3 = −8 When x = 1, y = −5.5 + 3 = −2.5

8

Write the coordinates of the two points of intersection.

The solutions are (2, −8) and (1, −2.5).

−0.5x2

+ 1.5x − 1 = 0

x2 − 3x + 2 = 0 (x − 2)(x − 1) = 0 x = 2 or x = 1

Worked example 32

a Solve the system of equations y = x2 − 8x + 12 and y = −6x + 11. b Illustrate the solution using a sketch graph. The turning point of the quadratic graph

is not required.

think

a 1 Write the original equations and label them.

Write/draW

a y = x2 − 8x + 12

[1] [2]

y = −6x + 11

x2 − 8x + 12 = −6x + 11

2

Put RHS [1] = RHS [2].

3

Collect terms on the side that makes the x2 term positive.

4

Combine like terms.

x2 − 2x + 1 = 0

5

Factorise if possible.

(x − 1)(x − 1) = 0

6

Solve for x.

7

Substitute x-values into the linear equation [2] to find the corresponding y-value.

If x = 1,

8

Write the solution coordinates.

Solution: (1, 5) only

b 1 Find intercepts for a sketch of the linear graph.

x2 − 8x + 12 + 6x − 11 = 0

x = 1 only y = −6x + 11 y = −6(1) + 11 y = −6 + 11 y=5

b For y = −6x + 11,

if x = 0, if y = 0,

y = 11 0 = −6x + 11 6x = 11 5 x = 11 6 or 1 6 Chapter 2 • Quadratic functions

85

2

Find intercepts for a sketch of the quadratic graph. (The turning point is not required here.)

3

Sketch the linear and quadratic graphs on the same axes, and include the points of intersection found in part a.

For y = x2 − 8x + 12, if x = 0, y = 12 if y = 0, 0 = x2 − 8x + 12 so 0 = (x − 6)(x − 2) and x = 6 or x = 2 y 12 11

Tangent point (1, 5) 11 — 6

2

x

6

Worked example 33

a Solve the system of equations y = −x2 + 3x + 18 and y = 4x + 22. b Illustrate the solution using a sketch graph. The turning point of the quadratic graph is not

required.

think

a 1 Write the original equations and label them.

Write/draW

a y = −x2 + 3x + 18 −x2

2

Put RHS [1] = RHS [2].

3

Collect terms on the side that makes the x2 term positive.

4

Combine like terms.

5

Try to factorise. No solution using the Null Factor Law is apparent. Check the value of the discriminant Δ of the quadratic in step 4.

(x )(x ) = 0? Δ = b2 − 4ac = (1)2 − 4(1)(4) = 1 − 16 = −15

6

Since Δ < 0, there is no solution.

Δ < 0, ∴ no solution; the graphs don’t meet.

b 1 Find intercepts for a sketch of the linear graph.

+ 3x + 18 = 4x + 22 0 = x2 − 3x − 18 + 4x + 22 x2 + x + 4 = 0

b For y = 4x + 22,

if x = 0, if y = 0,

y = 22 0 = 4x + 22 −22 = 4x x= x=

2

86

[1] [2]

y = 4x + 22

Find intercepts for a sketch of the quadratic graph. (The turning point is not required here.)

Maths Quest 11 Mathematical Methods CAS

− 22 4 − 11 2

1

or −5 2

For y = −x2 + 3x + 18, if x = 0, y = 18 if y = 0, 0 = −x2 + 3x + 18 x2 − 3x − 18 = 0 so (x − 6)(x + 3) = 0 and x = 6 or x = −3

3

y

Sketch the linear and quadratic graphs on the same axes. Note that the graphs do not intersect, indicating no solution.

22 18

— − 11 2

−3

6

x

Worked example 34

The graphs with equations y = x2 + 4 x + 33 and y = mx + 24 intersect once only. Find the possible values of m. think

Write

1

Write and label the equations.

y = x2 + 4x + 33 y = mx + 24

2

Put equation [1] = equation [2] and form a new quadratic equation [3].

x2 + 4x + 33 = mx + 24 + 4x − mx + 9 = 0 x2 + (4 − m)x + 9 = 0 [3]

3

For one solution only, we require Δ = 0.

Δ = (4 − m)2 − 4(1)(9) = 16 − 8m + m2 − 36 = m2 − 8m − 20 = 0 for one solution only.

4

Factorise and solve for m.

(m − 10)(m + 2) = 0 m = 10 or m = −2

5

State the rules for the two straight lines.

The two lines are y = 10x + 24 and y = −2x + 24.

[1] [2]

x2

Simultaneous quadratic and linear equations exercise 2l

1 We30, 31, 32, 33 For each of the following systems of equations: i solve to find any solution coordinates ii illustrate the solution (or lack of solution) using a sketch graph. The turning point of the quadratic

graph is not required. You may use a CAS calculator to verify solutions. a y = x2 + 6x + 5 and y = 11x − 1 c y = x2 + 9x + 14 and y = 3x + 5 e y = x2 − 2x − 3 and y = x − 6 g y = x2 + 5x − 36 and y = 15x − 61 i y = x2 − 2x − 24 and y = 4x + 3 k y = −x2 + 4x + 21 and y = x + 11 m y = −x2 + 4x + 12 and y = 9x + 16 o y = −x2 − 4x + 5 and y = −4x + 9

b y = x2 + 5x − 6 and y = 8x − 8 d y = x2 − 7x + 10 and y = −11x + 6 f h j l n p

y = x2 + 11x + 28 and y = 10x + 40 y = x2 − 6x − 16 and y = −4x − 17 y = x2 − 7x + 10 and y = −4x + 6 y = −x2 + 14x − 48 and y = 13x − 54 y = x2 + 7x + 12 and y = 20 y = x2 − 4x + 4 and y = 8x − 32

2 State how many points of intersection exist with each of the following pairs of simultaneous equations. a y = x2 + x − 6 and y = −9x − 31 b y = x2 − 16 and y = 6x + 11 c y = −x2 + 3x + 4 and y = −7x + 25 d y = x2 − 6x + 5 and y = 2x − 12 Chapter 2 • Quadratic functions

87

3 WE34 The graphs of equations y = x2 − 14x + 49 and y = mx + 48 intersect once only, at x = 1. Find

the value of m. 4 The system of equations y = x2 − 4x + c and y = −7x + 8 has two solutions, one at x = 1 and another at x = −4. Find the value of c. 5 The graphs of y = x2 + bx − 14 and y = −9x + c intersect at (−1, −8) and (−3, 10). Find the values of b and c. 6 Using a CAS calculator, find the points of intersection, in exact form, for each of the following systems of equations. a y = x2 + 5x − 3 and y = 2x − 2 b y = 2x2 + 7x − 8 and y = 3x + 4 c y = −3x2 − 2x + 12 and y = −4x + 8 d y = −5x2 − 2x − 5 and y = 43 x − 5 y

7 An engineer’s plans for a proposed road through a

mountain are shown at right. At what heights above sea level will the entrance and exit to the tunnel be, given the equations of the mountain profile and road path as shown on the plan?

y=−

x–2 +5 2

y = x–4 +2 Proposed road

Entrance of tunnel

x

8 A graphic designer draws a logo involving a parabola ‘sitting’ in a

y

V shape on a set of axes as shown at right. Find the equation of the parabola, given it is of the form y = kx2, and find the points of intersection of the V with the parabola. −1 −2 9 Use a CAS calculator to find the values of a in exact form such that the linear equation y =

and the quadratic equation y =

88 Maths Quest 11 Mathematical Methods CAS

−3 8

x 2 + 3 x − 5 do not intersect.

Sea level

y = kx 2

1

ax +a 2

x

10 Use a CAS calculator to help answer the following questions.

A pole 11 metres high is firmly secured to the ground. A parabolic arch is to be attached to the pole 3 metres above the ground as shown in the diagram below. A metal rod will run directly from the top of the pole straight to the ground so that it just touches the arch in one position only. Let y be the height above the ground in metres and x be the distance along the ground from the foot of the pole in metres. y (0, 11)

(0, 3) 0

A

B

x

Give all answers in exact form unless told otherwise. −1 a a The arch is to be modelled by the quadratic equation y = 8 x 2 + 4 x + 3, where a is a non-zero positive constant. Find where the arch meets the ground in terms of a, that is, find the x-intercept. b If it is decided that the distance from the foot of the pole to the point A is 12 metres, find the value of the constant a. c If the equation of the metal rod is modelled by the linear function y = 11 + bx, find the value of b. (Hint: Consider the discriminant.) d Find the coordinates of the point where the arch touches the metal rod. e Find the maximum height of the arch correct to 2 decimal places. f Find the distance from point A to point B both in exact form and correct to 2 decimal places.

Chapter 2 • Quadratic functions 89

Summary polynomials

1. A polynomial in x, sometimes denoted by P(x), is an expression containing only non-negative whole powers of x. 2. The degree of the polynomial is given by the highest power of the variable x. 3. The general form of a degree n polynomial is: P(x) = an x n + an − 1xn − 1 + . . . + a2 x 2 + a1x + a0 where n is a positive whole number and an, an − 1, . . ., a2, a1 and a0 are coefficients.

expanding quadratic expressions

• To expand: 2 1. First term × everything in the second brackets, then 1 2. Second term × everything in the second brackets. (4x + 9)(2x − 3) • Perfect squares (ax + b)2 = a2x2 + 2abx + b2 3 4 • Difference of squares (ax + b)(ax − b) = a2x2 − b2 • Expand brackets first, then multiply if there is an ‘external’ factor in expressions like k(ax + b)(cx + d).

Factorising quadratic expressions

• • • •

Look for a common factor first. General quadratics: Write down (_x )(_x ) and try factors of the constant term. Perfect squares: (ax + b)2 = a2x2 + 2abx + b2 Difference of squares: ( a x + b )( a x − b ) = ax 2 − b

Factorising by completing the square

• • • •

Use when whole number factors are not apparent. Halve and square the x-coefficient, then add and subtract this new term. Form a perfect square from three of the terms. Continue to factorise using a difference of squares.

Solving quadratic equations — null Factor law

• Factorise. • Set each factor equal to zero. • Solve two mini-equations.

Solving quadratic equations — completing the square

• Halve and square the x-coefficient. • Add and subtract the new term. • Form a perfect square and solve for x.

the quadratic formula

• If ax2 + bx + c = 0, then solutions may be found using x =

the discriminant

• • • •

If ax2 + bx + c = 0, then the discriminant Δ = b2 − 4ac. If Δ < 0, no real solutions exist. If Δ > 0, there are two real solutions. If Δ = 0, there is only one real solution.

Graphs of quadratic functions as power functions (turning point form)

• • • • •

Turning point form: y = a(x − b)2 + c. The turning point is at (b, c) and the dilation factor is a. If a > 0, the graph is a positive (∪) shape. If a < 0, the graph is a negative (∩) shape. To convert to turning point form, complete the square.

90

Maths Quest 11 Mathematical Methods CAS

−b ±

b 2 − 4 ac . 2a

y

(b, c) x

Graphs of quadratic functions (intercepts method)

• To sketch a quadratic graph of the form y = ax2 + bx + c: 1. Find the y-intercept (when x = 0) 2. Factorise if possible, and find the x-intercepts (when y = 0) It may be easier to use the quadratic y formula in some cases. If Δ < 0, there are no x-intercepts. If Δ = 0, one intercept only. y-intercept If Δ > 0, two intercepts. 3. Find the x-coordinate of the turning point −b using xt = or by completing the 2a square.

Axis of symmetry x-intercepts x

Turning point b b2 — — (− 2a , c − 4a )

4. Find the y-coordinate of the turning point by substituting xt into the equation for y, or by using b2 or by completing the square. yt = c − 4a 5. Combine all the information and sketch it. Functions with a positive x2 coefficient are ∪ shaped, and those with negative x2 coefficients are ∩ shaped. If the information you have gathered doesn’t seem to fit, check for calculation errors. Simultaneous quadratic and linear equations

• If given equations of the form y = ax2 + bx + c and y = mx + k: 1. Set ax2 + bx + c = mx + k. 2. Rearrange to form a new quadratic equation Ax2 + Bx + C = 0. 3. Solve to find any x-coordinates of intersection. 4. Substitute any x-coordinates into the linear equation to find the corresponding y-coordinates. 5. State the solutions (or state that there are none).

Chapter 2 • Quadratic functions

91

Chapter review S h ort anS Wer

1 Expand: a (7x + 8)(7x − 8) 2 Factorise: a 25x2 + 110x + 121

b (2x − 9)2

c (5 x − 3)(5 x + 3)

d (4x − 9)(2x − 13)

b 6x2 + 37x + 6

c 12x2 − 37x + 21

d 36x2 − 49

3 Factorise x2 − 6x − 14. 4 Solve the following. a (4x + 1)(3x − 9) = 0

b 3x2 − 40x − 75 = 0

5 Solve: a 4x2 − 5 = 0

b x2 − 14x + 42 = 0

6 Solve x2 + 6x − 2 = 0. 7 Use the quadratic formula to solve 5x2 − 7x + 1 = 0. 8 Evaluate the discriminant for 4x2 − 3x + 9 = 0. 9 Find the value(s) of k for which the equation 2x2 + 3kx + 6 = 0 has: a no solution b one solution c two solutions. 10 Sketch the following, showing the turning point and y-intercept in each case (x-intercepts not required). a y = −4(x + 1)2 + 2 b y = 6x2 − 1 11 Convert y = x2 + 8x + 3 to turning point form. 12 Sketch the following. a y = (x − 6)(x − 14)

b y = x2 + 2x − 80

c y = 3x2 − 26x + 48

d y = −2x2 − 5x − 3

13 Find any points of intersection of the line y = −2x − 6 and the parabola y = x2 + 7x + 12. 14 Solve 4kx2 − 5x + 3k = 0 for x. m U lt ip l e C h oiCe

1 Which of the following is the expansion of (3x − 7)(x + 5)? a 3x2 + 8x − 35 d 4x2 − 2x − 2

B 3x2 − 7x − 35 e 4x − 2

2 Which of the following is a perfect square quadratic expression? a x2 + 10x + 16 d (x + 6)(x − 6)

B x2 − 8x + 16 e (x2 + 7)2

3 The expression 9x2 − 64 is an example of: a a perfect square d a quadratic term

B a difference of squares e a factorised expression

4 Which of the following is equivalent to 36x2 − 49? a (6x + 7)2 d (36x + 1)(x − 49)

B (6x − 7)2 e (9x + 7)(4x − 7)

5 The expression (x − 2)2 + 8 can be factorised to:

C 3x2 − 2x − 35

C x2 − 25

C a negative quadratic

C (6x + 7)(6x − 7)

a (x + 6)2

B (x + 6)(x − 10)

d ( x − 2 + 2 2)( x − 2 + 2 2)

e The expression cannot be factorised using real numbers.

C (x − 2 + 4)(x − 2 − 4)

6 Which of the following gives all solutions of the equation (2x − 7)(x + 4) = 0? a x = 7, −4 d x=

−7 2

B x = −2, −1

,4

e

x=

C x = 2, 1

7 − , 4 2

7 Which of the following represent the full solution to x2 − 6x + 9 = 0? a x = −3 d x = −3, −6

B x = −3, +3 e No real solution

C x=3

B x = 5+

C x = 7+ 5

8 A solution to (x − 5)2 − 7 = 0 is: a x = −5− d x=

92

−7+

7 5

Maths Quest 11 Mathematical Methods CAS

e

x=

−5+

7 7

9 If the equation 3x2 = 9 + x is to be solved using the quadratic formula, which set of values should be

substituted into the formula? a a = 3, b = −1 and c = 0 1 d a = 1, b = 3 and c = 9

B a = 3, b = −1 and c = −9 e a = 9, b = 1 and c =

C a = 3, b = 9 and c = 1

−3

10 The quadratic formula, which can be used to solve equations of the type ax2 + bx + c = 0, is given by: a x= d x=

−b ±

− b2

b 2 − 4 ac 2a

± b − 4 ac 2a

B x=

e

x=

b ± b 2 − 4 ac 2a −b

C x = −b ±

b 2 − 4 ac 2a

± b − 4 ac 2a

11 The value of the discriminant in 2x2 + 3x + 4 = 0 is: a −23 d 32

B −7 e 41

C 25

12 For a quadratic equation to have at least one real solution, the discriminant could be: a negative d either positive or negative

B zero e a perfect square

13 The turning point of the graph of y = 5(x + 2)2 − 1 is at: a (1, 2) d (−2, 1)

B (1, −2) e (−2, −1)

C positive

C (2, −1)

14 If y = −3(x − 2)2 + 8, the maximum value of y is: a −3 C 2 e 20

B −4 d 8

y

15 The graph shown could be for the rule: a B C d e

y = + 5x − 30 y = x2 − 11x + 30 y = x2 + 11x + 30 y = −x2 + 11x + 30 y = −x2 + 11x − 30 x2

30 −6

−5

x

16 The turning point of y = x2 + bx + 40 is at x = −7. The value of b is: a −14

B −7

7 2

d 7

C

e 14

17 Using a CAS calculator, the approximate solutions to y = −x2 + x + 4 are: a B C d e

x = 0.5 and y = 4.25 x = 0 and y = 4 x = 2.56 and x = 0 x = −1.56 and x = 2.56 x = 1.56 and x = 0

18 The system of 2 simultaneous equations represented by the graphs

at right has: a no solutions for x B one solution for x C one positive and one negative solution for x d two negative solutions for x e insufficient information for us to find a solution

y

x

19 A solution to the system of equations y = 2x − 1 and y = x2 + 8x + 8 is: a (−3, −7) C (0, 8) e (2, 28)

B (0, −1) d (3, 5)

Chapter 2 • Quadratic functions

93

e x tended r e Sp onS e

1 A 100 m length of steel cable is threaded through a series of posts in order to construct

w

a fence around a paddock with four straight sides as shown at right. a Write an equation that links l and w. b Rearrange the equation in part a to write an expression for l in terms of w. l c Write an expression for the area of the paddock in terms of w. d Plot a graph of area against w. Plot area on the vertical axis and w on the horizontal axis. On the graph, label all intercepts and the turning point. e What is the maximum area of the paddock? f What values of l and w give the maximum area of the paddock? g Comment on the relationship between l and w, and state what type of shape the paddock is. 2 MacBurger’s restaurants have employed a mathematician to y design a new logo based on an ‘M’ made up of two parabolas Parabola 1 Parabola 2 as shown in the sketch at right. The parabolas both have the form y = −2x2 + bx + c. h a Give the coordinates of two points on each parabola. b Find the equation of each parabola. (−2, 0) (2, 0) x c Find the height, h, of the sign. d Which domain of x-values should each graph be restricted to so that the parabolas form the logo shown above? e Determine the equations of two parabolas that may be used to form the ‘W’ shape below, given they are both of the form y = x2 + bx + c. y 9

(−3, 0)

x

(3, 0)

3 Find the equation of a parabola that goes through the points listed in each case.

(Hint: Assume equations of the form y = ax2 + bx + c and form simultaneous equations by substituting coordinate values.) a (0, 1), (1, 0) and (2, 3) b (0, −1), (1, 4) and (2, 15) c (0, 5), (−1, 11) and (1, −3) d Find the equations of two parabolas that form a path similar to the one below depicting a proposed water slide, given the x2 coefficient of each curve is 1 or −1. y (8, 8) (4, 4) x

4 a Using written algebra, sketch the graphs of the quadratic equations y = x2 − 6x + 8 and diGital doC doc-9723 Test Yourself Chapter 2

94

y = 3x2 + 5x − 28, showing intercepts and turning points. b Why was one equation easier to sketch than the other? c Determine another quadratic function whose graph has integer intercepts and turning point coordinates. How can this be ensured?

Maths Quest 11 Mathematical Methods CAS

ICT activities Chapter opener diGital doC • 10 Quick Questions doc-9710: Warm up with ten quick questions on quadratic functions (page 39)

2a

polynomials

diGital doC • History of mathematics doc-9711: Learn about the life of Galois, a 19th century mathematician (page 40)

2B

expanding quadratic expressions

diGital doCS • SkillSHEET 2.1 doc-9712: Practise expanding perfect squares (page 43) • SkillSHEET 2.2 doc-9713: Practise expanding differences of squares (page 43)

2e

Solving quadratic equations — null Factor law

tUtorial • We 10 eles-1407: Watch how to apply quadratic algebra skills to determine the dimensions of a triangle (page 51) diGital doCS • doc-9714: Apply the Null Factor Law to solve quadratic equations (page 53) • WorkSHEET 2.1 doc-9715: Expanding brackets and factorising quadratic expressions (page 54)

2F Solving quadratic equations — completing the square tUtorial • We 12 eles-1408: Watch how to solve a quadratic equation giving solutions in exact form (page 55) diGital doCS • SkillSHEET 2.3 doc-9716: Practise solving equations in the complete square form (page 57) • SkillSHEET 2.4 doc-9717: Practise simplifying surds (page 57)

2h

the discriminant

tUtorial • We 19 eles-1409: Watch how to find the possible values of an unknown constant in a quadratic equation with zero, one and two solutions (page 63)

diGital doC • doc-9718: Observe the discriminant and the number of solutions to a quadratic equation (page 66)

2i Graphs of quadratic functions as power functions (turning point form) diGital doC • WorkSHEET 2.2 doc-9719: Solve quadratic equations, understanding the discriminant and sketching parabolas (page 70)

2J Graphs of quadratic functions (intercepts method) diGital doCS • doc-9720: Observe axial intercepts of the graph of a quadratic in general form (page 78) • doc-9721: Observe the turning point of the graph of a quadratic in turning point form (page 78)

2k Using technology to solve quadratic equations diGital doC • doc-9722: Calculate the solutions to linear and quadratic simultaneous equations (page 82)

2l Simultaneous quadratic and linear equations interaCtiVitY • Simultaneous quadratic and linear equations int-0261: Consolidate your understanding of how to solve simultaneous quadratic and linear equations (page 83)

Chapter review diGital doC • Test Yourself doc-9723: Take the end-of-chapter test to test your progress (page 94)

To access eBookPLUS activities, log on to www.jacplus.com.au

Chapter 2 • Quadratic functions

95

Answers CHAPTER 2 QUadratiC FUnCtionS exercise 2a

1 a d 2 a d 3 a b c 4 a d 5 a c e 6 a b c 7 a b c d

polynomials 3 b1 c 2 6 e 5 x bt c x x e u Polynomial 1b Polynomial 1c Polynomial 1a 5 N; bP c N; −3 x x − N; k 2 e N; 2x 3 bx −3 d5 Coefficient of the quadratic term 1 − 19 −2x2 + x + 2 6 t 0.9937; she is swimming 0.9937 m/s at 0.2 s. 1.818 m/s

exercise 2B expanding quadratic expressions 1 a 2x2 + 16x + 30 b 12x2 − 5x − 3 2 c 25x − 25x − 14 d 48x2 − 26x + 3 e 7x2 + 19x − 36 f x2 + 21x + 90 g −18x2 + 51x − 26 h x2 − 11x + 30 i 9x2 − 82x + 9 j 4x2 + 9x − 63 2 a 4x2 + 12x + 9 b 9x2 − 30x + 25 c 36x2 + 12x + 1 d 49x2 − 84x + 36 2 e x − 16x + 64 f x2 + 26x + 169 g 4x2 − 36x + 81 h 4x2 + 36x + 81 2 i 16 − 24x + 9x j 36 − 12x + x2 2 3 a 4x − 36 b 9x2 − 25 c 36x2 − 1 d 4x2 − 81 2 e 121x − 9 f x2 − 144 g x2 − 36 h 49 − 4x2 i 1 − x2 j 25x2 − 1 2 4 a 6x + 46x + 60 b 6x2 − 3x − 84 c −8x2 + 4x + 40 d 24x2 − 66x + 27 e 96x2 − 56x + 8 f −70x2 + 35x + 210 2 g 4x − 44x + 112 h 35x2 − 90x + 40 i 2ax2 − 3ax − 54a j 2bx2 − 32b 5 a 36x2 − 169 b 60x2 − x − 10 c −6x2 + 27x + 168 d 9x2 + 66x + 121 e 4x2 − 196 f 36x2 − 60x + 25 2 g x + 7x − 144 h −x2 + 16 i −100x2 + 120x − 36

j

20x2 − 245

2x2 − 7x − 34 12x2 + 25x − 55 4x2 − 24x − 62 g x2 − 3

6 a c e i 7 a b c d

96

4x2

+ 9x +

7 2

18x2 − 14x + 5 −2x2 − 18x − 159 23x2 − 114 h 12x2 + 2 x − 2 5 j 15x2 − 24x − b d f

4a2 + 8a + 2ab + 4b 3x2 − 10y2 + xy 49c2 − 56c + 12 u2 + 8uv + 16v2

3

e f g h

12r2 + 28rs − 5s2 9u2 − 4t2 2h2 − 128k2 −3m2 + 36mn − 108n2

exercise 2C

Factorising quadratic

expressions 1 a 5xy2(xy + 4) b −2ax(4x − 7) 2 c 7(p q − 3p + 1) d 11r2s(2r2s2 + 1) e 5(x2 + 2x + 15) f −3(x2 + 6x − 2) 2 a (x + 9)(x + 7) b (x − 11)(x − 6) c 2(x + 3)(x + 8) d −3(x − 2)(x + 4) e (8 − x)(x + 12) f (3x + 5)(2x − 7) g (2x − 9)(3x − 5) h −2(4x + 1)(5x − 7) 3 a (2x + 3)(2x − 3) b (9x + 5)(9x − 5) c 2(x + 4)(x − 4) d 18(x + 3)(x − 3) e (x + 3)(x − 1) f 3(x − 1)(3x − 5) g 2(x + 10)(4 − x) h −15(x − 1)(5x − 11) 4 a (x + 7)2 b (5x − 1)2 c 2(x + 3)2 d 3(x − 2)2 e −2(6x − 1)2 f (x + 3)2 2 g (2x − 5) h (2x − 3 2)2 5 a x(x + 6) b (x + 1)(x − 7) c −(2x + 1)(2x + 13) d (3x + 7)(3x − 11) e 2(x − 1)(x + 3) f 3x(8 − x) g 6(5x + 1)(15x + 17) h −(x − 5)(x − 7) i 3(x + 2)(10 − 3x) 6 a (x + 11)(x + 12) b (3x − 4)(2x − 5) c 2(2x − 1)(x + 3) d (9x − 31)(12x − 53) 7 a 3(x − 9)(x + 1) b (5x + 1)2 c (x − 14)2 d (x − 14)2 e 3 x ( 2 x − 2) f 5(2x + 1)(6x + 1) g 5x(12x − 1) h 9(1 − xy)(1 + xy) i 7x(5x − 4) j −2(6x + 7)(x − 7) 8 a C b E c A 9 k = 5, m = −4 exercise 2d

Factorising by completing

e

9 65 9 65 x + 2 + 2 x + 2 − 2

f

11 145 11 145 x + 2 + 2 x + 2 − 2

No real factors 5 17 5 17 x+ − h x+ + 2 2 2 2 g

i

1 13 1 13 x + 2 + 2 x + 2 − 2 3 E 4 E 5 B 6 C 7 a 3( x − 3 + 10)( x − 3 − 10) j

b c d

1 a

2 a

6, −12

b 4,

d

5, −11

e 3, −3

g

2 −2 , 5 5

h 0, 5

3 a

5, 9

b

d

d

( x + 1 + 2 2)( x + 1 − 2 2)

e

( x + 4 + 3)( x + 4 − 3)

f

( x − 2 + 5)( x − 2 − 5)

g

( x − 6 + 17)( x − 6 − 17)

h

( x − 1 + 6)( x − 1 − 6)

i

( x − 4 + 6)( x − 4 − 6)

j

( x − 3 + 13)( x − 3 − 13)

1 5 1 5 x − 2 + 2 x − 2 − 2

Maths Quest 11 Mathematical Methods CAS

6

−1

h 0, 1

j

d

1 b −1,

6 − , 6 7

( x + 3 + 2)( x + 3 − 2)

7 41 7 41 x − 2 + 2 x − 2 − 2

2

g

c

c

−3

−6,

e

g

5 37 5 37 x + 2 + 2 x + 2 − 2

− 1)(x + 3) or 5(1 − x)(x + 3) − x + 2)

−4(3x2

−12 − 2 , 7 9

( x + 5 + 5)( x + 5 − 5)

b

−5(x

d

b

3 5 3 5 x + 2 + 2 x + 2 − 2

5 17 5 17 2 x + + x+ − 2 2 2 2

exercise 2e Solving quadratic equations — null Factor law

the square 1 a ( x + 2 + 7)( x + 2 − 7)

2 a

13 173 13 173 x − 2 + 2 x − 2 − 2

1 −1 , 6 6 9 −9 4

,

4

e

2

,3

−1 2

−7 7 , 2 3 − −5, 1 3

h 0,

8 5

c 5, −2 f −7, 23

−5 c −2, 4

f

c f

1 4

−1 4 3 2

i 2,

−5 2

5 1 , 3 2

4 D 5 E 6 D 7 Length = 8 cm, width = 5 cm 8 a The area is doubled. b

Length = 20 2 + 33 + 1 cm Width = 20 2 + 33 − 1 cm

9 4 hours 10 a 30 11 a 42 b 16 12

9 2

b 13

or 4.5 kg

13 a 3.7 s b 2 5 14 a x = 1.998 using fixed point iteration.

(The correct answer is x = 2.)

b x = −0.562 15 a 0.228 b 1.140 c 0.268 d 0.863 16 Answers will vary. 17 Answers will vary. 18 Answers will vary.

exercise 2F Solving quadratic equations — completing the square

1 a

−4

c e

5± 7

2 a c e 3 a

b −6 ±

± 15

−2 ±

6

−3 ±

13

7 ± 29 2 − 11 ±

d 3±

5

f 3±

3

−5 ±

b

2

9 ± 89 2

f −2, 3

2 2± 5

b 3±

c

9 ± 73 2

e

3 ± 19

g

17

2

d

105

33

d

11

65

7±5 2

2 15 ± 201 f 2 h No solutions

4 a

± 14

b± 6

c

±2 3

d ±3

5 6 7 8

2

Solutions do not exist.

exercise 2G

the quadratic formula 1 a 1, 4, −3 b 1, −7, 9 − c 1, 4, 9 d −6, −3, 7 1 − e , 7, 5 f 1, 2, 1 2

4, −12, 9 7, −21, 4 −4, −5 8,

h j b

−3

−2,

0, 1 3, 0, 2 2, 8 −

−5

b

b − 4 ± 11

c

d − 5 ± 13 f 6, 1

9 113 ± 2 2

g

2± 6

h

i

3 17 ± 4 4

3 j ± 2 2

k l 4 a

x=

±

3k 2

+1 +1

k k + 1 ± k 2 − 30 k + 1 x= 4k −2 ±

7; k =

−1 2

8± 2 b ; k = 31 31

2 a d g 3 a b c d e f g h 4 D 5 a

1 4

d 3,

2

13 ± 2 2 −3 5 c ± 2 2 e 3± 7

3 a

exercise 2h

g

Solutions do not exist. 21 b x =2± 6 3 − c , 1

c

13

d

a

g i 2 a

12

1 a

E C D

d

7 8 9 10

7

−7 ±

6.317, −0.317 b 1.854, −4.854 d No real solutions 3.637, −0.137 f No real solutions 0.123, −8.123 1.472, −7.472 b 1.121, −3.121 − d 7.162, 0.838 0.225, −1.775 3.851, 0.649 f 7.772, −0.772 No real solutions h No real solutions 22.5 cm 11.3 hours 5.2 minutes a 1 metre b At t = 0.2 seconds c At t = 1.2 and 2.3 seconds a 1.2 m b She will need to buy an extra 5.25 m2 of pavers. c She will have 3 m2 of pavers left over. 0.711 m −2 2 a y= x + 4x 95 b 190 m c 190 m

5 a c e 6 a c e g

the discriminant 73 b 20 −71 e 37 229 h 193 0 b 2 1 e 1 2 h 2 a2 − 4 4 − 12a 36 − 4a b2 − 4a 4m2 − 4m m2 + 2m − 11 m2 + 4m + 16 k2 − 8k + 8 i iii i iii i iii

k > 4 or k < −4 4

6

Question

c 176 f −20 i c f i

k < 4.05 ii k > 4.05 e k<3 ii k>3 f k > 3 or ii k < −1.5 iii −1.5 < k < 3 ii g i No values of k iii No values of k h i k<1 ii iii k > 1 i i k < 3 or k > 3 ii iii No values of k 6 (See table bottom of page) d

v

60 0 2 1

k = ±4

ii

k=4

ii

k = ±1

Equation after substitution k = −1

k = 4.05 k=3 k = −1.5 or k = 3 All values of k k=1 k=3

exercise 2i Graphs of quadratic functions as power functions (turning point form) 1 a (5, 0) b (−7, 3) − d (1, 8) c (2, 7) f (2, 2) e (−3, −4) 2 a i (4, 2) ii Same iii Minimum y = 2 iv 18

−

ii

i iii i iii i

y

(0, 18)

(4, 2) x

b

−3)

i (2, iii Minimum y = −3 v y

ii Same iv 1

1 x

(2, −3)

Number of solutions

Reasoning Δ = k2 − 16 = −15

a

x2

−x+4=0

0

b

x2 − 4x − 1 = 0

2

Δ = 16 − 4k = 20

c

x2

− 4x + 4 = 0

1

Δ = 16k2 − 16 = 0

− 18x + 20 = 0

2

Δ = 324 − 80k = 404

x2 − 4x = 0

2

Δ = 12 − 4k = 16

− 4x + 2 = 0

0

Δ = 16k2 − 24k − 72 = −32

d e

−x2

f

6x2

g

−4x2

− 12x − 9 = 0

1

Δ = 144k2 − 144k2 =0

h

3x2 + 10x + 5 = 0

2

Δ = 20 − 20k = 40

2

Δ = k2 − 6k + 9 = 16

i

−2x2

+2=0

Chapter 2 • Quadratic functions

97

c

i ii iii iv v

(−5, −8) Same Minimum y = −8 17

h

i (3, 3) iii Maximum y = 3 y v

ii Thinner iv −60

y = (x − 2)2 + 4 y = −(x + 1)2 + 5 y = (x − 2)2 − 5 y = −(x + 3)2 + 1 y = −(x − 6)2 + 6 y = (x + 3)2 − 2 5 a 6 b −7 6 a (2, 5) minimum y = 5 b (3, 8) minimum y = 8 c (6, 1) minimum y = 1 d (−4, −3) minimum y = −3 e (2, −9) minimum y = −9 f (0, 7) minimum y = 7 g (−9, −81) minimum y = −81 h (3, 4) minimum y = 4 i (−2, 3) minimum y = 3 1 2 7 a i( , ) ii (0.33, 0.67) 3 3 4 a b c d e f

(3, 3)

x

y 17 −60

i

x (−5, −8)

d

i ii iii iv v

(1, −1) Same Minimum y = −1 0 y

−73

j

(1, −1)

i ii iii iv v

ii Same iv −73

x

(8, −9)

x

e

i (8, −9) iii Maximum y = −9 y v

(−5, −9) Thinner Minimum y = −9 41

i ii iii iv v

8 9 10 11 12

(1, 20) Same Minimum y = 20 21 y

y

(1, 20)

x

3 a

y

y

b

− − i ( 5 , 7) 4 8

ii (−1.25, −0.875)

c

− i ( 1 , 35 ) 4 8

ii (0.25, −4.375)

y = −(x − 5)2 + 9 C E B b n a −h d M e 2Mh

3 a d x

g

(−5, −9)

f

i (4, 4)

iv

(−1, 3)

28 3

x

x

c

d

y

c x

28 — 3

i

d 5 a

x

x

g

b

y (6, 0)

(4, 4)

(−1, −12)

iii Minimum y = −12 y v

iv

− 54 5

e

f

y

−1, 6 5 −a −b

e

7 2

±

−9 4

±

5

98

Maths Quest 11 Mathematical Methods CAS

2

,

4

73 4

No x-intercepts y

b

x

y

(7, −7)

— − 54

(−1, −12)

1 3 − 3 −1

21 2

1 3 (2, −1)

y

x

x

b , −2

3

(0, −4)

ii Thinner

5 − , 1 2

, A B 4 a No x-intercepts

(2, 5)

ii Wider

iii Minimum y = 4 v y

c M f Mh2 + n

exercise 2J Graphs of quadratic functions (intercepts method) 1 a 2 b −4 c 0 e 2 f 1 d 6 2 a 1, 6 b 3, −2 c −5, −1 d 4, 5 e 0, 2 f 3, 4 g −5, 5 h 0, −8 i −9 j 2, −10

21

41

b

1

(0, 2) x

(−1, 0)

x

c 9,

9 4

f 0, −4

y

c

y

i

d

8

y

3 −1

x

12

−12 −4 −2 (−3, −1)

1– 2

x — , − 169 —– ) (11 2

4

y

j

y

d

x

3– 2

(1, −1)

e

y 3

35

49

k

1– 2

x

(−7, 0)

x

3– 4

( 5– , − 1– ) 8

8

y

f

−7 −5 (−6, −1)

y 3– 7

x −3

64

x

−9

e

y x

(8, 0) 12

–— ) (− 9– , − 144

y

l

7

−17 2

x

9

x

6

7

y

g

(4, −4) 9

f

y

−153

(−4, −169)

−9

x

7

y

6 a

− 1– 5

−3

x

–— ) (− 8– , − 147 5

(−1, −64)

−63

−2

y

g

4

c

y 6

—) ( 1– , − 64 9

i

—– ) (− 9– , − 121 5

h

−7

x

y

y

3

−4 3

2

4

7– 2

x

x

−15 3 ( 5– , − 1– )

9

5

− 5–

2

x

1

9

−8

x

−1

− 7–

2– 5

−4

(− 3– , − 1– ) 2

−8

b

y

h

y

2 −2

x

4– 3

—) (− 1– , − 25 3 3

5

−28

x —) ( 2– , − 49 3 3

–—) (− 1– , − 225 4

8

Chapter 2 • Quadratic functions

99

y

j

y

e

y

f

( 3– , 4)

12

2

2 1– 2

−5

x

5– 2

− 2– 3

−1

x

−2

f

1 ) (− 5– , − — 6 12

—) (− 3– , 49 8

8

y

g 5

y

k

x

2

y

2

4

−5

x

0

—) (− 5– , − 25 2 4

y

8 a

5 x

−5 −9

x

1– 2

− 5–

3 x

− 3–

y

h

—) ( 3– , − 81 8

4

7 a

(−4, 49)

0

x

8

y (4, −16)

−25 33

i

y

b −11 −11

3

y

11 x 0

x

x

2

(1, −5) y

b

(1, 4)

−1

3

j

−121

3

y

c

x

y

(−3, 36)

1 −6

1 x

−1

x

0

y

c

k

(−9, 36) −15

d

x

−3

y

–—) ( 7– , 147 2

y

4

3

−45 3 x

− 3

0

d

y

y

e

(9, 0) x

−3

−81

100

Maths Quest 11 Mathematical Methods CAS

y

9 a 3

−18

x

x

7

−3.828

1.828

−7 (−1, −8)

x

y

b

−0.479

1

−0.268 −3.732

y

i

exercise 2l Simultaneous quadratic and linear equations

3.479 x

y

1 a

(3, 32)

−5

x

(2, 21) —) ( 3– , − 47

(−2, −3)

2

y

c

j −0.395

5

y

1 — 11

−1

−3 −2

(0, 17)

3 −7.606

4

x

x

y

b

(2, 8) 2.380 x

−2.380

k

(−4, −13)

d

−6

y

−6

y

—) ( 1– , 13 4 4

5– 4

0

3

−0.651

1.151 x y

(−9, 80)

−17.944

−0.056

−1 x

10 D 12 a 2 b 2 c 1 d 0 e 2 f 1 13 a C b A c D d B 14

f

25 8

x

−8

x

—) ( 5– , − 25 8

e

(1, 0)

y

c

16

11 B

14 5

−2

−7

x

(−3, −4)

d

y

(−2, 28)

3

units; up; y = 2x2 − 3x + 9 ; y = 2( x − 4 )2 8

15 16 units; up; y = −3x2 − 12x;

y = −3(x + 2)2 + 12

y

10 6

exercise 2k

1 0.382

2.618 ( 3– , − 5– ) 2

x

4

y

g

7.275 x

−0.275 −2 —) ( 7– , −57 2

4

y

h

4

−2.781

−0.719

—) (− 7– , − 17 4 8

x

Using technology to solve quadratic equations 1 a −3.732, −0.268 b −0.179, 11.179 c 3.414, 0.586 d −8.899, 0.899 e −11.196, −0.804 f 0.232, 1.434 2 a 0.377, −0.681 b −5.747, −0.665 c 4.712, −0.430 d −1.086, 1.600 e −153.441, −0.392 f 0.094, −500.094 3 4.355 h 4 56.4 h 5 5.7 m 6 Christine; her sum was 0.046 less than 10.25, whereas Harry’s was 0.167 more than 10.25. 7 a k=8 b x = 0.5

6 — 11

2

e

x

5

y

−1 −6

3

6

x

y

f

(3, 70)

40 28 (−4, 0) −7

−4

x

Chapter 2 • Quadratic functions

101

y

g

y

m

16

−2 — − 16 9

x

4

6

x

6.13 metres |AB| = 2.67 metres (approx) = 2 2 m (exact)

49x2 − 64 b 25x2 − 3 d (5x + 11)2 b (3x − 7)(4x − 3) d 3 (x − 3 + 23)(x − 3 −

8 x

(1, −21)

4 a 3,

x

−4 −3

−16

5 a ±

y

o

7 y 9– 4

(9, 39) −5

p

1

8

b 15,

4 5 2

b 7±

−4 3 3

b k=±

y

c k>

x

(6, 16)

4 3 3

10 a

or k <

(−1, 2)

−4 3 3

y

x

3 4 5

6 3– 2

2

5

x

11

(5, 16)

y

7 x

54 — 13

6 8 (3, −15)

−48

−54

(−2, −80)

102

d 0

b

y

b

x

− 3 − 13 − , 5 − 13 and 2 − 3 + 13 − , 5 + 13 2

−3

l

c 2

4 b = −5, c = −17

6 a

21 (−2, 9)

b2

−12

y

k

1

−2

x

2 4 (2, 0)

y

2 a

7

4 3 3

4

10

3

4 3 3

−24

j

−5

7 29 ± 10 10 −135

9 a

x

3 6

−1

4x2 − 36x + 81 8x2 − 70x + 117 (x + 6)(6x + 1) (6x + 7)(6x − 7) 23)

6 − 3 ± 11

9

− 3–4

e

d (8, 5)

4

1 a c 2 a c

12

−2

−4 (−3, −9)

b=

Short anSWer (1, 20)

y

i

c

Chapter reVieW

y

n

b a=5

−3

3

(−8, 20)

−17

x = a + a 2 + 24

(−4, −20)

−61

– −17 4

10 a

f

−36

h

3 − 3 < a <3 3 + 9

(−1, 7)

(5, 14)

−9

9

(0, −1)

11 y = (x + 4)2 − 13 y 12 a

(− 1 − 7, 1 − 3 7) and

84

(− 1 + 7, 1 + 3 7) c

1 − 13 4(5 + 13) 3 , and 3 1 + 13 4(5 − 13) 3 , 3

x

d

− 11

, 20

6 14 (10, −16)

b

y

− 433

and (0, −5) 80

8 x

−10

7 1.322 km and 2.553 km 8 y=

x2 , (2, 2) and (−2, 2) 2

Maths Quest 11 Mathematical Methods CAS

x

−80 (−1, −81)

c

y

7 10 13 16 19

48 8– 3 ( 13 —, 3

d

6

—) − 25 3

− 3–2

−1

8 11 14 17

B A D D

9 12 15 18

B C C D

extended reSponSe

x

1 a b c d

y

( − 5– , 1– ) 4 8

C A E E A

100 = 2l + 2w l = 50 − w A = −w2 + 50w Area

d e 3 a b c d 4 a

13

14 x =

0) and 5±

2 4 (3, −1)

6)

8k 2B 5E

3 B 6 E

(50, 0)

w

Area = 625 m2 l = 25 m, w = 25 m l = w: the paddock is a square. Parabola 1: (−2, 0) and (0, 0), parabola 2: (2, 0) and (0, 0) b y = −2x2 − 4x, y = −2x2 + 4x c h = 2 units

e f g 2 a

25 − 48 k 2

mUltiple ChoiCe

1A 4C

y

8

x

(0, 0)

(−6,

≤ x ≤ 0, 0 ≤ x ≤ 2 y = x2 + 6x + 9, y = x2 − 6x + 9 y = 2x2 − 3x + 1 y = 3x2 + 2x − 1 y = −x2 − 7x + 5 y = −x2 + 5x, y = x2 − 11x + 32

(25, 625)

−3

(−3,

−2

x

The turning point of the first graph has whole number (integer) coordinates. c The x-coordinate of the turning point − b , so is a whole number if b is a is 2a multiple of 2a. b

Chapter 2 • Quadratic functions

103

ChapTer 3

Cubic and quartic functions diGiTal doC doc-9724 10 Quick Questions

ChapTer ConTenTS 3a Expanding 3B Long division of polynomials 3C Polynomial values 3d The remainder and factor theorems 3e Factorising polynomials 3F Sum and difference of two cubes 3G Solving polynomial equations 3h Cubic graphs — intercepts method 3i Quartic graphs — intercepts method 3J Graphs of cubic functions in power function form 3k Domain, range, maximums and minimums 3l Modelling using technology 3m Finite differences

polynomials of degree 3 and 4 This chapter will deal mainly with polynomials of degree 3 (cubics). The general equation of a cubic polynomial is P(x) = a3x3 + a2x2 + a1x + a0, more commonly written as y = ax3 + bx2 + cx + d. Degree 4 polynomials (quartics) will also be considered. The general equation of a quartic polynomial is P(x) = a4 x4 + a3 x3 + a2 x2 + a1 x + a0, more commonly written as y = ax4 + bx3 + cx2 + dx + e.

3a

expanding

If we expand three linear factors, for example, (x + 1)(x + 2)(x − 7), we get a cubic polynomial (a polynomial of degree 3) as the following worked example shows. Worked example 1

Expand: a x(x + 2)(x − 3)

b (x − 1)(x + 5)(x + 2).

Think

a 1 Write the expression. 2

Expand two linear factors and simplify.

3

Multiply by the remaining factor.

b 1 Write the expression. 2

Expand two linear factors and simplify.

3

Multiply by the remaining factor and simplify.

WriTe

a x(x + 2)(x − 3)

= x(x2 − 3x + 2x − 6) = x(x2 − x − 6) = x3 − x2 − 6x

b (x − 1)(x + 5)(x + 2)

= (x − 1)(x2 + 2x + 5x + 10) = (x − 1)(x2 + 7x + 10) = x3 + 7x2 + 10x − x2 − 7x − 10 = x3 + 6x2 + 3x − 10

ChapTer 3 • Cubic and quartic functions

105

Note: Just as there is a shortcut for expanding perfect squares, there is also a shortcut for expanding cubes. We can find the shortcut by expanding (a + b)3 as usual. (a + b)3 = (a + b)(a + b)(a + b) = (a + b)(a2 + 2ab + b2) = a3 + 2a2b + ab2 + a2b + 2ab2 + b3 (a + b)3 = a3 + 3a2b + 3ab2 + b3 = a3 + 3a2b + 3ab2 + b3 (a − b)3 = a3 − 3a2b + 3ab2 − b3 Similarly, (a − b)3 = a3 − 3a2b + 3ab2 − b3. Worked example 2

Expand the perfect cube (x − 4)3 using the appropriate rule. Think

WriTe

1

Use the rule (a − b)3 = a3 − 3a2b + 3ab2 + b3. In this case a is x and b is 4.

(x − 4)3 = x3 − 3 × x2 × 4 + 3 × x × 42 − 43

2

Simplify.

(x − 4)3 = x3 − 12x2 + 48x − 64

Worked example 3

Expand −2x(x + 5)(x − 12). Think

WriTe −2x(x

+ 5)(x – 12)

1

Write the expression.

2

Expand the two linear factors and simplify.

= −2x(x2 − 12x + 5x − 60) = −2x(x2 − 7x − 60)

3

Multiply the remaining factor. No simplification is needed.

= −2x3 + 14x2 + 120

exercise 3a

expanding

1 We1a Expand each of the following. a x(x + 6)(x + 1) b x(x − 9)(x + 2) e −3x(x − 4)(x + 4) f 5x(x + 8)(x + 2) i (5x)(−6x)(x + 9) j −7x(x + 4)2

c x(x − 3)(x + 11) g x2(x + 4)

2 We1b Expand each of the following. a (x + 7)(x + 2)(x + 3) b (x − 2)(x + 4)(x − 5) d (x − 1)(x − 2)(x − 3) e (x + 6)(x − 1)(x + 1) g (x + 11)(x + 5)(x − 12) h (x + 5)(x − 1)2 j (x + 1)(x − 1)(x + 1) 3

d 2x(x + 2)(x + 3) h −2x2(7 − x)

c (x − 1)(x − 4)(x + 8) f (x − 7)(x + 7)(x + 5) i (x + 2)(x − 7)2

Expand each of the following. a (x − 2)(x + 7)(x + 8) d (5x + 3)(2x − 3)(x − 4) g −9x(1 − 2x)(3x + 8) j 2(7 + 2x)(x + 3)(x + 4)

b (x + 5)(3x − 1)(x + 4) e (1 − 6x)(x + 7)(x + 5) h (6x + 5)(2x − 7)2

c (4x − 1)(x + 3)(x − 3) f 3x(7x − 4)(x − 4) i (3 − 4x)(2 − x)(5x + 9)

4 We2 Expand the following using the appropriate rule for expanding cubes. a (x + 2)3 b (x + 5)3 c (x − 1)3 3 3 d (x − 3) e (2x − 6) f (3x + 4)3 5 We3 Expand each of the following. a (x + 5)(x − 11)(x + 2) b 3x(x + 6)(x − 1) d 106

(−x

+ 5)(x −

12)2

Maths Quest 11 Mathematical Methods CAS

e

−x(x

+

10)2

c f

6(x − 5)(−x + 15)(x + 8) + 13)3

−(x

3B

long division of polynomials

The reverse of expanding is factorising (expressing a polynomial as a product of its linear factors). Before learning how to factorise cubics, you must be familiar with long division of polynomials. You may remember in earlier levels doing long division questions. Consider 745 ÷ 3, or 3 745 The process used is as follows. 3 into 7 goes 2 times. Write 2 at the top. 2 × 3 = 6. Write down the 6. Subtract to get 1.

2 3 745 −6 14

Bring down the 4 to form 14. 3 into 14 goes 4 times. Write 4 at the top. 4 × 3 = 12. Write down the 12. Subtract to get 2.

Divisor

Bring down the 5 to form 25. 3 into 25 goes 8 times. Write 8 at the top. 8 × 3 = 24. Write down the 24. Subtract to get 1. Answer: 745 ÷ 3 = 248 remainder 1

248 3 745

Quotient Dividend

−6 14 − 12 25 − 24 1

Remainder

The same process can be used to divide polynomials by polynomial factors. Consider (x3 + 2x2 − 13x + 10) ÷ (x − 3)

or x − 3 x3 + 2x2 − 13x + 10

x into x3 goes x2 times (consider only the leading terms). Write x2 at the top. x2 × (x − 3) = x3 − 3x2 Write down the x3 − 3x2.

x−3

x3

x2 + 2x2 − 13x + 10

− (x3 − 3x2) 5x2 − 13x

Subtract. (x3 − x3 = 0, 2x2 − −3x2 = 5x2) Bring down the −13x. x into 5x2 goes 5x times. Write + 5x at the top. 5x × (x − 3) =

5x2

− 15x

Divisor

x2 + 5x + 2 + 2x2 − 13x + 10

Quotient Dividend

− (x3 − 3x2) 5x2 − 13x − (5x2 − 15x) 2x + 10 − (2x − 6) 16

Remainder

x−3

Write down the 5x2 − 15x. Subtract. Note: 5x2 − 5x2 = 0, −13x − −15x = +2x Bring down the 10.

x3

x into 2x goes 2 times. Write + 2 at the top. 2 × (x − 3) = 2x − 6 Write down the 2x − 6. Subtract to get 16. Answer: (x3 + 2x2 − 13x + 10) ÷ (x − 3) = x2 + 5x + 2 remainder 16

ChapTer 3 • Cubic and quartic functions

107

Worked example 4

Perform the following long divisions and state the quotient and remainder. a (2x3 + 6x2 − 3x + 2) ÷ (x − 6) b (x3 − 7x + 1) ÷ (x + 5) Think

WriTe

a 1 Write the question in long division format. a 2

Perform the long division process.

3

Write down the quotient and remainder.

b 1 Write the question in long division

x2

Perform the long division process.

3

Write down the quotient and remainder.

2 2 + 18xx + 105 Q 2x x −6 + 6x2 − 3x + 2 3 − (2 (2x − 12x 12 2) 18x2 − 3x − (18x2 − 108x) 105x + 2 − (105x − 630) 632 R The quotient is 2x2 + 18x + 105; the remainder is 632. 2 3 2x

b

format. Note that there is no term in this equation. Include 0x2 as a ‘place holder’. 2

TUTorial eles-1412 Worked example 4

x2 − 5x + 18 Q x + 5 + 0x2 − 7x + 1 − (x3 + 5x2) −5x2 − 7x − − ( 5x2 − 25x) 18x + 1 − (18x + 90) −89 R The quotient is x2 − 5x + 18; the remainder is −89. x3

Worked example 5

Find the quotient and remainder when x4 − 3x3 + 2x2 − 8 is divided by x + 2. Think 1

Write the question in long division format. Include 0x as a ‘place holder’.

2

Divide x into x4 and write the result above.

3

Multiply x3 by x + 2 and write the result underneath.

4

Subtract and then bring down the next term.

5

Continue to perform the long division process (as you did for cubic polynomials).

6

Write down the quotient and remainder.

WriTe

x3 − 5x2 + 12x − 24 x + 2 − 3x3 + 2x2 + 0x − 8 − (x4 + 2x3) −5x3 + 2x2 − − ( 5x3 − 10x2) 12x2 + 0x − (12x2 + 24x) −24x − 8 − (−24x − 48) 40 x4

The quotient is x3 − 5x2 + 12x − 24. The remainder is 40.

Worked example 6

Calculate the quotient and remainder when x3 − 4x2 − 7x − 5 is divided by x − 1. Think

108

1

Write the question in long division format.

2

Perform the long division process.

Maths Quest 11 Mathematical Methods CAS

WriTe

x2 − 3x − 10 x − 1 − 4x2 − 7x − 5 − (x3 − x2) −3x2 − 7x − (−3x2 + 3x) −10x − 5 − (−10x + 10) −15 x3

3

The quotient is x2 – 3x – 10. The remainder is –15.

Write the quotient and the remainder.

exercise 3B

long division of polynomials

1 We4a Perform the following long divisions, and state the quotient and remainder. 3 2 a (x + 6x + 3x + 1) ÷ (x + 3) b (x3 + 4x2 + 3x + 4) ÷ (x + 2) c (x3 + x2 + x + 3) ÷ (x + 1) 3 2 3 2 d (x + x + 4x + 1) ÷ (x + 2) e (x + 2x − 5x − 9) ÷ (x − 2) f (x3 + x2 − 9x − 5) ÷ (x − 2) 3 2 3 2 g (x − 5x + 3x − 8) ÷ (x − 3) h (x − 9x + 2x − 1) ÷ (x − 5) i 3x3 − x2 + 6x + 5, x + 2 3 2 3 2 j 4x − 4x + 10x − 4, x + 1 k 2x − 7x + 9x + 1, x − 2 l 2x3 + 8x2 − 9x − 1, x + 4 2

Divide the first polynomial by the second, and state the quotient and remainder. b 6x3 + 23x2 + 2x − 31, 3x + 4 c 8x3 + 6x2 − 39x − 13, 2x + 5 3 2 e 3x + 5x − 16x − 23, 3x + 2 f 9x3 − 6x2 − 5x + 9, 3x − 4

a 6x3 − 7x2 + 4x + 4, 2x − 1 d 2x3 − 15x2 + 34x − 13, 2x − 7

State the quotient and remainder for each of the following.

3 a c

− x3

− 6 x 2 − 7 x − 16 x +1

− 2x3

b

+ 9 x 2 + 17 x + 15 2x + 1

d

− 3x 3

+ 7 x 2 + 10 x − 15 x−3

4 x 3 − 20 x 2 + 23 x − 2 −2 x+3

4 We4b State the quotient and remainder for each of the following. a (x3 − 3x + 1) ÷ (x + 1) b (x3 + 2x2 − 7) ÷ (x + 2) − 3 d ( x − 7x + 8) ÷ (x − 1) e (5x2 + 13x + 1) ÷ (x + 3) − 3 g ( 2x − x + 2) ÷ (x − 2) h (−4x3 + 6x2 + 2x) ÷ (2x + 1)

c (x3 − 5x2 + 2x) ÷ (x − 4) f (2x3 + 8x2 − 4) ÷ (x + 5)

5 We5 Find the quotient and remainder for each of the following. a (x4 + x3 + 3x2 – 7x) ÷ (x – 1) b (x4 – 13x2 + 36) ÷ (x – 2) c (6x4 – x3 + 2x2 – 4x) ÷ (x – 3) 6 We6 Calculate the quotient and remainder for each of the following. a (x3 + 9x2 + 11x + 25) ÷ (x + 15) b (2x3 − 18x2 + 5x − 9) ÷ (x − 31) 3 c (12x + 32x − 9) ÷ (3x + 4) d (18x4 + 3x3 + 45) ÷ (2x + 7)

3C

polynomial values

Consider the polynomial P(x) = x3 − 5x2 + x + 1. The value of the polynomial when x = 3 is denoted by P(3) and is found by substituting x = 3 into the equation in place of x. That is, P(3) = (3)3 − 5(3)2 + (3) + 1 = 27 − 5(9) + 3 + 1 = 27 − 45 + 4 = −14.

Worked example 7

If P(x) = 2x3 + x2 − 3x − 4, determine: a P(1) b P(−2) Think

a 1 Write the expression. 2

Replace x with 1.

3

Simplify.

c P(a)

d P(x + 1).

WriTe

a P(x) = 2x3 + x2 − 3x − 4

P(1) = 2(1)3 + (1)2 − 3(1) − 4 =2+1−3−4 = −4 ChapTer 3 • Cubic and quartic functions

109

b 1 Write the expression. 2

Replace x with −2.

3

Simplify.

P(x) = 2x3 + x2 − 3x − 4 P(−2) = 2(−2)3 + (−2)2 − 3(−2) − 4 = 2(−8) + (4) + 6 − 4 = −16 + 4 + 6 − 4 = −10

c 1 Write the expression. 2

b

c P(x) = 2x3 + x2 − 3x − 4

P(a) = 2a3 + a2 − 3a − 4

Replace x with a. No further simplification is possible.

d 1 Write the expression. 2

Replace x with (x + 1).

3

Expand the right-hand side and collect like terms. Use the rules for expanding cubics and quadratics.

d

P(x) = 2x3 + x2 − 3x − 4 P(x + 1) = 2(x + 1)3 + (x + 1)2 − 3(x + 1) − 4 = 2(x3 + 3x2 + 3x + 1) + x2 + 2x + 1 − 3x − 3 − 4 = 2x3 + 6x2 + 6x + 2 + x2 − x − 6 = 2x3 + 7x2 + 5x − 4

Worked example 8

Determine the following, considering P(x) = 16x4 + 3x3 − 22x + 17. a P(−14) b P(y + 7) Think

WriTe

a 1 Write the expression.

a

P(x) = 16x4 + 3x3 – 22x + 17

2

Replace x with −14.

P(−14) = 16(–14)4 + 3(−14)3 – 22(−14) + 17

3

Simplify.

P(−14) = 614 656 – 8232 + 308 + 17 = 606 749

b 1 Write the expression.

b

P(x) = 16x4 + 3x3 – 22x + 17

2

Replace x with y + 7.

3

Expand each term.

16(y + 7)4 = 16y4 + 448y3 + 4704y2 + 21 952y + 38 416 3(y + 7)3 = 3y3 + 63y2 + 441y + 1029 −22(y + 7) = −22y − 154 17 = 17

4

Collect like terms.

P(y + 7) = 16y4 + 451y3 + 4767y2 + 22 371y + 39 308

exercise 3C

P(y + 7) = 16(y + 7)4 + 3(y + 7)3 − 22(y + 7) + 17

polynomial values

1 We7 If P(x) = 2x3 − 3x2 + 2x + 10, determine the following. a P(0) b P(1) d P(3) e P(−1) − h P(a) g P( 3) j P(x + 2) k P(x − 3)

c f i l

2 We8 Determine the following considering P(x) = 16x4 + 3x3 − 22x + 17. b P(102) a P(−11) c P(2x + 9) d P(x3 + 2) 110

Maths Quest 11 Mathematical Methods CAS

P(2) P(−2) P(2b) P(−4y)

3

Copy the following table. Column Column Column Column Column Column Column Column Column 1 2 3 4 5 6 7 8 9

P(x)

P(1)

P(2)

P(−1)

P(−2)

Rem. Rem. Rem. Rem. when when when when divided divided divided divided by by by by (x − 1) (x − 2) (x + 1) (x + 2)

a b c d

Complete columns 2 to 5 of the table for each of the following polynomials. a P(x) = x3 + x2 + x + 1 b P(x) = x3 + 2x2 + 5x + 2 3 2 c P(x) = x − x + 4x − 1 d P(x) = x3 − 4x2 − 7x + 3 4 Find the remainder when each polynomial in question 2 is divided by (x − 1) and complete column 6 of the table. 5 Find the remainder when each polynomial in question 2 is divided by (x − 2) and complete column 7 of

diGiTal doC doc-9725 Cubic valuer

the table. 6 Find the remainder when each polynomial in question 2 is divided by (x + 1) and complete column 8 of

the table. 7 Find the remainder when each polynomial in question 2 is divided by (x + 2) and complete column 9 of

the table. 8

Copy and complete the following sentences, using your answers to questions 3 to 7 to find the pattern. a A quick way of finding the remainder when P(x) is divided by (x + 8) is to calculate b A quick way of finding the remainder when P(x) is divided by (x − 7) is to calculate c A quick way of finding the remainder when P(x) is divided by (x − a) is to calculate

3d The remainder The remainder theorem

. . .

diGiTal doC doc-9726 WorkSHEET 3.1

and factor theorems

In the previous exercise, you may have noticed that: The remainder when P(x) is divided by (x − a) is equal to P(a). That is, R = P(a). This is called the remainder theorem. We could have derived this result as follows. If 13 is divided by 4, the quotient is 3, and the remainder is 1. That is, 13 ÷ 4 = 3 + 1 4

and

13 = 4 × 3 + 1. Similarly, if P(x) = is 15. That is,

x3

+

x2

+ x + 1 is divided by (x − 2), the quotient is x2 + 3x + 7 and the remainder

15 and x−2 (x3 + x2 + x + 1) = (x2 + 3x + 7)(x − 2) + 15.

(x3 + x2 + x + 1) ÷ (x − 2) = x2 + 3x + 7 +

In general, if P(x) is divided by (x − a), the quotient is Q(x), and the remainder is R, we can write R P(x) ÷ (x − a) = Q(x) + and ( x − a) P(x) = (x − a)Q(x) + R. ChapTer 3 • Cubic and quartic functions

111

Substituting x = a into this last expression yields P(a) = (a − a)Q(x) + R = 0 × Q(x) + R = R as before. Worked Example 9

Without actually dividing, find the remainder when x3 − 7x2 − 2x + 4 is divided by: a x − 3 b x + 6. Think

a

Write

a Let P(x) = x3 − 7x2 − 2x + 4

1

Name the polynomial.

2

The remainder when P(x) is divided by (x − 3) is equal to P(3).

b The remainder when P(x) is divided by

(x + 6) is equal to P(−6).

R = P(3) = 33 − 7(3)2 − 2(3) + 4 = 27 − 7(9) − 6 + 4 = 27 − 63 − 6 + 4 = −38 b R = P(−6)

= (−6)3 − 7(−6)2 − 2(−6) + 4 = −216 − 7(36) + 12 + 4 = −216 − 252 + 12 + 4 = −452

Worked Example 10

The remainder when x3 + kx2 + x − 2 is divided by (x − 2) is equal to 20. Find the value of k. Think

Write

1

Name the polynomial.

2

The remainder when P(x) is divided by (x − 2) is equal to P(2).

3

We are given R = 20. Put 8 + 4k = 20.

4

Solve for k.

Let P(x) = x3 + kx2 + x − 2. R = P(2) = 23 + k(2)2 + 2 − 2 = 8 + 4k Since R = 20. 8 + 4k = 20 4k = 12 k = 3

The factor theorem The remainder when 12 is divided by 4 is zero, since 4 is a factor of 12. Similarly, if the remainder (R) when P(x) is divided by (x − a) is zero, then (x − a) must be a factor of P(x). Since R = P(a), all we need to do is to find a value of a that makes P(a) = 0, and we can say that (x − a) is a factor. If P(a) = 0, then (x − a) is a factor of P(x). This is called the factor theorem. Imagine P(x) could be factorised as follows: P(x) = (x − a)Q(x), where Q(x) is ‘the other’ factor of P(x). Then we have P(a) = (a − a)Q(a) = 0 × Q(a) = 0. So if P(a) = 0, (x − a) is a factor. 112 Maths Quest 11 Mathematical Methods CAS

Worked example 11

Apply the factor theorem to determine which of the following is a factor of x4 − 4x3 − 43x2 + 58x + 240. a (x + 2) b (x − 1) Think

WriTe

a 1 Name the polynomial.

a Let P(x) = x4 − 4x3 − 43x2 + 58x + 240

2

To find the remainder when P(x) is divided by (x − a), find P(a).

P(−2) = (−2)4 − 4(−2)3 − 43(−2)2 + 58(−2) + 240 = 16 − 4(−8) − 43(4) − 116 + 240 = 16 + 32 − 172 − 116 + 240 =0

3

State the answer.

As P(−2) = 0, the remainder when P(x) is divided by (x + 2) is zero; therefore, (x + 2) is a factor.

b 1 To find the remainder when P(x) is

divided by (x − a), find P(a).

2

State the answer.

exercise 3d

b P(1) = (1)4 − 4(1)3 − 43(1)2 + 58(1) + 240

= 1 − 4 − 43 + 58 + 240 = 252

As P(1) = 252, the remainder when P(x) is divided by (x − 1) is 252; therefore, (x − 1) is not a factor.

The remainder and factor theorems

1 We9 Without actually dividing, find the remainder when x3 + 3x2 − 10x − 24 is divided by: a x−1 b x+2 c x−3 d x+5 e x−0 f x−k g x+n h x + 3c.

diGiTal doC doc-9725 Cubic valuer

2 Find the remainder when the first polynomial is divided by the second without performing

long division. a x3 + 2x2 + 3x + 4, x − 3 c x3 + 3x2 − 3x + 1, x + 2 e 2x3 + 3x2 + 6x + 3, x + 5 g x3 + x2 + 8, x − 5 i −x3 + 8, x + 3

b d f h j

x3 − 4x2 + 2x − 1, x + 1 x3 − x2 − 4x − 5, x − 1 −3x3 − 2x2 + x + 6, x + 1 x3 − 3x2 − 2, x − 2 x3 + 2x2, x − 7

The remainder when x3 + kx + 1 is divided by (x + 2) is −19. Find the value of k. The remainder when x3 + 2x2 + mx + 5 is divided by (x − 2) is 27. Find the value of m. The remainder when x3 − 3x2 + 2x + n is divided by (x − 1) is 1. Find the value of n. The remainder when ax3 + 4x2 − 2x + 1 is divided by (x − 3) is −23. Find the value of a. The remainder when x3 − bx2 − 2x + 1 is divided by (x + 1) is 0. Find the value of b. The remainder when −4x2 + 2x + 7 is divided by (x − c) is −5. Find a possible whole number value of c. g The remainder when x2 − 3x + 1 is divided by (x + d) is 11. Find the possible values of d. h The remainder when x3 + ax2 + bx + 1 is divided by (x − 5) is −14. When the cubic polynomial is divided by (x + 1), the remainder is −2. Find a and b.

3a b c d e f

We10

4 We11 Apply the factor theorem to determine which of the following are factors of x3 + 2x2 − 11x − 12. a (x − 1) b (x − 3) c (x + 1) d (x + 2) 5

Prove that each of the following are linear factors of x3 + 4x2 − 11x − 30 by substituting values into the cubic function: (x + 2), (x − 3), (x + 5).

6 Use the factor theorem to show that the first polynomial is exactly divisible by the second (that is, the

second polynomial is a factor of the first). a x3 + 5x2 + 2x − 8, x − 1 c x3 − 7x2 + 4x + 12, x − 2 e x3 + 3x2 − 9x − 27, x + 3 g −2x3 + 9x2 − x − 12, x − 4

b x3 − 7x2 − x + 7, x − 7 d x3 + 2x2 − 9x − 18, x + 2 f −x3 + x2 + 9x − 9, x − 1 h 3x3 + 22x2 + 37x + 10, x + 5 ChapTer 3 • Cubic and quartic functions

113

7 mC a When x3 + 2x2 − 5x − 5 is divided by (x + 2), the remainder is: a −5

B −2

C 0

d 2

e 5

a (x − 1)

B (x − 2)

C (x + 3)

d (x − 5)

e (x + 4)

a −3

B −2

C −1

d 0

e 1

d (x + 5)

e (x + 7)

b Which of the following is a factor of 2x3 + 15x2 + 22x − 15?

c When x3 − 13x2 + 48x − 36 is divided by (x − 1), the remainder is: d Which of the following is a factor of a (x − 2)

diGiTal doC doc-9727 SkillSHEET 3.1 reviewing the discriminant

B (x + 2)

x3

− 5x2 − 22x + 56? C (x − 5)

8 Find one factor of each of the following cubic polynomials. a x3 − 3x2 + 3x − 1 b x3 − 7x2 + 16x − 12 3 2 c x + x − 8x − 12 d x3 + 3x2 − 34x − 120 9 Consider the polynomial P(x) = 6x3 + 7x2 − x − 2. a Determine:

i P(−1)

ii P

( 12 )

iii P

( ) −2 3

i Factorise P(x) as the product of (x + 1) and a quadratic factor. ii Further factorise so P(x) is written as the product of three linear factors. c Explain how the other two linear factors relate to what you found in parts b and c. d Copy and complete the following: In general if (ax + b) is a factor, then P(…) = 0. b

3e Factorising Using long division

polynomials

Once one factor of a polynomial has been found (using the factor theorem as in the previous section), long division may be used to find other factors. Worked example 12

Use long division to factorise x3 − 19 x + 30. Think

WriTe

1

Name the polynomial. Note: There is no x2 term, so include 0x2.

P(x) = x3 − 19x + 30 P(x) = x3 + 0x2 − 19x + 30

2

Look at the last term in P(x), which is 30. This suggests it is worth trying P(5) or P(−5). Try P(−5). P(−5) = 0, so (x + 5) is a factor.

P(−5) = (−5)3 − 19 × (−5) + 30 = −125 + 95 + 30 =0 Therefore (x + 5) is a factor.

3

Divide (x + 5) into P(x) using long division to find a quadratic factor.

x2 − 5x + 6 x +5 + 0x2 − 19x + 30 3 − (x + 5x2) −5x2 − 19x −(−5x2 − 25x) 6x + 30 −(6x + 30) 0

4

Write P(x) as a product of the two factors found so far.

P(x) = (x + 5)(x2 − 5x + 6)

5

Factorise the second bracket if possible.

P(x) = (x + 5)(x − 2)(x − 3)

TUTorial eles-1413 Worked example 12

x3

Note: In this example, P(x) may have been factorised without long division by finding all three values of x that make P(x) = 0, and hence three factors, then checking that the three factors multiply to give P(x).

Using short division The process of long division can take a lot of time (and space). One short division method is shown here; it may take a little longer to understand, but it is quicker than long division once mastered. 114

Maths Quest 11 Mathematical Methods CAS

Consider P(x) = x3 + 2x2 − 13x + 10. Using the factor theorem, we can find that (x − 1) is a factor of P(x). So, P(x) = (x − 1)(ax2 + bx + c). Actually, we know more than this: as P(x) begins with x3 and ends with +10, we could write: P(x) = (x − 1)(x2 + bx − 10) Imagine expanding this version of P(x). Our x2 terms give −1x2 + bx2. Since P(x) = x3 + 2x2 − 13x + 10, we need +2x2. That is, we need −1x2 + 3x2. To get this, the bx must be 3x, as when x in the first bracket is multiplied by 3x in the second bracket, +3x2 results. That is, we have deduced P(x) = (x − 1)(x2 + 3x − 10). Factorising the second bracket gives P(x) = (x − 1)(x + 5)(x − 2) Worked example 13

Factorise the following, using short division where possible. a x3 − 5x2 − 2 x + 24 b x4 + x3 − 13 x2 − 25 x − 12 Think

a 1 Name the polynomial. 2

Look for a value of x such that P(x) = 0. Try P(−2). P(−2) does equal 0, so (x + 2) is a factor.

WriTe

a Let P(x) = x3 − 5x2 − 2x + 24

P(−2) = (−2)3 − 5 × (−2)2 − 2 × (−2) + 24 = −8 − 20 + 4 + 24 = −28 + 28 =0 So (x + 2) is a factor.

3

Write the original polynomial as the found factor multiplied by ax2 + bx + c. The first term in the brackets must be x2, and the last term must be 12.

P(x) = x3 − 5x2 − 2x + 24 P(x) = (x + 2)(ax2 + bx + c) = (x + 2)(x2 + bx + 12)

4

Imagine the expansion of the expression in step 3. We have 2x2, and require −5x2. We need an extra −7x2. So b = −7.

2x2 + bx2 = −5x2 b = −7 P(x) = (x + 2)(x2 − 7x + 12)

5

Factorise the second bracket if possible.

P(x) = (x + 2)(x − 3)(x − 4)

b 1 Name the polynomial.

b Let P(x) = x4 + x3 − 13x2 − 25x − 12

2

Look for a value of x such that P(x) = 0. Try P(−1).

P(−1) = 0 So (x + 1) is a factor.

3

It is difficult to factorise a quartic using short division, so we will use long division here.

x3 + 0x2 − 13x − 12 x + 1 + x3 − 13x2 − 25x − 12 − (x4 + x3) 0 − 13x2 − 25x − (−13x2 − 13x) −12x − 12 − − ( 12x − 12) 0

4

Name the cubic factor and try to find another factor using the factor theorem.

Let Q(x) = x3 − 13x − 12. Q(−3) = 0 So (x + 3) is a factor.

5

Factorise the cubic using short division.

Q(x) = x3 − 13x − 12 = (x + 3)(x2 − 3x − 4)

x4

ChapTer 3 • Cubic and quartic functions

115

6

Factorise the quadratic if possible.

Q(x) = (x + 3)(x − 4)(x + 1)

7

Write the original polynomial in factorised form.

P(x) = x4 + x3 − 13x2 − 25x − 12 = (x + 1)(x + 3)(x − 4)(x + 1) = (x + 1)2(x + 3)(x − 4)

Worked example 14

Factorise the expression 2 x3 − 9 x2 − 2 x + 24. Think

WriTe

Name the polynomial.

2

Look for values of x such that P(x) = 0.

x = 1: P(1) = 2 – 9 – 2 + 24 ≠ 0 x = 2: P(2) = 16 – 36 – 8 + 24 = 0 So x – 2 is a factor.

3

Write P(x) as the product of x – 2 and an unknown quadratic.

P(x) = (x – 2)(ax2 + bx + c)

4

Consider the x3 term (2x3). This must equal ax3.

(x)ax2 = ax3 = 2x3 Therefore a = 2.

5

Consider the constant term (24). This must equal –2c.

(–2)c = –2c = 2c Therefore c = –12.

6

Rewrite P(x).

P(x) = (x – 2)(2x2 + bx – 12)

7

Consider the x2 term from step 6. This must equal –9x2 from the original cubic.

+ bx2 = –9x2 bx2 = –5x2 Therefore b = –5.

8

Consider the x term from step 6. This must equal –2x from the original cubic. This confirms step 7.

–2bx

9

Write P(x).

P(x) = (x – 2)(2x2 – 5x – 12)

Factorise the quadratic term.

P(x) = (x – 2)(2x + 3)(x – 4)

10

exercise 3e 1 We12

–4x2

– 12x = –22 –2bx = 10x b = –5

Factorising polynomials

Use long division to factorise each dividend.

a x + 1 x3 + 10x2 + 27x + 18

b x + 2 x3 + 8x2 + 17x + 10

c x + 9 x3 + 12x2 + 29x + 18

d x + 1 x3 + 8x2 + 19x + 12

e x + 3 x3 + 14x2 + 61x + 84

f

x + 7 x3 + 12x2 + 41x + 42

g x + 2 x3 + 4x2 + 5x + 2

h x + 3 x3 + 7x2 + 16x + 12

i

x + 5 x3 + 14x2 + 65x + 100

k x x3 + 7x2 + 12x

l

x + 5 x3 + 10x2 + 25x

j

x x3 + 13x2 + 40x

m x + 1 x3 + 6x2 + 5x

116

P(x) = 2x3 – 9x2 – 2x + 24

1

Maths Quest 11 Mathematical Methods CAS

n x + 6 x3 + 6x2

2 We12,13 Factorise the following as fully as possible. a x3 + x2 − x − 1 b x3 − 2x2 − x + 2 3 2 d x + x − 8x − 12 e x3 + 9x2 + 24x + 16 3 2 g x + 2x − x − 2 h x3 − 7x − 6 3 2 j x +x +x+6 k x3 + 8x2 + 17x + 10 3 2 m x − x − 8x + 12 n x3 + 9x2 − 12x − 160 4 3 2 p x + 3x − 6x − 28x − 24 q x4 + 6x3 + 8x2 − 6x − 9

c f i l o r

x3 + 7x2 + 11x + 5 x3 − 5x2 − 4x + 20 x3 + 3x2 − 4 x3 + x2 − 9x − 9 x4 + 4x3 + 3x2 − 4x − 4 x4 − 5x3 − 17x2 + 21x

diGiTal doC doc-9728 polynomials zero search

3 We14 Use a CAS calculator to factorise the following as fully as possible. a 3x3 − x2 − 10x b 4x3 + 2x2 − 2x c 3x3 − 6x2 − 24x − 3 2 3 2 d 2x − 12x − 18x e 6x − 6x f −x3 − 7x2 − 12x − 3 2 − 3 2 g x − 3x + x + 3 h 2x + 10x − 12x i −6x3 − 5x2 + 12x − 4 − 3 2 − 5 4 3 2 j 5x + 24x − 36x + 16 k x − x + 21x + 49x − 8x − 60 l 24x4 − 53x3 − 71x2 + 152x + 20 4 a d g j

Factorise the following as fully as possible. 2x3 + 5x2 − x − 6 b 3x3 + 14x2 + 7x − 4 3 2 4x + 35x + 84x + 45 e 5x3 + 9x2 + 3x − 1 3 2 4x + 16x + 21x + 9 h 6x3 − 23x2 + 26x − 8 3 2 7x + 12x − 60x + 16 k 2x4 − x3 − 11x2 − 11x − 3

3F

c f i l

3x3 + 2x2 − 12x − 8 x3 + x2 + 4 10x3 + 19x2 − 94x − 40 6x4 + 11x3 − 22x2 − x + 6

Sum and difference of two cubes

Two special cases of cubic polynomials, called ‘sum of cubes’and ‘difference of cubes’, are discussed in this section. There are shortcuts for factorising such cubic expressions. Examples of each are shown in the table below. Sum of cubes x + 3

23

Difference of cubes x3 − 27

125 + 64b3

x3 − 81y3 1000

x3y3 + 1

w6 − 1

(2x + 1)3 + 8

216 − (uv)3

Consider the following expansions. (a + b)(a2 − ab + b2) and (a − b)(a2 + ab + b2) 3 2 2 2 2 3 = a − a b + ab + ba − ab + b = a3 + a2b + ab2 − ba2 − ab2 − b3 3 2 2 2 2 3 = a − a b + ab + a b − ab + b = a3 + a2b + ab2 − a2b − ab2 − b3 3 3 =a +b = a3 − b3 These expansions show that: a3 + b3 = (a + b)(a2 − ab + b2) and a3 − b3 = (a − b)(a2 + ab + b2). That is, we have two formulas that may be used to factorise sums and differences of cubes. Worked example 15

Factorise the following using the sum or difference of cubes formula. a x3 − 1000y3 b 2(x + 6)3 + 16 Think

a 1 Write the expression.

TUTorial eles-1414 Worked example 15

WriTe

a x3 − 1000y3

= x3 − (10y)3

2

Recognise a difference of cubes.

3

Identify a and b for use with the formula a3 − b3 = (a − b)(a2 + ab + b2).

4

Use the formula to factorise.

= (x − 10y)[x2 + x(10y) + (10y)2]

5

Simplify.

= (x − 10y)(x2 + 10xy + 100y2)

a = x, b = 10y

ChapTer 3 • Cubic and quartic functions

117

b 2(x + 6)3 + 16

b 1 Write the expression.

= 2[(x + 6)3 + 8] = 2[(x + 6)3 + 23]

2

Take out a common factor of 2 to produce a sum of cubes.

3

Identify a and b for use with the formula a3 + b3 = (a + b)(a2 − ab + b2).

4

Apply the sum of cubes formula.

= 2[(x + 6) + 2][(x + 6)2 − (x + 6)(2) + 22]

5

Simplify.

= 2(x + 8)(x2 + 12x + 36 − 2x − 12 + 4) = 2(x + 8)(x2 + 10x + 28)

exercise 3F

a = (x + 6), b = 2

Sum and difference of two cubes

1 Identify a and b (as used in the above sum and difference of cubes expressions) in each of the following

(do not factorise). a x3 + 63

b 8y3 + z3

d 1 − 64h3

e

2 We15a a x3 − 125

g j

+ 8p3

729 s3t3

+

− e3g3

f

t3 −

u3 216

Factorise the following using the sum or difference of cubes formula. b j3 + k3 c y3 − 8

d 27x3 + y3

x3

1 3 8c

c (x + 5)3 + 27

e

64t3 − 216u3

h 27r3 − 1

f

x3 − 1

i

(3k)3 − 1 8

g6

Factorise the following. a (a − + a3 b (x + 2)3 − 8 c (2x + 3)3 + 1 3 3 3 3 d (w − 5) − w e (2m + p) + (3m − p) f 27x3 − (x + 3)3 3 3 3 3 g (2y + 7) + (y − 2) h (3x + y) + (x − 4y) i (2 − 4p)3 − (p + 1)3 3 3 6 9 j (5x − 9) − (7 − x) k x +y l 2x3 − 54 3 2 3 m 3a + 3 n 6(x + 1) + 162 4 When mx3 − ny3 is fully factorised it gives (3x − y)(9x2 + 3xy + y2). What are the values of m and n? 5 a Write 3x3 + my3 in the form a3 + b3. b Identify the values of a and b. c Factorise using the rule for sum of cubes. 3 We15b

1)3

3G Solving polynomial equations making x the subject of the equation Cubic equations of the form a(x − b)3 + c = 0 may be solved by isolating x as follows. a(x − b)3 = −c −c (x − b)3 = a −

c a

x=b+

3

x−b=

3

−

c a

Unlike a square root, a cube root can be only positive or negative, not both; for example, 3 − 8

118

+= −2, 3 8 = 2.

Maths Quest 11 Mathematical Methods CAS

Worked example 16

Solve 3(x + 2)3 + 192 = 0 by making x the subject. Think

WriTe

3(x + 2)3 + 192 = 0

1

Write the equation.

2

Subtract 192 from both sides.

3

Divide both sides by 3.

4

Take the cube root of both sides.

5

Subtract 2 from both sides and simplify.

3(x + 2)3 = −192 (x + 2)3 = −64 x + 2 = −4 x = −4 − 2 = −6

Factorising to solve polynomial equations The Null Factor Law applies to cubic and quartic equations just as it does for quadratics. If P(x) = (x − a)(x − b)(x − c) = 0, then the solutions are x = a, x = b and x = c. If P(x) = k(lx − a)(mx − b)(nx − c) = 0, solutions are found by solving the following equations: lx − a = 0, mx − b = 0 and nx − c = 0 Worked example 17

Solve each of the following equations. a x4 = 16x2 b 2x3 − 11x2 + 18x − 9 = 0 Think

a 1 Write the equation.

WriTe

a

x4 = 16x2 x4 − 16x2 = 0

2

Rearrange so that all terms are on the left.

3

Take out a common factor of x.

4

Factorise the brackets using a difference of squares.

x2(x + 4)(x − 4) = 0

5

Use the Null Factor Law to solve.

x = 0, x + 4 = 0 or x − 4 = 0 so x = 0, x = −4 or x = 4

b 1 Name the polynomial.

x2(x2 − 16) = 0

b Let P(x) = 2x3 − 11x2 + 18x − 9.

2

Use the factor theorem to find a factor (search for a value a such that P(a) = 0). Consider factors of the constant term (that is, factors of 9 such as 1, 3). The simplest value to try is 1.

P(1) = 2 − 11 + 18 − 9 =0 So (x − 1) is a factor.

3

Use long or short division to find the other factors of P(x).

2 2 − 9x + 9 2x x − 1 2x 2 3 − 11x2 − 18x − 9 22x3 − 2x 2 2 −9x2 + 18x −9x2 + 9x 9x − 9 9x − 9 0

4

Factorise the brackets.

P(x) = (x − 1)(2x2 − 9x + 9) P(x) = (x − 1)(2x − 3)(x − 3)

5

Consider the factorised equation to solve.

For (x − 1)(2x − 3)(x − 3) = 0

6

Use the Null Factor Law to solve.

x − 1 = 0, 2x − 3 = 0 or x − 3 = 0 3

so x = 1, x = 2 or x = 3

ChapTer 3 • Cubic and quartic functions

119

Polynomial equations can also be solved using a CAS calculator, which is useful when the solutions are not rational. The solutions can be found by using the solve function on a calculator page, or by constructing a graph. Worked example 18

Use a CAS calculator to solve x3 + 3 x2 − 8 x − 5 = 0. Think

WriTe

1

Use the CAS calculator to solve the equation.

Enter solve (x3 + 3x2 − 8x − 5 = 0, x) into the CAS calculator.

2

Read the results supplied by the calculator.

x = −4.524 04 or x = −0.536 396 or x = 2.060 43

3

Write the answer.

The solutions of x3 + 3x2 − 8x − 5 = 0 are x = −4.52, −0.54 and 2.06, rounded to 2 decimal places.

exercise 3G 1 We16

Solving polynomial equations

Solve the following equations by making x the subject.

a 2(x − − 250 = 0 c (x − 4)3 − 1000 = 0 e −2(x − 5)3 − 2 = 0 g (2x + 3)3 − 27 = 0

b 3(x + 2)3 + 81 = 0 d (x + 7)3 − 8 = 0 f −(x + 3)3 + 1 = 0 h 4(3x − 1)3 + 500 = 0

1)3

i

2 3

(x + 1)3 + 18 = 0

j

k (x − 5)3 = 343

1 2

(5 − x)3 − 32 = 0

l 4−

4 5

(x + 8)3 = 104

Find all solutions of the following equations.

2 a c e g i k m

(x − 1)(x − 2)(x − 5) = 0 (x − 5)(x + 2)(x − 9) (x − 1) = 0 (3x + 12)(x − 4)(x + 4) = 0 (x + 5)(x − 8)2 = 0 x2(x + 1)(x − 1) = 0 (6 − x)2 (2 + x) = 0 x(5x − 6)(2x + 3) (6 − 7x) = 0

3 We17a a d g j m

b d f h j l n

(x + 3)(x + 4)(x + 7) = 0 (2x − 4)(x + 1)(x − 3) (x + 2) = 0 (2 − x)(x + 2)(1 − x)(1 + x) = 0 (x − 1)3 = 0 −3x(x − 9)3 = 0 x2(2x + 7) = 0 (3 − 4x)2(5x − 1) = 0

Solve the following equations.

x3 − 4x = 0 −3x4 + 81 = 0 −4x3 + 8x = 0 x4 − 5x3 + 6x2 = 0 9x2 = 20x + x3

b e h k n

x3 − 16x = 0 x3 + 5x2 = 0 12x3 + 3x2 = 0 x3 − 8x2 + 16x = 0 x3 + 6x = 4x2

c f i l

2x4 − 50x2 = 0 x3 − 2x2 = 0 4x3 − 20x4 = 0 x3 + 6x2 = 7x

4 We17b Use the Null Factor Law to solve the following equations. a x3 − x2 − 16x + 16 = 0 b x3 − 6x2 − x + 30 = 0 3 2 c x − x − 25x + 25 = 0 d x3 + 4x2 − 4x − 16 = 0 3 2 e x − 4x + x + 6 = 0 f x3 − 4x2 − 7x + 10 = 0 4 3 2 − g x − 3x − 7x + 15x = 18 h x4 + 2x3 − 13x2 + 10x 3 2 i 2x + 15x + 19x + 6 = 0 j −4x3 + 16x2 − 9x − 9 = 0 − 3 2 k 2x − 9x − 7x + 6 = 0 l 2x3 + 4x2 − 2x − 4 = 0 5 We18 Use a CAS calculator to find all solutions to the following equations. a x3 − 17x2 − 56x + 1153 = 0 b x3 + 12x2 − 49x − 588 = 0 − 3 2 c x + 17x + 65x − 1521 = 0 d x3 – 48x2 + 768x − 4096 = 0 120

Maths Quest 11 Mathematical Methods CAS

e x3 + 6x2 − 6x + 2 = 0 g 3x2 + 2x + 1 = x3 i x4 + 4x3 − 5x2 + 15 = 0

f x3 – 14x2 – 4x + 13 = 0 h −2x3 – 3x2 + 2x + 0.5 = 0 j 2x4 + 25x3 + 17x = 9

6 mC Which of the following is a solution to x3 − 7x2 + 2x + 40? a −5 d 1

B −4 e 2

C −2

7 mC A solution of x3 − 9x2 + 15x + 25 = 0 is x = 5. How many other (distinct) solutions are there? a 0 d 3

3h

B 1 e 4

C 2

diGiTal doC doc-9729 WorkSHEET 3.2

Cubic graphs — intercepts method

A good sketch graph of a function shows: 1. x- and y-intercepts 2. the behaviour of the function at extreme values of x, that is, as x approaches infinity (x → +∞) and as x approaches negative infinity (x → −∞) 3. the general location of turning points. The graphs below show the two main types of cubic graph. y

y Turning points

x

x

A positive cubic

A negative cubic

For positive cubic graphs, as positive values of x become larger and larger, y-values also become larger. For negative cubic graphs, as positive values of x become larger and y Point of inflection larger, y-values become smaller. Sometimes, instead of two turning points, there is a point of inflection, where the graph changes from a decreasing gradient to zero to an increasing gradient (or vice versa). When this occurs, there is only one x-intercept. Consider the general factorised cubic f (x) = (x − a)(x − b)(x − c). x The Null Factor Law tells us that f (x) = 0 when x = a or x = b or x = c. The y-intercept occurs when x = 0; that is, the y-intercept is f (0) = (0 − a)(0 − b)(0 − c) = −abc y

c

b

a

Cubic with a point of inflection

x

−abc

ChapTer 3 • Cubic and quartic functions

121

Worked Example 19

Sketch graphs of the following, showing all intercepts. a y = (x − 2)(x − 3)(x + 5) b y = (x − 6)2 (4 − x) c y = (x − 2)3 Think

Write/Draw

a 1 Note that the function is already

factorised and that the graph is a positive cubic.

a y = (x − 2)(x − 3)(x + 5)

2

The y-intercept occurs where x = 0. Substitute x = 0 into the equation.

y-intercept: if x = 0, y = (−2)(−3)(5) = 30 Point: (0, 30)

3

Use the Null Factor Law to find the x-intercepts. (Make each bracket equal to 0 and solve a mini-equation.)

x-intercepts: if y = 0, x − 2 = 0, x − 3 = 0 or x + 5 = 0 x = 2, x = 3 or x = −5 Points: (2, 0), (3, 0), (−5, 0)

4

Combine information from the above steps to sketch the graph.

y

30 −5 b 1 The graph is a negative cubic −x

(the in the last factor produces a negative x3 coefficient if the RHS is expanded).

2

3

x

b y = (x − 6)2(4 − x)

2

Substitute x = 0 to find the y-intercept.

y-intercept: if x = 0, y = (−6)2(4) = 144 Point: (0, 144)

3

Use the Null Factor Law to find the x-intercepts. (Make each bracket equal to 0 and solve a mini-equation.)

x-intercepts: if y = 0, x − 6 = 0 or 4 − x = 0 x = 6 or x=4 Points: (6, 0), (4, 0)

4

Combine all information and sketch the graph. Note the skimming of the x-axis indicative of a repeated factor, in this case the (x − 6)2 part of the expression.

y 144

4 c 1 Positive cubic.

2

Substitute x = 0 to find the y-intercept.

122 Maths Quest 11 Mathematical Methods CAS

6

x

c y = (x − 2)3

y-intercept: if x = 0, y = (−2)3 = −8

3

Use the Null Factor Law to find the x-intercept. (Make each bracket equal to 0 and solve a mini-equation.)

4

Combine all information and sketch the graph. The cubed factor, (x − 2), indicates a point of inflection and only one x-intercept.

x-intercept: if y = 0, x−2=0 x=2 y

x

2 −8

If a cubic function is not in the form f (x) = (x − a)(x − b)(x − c), we may try to factorise to find the x-intercepts. We can use the factor theorem and division of polynomials to achieve this. Worked example 20

Sketch the graph of y = 2x3 − 11x2 + 7x + 20, showing all intercepts. Think 1

Write the equation and name the polynomial P(x).

2

Note the graph is a positive cubic.

3

Let x = 0 to find the y-intercept. Note: All terms involving x are equal to zero.

4

Factorise P(x) to find x-intercepts. (x − 1) is not a factor.

TUTorial eles-1415 Worked example 20

WriTe/draW

y = P(x) = 2x3 − 11x2 + 7x + 20

y-intercept: if x = 0, y = 20 Point: (0, 20) P(1) = 2 × 13 − 11 × 12 + 7 × 1 + 20 = 2 − 11 + 7 + 20 = 18 ≠0 P(−1) = 2 × (−1)3 − 11 × (−1)2 + 7 × (−1) + 20 = −2 − 11 − 7 + 20 =0 So (x + 1) is a factor.

5

Use long or short division to factorise P(x). Here, short division has been used.

P(x) = (x + 1)(2x2 − 13x + 20) = (x + 1)(2x − 5)(x − 4)

6

Write down the x-intercepts (determined by making each bracket equal to 0 and solving for x).

x-intercepts: if y = 0, x = −1, 52 , 4

Points: (−1, 0), ( 52 , 0), (4, 0) 7

Use all available information to sketch the graph.

y 20

−1

5– 2

4

x

ChapTer 3 • Cubic and quartic functions

123

Worked example 21

Sketch the graph of y = 3x3 + x2 − 2x + 5 using a CAS calculator. Find all intercepts and stationary points. Think

WriTe/draW

1

Enter the rule for the function into the CAS calculator.

f1(x) = 3x3 + x2 − 2x + 5

2

Determine the y-intercept (when x = 0).

f1(0) = 3(0)3 + (0)2 − 2(0) + 5 =5 The y-intercept is (0, 5).

3

Use the CAS calculator to help sketch the graph.

A

y

(0, 5) B

x

(−1.51, 0)

4

Analyse the function with the CAS calculator to find zeros (x-intercepts). In this case there is only one x-intercept.

The x-intercept is (−1.51, 0).

5

Analyse the function with the CAS calculator to find the local maximum, point A.

The local maximum (A) is (−0.62, 5.9).

6

Analyse the function with the CAS calculator to find the local minimum, point B.

The local minimum (B) is (0.36, 4.6).

exercise 3h

Cubic graphs — intercepts method

1 We19 Sketch the following, showing all intercepts. a y = (x − 1)(x − 2)(x − 3) c y = (x + 8)(x − 11)(x + 1) e y = (4x − 3)(2x + 1)(x − 4) 2

Sketch the following.

a y = (2 − x)(x + 5)(x + 3) c y = x(x + 1)(x − 2) e y = 4x2(x + 8)

b y = (x + 8)(x − 8)(2x + 3) d y = 3(x + 1)(x + 10)(x + 5) f y = (6x − 1)2(x + 7)

Sketch each of the following as fully as possible.

3 We20 a c e g

b y = (x + 6)(x + 1)(x − 7) d y = (2x − 5)(x + 4)(x − 3) f y = (x − 3)2(x − 6)

y = + 2x2 − x − 2 y = x3 + 7x2 + 14x + 8 y = x3 − 3x2 − 25x − 21 y = 6x3 − 17x2 + 6x + 8 x3

b d f h

y = x3 + 6x2 + 11x + 6 y = x3 − x2 − 14x + 24 y = 3x3 + 17x2 + 28x + 12 y = −2x3 − 18x2

4 We21 Sketch the following using a CAS calculator. Find the intercepts and the coordinates

of all turning points, correct to 2 decimal places. y = −x3 − 8x2 − 5x + 14 y = −x3 + 8x2 + 13x − 140 y = 3x3 − 3x2 − 15x − 9 y = 7x3 + 29x2 + 32x + 4

a b c d 124

Maths Quest 11 Mathematical Methods CAS

5 mC Which of the following is a reasonable sketch of y = (x + 2)(x − 3)(2x + 1)? y

a

−3

1 – 2

−2

− 1– 2

−2

2 x

y

d

y

B

y 3

C

−2

y

e

1– 2

3 x

3 x

2

y

6 mC The graph shown on the right could be that of: a y= + 2) C y = (x − 2)(x + 2)2 e y = (x − 2)3

x2(x

x

− 1– 2

3 x

− 1– 2

diGiTal doC doc-9730 Cubic graphs — general form

B y = (x + d y = (x − 2)2(x + 2)

2)3

−2

2

x

−8 7 mC The graph below has the equation: y

−3

−1

2

x

−6 a y = (x + 1)(x + 2)(x + 3) d y = (x + 1)(x + 2)(x − 3)

B y = (x + 1)(x − 2)(x + 3) e y = (x − 1)(x − 2)(x − 3)

C y = (x − 1)(x + 2)(x + 3)

8 mC If a, b and c are positive numbers, the equation of the graph shown below is: a y = (x − a)(x − b)(x − c) d y = (x + a)(x + b)(x + c)

B y = (x + a)(x − b)(x + c) e y = (x − a)(x + b)(x − c)

C y = (x + a)(x + b)(x − c)

y

−b

c

a

x

9 mC Which of the following has only two distinct x-intercepts when graphed? a y = x(x + 1)(x + 2) d y = (x + 1)(x + 2)2

10 a b c d

B y = (x + 1)(x + 2)(x + 3) e y = x(x + 1)(x − 1)

C y = x3

Sketch the graph of y = x3 − x2 + 3x + 5 showing all intercepts. Factorise y = x3 − x2 + 3x + 5 expressing your answer in the form of y = (x + a)(x2 + bx + c). Hence, show that y = x3 − x2 + 3x + 5 has only one real solution. (Hint: Consider the discriminant.) If y = (x + a)(x2 + kbx + c) where k is a constant, find the values of k such that the cubic has: i two real solutions ii three real solutions. Note: a, b and c are the same values from part b. ChapTer 3 • Cubic and quartic functions

125

3I

Quartic graphs — intercepts method

Quartic functions are polynomial functions of degree 4. The graphs shown below are the main types of quartic graphs. y y = x4

y = (x − a)(x − b)(x − c)(x − d) y

y = (x − a)3(x − b) y

(2, 16) 0

2

x

a repeated factor (x − a)3

b

x a

b

d x

c

Negative quartics are reflected across the x-axis. Consider the general factorised quartic, f (x) = (x – a)(x – b)(x – c)(x – d). As for the cubic functions, the Null Factor Law tells us that f (x) = 0 (that is, an x-intercept occurs) when x = a, x = b, x = c or x = d. The y-intercept occurs when x = 0, therefore the y-intercept is f (0) = (0 – a)(0 – b)(0 – c)(0 – d) = abcd

y abcd

a

b

c

d x

Worked Example 22

Sketch the following graphs, showing all intercepts. a y = (x − 2)(x − 1)(x + 1)(x + 3) b y = (x − 3)2(x + 1)(x + 5) c y = (2x − 1)(x + 1)3 Think

a 1 The y-intercept occurs when x = 0.

Write/Draw

a y = (x − 2)(x − 1)(x + 1)(x + 3)

y = (−2)(−1)(1)(3) =6 Point: (0, 6)

2

Use the Null Factor Law to find the x-intercepts. (Make each bracket equal to 0 and solve a mini-equation.)

x-intercepts: if y = 0, x − 2 = 0, x − 1 = 0, x + 1 = 0 or x + 3 = 0 x = 2, x = 1, x = −1 or x = −3 Points: (2, 0), (1, 0), (−1, 0), (−3, 0)

3

The graph has a positive x4 coefficient, so large positive values for x result in large positive values for y.

Shape:

4

Combine information from steps 1 to 3 to sketch the graph.

y

6 −3

b 1 The y-intercept occurs when x = 0.

2

Use the Null Factor Law to find the x-intercepts. (Make each bracket equal to 0 and solve a mini-equation.)

126 Maths Quest 11 Mathematical Methods CAS

−1 0

1

2

x

b y = (x − 3)2(x + 1)(x + 5)

y = (−3)2(1)(5) = 45 Point: (0, 45)

x-intercepts: if y = 0, x − 3 = 0, x + 1 = 0 or x + 5 = 0 x = 3, x = −1 or x = −5 Points: (3, 0), (−1, 0), (−5, 0)

3

The graph has a positive x4 coefficient. (Large positive values for x result in large positive values for y.)

4

Combine all the information above to sketch the graph. Note that it touches the x-axis where there are repeated squared factors, (x − 3)2.

Shape:

y 45 −1 0

−5

c 1 The y-intercept occurs when x = 0.

2

Use the Null Factor Law to find the x-intercepts. (Make each bracket equal to 0 and solve a mini-equation.)

x

3

c y = (2x − 1)(x + 1)3

y = (−1)(1)3 = −1 Point: (0, −1)

x-intercepts: if y = 0, 2x − 1 = 0 or x + 1 = 0 x = 12 or x = −1 Points: (1, 0) and (−1, 0) 2

x4

3

The graph has a positive coefficient. (Large positive values for x result in large positive values for y.)

4

Combine all the information from above to sketch the graph. Note that the graph has a point of inflection where it crosses the x-axis with repeated cubic factors (x + 1)3.

Shape:

y

0

−1

1– 2

x

−1

Follow the instructions in worked example 21 to draw quartic graphs using a CAS calculator.

exercise 3i

Quartic graphs — intercepts method

1 We22 Sketch the following graphs, showing all intercepts. a y = (x − 3)(x − 2)(x + 1)(x + 2) b y = (x − 2)2(x + 1)(x + 2) 3 c y = (x + 5)(x − 1) d y = (x − 1)4 − e y = x(2x − 1)(x – 3)(x + 3) f y = (x − 2)2(x + 1)2 4 g y = (1 − 3x) h y = (x + 5)3(1 − x) 2 Compare your answers to question 1 to those found using a CAS calculator. 3 If the graph of y = (x + a)(x + 3)(x + 1)(x − 3) has four distinct x-intercepts and has a y-intercept at

(0, 45), find the value of a. 4 A quartic graph has only two x-intercepts, at x = a and x = b, and a y-intercept at the point (0, 81).

If a = −b: a find a possible equation for the quartic graph b sketch the graph, labelling all intercepts.

5 mC Which of the following has two distinct x-intercepts when graphed? a B C d e

y = x(x + 7)(x − 7)(x + 2) y = x2(x + 3)(x − 3) y = x3(x + 27) y = x(x + 7)2(x − 7) y = x4

ChapTer 3 • Cubic and quartic functions

127

y

6 MC If a, b and c are positive numbers, the equation of the graph

shown is: A y = (x + a)2(x + b)(x + c) C y = (x + a)2(x − b)(x + c) E y = (x − a)2(x + b)(x − c)

B y = (x − a)2(x − b)(x − c) D y = (x − a)(x + b)(x – c)

−b

−abc

a

c

x

7 MC For the graph of the quadratic equation y = (x − 2)(x + 1)(x + 3)2,

the y-intercept occurs at: A 6 b −6 C −12 D −18 E 18 8 Use a CAS calculator to help you sketch the following quartics, showing all intercepts and turning points correct to 2 decimal places. a y = 2x4 + x3 − 5x2 + 7x + 2 b y = 3x4 − 9x3 − 8x2 + 12x + 9 c y = −x4 − 2x3 + 5x2 + 4x d y = x4 − 5x3 − 45x2 + 8x + 120 e y = 3x4 − 10x2 − 3 f y = −8x4 − 10x3 + 120x2 + 15x + 358

Graphs of cubic functions in power function form 3J

Remember the power form or turning point form for quadratic graphs y = a(x − b)2 + c, which was related to transformations of the basic parabola? The same understanding of transformations can be used to sketch cubic functions. Cubic functions can be power functions. Power functions are functions of the form f (x) = xn, n ∈ R. The value of the power, n, determines the type of function. When n = 1, f (x) = x, and the function is linear. When n = 2, f (x) = x2, and the function is quadratic. When n = 3, f (x) = x3, and the function is cubic. When n = 4, f (x) = x4, and the function is quartic. Other power functions will be discussed later. Under a sequence of transformations of f (x) = xn, n ∈ R, the general form of a power function is f (x) = a(x − b)n + c (where a, b, c and n ∈ R). Although all linear and quadratic polynomials are also linear and quadratic power functions, this is not the case for cubic functions (or quartic functions). For example, a cubic power function in the form of f (x) = a(x − b)n + c has exactly one x-intercept and one stationary point of inflection. A cubic polynomial in the form f (x) = ax3 − bx2 + cx + d can have one, two or three x-intercepts and is therefore not always a power function. For example, the cubic function y = 2(x − 3)3 + 1 is a polynomial and a power function. It is the graph of y = x3 under a sequence of transformations. A cubic power function has a stationary point of inflection at (b, c). A stationary point of inflection is where a graph ‘levels off’ to have a zero gradient at one point with the same sign gradient either side. y

y = x3

y = −x3 y

x

x

Stationary point of inflection

Summary of transformations y = a(x − b)3 + c

y (b, c)

Dilation factor from the x-axis (y-stretch) 128 Maths Quest 11 Mathematical Methods CAS

x-translation

y-translation

b

c

Stationary point of inflection x

The effect of a is illustrated below. y

y = 2x3 y = x3 y = 1–2 x3

y = −2x3 y y = −x3 y = − 1–2 x3

x

x

Positive a

Negative a

intercepts Intercepts may be found by substituting x = 0 (to find the y-intercepts) and y = 0 (to find the x-intercepts) into the equation. Worked example 23

Sketch the graph of each of the following, showing the stationary point of inflection and intercepts. a y = 3(x − 2)3 + 3 b y = −2x3 + 54 c y = −2(1 − 2x)3 − 16 Think

a 1 Compare the equation with y = a(x − b)3 + c, which

WriTe/draW

a y = 3(x − 2)3 + 3

has a stationary point of inflection (b, c). 2

Note the values that match, namely a = 3, b = 2 and c = 3. State the stationary point of inflection (b, c).

Stationary point of inflection (2, 3)

3

Find the y-intercept.

If x = 0,

4

Find the x-intercept. (Note that the cube root of −1 is −1.)

If y = 0,

5

Note that the equation is for a positive cubic.

Shape:

6

Sketch, showing the stationary point of inflection and intercepts.

y = 3(0 − 2)3 + 3 y = 3(−8) + 3 y = −21 0 = 3(x − 2)3 + 3 − 2)3 = 3 (x − 2)3 = −1 x − 2 = −1 x=1

−3(x

y

(2, 3)

1

2

x

−21 b 1 Manipulate into y = a(x − b)3 + c form. 2

Note the graph is a negative cubic with stationary point of inflection (0, 54).

b y = −2x3 + 54

y = −2(x − 0)3 + 54

Stationary point of inflection (0, 54)

ChapTer 3 • Cubic and quartic functions

129

3

Find the y-intercept.

If x = 0,

4

Find the x-intercept.

If y = 0,

5

Sketch, showing the stationary point of inflection and intercepts.

y = −2(0) + 54 y = 54 0 = −2x3 + 54 2x3 = 54 x3 = 27 x=3

y (0, 54)

x

3

c 1 Manipulate into y = a(x − b)3 + c form.

c y = −2(1 − 2x)3 − 16

= −2[−2(x − 1)]3 − 16 2

= −2[−8(x − 12)3] − 16 = 16(x − 12)3 − 16 2

Note the graph is a positive cubic with stationary point 1 of inflection ( 2 , −16).

Stationary point of inflection ( 12 , −16)

3

Find the y-intercept.

If x = 0,

−

1

y = 16( 2 )3 − 16 = −2 − 16 = −18

4

Find the x-intercept. Use the result from step 1.

1

0 = 16( x − 2 )3 − 16

If y = 0,

1

16 = 16( x − 2 )3 1

1 = ( x − 2 )3 1=x − 3 2

5

Sketch, showing stationary point of inflection and intercepts.

=x

y

3– 2

−18

x

( 1–2, −16)

Graphs of cubic functions in power function form exercise 3J 1

130

Without sketching graphs for each of the following, state: i the dilation factor ii the coordinates of the stationary point of inflection. a y = 2(x − 1)3 + 3 b y = 3(x + 5)3 − 2 − 3 c y = 2(x − 6) − 8 d y = −7(x + 4)3 + 1 3 e y = (x − 9) + 4 f y = x3 − 7

Maths Quest 11 Mathematical Methods CAS

1 2

1

g y = −(x + 1)3 − 1

y= k y= i

−1 4 (x −1 3 x 3

h y = 2 (x + 2)3

− 3)3 + 2

j l

y = 4x3 y = −2x3 − 2

Sketch the graph of each of the following, showing the stationary point of inflection and intercepts. a y = 2(x − 2)3 + 2 b y = −3(x + 3)3 + 81 3 c y = 4(x − 4) − 32 d y = −5(x − 1)3 + 5 − 3 e y= x −8 f y = x3 − 1 1 3 g y = (x + 2) + 27 h y = (x + 5)3 − 32 2

2 We23

i

y=

k y= m y=

−1 3 3 (x − 3) − 9 1 (x + 2)3 + 25 5 5x3

−1 4

j

y=

(x + 1)3 + 2

l

y = −2x3

n y = 3x3 − 3

3 Sketch the following, showing the stationary point of inflection. Intercepts are not required. Use a

CAS calculator to verify answers. a y = (4 − x)3 + 1 c y = 2(4x − 1)3

b y = 3(5 − x)3 − 3 d y = 5(3 − 2x)3 + 1

y = − 17 (3 − 4x)3 − 2 h y = (9 − 5x)3 − 7 j y = −2(5 − 2x)3 − 1

2

e y = 5 (1 − x)3

f

g y = −(4 − x)3 + 3 i

diGiTal doC doc-9731 Cubic graphs — basic form

y = 83 (6 − x)3 + 4

4 mC The basic cubic graph y = x3 undergoes a dilation factor of 6 from the x-axis and is translated

right 4 units and down 3 units. The equation for this graph is:

a y = 6(x − 4)3 − 3 C y = −6(x − 3)3 − 4 e y = −4(x + 3)3 + 6

B y = 3(x − 4)3 − 6 d y = 4(x + 6)3 + 3

5 mC The graph of y = 5(2 − x)3 + 9 has a stationary point of inflection at: a (5, 2) d (2, −9)

B (5, 9) e (2, 9)

C (−2, 9)

6 Suggest a possible equation for each of the following, given that each is a cubic with a dilation factor of

1 or −1 from the x-axis.

a

y

y

b (1, 5)

(−2, 2) x

x y

c

y

d

(−3, 4)

(−3, 0)

x

x

7 Write an equation for a cubic with: a a dilation factor of 4 from the x-axis and a stationary point of inflection at (2, 3) b a dilation factor of –2 from the x-axis and a stationary point of inflection at (−5, 1) c a dilation factor of 1 from the x-axis and a stationary point of inflection at (1, −2) d a dilation factor of

4 −1 2

from the x-axis and a stationary point of inflection at (0, 4).

diGiTal doC doc-9732 Investigation Graphs of the form y = a (x − b )n + c

ChapTer 3 • Cubic and quartic functions

131

domain, range, maximums and minimums 3k

Range = [−8, 10]

The domain of a function is the set of x-coordinates of points on its graph. The range is the set of y-coordinates of points on the graph. Normally, the domain and range of a cubic function are the set of all real numbers, or R for short, as such graphs extend indefinitely in both positive and negative axis directions. The domain and range of a restricted cubic function may be a smaller set of numbers. Actual maximum (within given domain) y Local maximum 10

−5

3

x

−8 Domain = [−5, 3]

Range = [−8, 10]

The restricted graph has a domain of x-values between −5 and 3, denoted [−5, 3]. y The range is [−8, 10]. 10 Square brackets are used to indicate that an end value is included. We show this with a small coloured-in circle on the graph. If an end value is not included, a curved bracket is used. We show such points on a graph using a ‘hollow’ circle. −5

3

x

−8 Domain = [−5, 3]

Function notation

When we wish to convey information about the domain of a function, the following notation may be used: f : [−4, 1] → R, where f (x) = (x − 1)(x + 2)(x + 4)

}

} The ‘name’ of the function

The domain The co-domain. The range is within this set.

The rule for the function

Note: The range is not necessarily equal to R; the range is within R. Worked example 24

For the function f: [−4, 1] → R where f (x) = (x − 1)(x + 2)(x + 4), sketch the graph of f (x), showing intercepts and the coordinates of any local maximum or local minimum, and state the range. Think

132

WriTe/draW

1

Determine the y-intercept by setting x = 0.

f (x) = (x − 1)(x + 2)(x + 4) f (0) = (0 − 1)(0 + 2)(0 + 4) = −8 The y-intercept is (0, −8).

2

Determine the x-intercept by setting y = 0. Use the Null Factor Law.

0 = (x − 1)(x + 2)(x + 4) x − 1 = 0 or x + 2 = 0 or x + 4 = 0 x=1 x = −2 x = −4 − The x-intercepts are (1, 0), ( 2, 0) and (−4, 0).

Maths Quest 11 Mathematical Methods CAS

3

Enter the rule into a CAS calculator (including the restricted domain).

4

Sketch the graph over the restricted domain.

f (x) = (x − 1)(x + 2)(x + 4) | −4 ≤ x ≤ 1 y A (−2, 0)

(1, 0) x

(−4, 0)

B 5

Use the CAS calculator to determine the maximum at point A.

The maximum (A) is (−3.12, 4.06).

6

Use the CAS calculator to determine the minimum at point B.

The minimum (B) is (−0.21, −8.21).

7

Use the graph and the local maximum and minimum to determine the range.

The range is [−8.21, 4.06].

y

The maximum and minimum of a graph within a certain domain are not necessarily the values of the local maximum or minimum. Sometimes an extreme value is simply the y-coordinate of an end point of a graph.

Absolute maximum

Local maximum

Local minimum x Absolute minimum Worked example 25

Sketch f: [0, 7) → R where f (x) = x(x − 5)2, showing intercepts, end points, and the local maximum and minimum, and state the range. Think 1

Determine the y-intercept by setting x = 0.

2

Determine the x-intercept by setting y = 0. Use the Null Factor Law.

3

Calculate the value of the end points of the restricted domain.

4

Use a CAS calculator to sketch the graph over the restricted domain. Note the right end point (7, 28) should be shown as a hollow circle.

WriTe/draW

f (x) = x(x − 5)2 f (0) = 0(0 − 5)2 =0 The y-intercept is (0, 0). 0 = x(x − 5)2 x = 0 or (x − 5)2 = 0 → x = 5 The x-intercepts are (0, 0) and (5, 0). f (0) = 0 f (7) = 7(7 − 5)2 = 7 × 22 = 28 y (7, 28) A

(0, 0)

(0, 5)

x

ChapTer 3 • Cubic and quartic functions

133

5

Use the CAS calculator to determine the local maximum at point A.

The local maximum is (1.67, 18.52).

6

Use the graph to determine the range. Note the use of a curved bracket to indicate that the end value is not included in the range.

The range is [0, 28).

Domain and range will be discussed in more detail in chapter 4. A method of finding maximums and minimums without a calculator will be covered in the study of calculus later in this book.

domain, range, maximums and minimums exercise 3k

A graphics calculator is required for this exercise. State the domain and range of the sections of graph shown in each case.

1

y

a diGiTal doC doc-9733 SkillSHEET 3.2 interval notation

y

b

(−2, 5)

(1, 7)

(−5, 2)

(−2, 3)

(4, 0) x

x

(2, −2) c

(4, −2) d

y

(6, 10)

y

(−4, 2) x

(4, 1) x

(2, −3)

(−2, −2)

(−3, −5) y

e

(5, 5)

y

f

(2, 0)

(−1, 2)

(−3, 0)

x

(3, 0)

x

(−3, −8)

(4, −5)

(−1, −9) 2 We24,25 For each of the following, sketch the graph (showing local maximums and minimums, and

intercepts) and state the range. a f: [−1, 4] → R where f (x) = (x − 3)(x − 4)(x + 1) b f: [−5, 1] → R where f (x) = (x + 2)(x + 5)(x − 1) c f: [1, 3) → R where f (x) = (x − 2)2(x − 1) d f: (−3, 0] → R where f (x) = (x + 3)(x + 1)2 e f: [−8, 2) → R where f (x) = (2x − 3)(x + 1)(x + 7) f f: [0, 4] → R where f (x) = x3 − 6x2 + 9x − 4 g f: [−4, −1.442] → R where f (x) = x3 + 6x2 + 11x + 6 h f: (−2, 2.1) → R where f (x) = x3 + 2x2 − 5x − 6 i f: [−3, 5.1] → R where f (x) = −x3 + x2 + 17x + 15 1 j f: (−3, 2 ) → R where f (x) = 3x3 + 5x2 − 19x − 21 3 134

Maths Quest 11 Mathematical Methods CAS

3 mC The range of the function shown at right is: a

[−4.061,

18]

y (−0.786, 8.209)

B [−4.061, 8.209]

(4, 18)

6

C (0, 18] d (−2, 3] e (−2, 4]

x 1 3 (2.120, −4.061)

−2 y

4 mC Point A on the curve is: a B C d e

an intercept a local minimum an absolute minimum a local maximum an absolute maximum

A x y

5 A roller-coaster ride is modelled by the function

f (x) = 0.001(x − 10)(x + 20)(x − 40). a What is the height above ground level of the track at x = 50? b How far apart vertically are points A and B?

−20

20 A

Ground level 40

x

B

6 The course of a river as marked on a map follows the curve defined by the function

f (x) = 1.5x3 − 2.7x2 + x − 1. Find the coordinates of the southernmost point on the river between x = 0 and x = 2.

y

N

5 −1.5

1.5 −5

x diGiTal doC doc-9734 Investigation modelling the path of a roller-coaster

ChapTer 3 • Cubic and quartic functions

135

3l inTeraCTiViTY int-0262 modelling data using polynomials

modelling using technology

Scientists, economists, doctors and biologists often wish to find an equation that closely matches, or ‘models’, a set of data. For example, the wombat population of a particular island may vary as recorded in the following table. Year of study (x) Wombat population (W)

0

1

2

3

4

5

6

7

8

9

10

59

62

69

83

81

76

70

66

52

49

41

Wombat population

The graph below shows these data, with a possible model for the wombat population superimposed. 90 80 70 60 50 40 30 20 y = 0.0888x3 − 2.4598x2 + 14.196x + 55.063 10 0

2

4 6 Year

8 10

We will examine polynomial models up to degree 3, that is, models of the form: y y = a3x3 + a2x2 + a1x + a0 35 where a0, a1, a2 and a3 are constants. 30 Several technological options are available to assist in 25 Sum of squares of these obtaining models for data, including a CAS calculator, 20 lengths is minimised. spreadsheets and computer algebra systems such as 15 Mathcad. Many of these applications use a method 10 y = 2.6636x + 7.3182 5 involving minimising the sum of the squares of the vertical distances of the data points from the graph of the 0 2 4 6 8 10 12 x function — this is known as the least squares method.

modelling using a calculator A CAS calculator can be used to find a model or regression for a set of data. The following example employs cubic regression, but the general approach is the same for all types of regression. Worked example 26

Fit a cubic model to the following data using a CAS calculator. Write the equation and draw a rough sketch of the graph. 0 1 2 3 4 5 6 7 8 9 10 x 627 545 580 528 436 318 238 229 134 169 139 y Think 1 2 3

WriTe/draW

Use the CAS calculator to perform a cubic Enter the x- and y-values into two columns in regression. This fits a best-fit cubic to the given data. a spreadsheet on the CAS calculator. Write the equation. y = 1.37x3 + 19.14x2 + 8.55x + 607.48 Use the CAS calculator to sketch the graph over the even domain.

y 600

0

136

TUTorial eles-1416 Worked example 26

Maths Quest 11 Mathematical Methods CAS

10

x

modelling using technology

exercise 3l

Use a CAS calculator or other technology to answer the questions in this exercise. 1 We26 Find a linear model for each of the following sets of data, and draw a rough sketch of the graph. a

x

0

1

2

3

4

5

6

7

8

9

10

y

−30

0

5

−9

28

29

50

68

73

77

84

x

0

1

2

3

4

5

6

7

8

9

10

y

−15

−12

−26

−27

−12

−20

−39

−46

−50

−40

−67

c

x y

0 11

1 8

2 9

3 14

4 19

5 18

6 29

7 29

8 28

9 32

10 39

d

x y

0 53

1 44

2 39

3 42

4 35

5 32

6 30

7 29

8 23

9 27

10 19

b

diGiTal doC doc-9735 modelling

2 Find a quadratic model for each of the following sets of data, and draw a rough sketch of the graph. a

x y

0 19

1 4

2 48

3 60

4 36

5 88

6 126

7 116

8 159

9 168

10 122

b

x y

0 65

1 33

2 80

3 12

4 50

5 248

6 228

7 252

8 496

9 439

10 694

c

x y

0 −14

1 16

2 32

3 36

4 37

5 51

6 57

7 56

8 55

9 54

10 56

0

1

2

3

4

5

6

7

8

9

10

70

−27

2

−148

−327

−447

−639

−733

−910

−1204

d

x y

92

3 Find the cubic model for each of the following sets of data, and draw a rough sketch of the graph. a

x y

0 627

1 545

2 580

3 528

4 436

5 318

6 238

7 229

8 134

9 169

10 139

b

x y

0 21

1 28

2 91

3 182

4 81

5 203

6 345

7 397

8 730

9 873

10 1205

c

x

0

1

2

3

4

5

6

7

8

9

10

35

182

400

631

7 755

8 800

9 868

10 854

d

y

55

84

64

29

10

4

−17

x y

0 45

1 26

2 109

3 201

4 399

5 466

6 621

4 For the following data set, find and sketch: a a linear model b a quadratic model

x y

0 537

1 681

2 536

3 624

4 632

5 763

c a cubic model.

6 686

7 885

8 1090

9 1230

10 1451

5 Which of the models in question 4 fits best? 6 Use the model from question 1a to predict the value of y when x = 20. 7 Use the model from question 2a to predict the value of y to the nearest unit when x = 5.5. 8 Use the model from question 3a to predict the value of y to the nearest unit when x = 12. ChapTer 3 • Cubic and quartic functions

137

9 The value of shares in the company Mathsco is plotted by a sharemarket analyst over a

12-month period as shown. 0 J

Month Share price

1 J

2 A

3 S

4 O

5 N

6 D

7 J

8 F

9 M

10 A

11 M

0.50 0.58 0.53 0.76 1.00 1.50 1.55 2.20 3.06 3.83 4.79 4.40

a Find and sketch a quadratic model for the data. b Use your model to predict the share price 2 months later. c Give reasons why such a prediction may not be accurate. 10 The population of a colony of yellow-bellied sap-suckers on an isolated island is studied

over a number of years. The population at the start of each year is shown in the table below. Year Population

0

1

2

3

4

5

6

7

8

9

10

250

270

310

375

410

395

335

290

290

320

325

Find and sketch a cubic model for the population, and use it to estimate the population at the start of year 11.

3M

Finite differences

If pairs of data values in a set obey a polynomial equation, that equation or model may be found using the method of finite differences. Consider a difference table for a general polynomial of the form y = a3x3 + a2x2 + a1x + a0. We begin the difference table by evaluating the polynomial for x values of 0, 1, 2 etc. The differences between successive y-values (see table) are called the first differences. The differences between successive first differences are called second differences. The differences between successive second differences called the third differences. We will call the first shaded cell (nearest the top of the table) stepped cell 1, the second shaded cell stepped cell 2 and so on. x

y (= a3 x3 + a2 x2 + a1x + a0)

0

a0

First differences

Second differences

Third differences

a3 + a2 + a1 1

a3 + a2 + a1 + a0

6a3 + 2a2 7a3 + 3a2 + a1

2

8a3 + 4a2 + 2a1 + a0

6a3 12a3 + 2a2

19a3 + 5a2 + a1 3

27a3 + 9a2 + 3a1 + a0

6a3 18a3 + 2a2

37a3 + 7a2 + a1 4

64a3 + 16a2 + 4a1 + a0

6a3 24a3 + 2a2

61a3 + 9a2 + a1 5

125a3 + 25a2 + 5a1 + a0

If a3 ≠ 0, the above polynomial equation represents a cubic model, and the third differences are identical (all equal to 6a3). If a3 = 0, a2 ≠ 0 and the polynomial reduces to y = a2x2 + a1x + a0, that is, a quadratic model, and the second differences become identical (all equal to 2a2). 138 Maths Quest 11 Mathematical Methods CAS

If a3 = 0 and a2 = 0, the polynomial becomes y = a1x + a0, that is, a linear model, and the first differences are identical (all equal to a1). 1. Stepped cell 1 = a0 2. Stepped cell 2 = a1 + a2 + a3 3. Stepped cell 3 = 2a2 + 6a3 4. Stepped cell 4 = 6a3 Worked Example 27

Complete a finite difference table based on the data below, and use it to determine the equation for y in terms of x. x

0

1

2

3

4

5

y

−1

0

7

20

39

64

Think 1

2

3

Place the data in columns as shown, allowing space for 3 difference columns. Calculate the first differences and place them in the next column. The first differences are not constant, so we need to find the second differences. Calculate these and place them in the next column. The second differences are constant, so our table is complete. Showing the third differences is optional. The curve is a quadratic.

Write

x

y

0

−1

Differences 1st

2nd

3rd

1 1

0

6 7

2

0

7

6 13

3

0

20

6 19

4

39

5

64

0 6

25

4

Recall the stepped cell equations, and equate them to the shaded cells as shown: Stepped cell 1 = a0 Stepped cell 2 = a1 + a2 + a3 Stepped cell 3 = 2a2 + 6a3 Stepped cell 4 = 6a3 Here, * is used to denote solved values.

a0 = −1* [1] a1 + a2 + a3 = 1 [2] 2a2 + 6a3 = 6 [3] 6a3 = 0 [4]

5

[1] gives a0 = −1 and [4] gives a3 = 0. Substitute this information into [2] and [3].

a3 = 0* Sub a3 = 0 into [2]: a1 + a2 + 0 = 1 a1 + a2 = 1 Sub a3 = 0 into [3]: 2a2 + 6 × 0 = 6 2a2 = 6 a2 = 3* Sub a2 = 3 into [5]: So

[5]

6

Substitute a2 = 0 into [5] to find a1.

a1 + 3 = 1 a1 = −2*

7

Substitute your values for a0, a1, a2 and a3 into the equation y = a3x3 + a2x2 + a1x + a0.

y = a3x3 + a2x2 + a1x + a0 becomes y = (0)x3 + (3)x2 + (−2)x + (−1) y = 3x2 − 2x − 1

Chapter 3 • Cubic and quartic functions 139

The stepped equations work only if the finite differences table begins with x = 0 and x increases in steps of 1. It may be necessary on occasions to adjust the table to achieve this, as the following example shows. Worked Example 28

Complete a finite difference table based on these data and use it to determine the equation for y in terms of x. Think 1

2

3

4

Construct a difference table, leaving room for the x = 0 row. Calculate and fill in the first differences where possible. Note the first differences are constant, so the last two columns are optional. The relationship is linear. Working ‘backwards’, the first stepped cell must be −11 in order for the difference between it and the next cell to be 8.

1

x y

2 5

−3

3 13

4 21

5 29

Write

x

y

0

−11

Differences 1st

2nd

3rd

8 1

−3

0 8

2

5

0 0

8 3

13

0 0

8 4

21

0 0

8 5 5

Recall the stepped cell equations, and equate them to the shaded cells as shown: Stepped cell 1 = a0 Stepped cell 2 = a1 + a2 + a3 Stepped cell 3 = 2a2 + 6a3 Stepped cell 4 = 6a3 Here, an asterisk (*) is used to denote solved values.

6

[1] gives a0 = −11, [4] yields a3 = 0, and hence [3] yields a2 = 0. Substitute this information into [2].

7

Use the asterisked values to build the equation y = a3x3 + a2x2 + a1x + a0.

29

a0 = −11* [1] a1 + a2 + a3 = 8 [2] 2a2 + 6a3 = 0 [3] 6a3 = 0 [4]

a3 = 0* and a2 = 0* Sub a2 = 0 and a3 = 0 into [2]: a1 + 0 + 0 = 8 a1 = 8* y = a3x3 + a2x2 + a1x + a0 becomes y = (0)x3 + (0)x2 + (8)x + (−11) y = 8x − 11

So

Using simultaneous equations to find a polynomial model The method of fitting a polynomial to a set of data using finite differences requires the data to be sequential. Often this is not the case. Simultaneous equations can be used to find a polynomial model when the data are not sequential. 140 Maths Quest 11 Mathematical Methods CAS

The number of simultaneous equations required to find the rule of a degree n polynomial is n + 1. For example, to find a quadratic model, 2 + 1 = 3 points are required as a quadratic is a degree 2 polynomial. Each of the points are substituted into the general equation of the quadratic polynomial, y = ax2 + bx + c, to generate 3 simultaneous equations. These can be solved using elimination or by using a CAS calculator. Worked example 29

Using simultaneous equations, find a quadratic model for the points (−2, −9), (3, 1) and (1, 9). Think 1

Write down the general rule of a quadratic.

2

Substitute each point into the general equation to get three simultaneous equations.

3

WriTe

y = ax2 + bx + c a(−2)2 + b(−2) + c = −9 ∴ 4a − 2b + c = −9 a(3)2 + b(3) + c = 1 ∴ 9a + 3b + c = 1 a(1)2 + b(1) + c = 9 ∴a+b+c=9

Solve equations [1], [2] and [3] using elimination. Equation [1] − [2] to eliminate c. Equation [2] − [3] to eliminate c. Equation [4] × 8 Equation [5] × 5 Add equations [6] and [7] to eliminate a and solve for b. Substitute b = 4 into equation [4] to find a.

−5a

− 5b = −10 8a + 2b = −8 −40a − 40b = −80 40a + 10b = −40 −30b = −120 ∴b=4 −5a − 5(4) = −10 −5a = 10 ∴ a = −2 −2 + 4 + c = 9 ∴c=7

Substitute a = −2 and b = 4 into equation [3] to find c. 4

Write the rule.

[1] [2] [3] [4] [5] [6] [7]

y = −2x2 + 4x + 7

Worked example 30

Using simultaneous equations, find a cubic model for the points (−2, −10), (1, 2), (3, −20) and (6, 22). Use a CAS calculator to solve the simultaneous equations. Think

WriTe

1

Write the general rule for a cubic.

y = ax3 + bx2 + cx + d

2

Substitute each point into the general equation to get 4 simultaneous equations.

a(−2)3 + b(−2)2 + c(−2) + d = −10 −8a + 4b − 2c + d = −10 3 a(1) + b(1)2 + c(1) + d = 2 a+b+c+d=2 a(3)3 + b(3)2 + c(3) + d = −20 27a + 9b + 3c + d = −20 3 a(6) + b(6)2 + c(6) + d = 22 216a + 36b + 6c + d = 22

[1] [2] [3] [4]

3

Use the CAS calculator’s solve feature to determine a, b, c and d.

The values are a = 1, b = −5, c = −4 and d = 10.

4

Write the rule.

y = x3 − 5x2 − 4x + 10

ChapTer 3 • Cubic and quartic functions

141

exercise 3m

Finite differences

For each of the following, complete a finite difference table based on the data below and use it to determine the equation of y in terms of x.

1 We27,28 a

b

c

d

x

0

1

2

3

4

5

y

6

17

28

39

50

61

x

0

1

2

3

4

5

y

100

74

48

22

−4

−30

x

0

1

2

3

4

5

y

−4

7

32

71

124

191

x y

e

f

g

h

i

j

k

l

2

1

2

3

4

5

1

−9

−13

−11

−3

x

0

1

2

3

4

5

y

−7

−10

−1

26

77

158

x

0

1

2

3

4

5

y

16

17

20

31

56

101

x

0

1

2

3

4

5

y

−23

−11

5

25

49

77

x

1

2

3

4

5

y

28

27

32

31

12

x

0

1

2

3

4

5

y

−27

−18

−9

0

9

18

x y

1

2

3

4

5

−7

−3

−3

−7

−15

x

0

1

2

3

4

5

y

−66

−5

58

99

94

19

x

0

1

2

3

4

5

y

43

35

27

19

11

3

Triangular numbers may be illustrated as shown at right. If x is the number of dots on the base of each diagram, and y is the total number of dots: a complete the table below b find an equation linking x and y x (base dots)

0

1

2

3

4

5

y (total dots) c find the total number of cans in the supermarket display shown at right using the equation found in b, and check your answer by counting the cans.

142

Maths Quest 11 Mathematical Methods CAS

x=1

x=2

x=3

x=4

3

The diagonals in polygons of various types are shown at right in red. Find the relationship between the number of dots (x) and the number of diagonals (n). (Hint: Continue patterns in a difference table so that it is completed back to x = 0.)

4

If n is the number of different squares that can be found within a square grid of edge length x, find an equation for n in terms of x and use this equation to find the number of different squares on a chessboard.

x=1 x=2 n=1 n=5 5 Find a linear model for the following sets of points. a (2, 1) (−1, −23)

x=3 n=?

b (−4, 6) (8, −3)

6 We29 Using simultaneous equations, find a quadratic model for the following sets of points. a (−2, −13) (6, −37) (−4, −57) b (−1, 4) (1, −2) (4, 19) − c (4, 8) (0, 8) ( 4, 24) d (−5, −360) (−2, −96) (6, −272) 7 We30 Using simultaneous equations, find a cubic model for the following sets of points. Use a CAS

calculator to solve the simultaneous equations.

a (−6, 3) (−3, −27) (3, −33) (2, 3) c (4, −10) (6, 90) (8, 302) (−2, 2)

b (−2, −39) (1, 6) (4, 141) (−3, −118) d (−1, −4) (1, −8) (4, −314) (0, −6)

8 Using simultaneous equations, find a quartic model for the following sets of points. Use a

CAS calculator to solve the simultaneous equations. a (1, 2) (−3, 354) (4, 1313) (2, 79) (−1, −2) b (−4, 73) (0, 1) (2, −11) (−2, 13) (6, −707)

ChapTer 3 • Cubic and quartic functions

143

Summary expanding

• When expanding three linear factors: 1. expand two factors first, then multiply the result by the remaining linear factor 2. collect like terms at each stage 3. (x + 2)3 may be written as (x + 2)(x + 2)(x + 2) 4. (a + b)3 = a3 + 3a2b + 3ab2 + b3 (a − b)3 = a3 − 3a2b + 3ab2 − b3

long division of polynomials

• Long division of polynomials is similar to long division with numbers. • The highest power term is the main one considered at each stage. • The key steps are: 1. How many? 2. Multiply and write the result underneath. 3. Subtract. 4. Bring down the next term. 5. Repeat until no variables remain to be divided. 6. State the quotient and the remainder.

polynomial values

• P(a) means the value of P(x) when x is replaced by a and the polynomial is evaluated.

The remainder and factor theorems

• The remainder R = P(a) when P(x) is divided by x − a. • If P(a) = 0, then (x − a) is a factor of P(x).

Factorising polynomials

• To factorise a polynomial: 1. let P(x) = the given polynomial 2. use the factor theorem to find a linear factor 3. use long or short division to find another factor 4. repeat steps 2 and 3, or factorise by inspection if possible. Alternatively, use the factor function on a calculator.

Sum and difference of two cubes

• a3 + b3 = (a + b)(a2 − ab + b2) • a3 − b3 = (a − b)(a2 + ab + b2)

Solving polynomial equations

• To solve a polynomial equation: 1. rewrite the equation so it equals zero 2. factorise the polynomial as much as possible 3. let each linear factor equal zero and solve for x in each case.

Cubic graphs — intercepts method

y • To sketch a cubic function of the form f (x) = Ax3 + Bx2 + Cx + D: 1. determine if the expression is a positive or negative cubic (that is, 20 if A is positive or negative) 2. find the y-intercept (let x = 0) 3. factorise if necessary or possible; for example, obtain an expression in the form f (x) = (x − a)(x − b)(x − c) 4. find the x-intercepts (let factors of f (x) equal 0) −1 5. use all available information to sketch the graph.

Quartic graphs — intercepts method

Maths Quest 11 Mathematical Methods CAS

4

x

y

• To sketch a quartic function in the form f (x) = (x – a)(x – b)(x – c) (x – d): 1. find the y-intercept (f (0) = abcd) 2. find the x-intercepts (let factors of f (x) = 0) 3. use all available information to sketch the graph.

840

−8

144

5– 2

−5

0

3

7

x

Graphs of cubic functions in power function form

• A cubic function in the form y = a(x − b)3 + c has a y-dilation of a and a stationary point of inflection at (b, c). y

y

a>0

a<0

(b, c)

(b, c) x

x

Positive a

Negative a

domain, range, maximums and minimums

• The absolute maximum or minimum is either the y-value at a local maximum or minimum, or the y-value at an end of the domain.

modelling using technology

• An equation to model data may be obtained using a calculator. 1. Enter data as lists. 2. Find the regression equation (linear, quadratic etc.).

Finite differences

• To use the method of finite differences: 1. Set up a table as shown and find differences by subtracting successive values (value – previous value). Circle or shade the stepped cells. x

y

0

Stepped cell 1

First differences

Second differences

Third differences

Stepped cell 2 1

Stepped cell 3 Stepped cell 4

2

3

Etc. Etc.

4

Previous value Value – previous value

5

Value

2. Use the following equations to determine the polynomial model’s coefficients. 1. Stepped cell 1 = a0 2. Stepped cell 2 = a1 + a2 + a3 The equation of the polynomial model is 3. Stepped cell 3 = 2a2 + 6a3 y = a3x3 + a2x2+ a1x + a0. 4. Stepped cell 4 = 6a3

}

ChapTer 3 • Cubic and quartic functions

145

Chapter review S h orT anS Wer

1 Expand: a (x − 2)2(x + 10) c (x − 7)3

b (x + 6)(x − 1)(x + 5) d (5 − 2x)(1 + x)(x + 2).

2 Find the quotient and remainder when the first polynomial is divided by the second in each case. a x3 + 2x2 − 16x − 3, x + 2 b x3 + 3x2 − 13x − 7, x − 3 c −x3 + x2 + 4x − 7, x + 1 3 If P(x) = −3x3 + 2x2 + x − 4, find: a P(1)

b P(−4)

+ − 29x − 42.

4 Without dividing, find the remainder when 5 Show that x + 3 is a factor of

x3

−

2x2

c P(2a).

x3

3x2

− 16x + 5 is divided by x − 1.

6 Factorise x3 + 4x2 − 100x − 400. 7 Factorise: a 1 − 125x3

b (x − 2)3 + (x + 3)3.

8 Solve: a 5(x + 5)3 + 5 = 0

b (2x + 1)2 (x − 3)2 = 0

c x3 − 9x2 + 26x − 24 = 0.

9 Sketch: a y = x(x − 2)(x + 11)

b y = x3 + 6x2 − 15x + 8

c y = −2x3 + x2.

b y = (2x – 1)(x + 1)(x + 4)2

c y = −x(x + 5)3.

10 Sketch: a y = x(x – 7)(x – 2)(x + 4) 11 Sketch y =

−1 8

(x + 1)3 + 8.

12 Find the range of f : [−6, 3] → R, where f (x) = (x + 1)(2 − x)(x + 5). 13 Complete a finite difference table, and use it to determine the equation for y in terms of x for the

following data set. x

0

1

2

3

4

5

y

8

7

8

17

40

83

14 The following series of diagrams show the maximum number of regions

produced by drawing chords in a circle. x=0 x=1 x=2 x=3 Find a relationship between the number of chords (x) and the maximum r = 1 r = 2 r = 4 r = 7 number of regions (r). 15 Using simultaneous equations, find a cubic model for the points (−1, −10), (2, −4), (−3, −104) and (0, −2). mUlTip l e Ch oiCe

1 The expansion of (x + 5)(x + 1)(x − 6) is:

2

a x3 − 30

B x3 − 6x2 + 5x − 6

d x3 − 31x − 30

e x3 + 5x2 − 36x − 30

x3

+

5x2

+ 3x − 9 is the expansion of:

a (x + 3)3

C x3 + 12x2 − 31x + 30

B x(x + 3)(x − 3)

C (x − 1)(x + 3)2

d (x − 1)(x + 1)(x + 3)

e (x + 1)(x + 2)(x − 3) Questions 3 and 4 refer to the following long division.

x2 + x + 2 x + 4)x3 + 5x2 + 6x − 1 x3 + 4x2 x2 + 6x x2 + 4x 2 −1 2x 2 +8 2x −9 3 The quotient is: a −9

146

B 4

Maths Quest 11 Mathematical Methods CAS

C x+4

d x2 + x + 2

e x3 + 5x2 + 6x − 1

4 The remainder is: a −9

B 2

C 4

d 2x − 1

e 2x + 8

d 7

e 35

d 7

e 8

d (x + 3)

e (2x − 1)

5 If P(x) = x3 − 3x2 + 7x + 1, then P(−2) equals: a −34

B −33

C −9

− 7x is divided by x − 1 is: C 6 3 7 Which of the following is a factor of x − 3x2 − 18x + 40? a (x − 4) B (x − 2) C (x + 1) 6 The remainder when a −6

8

9

x3

+

x3

B 1

6x2

− 15x + 8 factorises to:

a (x − 1)2(x + 8)

B (x + 1)2(x + 8)

d (x + 1)(x + 2)(x + 4)

e (x − 1)(x + 2)(x + 4)

−

64x3

y3

C (x + 2)3

factorises to:

a (4x − y)(16x2 + 4xy + y2)

B (4x − y)(16x2 − 4xy + y2)

d (4x + y)(16x2 − 8xy + y2)

e (4x + y)(16x2 − 4xy + y2)

C (4x − y)(16x2 + 8xy + y2)

10 Which of the following is the solution to −(x − 4)3 − 2 = 6? a −6

B −2

C 2

d 4

e 6

11 Which of the following is a solution to (x − 11)(3x + 5)(7 − 3x)(2x + 5) = 0? a −11

B

−3

3

d 3

12 The equation for this graph could be: a B C d e

7

5

C 7

5

e 3

y

y = (x − 5)(x + 1)(x + 3) y = (x − 3)(x − 1)(x + 5) y = (x − 3)(x + 1)(x + 5) y = (3x − 1)(x + 1)(x − 5) y = (5 − x)(1 + x)(3 + x)

−3

−1

5

x

y

13 The equation for the graph shown below could be: a B C d e

y = (x – 3)2(x + 3)2 y = (x – 3)(x + 3)3 y = (x – 3)3(x + 3) y = (x – 3)4 y = (x + 3)4

0

−3

3

x

−81

14 Which of the following shows the graph of y = −2(x + 5)3 − 12? y y a B x

C

y x

(−5, 12)

(−5, −12)

(5, −12) x

d

y

y

e

(5, 12) x

(−5, −12)

x

ChapTer 3 • Cubic and quartic functions

147

Questions 15 and 16 refer to the following graph (below). y (−7, 252) (4.813, 60.370) −5

2

7 x

(−2.147, −108.222)

15 The domain of the graph is: a [−108.222, 252)

B (−2.147, 4.183]

d (−7, 7]

e (−6, 7]

16 The range of the graph is: a [−108.222, 252)

B [−2.147, 4.183]

d [7, 252)

e [0, 252)

C (−108.222, 60.370]

C [−108.222, 60.370]

17 The data below obey which type of relationship?

x

0

1

2

3

4

5

y

0

4

16

66

208

520

a Linear

B Quadratic

C Cubic

d Quartic

e None of the above

18 Which of the following points lies on the curve of the quadratic model that fits the points (1, 0), (0, −7)

and (2, 11)? a (3, −4)

e x T ended r e SponS e

C (−2, −7)

B (0, 7)

d (−1, −10)

e (−1, −14)

1 For P(x) = 5x3 − 3x2 − 6x − 22, find P(3) and P(−x). 2 Find the value of m if x + 3 is a factor of 2x3 − 15x2 + mx − 21. 3 Factorise x3 − 2x2 − 9x + 18. Sketch the graph of f (x) = x3 − 2x2 − 9x +18. 4 Factorise (3x − 2)3 + (x + 5)3. 5 Determine the x- and y-intercepts of the cubic graph y = (2 − 3x)(4x + 1)(2x − 7). Hence, sketch the

graph. 6 The graph y = x3 has been moved parallel to the x-axis 5 units to the left and moved upwards 2 units

from the x-axis. What is the equation of the translated graph and what are the coordinates of the point of inflection? Sketch the translated graph. 7 Sketch the graph of y = (x + 2)2(x − 3)(x − 4), showing all intercepts. 8 The polynomial P(x) = x3 + ax2 + bx + 54 is exactly divisible by x − 9 and also exactly divisible by x − 6. a Find the values of a and b. b Find the third factor. c Hence, sketch the graph of the polynomial y = x3 + ax2 + bx + 54. 9 Factorise x3 − 2x2 − 3x + 6 over the real number field. Sketch the graph of y = x3 − 2x2 − 3x + 6. 10 Find the points of intersection between y = x3 − x2 − 19x − 13 and 3x + y − 7 = 0. 11 Use the method of finite differences to fit a polynomial model to the following data.

x y

148

0 4

1 16

Maths Quest 11 Mathematical Methods CAS

2 25

3 30

4 30

12 A diagram of a proposed waterslide based on a cubic function appears below.

Find: a the height, h1, of the top of the slide b the coordinates of point A (where the slide enters the water) c the length, L, of the ladder d the height, h2, of the ‘mini-hump’ to the nearest centimetre. y

y = −0.008(x3 − 30x2 + 285x − 900)

L

h1 h2

(−5, 0)

A

(8, 0.22)

x

13 An innovative local council decides to put a map of the district on a website. Part of the map involves

two key features — the Cubic River and the Linear Highway. A mathematically able website designer has found the following equations for these features: Cubic River: y = x3 + x2 − 4x − 4 Linear Highway: y = 5x + 5. a Sketch the river and highway, showing x- and y-axis intercepts. b Find the coordinates of the points of intersection of the highway and the river. c A fun-run organiser wishes to arrange checkpoints at the closest points of intersection. Find the distance between the proposed checkpoints. 14 A cubic function in the form f (x) = ax3 + bx2 + cx + d has the following values. x y

0 42

1 36

2 20

3

4

5

0

−18

−28

Use finite differences to find the values of a, b, c and d. State one factor of f (x), giving your reasoning. Using long or short division, factorise f (x). Sketch the graph of f (x), labelling all intercepts. 15 The height (in centimetres) of a wave above a 1-metre pole is measured over an interval of 8 seconds. The wave’s height above the 1-metre pole has been found to approximate the function H1 = t3 − 13t2 + 48t. a Find the initial height of the wave above the 1-metre pole. b Using a CAS calculator, sketch the function and find the local maximum and minimum height of the wave. c The height of a later wave above the 1-metre pole is found to approximate the function H2 = t3 − 14t2 + 53t − 40. Show at what times the height of this wave is exactly the same height as the pole. a b c d

diGiTal doC doc-9737 Test Yourself Chapter 3

ChapTer 3 • Cubic and quartic functions

149

ICT activities Chapter opener diGiTal doC • 10 Quick Questions doc-9724: Warm up with ten quick questions on cubic and quartic functions (page 105)

3B

long division of polynomials

TUTorial • We4 eles-1412: Watch how to perform long division of polynomials (page 108)

3C

polynomial values

diGiTal doCS • doc-9725: Investigate solutions to cubic equations (page 111) • WorkSHEET 3.1 doc-9726: Review the discriminant (page 111)

3d

The remainder and factor theorems

diGiTal doCS • doc-9725: Investigate solutions to cubic equations (page 113) • SkillSHEET 3.1 doc-9727: Practise calculating and using the discriminant (page 114)

3e

Factorising polynomials

diGiTal doC • doc-9728: Investigate zeros of cubics (page 117) TUTorial • We12 eles-1413: Use long division to factorise a cubic (page 114)

3F

Sum and difference of two cubes

TUTorial • We15 eles-1414: Watch how to factorise expressions using the sum or difference of two cubes formulae (page 117)

3G

Solving polynomial equations

diGiTal doC • WorkSHEET 3.2 doc-9729: Factorising cubics and quartics using long division, applying the Null Factor Law to determine x-intercepts and sketching cubics and quartics (page 121)

150

Maths Quest 11 Mathematical Methods CAS

3h

Cubic graphs — intercepts method

TUTorial • We20 eles-1415: Sketch the graph of a cubic showing axial intercepts (page 123) diGiTal doC • doc-9730: Investigate the effect of changing coefficients of cubics in general form on its graph (page 125)

3J Graphs of cubic functions in power function form diGiTal doCS • doc-9731: Investigate the graphs of cubic functions in power form (page 131) • Investigation doc-9732: Graphs of the form y = a(x − b)n + c (page 131)

3k

domain, range, maximums and minimums

diGiTal doCS • SkillSHEET 3.2 doc-9733: Practise expressing intervals using varying notation (page 134) • Investigation doc-9734: Modelling the path of a roller-coaster (page 135)

3l

modelling using technology

inTeraCTiViTY • Modelling data using polynomials int-0262: Use the interactivity to consolidate your understanding of how to fit a polynomial model to data (page 136) TUTorial • We26 eles-1416: Watch how to fit a cubic model to a set of data using a CAS calculator (page 136) diGiTal doCS • doc-9735: Investigate the best model for a set of data (page 137)

Chapter review diGiTal doC • Test Yourself doc-9737: Take the end-of-chapter test to test your progress (page 149)

To access eBookPLUS activities, log on to www.jacplus.com.au

Answers CHAPTER 3 CUBiC and QUarTiC FUnCTionS exercise 3a

1 a b c d e f g h i j 2 a b c d e f g h i j 3 a b c d e f g h i j 4 a b c d e f 5 a b c d e f

expanding x3 + 7x2 + 6x x3 − 7x2 − 18x x3 + 8x2 − 33x 2x3 + 10x2 + 12x 48x − 3x3 5x3 + 50x2 + 80x x3 + 4x2 2x3 − 14x2 − 30x3 − 270x2 −7x3 − 56x2 − 112x x3 + 12x2 + 41x + 42 x3 − 3x2 − 18x + 40 x3 + 3x2 − 36x + 32 x3 − 6x2 + 11x − 6 x3 + 6x2 − x − 6 x3 + 5x2 − 49x − 245 x3 + 4x2 − 137x − 660 x3 + 3x2 − 9x + 5 x3 − 12x2 + 21x + 98 x3 + x2 − x − 1 x3 + 13x2 + 26x − 112 3x3 + 26x2 + 51x − 20 4x3 − x2 − 36x + 9 10x3 − 49x2 + 27x + 36 −6x3 − 71x2 − 198x + 35 21x3 − 96x2 + 48x 54x3 + 117x2 − 72x 24x3 − 148x2 + 154x + 245 20x3 − 19x2 − 69x + 54 4x3 + 42x2 + 146x + 168 x3 + 6x2 + 12x + 8 x3 + 15x2 + 75x + 125 x3 − 3x2 + 3x − 1 x3 − 9x2 + 27x − 27 8x3 − 72x2 + 216x − 216 27x3 + 108x2 + 144x + 64 x3 − 4x2 − 67x − 110 3x3 + 15x2 − 18x −6x3 + 72x2 + 510x − 3600 − 3 x + 29x2 − 264x + 720 −x3 − 20x2 − 100x −x3 − 39x2 − 507x − 2197

exercise 3B

1 a b c d e f g h i j k l 2 a b

long division of polynomials 2 x + 3x − 6, 19 x2 + 2x − 1, 6 x2 + 1, 2 x2 − x + 6, −11 x2 + 4x + 3, −3 x2 + 3x − 3, −11 x2 − 2x − 3, −17 x2 − 4x − 18, −91 3x2 − 7x + 20, −35 4x2 − 8x + 18, −22 2x2 − 3x + 3, 7 2x2 − 9, 35 3x2 − 2x + 1, 5 2x2 + 5x − 6, −7

3

4

5

6

4x2 − 7x − 2, −3 x2 − 4x + 3, 8 x2 + x − 6, −11 3x2 + 2x + 1, 13 −x2 − 5x − 2, −14 − 2 3x − 2x + 4, −3 −x2 + 5x + 6, 9 −2x2 + 7x − 1, 1 x2 − x − 2, 3 x2, −7 x2 − x − 2, −8 −x2 − x − 8, 0 5x − 2, 7 2x2 − 2x + 10, −54 −2x2 − 4x − 9, −16 −2x2 + 4x − 1, 1 x3 + 2x2 + 5x – 2, −2 x3 + 2x2 – 9x – 18, 0 6x3 + 17x2 + 53x + 155, 465 The quotient is x2 − 6x + 101; the remainder is −1490. b The quotient is 2x2 + 44x + 1369; the remainder is 42 430. 16 x 160 c The quotient is 4x2 − + ; − 721 3 9 the remainder is 9 .

exercise 3d The remainder and factor theorems b 0 c 0 1 a −30 d −24 e −24 f k3 + 3k2 − 10k − 24 g −n3 + 3n2 + 10n − 24 h −27c3 + 27c2 + 30c − 24 2 a 58 c 11 b −8 f 6 d −9 e −202 g 158 i 35 h −6 j 441 3 a 6 b 3 c 1 e 2 f 2 d −2 g −5, 2 h a = −5, b = −3 4 a Not a factor b Is a factor c Is a factor d Not a factor 5 Show P(−2) = 0, P(3) = 0 and P(−5) = 0. 6 For example, for a show P (1) = 0. 7 a E b C c D d A 8 a (x − 1) b (x − 3) or (x − 2) c (x − 3) or (x + 2) d (x − 6) or (x + 4) or (x + 5) 9 a i 0 ii 0 iii 0 b i (x + 1)(6x2 + x − 2) ii (x + 1)(2x − 1)(3x + 2)

c d e f a b c d a b c d e f g h a b c a

d The quotient is the remainder is exercise 3C

1 a 10 d 43 g h i j k l 2 a b c

−77

9x3

−

30x2

+ 105x −

5235 2 .

c P 1 = 0 and (2x − 1) is a factor. 2

735 2 ;

1

2x − 1 = 0 gives x = 2 . −2 P( 3 )

= 0 and (3x + 2) is a factor.

3x + 2 = 0 givesx =

polynomial values b 11 c 18 e 3 f −22

−2

3.

d In general if (ax + b) is a factor, then

− b = 0. P a

2a3 − 3a2 + 2a + 10 16b3 − 12b2 + 4b + 10 2x3 + 9x2 + 14x + 18 2x3 − 21x2 + 74x − 77 −128y3 − 48y2 − 8y + 10 P(−11) = 230 522 P(102) = 1 735 072 853 P(2x + 9) = 256x4 + 4632x3 + 31 428x2 + 94 726x + 106 982 d P(x3 + 2) = 16x12 + 131x9 + 402x6 + 526x3 + 253

exercise 3e

1 a b c d e f g h i j

3–7

Factorising polynomials (x + 1)(x + 3)(x + 6) (x + 1)(x + 2)(x + 5) (x + 1)(x + 2)(x + 9) (x + 1)(x + 3)(x + 4) (x + 3)(x + 4)(x + 7) (x + 2)(x + 3)(x + 7) (x + 1)2(x + 2) (x + 2)2(x + 3) (x + 4)(x + 5)2 x(x + 5)(x + 8)

Column Column Column Column Column Column Column Column Column 1 2 3 4 5 6 7 8 9

P (x)

P (1)

P (2)

P (−1)

P (−2)

Rem when divided by (x − 1)

a

4

15

0

−5

4

15

0

−5

b

10

28

−2

−8

10

28

−2

−8

c

3

11

−7

−21

3

11

−7

−21

d

−7

−19

5

−7

−7

−19

5

−7

8 a

P (−8)

b P (7)

Rem when divided by (x − 2)

Rem when divided by (x + 1)

Rem when divided by (x + 2)

c P (a)

ChapTer 3 • Cubic and quartic functions

151

k x(x + 3)(x + 4) l x(x + 5)2 m x(x + 1)(x + 5) n x2(x + 6) 2 a (x − 1)(x + 1)2 b (x − 2)(x − 1)(x + 1) c (x + 1)2(x + 5) d (x − 3)(x + 2)2 e (x + 1)(x + 4)2 f (x − 5)(x − 2)(x + 2) g (x − 1)(x + 1)(x + 2) h (x − 3)(x + 1)(x + 2) i (x − 1)(x + 2)2 j (x + 2)(x2 − x + 3) k (x + 1)(x + 2)(x + 5) l (x − 3)(x + 1)(x + 3) m (x − 2)2(x + 3) n (x − 4)(x + 5)(x + 8) o (x − 1)(x + 1)(x + 2)2 p (x − 3)(x + 2)3 q (x − 1)(x + 1) (x + 3)2 r x(x − 1)(x − 7)(x + 3) 3 a x(x − 2)(3x + 5) b 2x(x + 1)(2x − 1) c 3x(x − 4)(x + 2) d −2x(x + 3)2 e 6x2(x − 1) f −x(x + 4)(x + 3) g −(x − 1)(x + 1)(x + 3) h −2x(x − 3)(x − 2) i −(x + 2)(2x − 1)(3x − 2) j −(x − 2)2(5x − 4) k −(x − 1)(x + 3)(x − 5)(x + 2)2 l (x − 2)2(3x + 5)(8x + 1) 4 a (2x + 3)(x − 1)(x + 2) b (3x − 1)(x + 1)(x + 4) c (3x + 2)(x − 2)(x + 2) d (4x + 3)(x + 3)(x + 5) e (5x − 1)(x + 1)2 f (x + 2)(x2 − x + 2) g (x + 1)(2x + 3)2 h (x − 2)(2x − 1)(3x − 4) i (x + 4)(2x − 5)(5x + 2) j (7x − 2)(x − 2)(x + 4) k (x − 3)(x + 1)2(2x + 1) l (x − 1)(x + 3)(2x + 1) (3x − 2) exercise 3F

Sum and difference of

two cubes 1 a a = x, b = 6 b a = 2y, b = z c a = x + 5, b = 3 d a = 1, b = 4h c u e a= f a = t, b = , b = eg 2 6 2 a (x − 5)(x2 + 5x + 25) b ( j + k)( j2 − jk + k2) c (y − 2)( y2 + 2y + 4) d (3x + y)(9x2 − 3xy + y2) e 8(2t − 3u)(4t2 + 6tu + 9u2) f (x − 1)(x2 + x + 1) 2 x x − 2 xp + 4 p2 g + 2 p 9 81 9

3 a (2a − 1)(a2 − a + 1) b x(x2 + 6x + 12) c 2(x + 2)(4x2 + 10x + 7) d −5(3w2 − 15w + 25) e 5m(7m2 − 3mp + 3p2) f (2x − 3)(13x2 + 15x + 9) g (3y + 5)(3y2 + 21y + 67) h (4x − 3y)(7x2 + 9xy + 21y2) i (1 − 5p)(13p2 − 16p + 7) j 2(3x − 8)(21x2 − 60x + 67) k (x2 + y3)(x4 − x2y3 + y6) l 2(x − 3)(x2 + 3x + 9) m 3(a + 1)(a2 − a + 1) n 6(x2 + 4)(x4 − x2 + 7) 4 m = 27, n = 1

exercise 3G

152

−6 y

b −6

7 x

−1 −42

b d f

g 0

h

i −4

c e g

−2,

k

−

−4,

2, 6

−3

l 6 6

3 a −2, 0, 2 −3,

3

−5 −2 −4

y

d

3

60

2, 0, 2

m 0, 4, 5 4 a −4, 1, 4 −4, −2,

2

g −2, −1, 3 −1 2

,

3 2,

3

−

5– 2

−2, −1,

−7

2,

−4

e

y 12 − 1–

0

3 3– 4

4

2

x

x

n 5, 4 b −4, 0, 4 e

j 0, 2, 3

j

x

1 3

m 2 , 0, 7 , 5

d

11

−5

3, −4, −7 2, 3 −2, −1, 1, 2 1 j 0, 9

b d f h

1, 5, 9 4 −5, 8 i −1, 0, 1

g

−1 −88

l −13

2 a 1, 2, 5

−

−8

j 1

k 12

d

y

c

Solving polynomial

equations 1 a 6 c 14 e 4

6 C 7 B

3k

x

1 2 3

c (3 3x + 3 m y) ((3 3 x) 2 − 3 3mxy + ( 3 my)2 )

j (st + g2)(s2t 2 − stg2 + g4)

1

y

1 a

b a = 3 3x , b = 3 m y

1

2 + i 3k − 9 k + 2 2 4

Cubic graphs — intercepts

5 a 3x3 + my3 = ( 3 3 x )3 + ( 3 m y)3

5 a b c d e f g h i j

h (3r − 1)(9r2 + 3r + 1)

exercise 3h

method

h

−5,

c −5, 0, 5

0

f 0, 2

,0 4

i 0, 5

−1

k 0, 4

l −7, 0, 1

n 0 b −2, 3, 5

c −5, 1, 5

e

−1,

2, 3

h −5, 0, 1, 2 k −3, −2,

1 2

−1

2,

6 x

3

f −2, 1, 5 i −6,

y

f

1

−54 −1

y 30

2 a

l −2, −1, 1

−

8.002, 9.008, 15.994 12, −7, 7 − 9, 13 16 − 6.91 14.22, −1.071, 0.854 3.627 0.646, −1.948, −0.199 −4.894, −1.325 −12.556, 0.418 −

Maths Quest 11 Mathematical Methods CAS

−5

−3

2

x

8

x

y

b −8

− 3– 2

−192

y

e

y

c −1

−3 −1 −21

x

2

5D 8 E

x

7

6 C 9 D

10 a

7 B

y

5 y

f

y

d

−1

x

12

150

−10

−1

−5

x

−3 −2

−1

4 3

2

−8

x

x

2

y

h

x

−9

y

f

factor, Δ = (−2)2 − 4 × 1 × 5 = 4 − 20 = −16. ∴ As Δ < 0, the quadratic factor has no real solutions, so y = x3 − x2 + 3x + 5 has only one real solution at x = −1. d Find the discriminant of the quadratic factor, Δ = (−2k)2 − 4 × 1 × 5 = 4k2 − 20. i For 2 real solutions, the quadratic factor must have 1 solution. ∴ Δ = 0. Solve 4k2 − 20 = 0. ∴ If k = ± 5 , the cubic will have 2 real solutions. ii For 3 real solutions, the quadratic factor must have 2 solutions. ∴ Δ > 0. Solve 4k2 − 20 > 0.

x

y 8

g

y

e

b y = (x + 1)(x2 − 2x + 5) c Find the discriminant for the quadratic

−2 3

7 −7

x

1– 6

4 a

y (−0.33, 14.81) 14

y

3a

−7

−2

−1

−2

exercise 3i

1 a

12 (−5, −36)

−2

b y

b

y

x

1

Quartic graphs — intercepts

method

x

1

∴ If k < − 5 or k > 5 , the cubic will have 3 real solutions.

0 −2 −1

y (6.04, 10.02) −4 5 7 x

b

2 3

y

6

8 −2 −1 0

−3

−2

−140 (−0.71, −144.84)

−1

x

c

y

c

−1

3

−1

x

d

−2

1

y

0

e

x

(3, 16) x

1 y

−1 7

(−2, 0)

2 3

x

y

(1.67, −28.44)

4 24

−4

0 1 −5

−9

y

d

x

y −5

x

d −2

2

c

y

(−1, 0)

8

−4

x

(−0.76, −6.64)

x

−3

0

1– 2

3 x

ChapTer 3 • Cubic and quartic functions

153

6 E 7 D 8 a

y

f

4

i i y

−1

2

−2.23 −0.24 2

x

0

1– 3

h

x

b

2 a

ii (0, −2) y (2, 2)

(0.46, 12.09) −14 9

125

1.30

3.36 x

(−0.83, 0.10)

(−3, 81) (2.61, −34.98)

3 a = −5 4 a There are many possible quartic

equations, including y = −(x + 3)(x – 3)3 or y = −x4 + 81 or y = −(x − 3)(x + 3)3. b y = −(x + 3)(x – 3)3

y c (−2.38, 13.68)

x (1.22, 6.47)

−3.18 −0.68 0

y

81

d

1.86

c

0

3

y (0.09, 120.35) 9.51

−288

10

(1, 5)

y

e

2 y

−1.90 (0, −3) −3

81

1.90

x

0

−2

− 3)(x +

3

(−3.22, 1027.74)y

−4.75

y

y

f

358

3)3

x

−8

(−0.06, 357.53) (2.35, 682.19)

x

x

y

e

(−1.29, −11.33) (1.29, −11.33)

−3

y

d

f

x

x

(6.95, −1343.39)

y = −x4 + 81

6 (4, −32)

(−3.28, −98.19)

x

y

x

120 1.62 −4.33 −1.79 −3

y

b

0 1 x

−5

x

1

y

y

y=

ii (0, 0)

3

(−1.55, −13.04)

1

−(x

ii (0, 0)

−1

l i −2

x

y

g

ii (3, 2)

4

j i 4 k i

0

−1

3.74

1

x

x

−1 81

−3

5 C

154

0

3

x

exercise 3J Graphs of cubic functions in power function form 1 a i 2 ii (1, 3) b i 3 ii (−5, −2) c i −2 ii (6, −8) d i −7 ii (−4, 1) e i 1 ii (9, 4) f i 1 ii (0, −7) g i −1 ii (−1, −1) 1 h i ii (−2, 0) 2

Maths Quest 11 Mathematical Methods CAS

y

g

(−2, 27)

−5

35

x

y

h

b

1

y

y

j

30 –2 x

x

4

−1

( 5– , −1) 2

(5, −3)

x

(−5, −32) y

i

y

c x

x

( 1– , 0) 4

(3, −9) y

j (−1, 2)

y

d

1

c y = 4 (x − 1)3 − 2

1 3–4 1

d y=

( 3– , 1) 2 x

x

y

k (−2, 25)

y

e

26 3–5

(1, 0) x x

−7

4 A 5 E 6 a y = (x − 1)3 + 5 b y = (x + 2)3 + 2 c y = −(x + 3)3 + 4 d y = −(x + 3)3 7 a y = 4(x − 2)3 + 3 b y = −2(x + 5)3 + 1

−1

3 2x

+4

exercise 3k domain, range, maximums and minimums 1 a Domain [−5, 4], range [−2, 5] b Domain [−2, 4], range [−2, 7] c Domain [−4, 2], range [−5, 2] d Domain [−2, 6), range [−2, 10) e Domain (−3, 5), range [0, 5) f Domain (−3, 4], range [−9, 0]

2 a

y 12

(0.472, 13.129)

y

l

y

f

y

4 (3.528, −1.128)

( 3– , −2) 4

(1, −2)

m

−1

x

x

x

Range [−1.128, 13.129] y

b (−3.732, 10.392) (1, 5)

y

g

(4, 3)

−5 x

x

1 x

−2

(−0.268, −10.392)

−10

Range [−10.392, 10.392] y

n

c

y

h x

1

y

x

−3

(9–5 , −7)

(1.333, 0.148) 1

3 a

y

i

(4, 1) 5

Range [0, 2)

(3, 2)

y 3

d

y

x

2

Range [0, 3]

(−2.333, 1.185)

(6, 4) x x

−3

−1

x

ChapTer 3 • Cubic and quartic functions

155

e (−4.689, 105.526)

y

−7

b 3– (2, 27) 2

−1

x −21 (0.355, −22.822)

y 0 −10 −20 −30 −40 −50 −60

600 500 400 300 200 100 0

x

2 4 6 8 10

y = −4.6273x − 9.0455

1

−4

4

40 30 20 10 0

x

d

(3, −4)

g (−2.577, 0.385) y x −2 (−1.422, −0.385)

−3

2 a (−4, −6)

Range [−6, 0.385] h (−2, 4) y (2.1, 1.581) −1 −6

b x

2

(0.786, −8.209)

Range [−8.209, 4) y

i

(2.737, 48.517)

−3 −1 (−2.070, −7.035)

5 (5.1, −4.94)

−3

0 2 4 6 8 10

−21

Range [−32, 13.169]

3 A 5 a 28 m 6 (0.971, −1.201) exercise 3l

1 a

156

4 B b 20.785 m

modelling using technology

y y = 11.436x − 23.091 100 80 60 40 20 0 −20 2 4 6 8 10 x

2 4 6 8 10

y y = 12.953x2 − 47.679x + 613.66

0

c

0 2 4 6 8 10

b

y y = 1.5056x3 − 6.6906x2 + 35.465x + 20.615 1200 1000 800 600 400 200 0 0 2 4 6 8 10 x

Maths Quest 11 Mathematical Methods CAS

x

5 The cubic model 6 205.629 7 99 8 330 9 a y

x

y y = 1.3747x3 − 19.138x2 + 8.5536x + 607.48 600 500 400 300 200 100 0 0 2 4 6 8 10 x

x

0 2 4 6 8 10

y y = 10.006x3 − 2.1369x2 + 9.8652x + 577.45 1400 1200 1000 800 600 400 200 0

x

y = −11.124x2 − 15.655x + 79.329

x

0 2 4 6 8 10

1400 1200 1000 800 600 400 200

6 5 4 3 2 1 0

y −200 −400 −600 −800 −1000 −1200

3 a

b

y y = −1.1014x2 + 16.75x − 5.5664

0

(1, −32)

x

x

y y = 81.855x + 419.36 1400 1200 1000 800 600 400 200 0

y y = 8.2459x2 − 21.777x + 56.371 800 600 400 200 0 0 2 4 6 8 10 x

2 4 6 8 10

0 2 4 6 8 10

4a

y y = −0.5291x2 + 20.8x + 0.5175 250 200 150 100 50 0 0 2 4 6 8 10 x

d

2.333 x

x

0 2 4 6 8 10

50 40 30 20 10 0 −10

y −1

0

c

x

Range [−7.035, 48.517] j (−2.111, 13.169)

300

y y = −2.8818x + 48.318 50 40 30 20 10 0

Range [−4, 0]

600

Share price ($)

y

f

x

900

y y = 3.0273x + 6.3182

2 4 6 8 10 12 Time (months)

x

b $6.78 c Anything can happen with prices in the

sharemarket. The last data point may be the beginning of a share price crash!

10

y y = 1.048x3 − 19.309x2 + 98.296x + 220.7 Population

c

2 4 6 8 10

d y y = −2.2988x3 + 32.575x2 −14.094x + 30.154

(−8, −133)

Range [−133, 105.526]

y y = 2.5307x3 − 23.29x2 + 38.808x + 59.448

c

400 300 200 100 0

0 2 4 6 8 10 Year

x

Year 11 estimate: 360 birds exercise 3m

1 a b c d e

Finite differences y = 11x + 6 y = −26x + 100 y = 7x2 + 4x − 4 y = 3x2 − 19x + 17 y = x3 + 3x2 − 7x − 7

f y = x3 − 2x2 + 2x + 16 g y = 2x + 10x − 23

exTended reSponSe

y

9a

2

1 P(3) = 68

P(−x) = −5x3 − 3x2 + 6x − 22

h y = −2x3 + 15x2 − 32x + 47

2 m = −70 3 Factors: (x − 3)(x + 3)(x − 2)

i y = 9x − 27

j y = −2x2 + 10x − 15

y

k y = −4x3 + 13x2 + 52x − 66 l y= 2 a

−8x

−11

+ 43

x (base dots)

0

1

2

3

y (total dots)

0

1

3

6 10 15

b y=

1 2

4

b

5

18

y

8

1

−8

x (dots)

0

n (diagonals) 0 1

−1 −1

−1 2 1 , ,3 4 3 2 y-intercepts: y = −14

5 x-intercepts: x =

0 2 5 9

1

1

a chessboard

b y = 4x + 3

c y= d y=

d y=

2

7

x

− 1 0 2 3 1– 4 3 2

x

−14 y

b

1 2 x − 2x + 8 2 −10x2 + 18x –

3 3 1 x + 2 x 2 − 12 x 4 −5x3 + 3x – 6

0

−4

−5 2 2 x + 7x + 11 2x2 − 3x − 1

6 Equation: y = (x + 5)3 + 2

20 7 a y = −x3 − 5x2 + 8x + 15 b y = 3x3 − 4x2 + 2x + 5 c y=

y

10 a

−3

b y=

x

1– 2

+ 2 x 2 + 6 x, 204 squares on

5 a y = 8x − 15 6 a y=

}

y

3

1 3 x 3

x

2 3

4 (4x + 3)(7x2 − 15x + 39)

y

n = 2 x2 − 2 x 4 n=

x

1

c

1 2 3 4 5 6

0

−3

x2 + 2 x

c 28 3

x

2

−4

Point of inflection: (−5, 2)

1– x 2

−1 0

y

−16

− 18

(−5, 2) y

c

2

8 a y = 5x4 + x3 − 2x2 + x − 3

x

0

−5

−1

b y = 4 x4 − 2x3 + x2 + 2x + 1

0

−5

x

ChapTer reVieW ShorT anSWer

1 a x3 + 6x2 − 36x + 40 b x3 + 10x2 + 19x − 30 c x3 − 21x2 + 147x − 343 d −2x3 − x2 + 11x + 10 2 a x2 − 16, 29 b x2 + 6x + 5, 8 c −x2 + 2x + 2, −9 3 a −4 b 216 c −24a3 + 8a2 + 2a − 4 4 −7 5 Show P (−3) = 0. 6 (x − 10)(x + 4)(x + 10) 7 a (1 − 5x)(1 + 5x + 25x2) b (2x + 1)(x2 + x + 19) 8 a −6 b

−1

2,

3

c 2, 3, 4

y

7 y

11

7

(−1, 8)

48

7 –8

3

−2 0

x

12 [−32, 40] 13 y = x3 – 2x2 + 8 1

3

x

4

8 a a = −14, b = 39 b The third factor is x + 1. y c

1

14 r = 2 x 2 + 2 x + 1 15 y = 2x3 – 5x2 + x – 2 mUlTiple ChoiCe

1D 4A 7B 10 C 13 B 16 A

2C 5B 8A 11 E 14 A 17 D

54

3D 6A 9A 12 E 15 D 18 D

−10

6 9

x

ChapTer 3 • Cubic and quartic functions

157

9 y = ( x − 2)( x −

3)( x + 3)

y

c 8.766 m d 58 cm 13 a

(3, 20)

y

6

14 a a = 1, b = −8, c = 1, d = 42 b (x – 3), f (3) = 0 c (x – 7)(x – 3)(x + 2) d y 42

0− 3

− 3

2

x

−3

10 Points of intersection are (−2, 13) and (5, −8).

158

−2

−1

(−3, −10)

− 3

79 x − x2 + x + 4 6 6 12 a 7.2 m b (15, 0) 11 y =

−2

Linear Highway: 5 y = 5x + 5 0 −4

1

3

x

Cubic River: y = x3 + x2 − 4x − 4

b (−3, −10), (−1, 0), (3, 20) c 10.198 km

Maths Quest 11 Mathematical Methods CAS

2

0

3

7

x

15 a Initial height = 0 cm b y 54.5185 (2 2–3 , 54.5185) (8, 64)

64

(6, 36)

36 0

22–3

c 1 s, 5 s and 8 s

6

8 x

Exam practice 1 CHAPTERS 1–3 1 Determine the exact values for x for which x2 − 5 = 3x.

[−3,

3 marks

2 Let f : 1] → R, where f (x) = + − 4x − 3. a Show that (x − 1) is a linear factor. b When factorised, f (x) = (x − 1)(x + 3)(ax + b). Determine the values of a and b. c Hence, sketch the graph of f (x).

2x3

5x2

2 marks

S ho rT a n S W er 25 minutes

3 marks 3 marks

3 The line y = ax + b passes through the point (2, 1) and is parallel to the line y − 4x + 3 = 0. Determine

the values of a and b. [−1,

2 marks

4 Let f : 3] → R, f (x) = − 4x − 3. a Express f in the form f (x) = a(x − b)2 + c. Hence, state the coordinates of the turning point. b State the range for f. c Determine the exact values of the x-intercepts. d Sketch f on a set of axes. Label all key features.

2x2

3 marks 2 marks 3 marks 3 marks

1 A straight line passes through the points (2, 4) and (−1, −5). Its equation would be: a y=6−x

B y=x+2

C 3y = x + 10

e y = −3x + 10

d y = 3x − 2

d

B (3x + 4)(x − 2)

− 13 13 or 10 10

e

−1

2

or

C

−4

3

12 minutes

each question is worth one mark.

2 The exact values of x for which 3(2x − 1)2 + 2(2x − 1) − 8 = 0 are: a (2x + 1)(6x − 7)

m U lTip l e C ho iC e

or 2

7 6

3 If (x − a) is a linear factor of the function f (x) = x3 + 2x2 − 11x − 12, then the possible value for a is: a −3

B −1

C 0

d 1

e 4

4 A cubic function has the following axis intercepts: x-intercepts 1, 3 and −4, and y-intercept 24. The

equation that would best describe this function would be: B (x + 1)(x + 3)(x − 4) + 24 e 2(x + 1)(x + 3)(x − 4)

a 24(x + 1)(x + 3)(x − 4) d 2(x − 1)(x − 3)(x + 4)

C (x − 1)(x − 3)(x + 4) + 24

5 Three linear functions are defined as:

L1: 4y − 3x − 8 = 0 L2: 3y + 4x + 1 = 0 L3: 3y − 4x − 2 = 0 Which one of the following statements is correct? a L1 and L2 are parallel. B L1 and L3 are parallel. d L1 and L3 are perpendicular. e L1, L2 and L3 are parallel.

6

C L1 and L2 are perpendicular.

y

−a

b

x

Which one of the following rules best describes the graph above? a y = (x + a)2(b − x) B y = (x − a)2(x − b) d y = (x − b)2(x + a) e y = (x − a)2(x − b)

C y = (x + a)(x − b)

7 A polynomial of degree 3 passes through the origin and has x-intercepts at −2 and 3. The equation for

this polynomial would best be described by which one of the following? a y = (x − 2)(x + 3) B y = (x + 2)(x − 3) C y = x(x − 2)(x + 3) d y = x(x + 2)(x − 3) e y = (x − 1)(x − 2)(x + 3) 8 A linear function L(x) = ax + 1 intersects the quadratic function Q(x) = 2x2 − x − 1 at the point (−0.5, 0).

The coordinates of the second point of intersection would be closest to: a (−1, 1) B (−1, 2) C (−1, 4) d (2, 3) e (2, 5) Exam practice 1

159

e xT ended r e S p onS e 40 minutes

1 Points A (−2, 1), B (1, 5) and C (5, 2) are vertices on a triangle. a Show that the triangle is an isosceles triangle. b Determine the equation of the perpendicular bisector of the line AC. c Show that the perpendicular bisector found in part b passes through vertex B.

3 marks 4 marks 2 marks

2 Seng and Victor are training for a 1000-metre race. The distance in metres they each ran was recorded

in 1-minute intervals. Victor was delayed for 1 minute because he was tying up his shoelaces when he was 110 m from the starting line. The individual distances are recorded in the table below. Time (minutes) Seng’s distance (m) Victor’s distance (m)

0 0 −110

1 200 0

2 380 200

3 680 560

4 880 875

5 1040 1050

a The relationship between distance and time for Victor can be modelled using a quartic function.

diGiTal doC doc-10161 Solutions exam practice 1

Determine this quartic function, Qv. Write your function in terms of Qv (distance in metres) and t (minutes). Express coefficients correct to 2 decimal places. 2 marks b Seng’s distance and time relationship can be modelled using a cubic function. Determine this cubic function, Cs. Write your function in terms of Cs (distance in metres) and t (minutes). Express coefficients correct to 2 decimal places. 2 marks c i State the domain of Qv. 1 mark ii State the domain of Cs. 1 mark iii Sketch both Qv and Cs on the same set of axes. 3 marks d Using your functions Qv and Cs, determine the time first Victor passes Seng. Write your answer in minutes and seconds. 2 marks 3 A children’s playground in the shape of a rectangle is to be constructed at a local park. The longer side is 5 metres longer than the shorter side. a If x is the shorter side, write down an expression for the longer side in terms of x. 1 mark b Write down an equation for the area, A(x), of the playground in terms of x. 2 marks c If the maximum area of the playground is 150 m2, determine the y dimensions of the playground. 3 marks A miniature bike path will be built within the playground. It will travel from the gate to the seesaws, slides and swings. The bike path can be modelled by the cubic function B(x) = x3 − 6x2 +10x, where x is the horizontal distance, in metres, and B(x) is the vertical distance, in metres, from the gate. d Assuming that the slides are on the path, determine the vertical distance the slides are from the gate if their measurement is 3 metres horizontally from the gate. 2 marks e If the seesaws are on the bike path, show that the seesaws are located 1 metre horizontally and 5 metres vertically from the gate. 2 marks f Determine the shortest distance, in metres, between the seesaws and the slides. Write your answer in exact form. 2 marks 0

160

Maths Quest 11 Mathematical Methods CAS

x

Chapter 4

Relations, functions and transformations Chapter ContentS 4a 4b 4C 4d 4e 4F 4G 4h 4i 4J

Set notation Relations and graphs Domain and range Types of relations (including functions) Power functions (hyperbola, truncus and square root function) Function notation Special types of function (including hybrid functions) Inverse relations and functions Circles Functions and modelling

4a

diGital doC doc-9738 10 Quick Questions

Set notation

Set notation is used in mathematics in the same way as symbols that are used to represent language statements.

definitions 1. A set is a collection of things. 2. The symbol {. . .} refers to a set. 3. Anything contained in a set, that is, a member of a set, is referred to as an element of the set. (a) The symbol ∈ means ‘is an element of’, for example, 6 ∈ {2, 4, 6, 8, 10}. (b) The symbol ∉ means ‘is not an element of’, for example, 1 ∉ {2, 4, 6, 8, 10}. Chapter 4 • Relations, functions and transformations

161

4. A capital letter is often used to refer to a particular set of things. 5. The symbol ⊂ means ‘is a subset of’, so, if B ⊂ A, then all of the elements of set B are contained in set A. For example, {2, 4} ⊂ {2, 4, 6, 8, 10}. 6. The symbol ⊄ means ‘is not a subset (or is not contained in)’, for example {2, 3} ⊄ {2, 4, 6, 8, 10}. 7. The symbol ∩ means ‘intersection’, so, A ∩ B is the set of elements common to sets A and B. For example, {1, 2, 3} ∩ {2, 4, 6} is {2}. 8. The symbol ∪ means ‘union’, so, A ∪ B is the set of all elements belonging to either set A or B or both. For example, {1, 2, 3} ∪ {2, 4, 6} is {1, 2, 3, 4, 6}. 9. The symbol A \ B denotes all of the elements of A that are not elements of B. 10. The symbol ∅ means the null set. It implies that there is nothing in the set, or that the set is empty. Worked exaMple 1

If A = {1, 2, 4, 8, 16, 32}, B = {1, 2} and C = {1, 2, 3, 4}, find: a A∩B b A∪C c A\B d {3, 4} ∩ B e whether or not: i 8 ∈ A ii B ⊂ A iii C ⊂ A. think

Write

a The elements that A and B have in common are 1 and 2.

a {1, 2}

b The elements that belong to either A or C are 1, 2, 3, 4, 8, 16 and 32.

b {1, 2, 3, 4, 8, 16, 32}

c The elements of A that are not elements of B are 4, 8, 16 and 32.

c {4, 8, 16, 32}

d {3, 4} and B have no common elements.

d {∅}

e iii 8 is an element of A. ii All elements of B belong to A. iii 3 is an element of C but not A.

e i Yes. 8 ∈ A ii Yes. B ⊂ A iii No. C ⊄ A

Sets of numbers Certain letters are reserved for important sets that arise frequently in the study of mathematics. 1. R is the set of real numbers, that is, any number you can think of. 2. N is the set of natural numbers, that is, {1, 2, 3, 4, 5, . . .}. 3. Z is the set of integers, that is, {. . ., −3, −2, −1, 0, 1, 2, 3, . . .}. a 4. Q is the set of rational numbers (that is, numbers that can be expressed as fractions in the form b where a and b are integers and b ≠ 0). 5. Q′ is the set of numbers that are not rational (that is, cannot be expressed as a ratio of two whole numbers). These numbers are called irrational, for example, π, 3, etc. Note that N ⊂ Z ⊂ Q ⊂ R, that is, R

Q'

Q 1– 2 3– 4 2– 3

Z N 0 −1 −2 −3 ...

exercise 4a

3 π ...

1 2 3 4 ...

7– 5 33 — ... 51

Set notation

1 We1 If A = {2, 4, 6, 8, 10, 12, 14}, B = {1, 3, 5, 7, 9, 11, 13}, C = {4, 5, 6, 7} and D = {6, 7, 8}, find: a A∩B b A∩C c A∩C∩D d A∪B f A\C g C \ D. e C∪D 162

Maths Quest 11 Mathematical Methods CAS

2 If A = {−3, −2, −1, 0, 1, 2, 3}, B = {0, 1, 2, 3} and C = {−3, 2, 3, 4}, find: b A \ B a A ∩ B ∩ C c A \ (B ∪ C) d A \ (B ∩ C) e A ∪ C. 3 If F = {a, e, i, o, u}, G = {a, b, c, d, e, f, g, h, i} and H = {b, c, d, f, g, h}, find: c G \ H a F ∩ G ∩ H b G ∩ H d H \ F e (F ∪ H) \ G. 4 MC Given that A ⊂ B, then A ∩ B is equivalent to: b ∅ e A

A B d A ∪ B

c {1, 2}

5 MC Given that C ⊂ B ⊂ A, then it follows that: a A ∪ B ∪ C is equivalent to: A B

B C

C A

D A ∪ B

E B ∪ C

A B

B ∅

C C

D A ∩ B

E B \ C

b (A \ B) ∩ C is equivalent to:

6 Answer true (T) or false (F) to each of the following statements relating to the number sets N, Z, Q

and R. a 7 ∈ R d 5 ∈ Q g {5, 10, 15, 20} ⊂ N j Q ⊂ N

4B

b −4 ∈ N e 1.5 ∈ Z

c 6.4217 ∈ Q f {5, 10, 15, 20} ⊂ Z

h Z \ N = {. . ., −3, −2, −1} k Q ∩ Z = ∅

i Z ∩ N = N l (Z ∪ Q) ⊂ R

Relations and graphs

A relation is a set of ‘ordered pairs’ of values or ‘variables’. Consider the following relation. The cost of hiring a trailer depends on the number of hours for which it is hired. The table below outlines this relation. Number of hours of hire Cost ($)

3 50

4 60

5 70

6 80

7 90

8 100

Cost of trailer hire ($)

Since the cost depends upon the number of hours, the cost is said to be the dependent variable, and the number of hours is called the independent variable. The information in the table can be represented by a graph, which usually gives a better indication of how two variables are related. When graphing a relation, the independent variable is displayed on the horizontal (or x) axis and the dependent variable is displayed on the vertical (or y) axis. So we can plot the set of points {(3, 50), (4, 60), (5, 70), (6, 80), (7, 90), (8, 100)}. The points are called (x, y) ordered pairs, where x is the first element and y is the second element. y 100 90 80 70 60 50 40 0 1 2 3 4 5 6 7 8 x Number of hours

This graph clearly shows that the cost increases as the number of hours of hire increases. The relation appears to be linear. That is, a straight line could be drawn that passes through every point. However, the dots are not joined as the relation involves ‘integer-valued’ numbers of hours and not minutes or seconds. The number of hours can be referred to as a ‘discrete dependent variable.’ Discrete variables include names and numbers of things; that is, things that can be counted (values are natural numbers or integers). Chapter 4 • Relations, functions and transformations 163

Some variables are referred to as continuous variables. Continuous variables include height, weight and volume; that is, things that can be measured (values are real numbers). If a relationship exists between the variables, we may try to find a rule and then write this rule in mathematical terms. In our example, the relationship appears to be that for each extra hour of hire the cost increases by $10 after an initial cost of $20. Cost = 10 × number of hours + 20 Using x and y terms, this is written as y = 10x + 20 Worked Example 2

Sketch the graph by plotting selected x-values for the following relations and state whether each is discrete or continuous. a y = x2, where x ∈ {1, 2, 3, 4} b y = 2x + 1, where x ∈ R Think

a 1 Use the rule to calculate y and state the

ordered pairs by letting x = 1, 2, 3 and 4.

2

Plot the points (1, 1), (2, 4), (3, 9) and (4, 16) on a set of axes.

Write/draw

a When x = 1, y = 12

= 1 x = 2, y = 22 = 4 x = 3, y = 32 = 9 x = 4, y = 42 = 16

(1, 1) (2, 4) (3, 9) (4, 16)

y 16 12 8 4 0

3

Do not join the points as x is a discrete variable (whole numbers only).

b 1 Use the rule to calculate y. Select values of x,

say x = 0, 1 and 2 (or find the intercepts). State the ordered pairs.

2

Plot the points (0, 1), (1, 3) and (2, 5) on a set of axes.

1

2

Join the points with a straight line, continuing in both directions as x is a continuous variable (any real number).

164 Maths Quest 11 Mathematical Methods CAS

4

x

It is a discrete relation as x can be only whole number values. b When x = 0, y = 2(0) + 1

= 1 x = 1, y = 2(1) + 1 = 3 x = 2, y = 2(2) + 1 = 5

(0, 1) (1, 3) (2, 5)

y y = 2x + 1 5 4 3 2 1 −2 −1 −10 −2 −3

3

3

1

2

x

It is a continuous relation as x can be any real number.

Worked Example 3

The pulse rate of an athlete, R beats per minute, t minutes after the athlete finishes a workout, is shown in the table below. 0 180

t R

2 150

4 100

6 80

8 70

a Plot the points on a graph. b Estimate the athlete’s pulse rate after 3 minutes.

Think

Write/draw

a 1 Draw a set of axes with t on the horizontal axis

a&b

and R on the vertical axis because heart rate is dependent on the time.

R (beats/min)

2

180 160 140 120 100 80 60 40 20

Plot the points given in the table.

b 1 Join the points with a smooth curve as t (time)

is a continuous variable. 2

Construct a vertical line up from t = 3 until it touches the curve.

3

From this point draw a horizontal line back to the vertical axis.

4

Estimate the value of R where this line touches the axis.

0 1 2 3 4 5 6 7 8 t (min)

When t = 3, the pulse rate is approximately 125 beats per minute.

Note: In any defined domain, for example, −3 ≤ x ≤ 3, the variable (x) is assumed continuous in that domain unless otherwise stated.

Exercise 4B

Relations and graphs

Questions 1, 2, and 3 refer to the following information. A particular relation is described by the following ordered pairs: {(0, 4), (1, 3), (2, 2), (3, 1)}. 1 MC The graph of this relation is represented by: A y B y 4 4

C

y 4

3

3

3

2

2

2

1

1

1

0 D y 4

1

2

3

4 x

0 E y 4

3

3

2

2

1

1

0

1

2

3

4

x

0

1

2

3

4 x

1

2

3

4 x

0

1

2

3

4 x

Chapter 4 • Relations, functions and transformations 165

2 MC The elements of the dependent variable are: A {1, 2, 3, 4} d {0, 1, 2, 3}

b {1, 2, 3} E {1, 2}

c {0, 1, 2, 3, 4}

3 MC The rule for the relation is correctly described by: A y = 4 − x, x ∈ R c y = 4 − x, x ∈ N E y = 4 − x, x ∈ {0, 1, 2, 3}

b y = x − 4, x ∈ N d y = x − 4, x ∈ Z

4 MC During one week, the number of people travelling on a particular train, at a certain time,

Number of people

D

M T W T F

0

M T W T F

M T W T F

Number of people

0 E

0

M T W T F

Number of people

0 C

B

Number of people

A

Number of people

progressively increases from Monday through to Friday. Which graph below best represents this information?

0

M T W T F

5 State whether each of the following relations has discrete (D) or continuous (C) variables. a {(−4, 4), (−3, 2), (−2, 0), (−1, −2), (0, 0), (1, 2), (2, 4)} b The relation that shows the air pressure at any time of the day y y c d

0

x

x

0

e The relation that shows the number of student absences per day during term 3 at your school f The relation describing the weight of a child from age 3 months to one year 6 WE2 Sketch the graph representing each of the following relations, and state whether each is discrete

or continuous. a

Day Cost of petrol (c/L)

b c d e f

Mon

Tues

Wed

Thur

Fri

Sat

Sun

68

67.1

66.5

64.9

67

68.5

70

{(0, 0), (1, 1), (2, 4), (3, 9)} y = −x2, where x ∈ {−2, −1, 0, 1, 2} y = x − 2, where x ∈ R y = 2x + 3, where x ∈ Z y = x2 + 2, where −2 ≤ x ≤ 2

166 Maths Quest 11 Mathematical Methods CAS

7 WE3 The table at right shows the temperature of a cup

t (min) 0 2 4 6 8 of coffee, T °C, t minutes after it is poured. 80 64 54 48 44 T (°C) a Plot the points on a graph. b Join the points with a smooth curve. c Explain why this can be done. d Use the graph to determine how long it takes the coffee to reach half of its initial temperature.

8 A salesperson in a computer store is paid a base salary of $300 per week plus $40 commission for each

computer she sells. If n is the number of computers she sells per week and P dollars is the total amount she earns per week, then: a copy and complete the table below. n

0

1

2

3

4

5

6

P b plot the information on a graph. c explain why the points cannot be joined together. 9 The speed of an aircraft, V km/h, t seconds after it starts to accelerate down the runway, is shown in the

following table. t

0

1

2

3

4

5

V

0

30

80

150

240

350

a Plot a graph that represents the information shown in the table. b Use the graph to estimate the speed after: i 2.5 s ii 4.8 s. 10 The cost, C dollars, of taking n students on an excursion to the zoo is $50 plus $6 per student. a Complete a table using 15 ≤ n ≤ 25. b Plot these points on a graph. c Explain why the dots can or cannot be joined.

4C Domain and Domain and range

range

A relation can be described by: 1. a listed set of ordered pairs 2. a graph 3. a rule. The set of all first elements of a set of ordered pairs is known as the domain, and the set of all second elements of a set of ordered pairs is known as the range. Alternatively, the domain is the set of independent values, and the range is the set of dependent values. If a relation is described by a rule, it should also specify the domain. For example: 1. the relation {(x, y): y = 2x, x ∈ {1, 2, 3}} describes the set of ordered pairs {(1, 2), (2, 4), (3, 6)} 2. the domain is the set X = {1, 2, 3}, which is given 3. the range is the set Y = {2, 4, 6}, and can be found by applying the rule y = 2x to the domain values. If the domain of a relation is not specifically stated, it is assumed to consist of all real numbers for which the rule has meaning. This is referred to as the implied domain of a relation. For example: {(x, y): y = x3} has the implied domain R. {(x, y): y = x } has the implied domain x ≥ 0.

Interval notation If a and b are real numbers and a < b, then the following intervals are defined with an accompanying number line. (a, b) implies a < x < b or (a, b] implies a < x ≤ b or a

b

x

a

b

x

Chapter 4 • Relations, functions and transformations 167

(a, ∞) implies x > a or

[a, ∞) implies x ≥ a or x

a

(−∞, b) implies x < b or

(−∞, b] implies x ≤ b or x

b

x

a

b

[a, b) implies a ≤ x < b or

x

[a, b] implies a ≤ x ≤ b or x

x

a b A closed circle indicates that the number is included and an open circle indicates that the number is not included. a

b

Worked Example 4

Describe each of the following subsets of the real numbers using interval notation. a b c −4

0

x

2

−3

5 x

0

Think

0 1

3

Write

a The interval is x < 2 (2 is not included).

a (−∞, 2)

b The interval is −3 ≤ x < 5 (−3 is included).

b [−3, 5)

c The interval is both 1 ≤ x < 3 and x ≥ 5 (1 is included,

c [1, 3) ∪ [5, ∞)

3 is not). Worked Example 5

Illustrate the following number intervals on a number line. a (−2, 10] b [1, ∞) Think

Write/draw

a The interval is −2 < x ≤ 10 (−2 is not included, 10 is).

a −2

b The interval is x ≥ 1 (1 is included).

10 x

0

b 0

x

1

Worked Example 6

State the domain and range of each of the following relations. a {(1, 2), (2, 5), (3, 8), (4, 11)} b Weight (kg) 10 15 20 25 30 Cost per kg ($) 3.5 3.2 3.0 2.8 2.7 c

d

y

y 4

−4 0

168 Maths Quest 11 Mathematical Methods CAS

0

x −4

4

x

5

x

think

Write

a Domain = {1, 2, 3, 4}

a 1 The domain is the set of first elements of the

ordered pairs. 2

Range = {2, 5, 8, 11}

The range is the set of second elements of the ordered pairs.

b 1 The domain is the set of independent values in the

table, that is, the weight values. 2

Range = {2.7, 2.8, 3.0, 3.2, 3.5}

The range is the set of dependent values in the table, that is, the cost values.

c Domain = R

c 1 The domain is the set of values that the graph

covers horizontally. 2

Range = [0, ∞)

The range is the set of values that the graph covers vertically.

d Domain = [−4, 4]

d 1 The domain is the set of values that the graph

covers horizontally. 2

b Domain = {10, 15, 20, 25, 30}

Range = [−4, 4]

The range is the set of values that the graph covers vertically.

Worked exaMple 7

For each relation given, sketch its graph and state the domain and range using interval notation. a {(x, y): y = x − 1} b {(x, y): y = x2 − 4, x ∈ [0, 4]} think

a 1 The rule has meaning for x ≥ 1 because

tUtorial eles-1461 Worked example 7

Write/draW

a

if x < 1, y = negative number. 2

3

Calculate the value of y when x = 1, 2, 3, 4 and 5, and state the coordinate points.

Plot the points on a set of axes.

When x = 1, y = 0 =0 x = 2, y = 1 =1 x = 3, y = 2 x = 4, y = 3 x = 5, y = 4 =2 y

(2, 1) (3, 2) (4, 3) (5, 2)

y= x−1

Join the points with a smooth curve starting from x = 1, extending it beyond the last point. Since no domain is given we can assume x ∈ R (continuous).

2

5

Place a closed circle on the point (1, 0) and put an arrow on the other end of the curve.

−1

6

The domain is the set of values covered horizontally by the graph, or implied by the rule.

Domain = [1, ∞)

7

The range is the set of values covered vertically by the graph.

Range = [0, ∞)

4

(1, 0).

1 0

1

2

3

4 5

x

Chapter 4 • Relations, functions and transformations

169

b 1 Calculate the value of y when x = 0, 1, 2, 3 and 4,

b When x = 0, y = 02 − 4

= −4 x = 1, y = 12 − 4 = −3 x = 2, y = 22 − 4 =0 x = 3, y = 32 − 4 =5 x = 4, y = 42 − 4 = 12

say, as the domain is [0, 4]. State the coordinate points.

2

Plot these points on a set of axes.

3

Join the dots with a smooth curve from x = 0 to x = 4.

4

Place a closed circle on the points (0, −4) and (4, 12).

(0, −4) (1, −3) (2, 0) (3, 5) (4, 12)

y y = x2 − 4, x ∈ [0, 4] 12 10 8 6 4 2 0 −2 −4

1 2

3 4

5

The domain is the set of values covered by the graph horizontally.

Domain = [0, 4]

6

The range is the set of values covered by the graph vertically.

Range = [−4, 12]

x

Verify that the graphs are correct using a graphics calculator or other technology.

domain and range

exercise 4C

1 We4 Describe each of the following subsets of the real numbers using interval notation. b a −2

0

1

0

c

5

d −3

0

−8

4

e

0

9

f −1

0

0

g

1

h −5

−2

0

−3

3

0 1 2

2 We5 Illustrate each of the following number intervals on a number line. a [−6, 2) b (−9, −3) − c ( ∞, 2] d [5, ∞) e (1, 10] f (2, 7) g (−∞, −2) ∪ [1, 3) h [−8, 0) ∪ (2, 6] i R \ [1, 4] j R \ (−1, 5) k R \ (0, 2] l R \ [−2, 1) 3 Describe each of the following sets using interval notation. a {x: −4 ≤ x < 2} b {x: −3 < x ≤ 1} −1 c {y: −1 < y < 3} d {y:

{x: x > 3} R R \ {1} R \ {x: 2 ≤ x ≤ 3}

Maths Quest 11 Mathematical Methods CAS

2

f h j l

2

{x: x ≤ −3} R+ ∪ {0} R \ {−2} R \ {x: −2 < x < 0}

4

4 MC Consider the set described by R \ {x: 1 ≤ x < 2}. a It is represented on a number line as: a

b

0 1 2

0 1 2 C

d

0 1 2

0 1 2

e

0 1 2 b It is written in interval notation as: a b C d e

(−∞, 1) ∪ (2, ∞) (−∞, −1) ∪ [2, ∞) (−∞, 1) ∪ (2, ∞] (−∞, 1] ∪ (2, ∞) (−∞, 1) ∪ [2, ∞)

y 4

5 MC The domain of the relation graphed at right is: a b C d e

[−4, 4] (−4, 7) [−1, 7] (−4, 4) (−1, 7)

−1 0

diGital doC doc-9739 SkillSHEET 4.1 domain and range

7 x

3

−4 6 MC The range of the relation {(x, y): y = 2x + 5, x ∈ [−1, 4]} is: a [7, 13] C [3, ∞) e R \ (7, 13)

b [3, 13] d R

7 We6a, b State i the domain and ii the range of each of the following relations. a {(3, 8), (4, 10), (5, 12), (6, 14), (7, 16)} b {(1.1, 2), (1.3, 1.8), (1.5, 1.6), (1.7, 1.4)} c

Time (min)

Distance (m) d

Day Cost ($)

3

4

5

6

110

130

150

170

Monday

Tuesday

Wednesday

Thursday

Friday

25

35

30

35

30

e y = 5x − 2, where x is an integer greater than 2 and less than 6 f y = x2 − 1, x ∈ R 8 We6c, d State the domain and range of each of the following relations. Use a CAS calculator to view

more of each graph if required. a

b

y 2 −3

d

y

0 1

0

x

y= x−1

0 e

c

y y = 2ex 2

y 4

y 2

x

2

y = 4e−x

x

−2 f

0

2x

y 0

x

−3 0

x

Chapter 4 • Relations, functions and transformations

171

y

g

y= 0

diGital doC doc-9740 WorkSHEET 4.1

1– x

y

h

y

i

1

x

0

x

0 −2

x

9 We7 For each relation given, sketch its graph and state the domain and range using interval notation. a {(x, y): y = 2 − x2} b {(x, y): y = x3 + 1, x ∈ [−2, 2]} c {(x, y): y = x2 + 3x + 2} d {(x, y): y = x2 − 4, x ∈ [−2, 1]} e {(x, y): y = 2x − 5, x ∈ [−1, 4)} f {(x, y): y = 2x2 − x − 6}

Verify that the graphs are correct with a CAS calculator.

diGital doC doc-9741 Investigation interesting relations

10 State the implied domain for each relation defined by the following rules. a y = 10 − x b y=3 x −

16 − x 2 c y= 2 d y=x +3 1 x f y = 10 − 7x2 e y=

types of relations (including functions) 4d

one-to-one relations

y

A one-to-one relation exists if, for any x-value, there is only one corresponding y-value and vice versa. For example:

0

x

{(1, 1), (2, 2), (3, 3), (4, 4)}

one-to-many relations y

A one-to-many relation exists if there is more than one y-value for any x-value but for any y-value there is only one x-value. For example:

x

0

{(1, 1), (1, 2), (2, 3), (3, 4)}

Many-to-one relations A many-to-one relation exists if there is more than one x-value for any y-value but for any x-value there is only one y-value. For example:

y

{(−1, 1), (0, 1), (1, 2)} 0

Many-to-many relations A many-to-many relation exists if there is more than one x-value for any y-value and vice versa. For example: {(0, −1), (0, 1), (1, 0), (−1, 0)}

172

Maths Quest 11 Mathematical Methods CAS

y

y

0

x

x

0

x

Worked exaMple 8

What type of relation does each graph represent? a

b

y

0

c

y

x

y

think

x

0

x

0

tUtorial eles-1462 Worked example 8

Write

a 1 For some x-values there is more than one

a One-to-many relation.

y-value. A line through some x-values shows that two y-values are available: y x = −1

x

0

2

For any y-value there is only one x-value. A line through any y-value shows that only one x-value is available: y y=1 x

0

b 1 For any x-value there is only one y-value. 2

For any y-value there is only one x-value.

c 1 For any x-value there is only one y-value. 2

b One-to-one relation.

c Many-to-one relation.

For some y-values there is more than one x-value.

Functions Relations that are one-to-one or many-to-one are called functions. That is, a function is a relation where for any x-value there is at most one y-value. For example: y

0

y

x

0

x

Vertical line test A function is determined from a graph if a vertical line drawn anywhere on the graph cannot intersect with the curve more than once. Chapter 4 • Relations, functions and transformations

173

Worked Example 9

State whether or not each of the following relations are functions. a {(−2, 1), (−1, 0), (0, −1), (1, −2)} y b c y

x

0

x

0 Think

Write

a For each x-value there is only one y‑value. (Or, a plot of

a Function

the points would pass the vertical line test.) b It is possible for a vertical line to intersect with the curve

b Not a function

more than once. c It is not possible for any vertical line to intersect with

c Function

the curve more than once.

Exercise 4D

Types of relations (including functions)

1 WE8 What type of relation does each graph represent? y y a b

0

d

h

y

x

k

y

0

x

0

y

x

y

x

0

l

x

x

0

i

y

x

0

f

y

0

y

x

y

0

x

0

j

0

e

y

0

g

x

c

x

y

0

x

2 WE9 Use the vertical line test to determine which of the relations in question 1 are functions. 174 Maths Quest 11 Mathematical Methods CAS

3 MC Which of the following relations is not a function? a {(5, 8), (6, 9), (7, 9), (8, 10), (9, 12)} y C y2 = x b

d y = 8x − 3

x

0

y

0

4 MC Consider the relation y ≥ x + 1. a The graph that represents this relation is: a y

b

x

y 1

1

d

y

x

−1 0

x

−1 0

C

e

y 1

1 0 e

0

1

x

x

1

Note: The shaded side indicates the region not required.

y

1 −1 0

x

b This relation is:

a one-to-one C many-to-one e a function

b one-to-many d many-to-many

c The domain and range are respectively:

R and R+ R and R R and R− R+ and R R− and R 5 Which of the following relations are functions? State the domain and range for each function. a {(0, 2), (0, 3), (1, 3), (2, 4), (3, 5)} b {(−3, −2), (−1, −1), (0, 1), (1, 3), (2, −2)} − − − − d {(1, 2), (1, 0), (2, 1), (3, 2), (4, 3)} c {(3, 1), (4, 1), (5, 1), (6, 1)} e {(x, y): y = 2, x ∈ R} f {(x, y): x = −3, y ∈ Z} g y = 1 − 2x h y>x+2 2 2 i x + y = 25 j y = x + 1, x ≥ − 1 3 k y=x +x l x = y2 + 1 a b C d e

power functions (hyperbola, truncus and square root function) 4e

Power functions are functions of the form f (x) = xn, n ∈ R. The value of the power, n, determines the type of function. We saw earlier that when n = 1, f (x) = x, and the function is linear. When n = 2, f (x) = x2, and the function is quadratic. When n = 3, f (x) = x3, and the function is cubic. When n = 4, f (x) = x4, and the function is quartic.

interaCtiVitY int-0263 domain and range

Chapter 4 • Relations, functions and transformations

175

Other power functions are: − • when n = −1, f (x) = x 1, and the power function produces the graph of a hyperbola − − • when n = 2, f (x) = x 2, and the power function produces the graph of a truncus 1 • when n = 2 , f ( x ) = x , and the function is the square root function. Under a sequence of transformations of f (x) = xn, n ∈ R, the general form of a power function is f (x) = a(x − b)n + c (where a, b, c and n ∈ R). 1 2

The hyperbola

1

• The graph shown is called a hyperbola, and is given by the equation y = x . − • This can also be represented as the power function y = x 1. y

y=0 0

x

x=0

The graph exhibits asymptotic behaviour. • That is, as x becomes very large, the graph approaches the x‑axis but never touches it. As x becomes very small (approaches 0), the graph approaches the y‑axis, but never touches it. • So the line x = 0 (the y‑axis) is a vertical asymptote, and the line y = 0 (the x‑axis) is the horizontal asymptote. • Both the domain and the range of the function are all real numbers, except 0; that is, R \ {0}. 1 • The graph of y = can be subject to a number of transformations. x a − • Consider y = + c or y = a( x − b) 1 + c. x−b y

Dilation

3 2 1

• The value a is a dilation factor. It dilates the graph from the x‑axis. 3 • For example, the graph of y = compared to the basic graph of x 1 y = is shown at right. x

0 −3 −2 −1 −1

y= (1, 3)

(1, 1)

1 2 3

3– x

y=

1– x

x y=0

−2 −3

x=0

Reflection

• If a is negative, the graph of the basic hyperbola is reflected in the x‑axis. • If x is replaced with −x, the graph of the basic hyperbola is reflected in the y‑axis. 1 1 and y = − are reflections of each other across the y‑axis. • For example, the graphs of y = x +1 x+1 y 2 1 0 −2 −1 −1 −2

y y = x 1+ 1

1 2

3 2 1 x

0 −2 −1 −1 1

2

x

−2

y = −x1+ 1 x = −1

176 Maths Quest 11 Mathematical Methods CAS

x=1

translation horizontal translation

• The value b translates the graph b units horizontally, that is, parallel to the x-axis. If b > 0, the graph is translated to the right, and if b < 0, the graph is translated to the left. 1 is a basic hyperbola translated one unit to the left, since • For example, the graph with equation y = x +1 b = −1. This graph has a vertical asymptote of x = −1 and domain R \ {−1}, and a horizontal asymptote of y = 0. 1 , with a vertical asymptote • If a basic hyperbola is translated one unit to the right, it becomes y = x −1 of x = 1 and domain R \ {1}. • Hence, the equation of the vertical asymptote is x = b and the domain is R \ {b}. The horizontal asymptote and the range remain the same, that is, x = 0 and R \ {0} respectively. Vertical translation

• The value c translates the graph c units vertically, that is, parallel to the y-axis. If c > 0, the graph is translated upward, and if c < 0, the graph is translated c units downward. 1 • The graph with equation y = − 1 is a basic hyperbola translated one unit down. This graph has a x horizontal asymptote of y = −1, a range of R \ {−1} and a vertical asymptote of x = 0. 1 • If a basic hyperbola is translated two units up, it becomes y = + 2, with a horizontal asymptote of x y = 2 and a range of R \ {2}. Hence, the equation of the horizontal asymptote is y = c, with a vertical asymptote of x = 0, and the range is R \ {c}. y y=c

y=

a x−b

+c

c 0

b

x

x=b

a − + c or y = a(x − b) 1 + c shows the combination of these transformations. The graph of y = x−b Worked exaMple 10

3 Sketch the graph of y = + 2, clearly showing the intercepts with the axes and the position of +1 x the asymptotes. think

Write/draW

a Compare the given equation with y = + c, x−b and state the values of a, b and c.

a = 3, b = −1, c = 2

2

Write a short statement about the effects of 1 a, b and c on the graph of y = . x

The graph of y =

3

Write the equations of the asymptotes. The horizontal asymptote is at y = c. The vertical asymptote is at x = b.

Asymptotes: x = −1; y = 2

4

Find the value of the y-intercept by letting x = 0.

y-intercept: x = 0

1

1 is dilated by the factor of 3 x from the x-axis (a = 3), translated 1 unit to the left (b = −1) and 2 units up (c = 2).

3 +2 0 +1 =3+2 =5

y=

Point (0, 5)

Chapter 4 • Relations, functions and transformations

177

5

Find the value of the x-intercept by making y = 0.

x-intercept: y = 0

3 +2 x +1 3 − 2= x +1 − 2(x + 1) = 3 −2x − 2 = 3 −2x = 5 0=

x= Point 6

To sketch the graph: (a) draw the set of axes and label them (b) use dotted lines to draw the asymptotes The asymptotes are x = −1 and y = 2. (c) mark the intercepts with the axes −5 The intercepts are y = 5 and x = 2 . (d) treat the asymptotes as your new set of axes, sketch the graph of the hyperbola. (As a is positive, the graph is not reflected in the x-axis.)

(

−5

2

,0

−5 2

) y

y= 6 (0, 5) 4 2 0 −3 −2 −1 −2 (− 5 , 2

0)

3 x+1

+2

y=2 x

1 2 3

−4 −6

x = −1

the truncus • The graph shown is known as a truncus. The equation of the graph is given by: 1 y= 2 x −

• This can also be represented as the power function y = x 2. • The function is undefined for x = 0. Hence, the equation of the vertical asymptote is x = 0 and the domain of the function is R \{0}. • We can also observe that the graph approaches the x-axis very closely but never touches it. So y = 0 is the horizontal asymptote. • Since the whole graph of the truncus is above the x-axis, its range is R+ (that is, all positive real numbers). • Similarly to the graphs of the functions, discussed in the previous 1 sections, the graph of y = 2 can undergo various transformations. x a + c or y = a(a − b)−2 + c. • Consider the general formula y = ( x − b) 2

dilation • The value a is the dilation factor. It dilates the graph from the x-axis. The dilation factor does not affect the domain, range or asymptotes. 4 • Consider the graph of y = 2 . x

y y= y=0

1 x2

x

0 x=0

y 6 4 2 −3 −2 −1 0

y= 1 2 3

4 x2

x y=0

x=0

reflection • If a is negative, the graph of a basic truncus is reflected in the x-axis. The range becomes R− (that is, all negative real numbers). • If x is replaced with −x, the graph of the basic truncus is reflected in the y-axis. The effect of this reflection cannot be seen in the basic graph, but it becomes more obvious if the graph has been translated horizontally first. 178

Maths Quest 11 Mathematical Methods CAS

1 1 and y = − are reflections across the y-axis. 2 ( x − 2)2 ( x − 2) − The vertical asymptote changes from x = 2 to x = 2, and the domain changes from R \{2} to R \{−2}.

• For example, the graphs of y =

y

y 3 2 1 0 −2 −1 −1

y=

1

2

1 (x − 2)2

y=

x

1 (−x − 2)2

x

−3 −2 −1 0

−2

x = −2 x=2

translation horizontal translation

• The value b translates the graph b units horizontally. If b > 0, the graph is translated to the right, and if b < 0, the graph is translated left. 1 • For example, the graph of the equation y = results from translating a basic truncus 3 units to ( x − 3)2 the right. The vertical asymptote is x = 3 and the domain is R \ {3}. 1 , where the vertical asymptote • If a basic truncus is translated 2 units to the left, it becomes y = ( 2)2 x + is x = −2 and the domain is R \ {−2}. • Hence, the equation of the vertical asymptote is x = b, and the domain is R \ {b}. The range is still R+, and the equation of the horizontal asymptote is y = 0. Vertical translation

• The value c translates the graph c units vertically. If c > 0, the graph is translated upward, and if c < 0, the graph is translated c units downward. 1 • For example, the graph with equation y = 2 + 1 results when a basic truncus is translated 1 unit x upward. The horizontal asymptote is y = 1, and the range is (1, ∞). 1 • If a basic truncus is translated 1 unit down, it becomes y = 2 − 1, with y = −1 as the horizontal x asymptote and (−1, ∞) as the range. • Hence, the equation of the horizontal asymptote is y = c, and the range is (c, ∞). Note: If a is positive (see graph below), the whole graph of the truncus is above the line y = c (the horizontal asymptote), and hence its range is y > c, or (c, ∞). y • If a is negative, the whole graph is below its horizontal asymptote, y= − and therefore the range is y < c, or ( ∞, c). a + c or y = a(x − b)−2 + c shows the • The graph of y = ( x − b) 2 c combination of these transformations. 0

b

a (x − b)2

+c

y=c x

x=b

Worked exaMple 11

1 + 3, clearly showing the position of the asymptotes and the ( x − 2) 2 intercepts with the axes. Sketch the graph of y = think

Write

a +c ( x − b) 2

1

Write the general formula for the truncus.

y=

2

Identify the values of a, b and c.

a = 1, b = 2, c = 3

Chapter 4 • Relations, functions and transformations

179

1 The graph of y = 2 is translated 2 units to the right and 3 units up. x

3

Write a short statement about the 1 transformations the graph of y = 2 should undergo in order to be changed x into the one in question.

4

Write the equations of the asymptotes (y = c and x = b).

Asymptotes: x = 2 and y = 3

5

Find the x-intercept. As you cannot get the square root of a negative number, there is no solution and therefore no x-intercepts. An inspection of the equation of the graph would also have revealed this.

x-intercept: y = 0 1 0= +3 ( x − 2)2 1 = −3 ( x − 2)2 ( x − 2)2 =

−1 3 −1

( x − 2) = ±

3

y-intercept: x = 0 1 y= +3 (0 − 2)2 1 = +3 4 13 = 4

6

Find the y-intercept.

7

To sketch the graph: (a) draw the set of axes and label them (b) use dotted lines to draw asymptotes (c) mark the x- and y-intercepts (d) treating the asymptotes as the new set of axes, draw the basic truncus curve (e) make sure it intersects the axes in the right places.

y (0,

13 ) 4

6 4 2

−2 −1 0

y=3 1 2 3 4x x=2

the square root function • • • • •

The square root function is given by y = x . This can be written as the power function y = x . The function is defined for x ≥ 0; that is, the domain is R+ ∪ {0}, or [0, ∞). As can be seen from the graph, the range of the square root function is also R+ ∪ {0}, or [0, ∞). Throughout this section we will refer to the graph of y = x as ‘the basic square root curve’. 1 2

y y= x x

0

• Let us now investigate the effects of various transformations on the basic square root curve. • Consider the function y = a x − b + c or y = a( x − b) + c. 1 2

dilation • The value a is a dilation factor; it dilates the graph from the x-axis. The domain is still [0, ∞). 180

Maths Quest 11 Mathematical Methods CAS

Reflection • If a is negative, the graph of a basic square root curve is reflected in the x‑axis. The range becomes (−∞, 0]. The domain is still [0, ∞). • If x is replaced with −x, the graph is reflected in the y‑axis. For example, the graphs with equations y = x and y = − x are reflected across the y‑axis. • The domain becomes (−∞, 0] and the range is [0, ∞). y

y y= x

0

1 2

y = −x x

−2 −1 0

1 2

x

Translation Horizontal translation

• The value h translates the graph horizontally. If b > 0, the graph is translated to the right, and if b < 0, the graph is translated to the left. • The graph with the equation y = x − 1 results when the basic curve is translated 1 unit to the right. This translated graph has domain [1, ∞) and range [0, ∞). • If the basic curve is translated 3 units to the left, it becomes y = x + 3 and has domain [−3, ∞) and range [0, ∞). y y= x+3 2 1

3

−3 −2 −1 0

1 2 3

x

• The domain of a square root function after a translation is given by [b, ∞). Vertical translation

• The value c translates the graph vertically. If c > 0, the graph is translated vertically up, and if c < 0, the graph is translated vertically down. • If y = x is translated 3 units vertically up, the graph obtained is y = x + 3, with domain [0, ∞) and range [3, ∞). y 5 4 3 2 1

y= x+3

0 −2 −1 −1

1 2 3 4x

• If the basic curve is translated 2 units down, it becomes y = x − 2, with domain [0, ∞) and range [−2, ∞). • The range of the square root function is [c, ∞) for a > 0. • The graph of y = a x − b + c or y = a( x − b) + c shows the combination of these transformations. 1 2

y y=a b−x+c

(−b, c)

y=a x+b+c

(b, c) x

Chapter 4 • Relations, functions and transformations 181

Worked Example 12

Sketch the graph of y = 3 x − 1 + 2, clearly marking intercepts and the end points. Think

Write/draw

1

Write the equation.

y = 3 x − 1 + 2,

2

Write the coordinates of the end point.

End point: (1, 2)

3

State the shape of the graph.

Shape:

4

Inspection of the equation reveals that there is no y-intercept.

There is no x-intercept.

5

Inspection of the equation reveals that there is no y-intercept.

There is no y-intercept.

6

To help sketch the graph, determine the coordinates of a second point. Let x = 3.

x = 3: y = 3 3−1 + 2

7

.

=3× 2+2 Point: (3, 3 2 + 2)

Sketch the graph by plotting the end point, showing the second point, and drawing the curve so that it starts at the end point and passes through the second point.

y 7 6 5 4 3 2 1 0 −2 −1 −1

y=3 x−1+2 (3, 3 2 + 2)

(1, 2) 1 2 3 4x

Power functions (hyperbola, truncus and square root function)

Exercise 4E

1 State the dilation factor and the vertical and horizontal translations or reflections for each of the

following. Write a short statement about the effects each has on the basic graph of that function. 3 +1 a y= x−5 2 b y = 2 +1 x 1 −2 c y= 4−x d y= 2 x+4 e y= f

1 (3 − x )2

y = −4 x −5

2 WE10, 11, 12 Sketch a graph for each of the following functions, clearly showing x- and y‑intercepts

and any asymptotes. 3 a f (x) = 2 + 1 x 2 c f (x) = +5 x −1 182 Maths Quest 11 Mathematical Methods CAS

b f (x) =

3− x −1

d f (x) = 2 −

2 ( x − 1)2

3 Assuming the dilation factor is 1, write the equation of the graph. y 9 8 7 6 5 4 y=3 3 5– 2 2 1 1 2 3 4x

−1 0 5– 3

x=2

4F

Function notation

Consider the relation y = 2x, which is a function. The y-values are determined from the x-values, so we say ‘y is a function of x’, which is abbreviated to y = f (x). So, the rule y = 2x can also be written as f (x) = 2x. If x = 1, then y = f (1) If x = 2, then y = f (2) =2×1 =2×2 =2 = 4, and so on.

evaluating functions For a given function y = f (x), the value of y when x = 1 is written as f (1), the value of y when x = 5 is written as f (5) and so on. Worked exaMple 13

If f (x) = x2 − 3, find: a f (1) b f (−2)

c f (a)

d f (2a).

think

a 1 Write the rule. 2

Substitute x = 1 into the rule.

3

Simplify.

b 1 Write the rule. 2

Substitute x = −2 into the rule.

3

Simplify.

c 1 Write the rule. 2

Substitute x = a into the rule.

d 1 Write the rule. 2

Substitute x = 2a into the rule.

3

Simplify the expression if possible.

Write

a f (x) = x2 − 3

f (1) = 12 − 3 =1−3 = −2 b f (x) = x2 − 3

f (−2) = (−2)2 − 3 =4−3 =1 c

f (x) = x2 − 3 f ( a) = a2 − 3

d f (x) = x2 − 3

f (2a) = (2a)2 − 3 = 22a2 − 3 = 4a2 − 3

Chapter 4 • Relations, functions and transformations

183

Fully defining functions To fully define a function: 1. define the domain 2. state the rule. That is, if a function f (x) has domain X, the function may be defined as follows: f : X → Y, f (x) = . . . . . . Domain Co-domain Rule Y is not necessarily the range but is a set that contains the range, called the co‑domain. The co-domain gives the set of possible values that contains y. It is usually R (the set of real numbers). The actual values that y can be — the range — is determined by the rule. When using function notation the domain can be abbreviated as dom f and the range as ran f. For example, the function defined by {(x, y): y = 2x, x ∈ [0, 3]} can be expressed in function notation as f : [0, 3] → R, f (x) = 2x. For this function we can write dom f = [0, 3]. The co-domain = R. Also, ran f = [0, 6] (x = 0 gives y = 0 and x = 3 gives y = 6, which are the minimum and maximum values of y). R

R –5.1

– 7–8

– 2

1– 3

0.6

[0, 3]

3 — – 11 3

f : domain

0

etc. 2– 2 1.2 3 46 2 3 etc. 0 Range

1 2 3

–10

etc. Domain

[0, 6]

Co-domain co-domain , f(x) = rule

The graph of this function is shown below. y 6 5 4 3 2 1 0

f(x)

1

2

3x

The maximal domain of a function is the largest possible set of values of x for which the rule is defined. The letters f, g and h are usually used to name a function, that is, f (x), g(x) and h(x). Note: If a function is referred to by its rule only, then the domain is assumed to be the maximal domain. Worked Example 14

Express the following functions in function notation with maximal domain. 1 a {(x, y): y = x2 − 4} b y = 3x − 4, −2 ≤ x ≤ 5 c y = x Think

Write

a The rule has meaning for all values of x (it is a

quadratic), so the domain of the function is R. b The rule has meaning for all values of x in the

given domain

[−2,

5].

c The rule has meaning for all values of x except 0.

184 Maths Quest 11 Mathematical Methods CAS

a f : R → R, f (x) = x2 − 4 b f : [−2, 5] → R, f (x) = 3x − 4

c f : R \ {0} → R, f (x) =

1 x

Worked exaMple 15

State i the domain, ii the co-domain and iii the range for each of the following functions. 1 a f : R → R, f (x) = 5 − x b g : R+ → R, g(x) = x think

Write/draW

a 1 The domain is given as R. 2

The co-domain is given as R.

3

Use a CAS calculator to obtain the graph of the function, or sketch it.

a i dom f = R ii The co-domain is R. y 5

0

4

From the graph the range is observed to be R.

b 1 The domain is given as R+. 2

The co-domain is given as R.

3

Use a CAS calculator to obtain the graph of the function, or sketch it.

5

iii ran f = R b i dom g = R+ ii The co-domain is R. y

1 0 4

The range is observed from the graph to be R+.

x f(x)

g(x) x

1

iii ran g = R+

Worked exaMple 16

State i the maximal domain and ii the range for the function defined by the rule: 1 . a y = x+1 b y = x+2 think

Write/draW

a 1 The rule has meaning for all x if x + 1 ≥ 0

(that is, contents of

tUtorial eles-1463 Worked example 16

a Require x + 1 ≥ 0

are positive).

2

Solve this inequation.

So x ≥ −1

3

State the maximal domain.

i Maximal domain = [−1, ∞)

4

To confirm the maximal domain and see the range, use a CAS calculator or other technology to obtain the graph of the function, or sketch it by plotting selected points.

y y= x+1 1 (−1, 0) 0

5

The range is observed from the graph to be [0, ∞).

x

ii Range = [0, ∞)

Chapter 4 • Relations, functions and transformations

185

b 1 The rule exists for all x, except when x + 2 = 0. 2

Therefore x ≠ −2.

x ≠ −2

3

State the maximal domain.

i Maximal domain = R \ {−2}

4

Use a CAS calculator to obtain the graph of the function, or sketch it by plotting selected points to confirm the maximal domain and see the range.

5

y 1 y = —— x+2

1 −2 −1 0

x

ii Range = R \ {0}

The range is observed from the graph to be R \ {0}.

exercise 4F

Function notation

1 We13 a If f (x) = 3x + 1, find: diGital doCS doc-9742 SkillSHEET 4.2 Substitution doc-9743 SkillSHEET 4.3 transposition of equations

bx+2≠0

b If g(x) =

x + 4 , find: 1 c If g(x) = 4 − , find: x d If f (x) = (x + 3)2, find: 24 e If h (x) = , find: x

i f (0),

ii f (2),

iii f (−2) and

iv f (5).

i g(0),

ii g(−3),

iii g(5) and

iv g(−4),

i g(1),

ii g 1 ,

iii g

i f (0),

ii f (−2),

iii f (1) and

iv f (a).

i h (2),

ii h (4),

iii h (−6) and

iv h (12).

2

− 1 , 2

− 1 , 5

and

iv g

2 Find the value (or values) of x for which each function has the value given. a f (x) = 3x − 4, f (x) = 5 b g(x) = x2 − 2, g(x) = 7

1 , f (x) = 3 x e g(x) = x2 + 3x, g(x) = 4 c f (x) =

10 3 Given that f ( x ) = − x , find: x a f (2) d f (x2)

d h (x) = x2 − 5x + 6, h (x) = 0 f

f (x) =

b f (−5) e f (x + 3)

8 − x , f (x) = 3 c f (2x) f f (x − 1)

4 We14 Express the following functions in function notation with maximal domain. b y=

x−6 2 d y= x +1 f y = x2 + 3x, where x ≥ 2 h y = x2 + x

a {(x, y): y = 4x + 1}

1 x −1 e y = (x + 2)2, where x ∈R+ g y = 8 − x, where x ≤ 0 c y=

5 We15 For each of the following functions, state: i the domain ii the co-domain iii the range. diGital doC doc-9744 Square root graphs

a f : {0, 1, 2, 3} → Z, f (x) = 3x − 7 c f : {2, 4, 6, 8, 10} → N, f (x) =

x 2

e g: R+ → R, g(x) = x2 − 2

d f : (−∞, 0) → R, f (x) = f

6 We16 State i the maximal domain and

186

b g: (0, 10] → R, g(x) =

3 x 1 −x

h: [−3, 3] → R, h(x) = 9 − x 2

ii the range for the function defined by the rule:

a f (x) = 3 − x

b f (x) = 5 x

c y = x3 + 2

d y = 5 − 3x2

e y=

f

Maths Quest 11 Mathematical Methods CAS

x−4

y=

1 x−3

Special types of function (including hybrid functions) 4G

one-to-one functions

As we have already seen, one-to-one relations and many-to-one relations are functions. A one-to-one function has, at most, one y-value for any x-value and vice versa. The graph of a relation is a function if any vertical line crosses the curve at most once. Similarly, a one-to-one function exists if any horizontal line crosses the curve at most once. For example: y

0

y

x

0

A function that is not one-to-one; this passes the vertical line test but not the horizontal line test.

eleSSon eles-0077 hybrid functions

x

A one-to-one function; this passes the vertical line test and the horizontal line test.

Worked exaMple 17

Which of the following functions are one-to-one? a {(0, 1), (1, 2), (2, 3), (3, 1)} b {(2, 3), (3, 5), (4, 7)} c f (x) = 3x think 1

Write/draW

a When x = 0 and x = 3, y = 1.

Check whether each function has, at most, one y-value for any x-vaue and vice versa.

It is not a one-to-one function. b There is only one x-value for each y-value.

2

Sketch the graph of f (x) = 3x. Check whether both a vertical line and a horizontal line crosses only once.

y

c

f(x)

3 0

1

x

It is a one-to-one function. 3

Write a statement to answer the question.

The functions are one-to-one for b and c.

Worked exaMple 18

Which of the following graphs show a one-to-one function? a

b

y

0

x

c

y

0

x

think

If a function is one-to-one, any vertical or horizontal line crosses the graph only once.

y

0

x

Write

Only b is a one-to-one function.

Chapter 4 • Relations, functions and transformations

187

Restriction of functions Restrictions can be placed on a function through its domain. If we have one relation, for example f (x) = x2, we can create several different functions by defining different domains. For example: y

f(x)

x

0

y

y

h(x)

g(x) −1 0

1x

x

0

f : R → R, f (x) = x2 g: [−1, 1] → R, g(x) = x2 h: R+ → R, h(x) = x2 The restriction imposed on the function f to produce the function h has created a one-to-one function.

Worked Example 19

For each function graphed below state two restricted, maximal (largest possible) domains that make the function one-to-one. y y = (x − 2)2 b a y 1 y=— 2

4

0

x

2

Think

a 1 One-to-one functions will be formed if the

x

0

x

Write/draw

a

curve is split into two through the vertical line x = 2.

y

y

4

0 2

State the required domains.

b 1 One-to-one functions will be formed if the

curve is split into two through the line x = 0.

2

x

0

x

2

For the function to be one-to-one, the domain is (−∞, 2] or [2, ∞). b

y

0

y

x

x

0

2

State the required domains.

For the function to be one-to-one, the domain is (−∞, 0) or (0, ∞).

Hybrid functions A hybrid, mixed, or piecewise defined function is a function that has different rules for different subsets of the domain. For example: x + 1, for x ≤ 0 f (x) = 2 is a hybrid function that obeys the rules y = x + 1 if x , for x > 0 x ∈ (−∞, 0] and y = x2 if x ∈ (0, ∞). The graph of f (x) is shown at right. Note that the domains do not overlap. 188 Maths Quest 11 Mathematical Methods CAS

y

f(x)

1 −1 0

x

Worked exaMple 20

x, a Sketch the graph of f ( x ) = x + 1, 5 − x, b State the range of f.

x<0 0≤x<2

tUtorial eles-1464 Worked example 20

x≥2

think

Write/draW

a If x = −1, y = x

a Calculate and plot points as shown.

= −1

1

Sketch the graph of y = x for the domain (−∞, 0).

If x = 0, y = x =0

2

On the same axes sketch the graph of y = x + 1 for the domain [0, 2).

If x = 0, y = x + 1 =1 If x = 2, y = x + 1 =3

3

On the same axes sketch the graph of y = 5 − x for the domain [2, ∞).

If x = 2, y = 5 − x =3 If x = 5, y = 5 − x =0 y 3 y=x+1 2 1

f(x) y=5−x

0 1 2 3 4 5 x y = x−1−1

b ran f = (−∞, 3]

b The range is made up of (or is the union of) two

sections, (−∞, 0) with (−∞, 3].

Special types of function (including hybrid functions) exercise 4G

1 We17 Which of the following functions are one-to-one? Use a CAS calculator or other technology to

obtain the graph of the function where appropriate.

a {(1, −1), (2, 1), (3, 3), (4, 5)} c {(x, y): y = x2 + 1, x ∈ [0, ∞)} e {(x, y): y = 3 − 2x2}

b {(−2, 1), (−1, 0), (0, 2), (1, 1)} d {(x, y): y = 3 − 4x} f f (x) = x3 − 1

g y = x2, x ≤ 0

h g(x) = 1 − x 2

2 We18 Consider the relations below and state: i which of them are functions ii which of them are one-to-one functions. a b y y

0

x

0

c

x

y

0

x

Chapter 4 • Relations, functions and transformations

189

y

d

h

y

j

y

x

l

y

0

x

0

x

y

x

0

x

0

i

0

k

y

x

y

x

0

y

f

0

x

0

g

y

e

x

0

3 WE19 For each function below, state two restricted, maximal domains that make the function

one-to-one. a

−1 0 d

b

y

y

g f (x) = 1 − x2

3 x

−3 e

x

y −3 0

x

2

0

x

(3, 4)

0

c

y

f

y 0

−4

(−1, 4)

y

x

(−2, −2)

0 (1, 0)

4 − x 2 , x ∈ [−2, 2] j f (x) = (x + 3)2 h g(x) =

x

1 , x ∈ R \ {0} x2 4 MC Use the graph of the relation y2 = x − 1, shown below, to answer the following questions. a A one-to-one function can be formed by: y A restricting the domain to R+ B restricting the domain to [1, ∞) C restricting the domain to (1, ∞) x 0 1 D restricting the range to [0, ∞) E restricting the range to R \ {0} i g(x) =

b A rule that describes a one-to-one function derived from the relation y2 = x − 1 is: a y2 = x − 1 b y = ±

x −1

−

x −1

c y = d y =

x −1

e y =

x −1

190 Maths Quest 11 Mathematical Methods CAS

5 MC Consider the following hybrid function.

− x, x < 1 f (x) = x , x ≥ 1

a The graph that correctly represents this function is: a b y y

d

y

1

1 −1

C

0 1

2

x

0 −1

e

y

1

2

1

x

0 −1

x

1

y

1 0 −1

1

x

1

0 −1

x

1

b The range of this hybrid function is:

b R \ {−1} e R+

a R

d [0, ∞)

C (−1, ∞)

6 We20 a Sketch the graph of the following function.

1 x<0 , f (x) = x x + 1, x ≥ 0 b State the range of f. 2 x + 1,

7 a Sketch the graph of the function g( x ) = 2 − x , b State the range of g. c Find: i g(−1) ii g(0) iii g(1).

x≥0 x<0

.

x − 2, x < − 2 8 a Sketch the graph of the function f ( x ) = x 2 − 4, − 2 ≤ x ≤ 2 . x + 2, x > 2 b State the range of z. c Find: i f (−3)

ii f (−2)

iii f (1)

iv f (2)

v f (5).

y 3 2 1

9 Specify the rule for the function represented by the graph at right.

−2 −1 0 −1 −2 10 The graph of the relation {(x, y): x2 + y2 = 1, x ≥ 0} is shown at right.

From this relation, form two one-to-one functions and state the range of each.

f(x)

1

x

1

x

y 1 0

−1 11 a Sketch the graph of the function f : R → z, f (x) = (x − 3)2. b By restricting the domain of f, form two one-to-one functions that have the same rule as f (use the

largest possible domains). 12 a Sketch the graph of the function g : R → R, g(x) = x2 + 2x + 1. b By restricting the domain of g, form two one-to-one functions that have the same rule as g

(use maximal domains).

diGital doC doc-9745 WorkSHEET 4.2

Chapter 4 • Relations, functions and transformations

191

4H

Inverse relations and functions

A relation is a set of ordered pairs that can be graphed or described by a rule. The inverse of a set of ordered pairs is obtained simply by interchanging the x and y elements. So, the inverse of {(1, 5), (2, 6), (3, 7)} is {(5, 1), (6, 2), (7, 3)}. If these points are plotted on a set of axes, it can be seen that when each original point is reflected across the line y = x, the inverse points are obtained. Similarly, if the graph of a function is given, then its inverse function can be sketched by reflecting the original function across the line y = x. Only functions that are one-to-one have inverses.

Worked Example 21

Sketch the graph of the following and then sketch the inverse. a {(3, −6), (4, −4), (5, −2), (6, 0)} b y

c y = x2 for x ≥ 0

x Think

a 1 Plot the points on a set of axes. 2

Interchange the x- and y-values and plot them. Alternatively, reflect the original points across the line y = x. So, (3, −6) → (−6, 3) (4, −4) → (−4, 4) (5, −2) → (−2, 5) (6, 0) → (0, 6)

b 1 Re-draw the given graph. 2

On the same set of axes, plot the line y = x.

3

Sketch a reflection of the original graph across the line y = x.

Write/draw

a

y y=x 6 5 4 3 2 1 0 −6−5−4−3−2−1 −1 1 2 3 4 5 6 x −2 −3 −4 −5 −6

b y

y=x

x c Sketch the original function. The graph is a

c y

parabola with turning point (0, 0). The domain is restricted.

y = x2

y=x

1 0

192 Maths Quest 11 Mathematical Methods CAS

1

x

inverse relations and functions

exercise 4h

1 We21a Sketch the graph of the following and then sketch the inverse. a (1, 7), (2, 5), (3, 3)} b (1, 3), (2, 6), (3, 9)} c (−2, 11), (0, 6), (2, 1)} 2 We21b Sketch the graph of the following and then sketch the inverse. (Assume each set of axes has

the same scale for x and y.) a

y

b

y

x

x c

y

d

y

x x y

e

y

f

x x 3 We21c Sketch the graph of the following and then sketch the inverse. a y = 4x

4i

b y = x2 + 3, x ≥ 0

1

c y = 2x + 1

d y = x3 + 4

Circles

y r

A circle is a many-to-many relation. The rule that defines a circle with its centre at (0, 0) and of radius r is −r

x2 + y2 = r 2 The graph of this circle is shown at right. The vertical-line test clearly verifies that the circle graph is not a function. Solving the equation for y we have y2 = r 2 − x2, so y = r 2 − x 2 or y = − r 2 − x 2 . These two relations represent two semicircles that together make a complete circle: • y = r 2 − x 2 is the ‘upper semicircle’ (above the x-axis).

r x −r

diGital doC doc-9746 Investigation a special relation

• y = − r 2 − x 2 is the ‘lower semicircle’ (below the x-axis). y r −r

y y = r 2− x2

r

x

−r

r x −r y = − r 2 − x 2

Both these relations are many-to-one functions. Chapter 4 • Relations, functions and transformations

193

Worked Example 22

Sketch the graphs of the following relations. a x2 + y2 = 16 b x2 + y2 = 9, 0 ≤ x ≤ 3 c y = 8 − x 2 Think

draw

a 1 This relation is a circle of centre (0, 0) and radius = 16 = 4. 2

On a set of axes mark x- and y‑intercepts of −4 and 4.

3

Draw the circle.

y

a

4 −4

4 x

0 −4

b 1 This relation is part of a circle of centre (0, 0) and radius = 2

Since the domain is [0, 3], on a set of axes mark y-intercepts −3 and 3 and x‑intercept 3.

3

Draw a semicircle on the right-hand side of the y-axis.

−3

3

Draw a semicircle above the x-axis.

y

c

and radius = 8.

On a set of axes mark the x‑intercepts of y‑intercepts of 8 .

3 x

0

c 1 This relation is an ‘upper semicircle’ (as y > 0) of centre (0, 0)

2

y 3

b

9 = 3.

8 −

8 and 8 and − 8

General equation of a circle

y

The general equation of a circle with centre (h, k) and radius r is (x − h)2 + (y − k)2 = r 2. The domain is [h − r, h + r]. The range is [k − r, k + r].

k+r Range

8x

0

(x − h)2 + (y − k)2 = r2

k

(h, k)

k−r 0 h−r

h Domain

Worked Example 23

Sketch the graphs of the following circles. State the domain and range of each. a x2 + (y − 3)2 = 1 b (x + 3)2 + (y + 2)2 = 9 Think

a 1 This circle has centre (0, 3) and radius 1. 2

On a set of axes mark the centre and four points; 1 unit (the radius) left and right of the centre, and 1 unit (the radius) above and below the centre.

Write/draw

a

y 4

x 2 + (y − 3)2 = 1

3 2 −1 0

1

x

3

Draw a circle that passes through these four points.

4

State the domain.

The domain is [−1, 1].

5

State the range.

The range is [2, 4].

194 Maths Quest 11 Mathematical Methods CAS

h+r x

b 1 This circle has centre (−3, −2) and radius 3. 2

3

b

y 1

(−3, 1)

On a set of axes mark the centre and four points; 3 units left and right of the centre, and 3 units above and below the centre.

−6 −5 −4 −3 −2 −1 0 −1 (−6, −2)

Draw a circle that passes through these four points.

(0, −2)

−2

(−3, −2)

1 x

−3 −4 −5

(−3, −5) 4

State the domain.

The domain is [−6, 0].

5

State the range.

The range is [−5, 1].

Note: When using a CAS calculator to plot circle graphs, ensure that the upper and lower values are entered as separate equations on the Graphs & Geometry page; for example, f1(x) = − 16 − x 2 and f2(x) = 16 − x 2 .

exercise 4i

Circles

1 State the equation of each of the circles graphed below. y y a b 3 1 −3

−1 0 −1

3x

0 −3

d

e

y 10 −10

c

5 −5

x

1

10 x

f

y

y

−10

2 2 −2 2

6 x

0

− 6

5 x

0 −5

6

0

diGital doC doc-9747 Circle graphs

y

2 2x

0

−2 2

− 6 h

y

g

3 −3

y −4

3x

0

0

4 x

−4

2 State the domain and range of each circle in question 1. 3 We22 Sketch the graphs of the following relations. a x2 + y2 = 4 b x2 + y2 = 16 d x2 + y2 = 7

c x2 + y2 = 49

e x2 + y2 = 12

f

x2 + y2 =

1 4

4 Sketch the graph of each of the following relations and state whether it is a function or not. a y = ± 81 − x 2

b y=

4 − x2

c y = − 1− x2

d y=

1 9

e y=−

1 4

− x2

g y = ± 10 − x 2

f

− x2

y = 5 − x2

h x2 + y2 = 3,

−

3≤x≤0

Chapter 4 • Relations, functions and transformations

195

5 MC Consider the circle below.

y 2 0

2

4 x

−2 a The equation of the circle is: A x2 + (y − 2)2 = 4 c (x + 2)2 + y2 = 16 e (x + 2)2 + y2 = 4

b The range of the relation is: A R d [2, 4]

b (x − 2)2 + y2 = 16 d (x − 2)2 + y2 = 4

b [−2, 2] e [−2, 1]

6 MC Consider the equation (x + 3)2 + (y − 1)2 = 1. a The graph that represents this relation is: y B y A 4 2 1 1 0 2 3 4 0 x −3 −6 −2 D

y −1 −2

c [0, 4]

2 1 x

2

3

4

−4 −3 −2 y 2

E

0

y

C

x 1 0

−3.5−3−2.5 b The domain of the relation is: A [−3.5, −2.5] D [2, 4]

B E

x

(−4, −2) [−4, −2]

C R

7 WE23 Sketch the graph of the following circles. State the domain and range of each. a x2 + (y + 2)2 = 1 b x2 + (y − 2)2 = 4 c (x − 4)2 + y2 = 9 d (x − 2)2 + (y + 1)2 = 16 e (x + 3)2 + (y + 2)2 = 25 f (x − 3)2 + (y − 2)2 = 9 2 2 9 3 1 g (x + 5) + (y − 4) = 36 h (x − )2 + (y + )2 = 2

8 Express the relation x2 + y2 = 36 as two

functions, and state the largest domain and range of each. 9 Express the relation x2 + (y − 2)2 = 9 as

two functions, and state the largest domain and range of each. 10 Circular ripples are formed when a

water drop hits the surface of a pond. If one ripple is represented by the equation x2 + y2 = 4 and then 3 seconds later by x2 + y2 = 190, where the length of measurements are in centimetres: a find the radius (in cm) of the ripple in each case b calculate how fast the ripple is moving outwards. (State your answers to 1 decimal place.) 196 Maths Quest 11 Mathematical Methods CAS

2

4

0

x

4J

Functions and modelling

When using functions to model rules in real-life situations the domain usually has practical restrictions imposed on it. For example, the area of a circle is determined by the function A(r) = π 2. For a circle to be drawn the radius needs to be a positive number. Hence the domain is (0, ∞) or R+.

Worked exaMple 24

The table describes hire rates for a removal van. a Express the cost as a hybrid function. b Sketch the graph of the function. Hours of hire (h)

Cost ($C)

Up to 3

200

Over 3 up to 5

300

Over 5 up to 8

450

think

Write/draW

a 1 The cost is $200 if 0 < h ≤ 3. 2

The cost is $300 if 3 < h ≤ 5.

3

The cost is $450 if 5 < h ≤ 8.

4

tUtorial eles-1465 Worked example 24

a

200, C (h) = 300, 450,

State the cost function C(h).

b Sketch a graph with 3 horizontal lines over the

appropriate section of the domain.

0

b C ($) 450 400 350 300 250 200 150 100 50 0 1 2 3 4 5 6 7 8 h (hours)

exercise 4J

Functions and modelling

1 We24 The cost of hiring a paper recycling removalist is described in the following table:

Hours of hire

Cost

Up to 1

$40

Over 1 up to 2

$70

Over 2 up to 4

$110

Over 4 up to 6

$160

a State the cost function, $C, in terms of the time, t hours, for hiring up to 6 hours. b Sketch the graph of the function. Chapter 4 • Relations, functions and transformations

197

2 The charge for making a 10-minute STD call on the weekend is listed below.

Distance d (km) Cost $C

Up to 50 km 0.40

50 to 100 km 0.60

100 to 200 km 0.80

200 to 700 km 1.70

Over 700 km 2.00

a State the cost function in terms of the distance. b Sketch the graph of the function. 1 2

3 A car travels at a constant speed of 60 km/h for 1 hours, stops for half an hour, then travels for another

2 hours at a constant speed of 80 km/h until it reaches its destination. a Construct a function that describes the distance travelled by the car, d (km), at time, t hours. b State the domain and range of this function. c Calculate the distance travelled after: i 1 hour ii 3 hours. 4 At a fun park, a motorised toy boat operates for 5 minutes for every dollar coin placed in a meter. The meter will accept a maximum of 120 one-dollar coins. a Write a rule that gives the time of boat operation, B hours, in terms of the number of dollar coins, n. b Sketch the graph of the function and state the domain and range. c How much is in the meter when the boat has operated for 450 minutes? 5 The tax for Australian residents who earn a taxable income between $37 000 and $80 000 is $4650 plus 30 cents for every dollar earned over $37 000. a Write a rule for the tax payable, $T, for a taxable income, $x, where 37 001 ≤ x ≤ 80 000. b Sketch a graph of this function. c Calculate the tax paid on an income of $42 000. 6 The maximum side length of the rectangle shown is 10 metres. (x + 4) m (x − 1) m a Write a function that gives the perimeter, P metres, of the rectangle. b State the domain and range of this function. 7 A rectangular swimming pool is to have a length 4 metres greater than its width. a Write a rule for the area of the pool, A m2, as a function of the width, x metres. b State the domain and range if the maximum side length is 12 metres. 8 Timber increases in value (appreciates) by 2% each year. If a consignment of timber is currently

worth $100 000: a express the value of the timber, P dollars, as a function of time, t, where t is the number of years from now. b what will be the value of the timber in 10 years?

198 Maths Quest 11 Mathematical Methods CAS

9 The number of koalas remaining in a parkland t weeks after a virus strikes is given by the function

96 koalas per hectare. t+3 How many koalas per hectare were there before the virus struck? How many koalas per hectare are there 13 weeks after the virus struck? How long after the virus strikes are there 23 koalas per hectare? Will the virus kill off all the koalas? Explain why.

N (t ) = 15 + a b c d

10 A school concert usually attracts 600 people at a cost of $10 per person. On average, for every $1 rise

in admission price, 50 less people attend the concert. If T is the total amount of takings and n is the number of $1 increases: a write the rule for the function that gives T in terms of n b sketch the graph of T versus n c find the admission price that will give the maximum takings.

Chapter 4 • Relations, functions and transformations 199

Summary Set notation

• {. . .} refers to a set. • ∈ means ‘is an element of’. • ∉ means ‘is not an element of’. • ⊂ means ‘is a subset of’. • ⊄ means ‘is not a subset (or is not contained in)’. • ∩ means ‘intersection with’. • ∪ means ‘union with’. • \ means ‘excluding’. • ∅ refers to ‘the null, or empty set’. • {(a, b), (c, d), . . .} is a set of ordered pairs. • A relation is a set of ordered pairs. • N refers to the set of natural numbers. • Z refers to the set of integers. • Q refers to the set of rational numbers. • R refers to the set of real numbers.

relations and graphs

• The independent variable (domain) is shown on the horizontal axis of a graph. • The dependent variable (domain) is shown on the vertical axis of a graph. • Discrete variables are things that can be counted. • Continuous variables are things that can be measured.

domain and range

• The domain of a relation is the set of first elements of a set of ordered pairs. • The range of a relation is the set of second elements of a set of ordered pairs. • The implied domain of a relation is the set of first element values for which a rule has meaning. • In interval notation a square bracket means that the end point is included in a set of values, whereas a curved bracket means that the end point is not included. (a, b] a

types of relations (including functions)

b

• A function is a relation that does not repeat the first element in any of its ordered pairs. That is, for any x-value there is only one y-value. • The graph of a function cannot be crossed more than once by any vertical line. y

y

0 x

0

Function power functions (hyperbola, truncus and square root function)

200

1

x

Not a function

1. The graph of y = x is called a hyperbola. a + c is the graph of the basic hyperbola, dilated by the factor of a in the • The graph of y = x−b y-direction, translated b units horizontally (to the right if b > 0 or to the left if b < 0) and c units vertically (up if c > 0 or down if c < 0). If a < 0, the graph is reflected in the x-axis. The equations of the asymptotes are: x = b and y = c. The domain of the function is R \ {b} and its range is R \ {c}.

Maths Quest 11 Mathematical Methods CAS

2. The graph of y = 12 is called a truncus. x a • The graph of y = + c is the basic truncus curve, dilated by a factor of a in the y-direction ( x − b) 2 and translated b units along the x-axis (to the right if b > 0 or to the left if b < 0) and c units along the y-axis (up if c > 0 or down if c < 0). If a is negative, the graph is reflected in the x-axis. The vertical asymptote is x = b. The horizontal asymptote is y = c. The domain is R \ {b}. The range is y > c if a > 0, or y < c if a < 0. 3. The graph of the function y = a x − b + c is the graph of y = x , dilated by the factor of a in the y-direction and translated b units along the x-axis and c units along the y-axis. • If a < 0, the basic graph is reflected in the x-axis. • The end point of the graph is (b, c). • The domain is x ≥ b. • The range is y ≥ c for a > 0, or y ≤ c for a < 0. • If y = a b − x + c, the domain is x ≤ b; the graph of y = a x is reflected in the y-axis. Function notation

• f (x) = . . . is used to describe ‘a function of x’. To evaluate the function, for example, when x = 2, find f (2) by replacing each occurrence of x on the RHS with 2. • Functions are completely described if the domain and the rule are given. • Functions are commonly expressed using the notation f : X → Y, f (x) = . . . . . . Domain

Co-domain

Rule

• dom f is an abbreviation for the domain of f ( x). • ran f is an abbreviation for the range of f ( x). • The maximal domain of a function is the largest domain for which the function will remain defined. Special types of function (including hybrid functions)

• A function is one-to-one if for each x-value there is at most one y-value and vice versa. • A one-to-many function may be ‘converted to’ a one-to-one function by restricting the domain. • A hybrid function obeys different rules for different subsets of the domain.

inverse relations and functions

• An inverse relation is obtained by interchanging the x- and y-values of the original relation. • The graph of a function and its inverse are reflections of each other across the line y = x. • Only one-to-one functions have inverses.

Circles

• The general equation of a circle with centre (h, k) and radius r is (x − h)2 + (y − k)2 = r2 • An ‘upper semicircle’ with centre (0, 0) and radius r is y = r 2 − x 2 . • A ‘lower semicircle’ with centre (0, 0) and radius r is y = − r 2 − x 2 .

Functions and modelling

• When using functions to model situations: 1. form an equation involving one variable and sketch a graph 2. use the graph to determine domain and range.

Chapter 4 • Relations, functions and transformations

201

Chapter review S h ort anS Wer

1 The total number of cars that have entered a car park during the first 5 hours after opening is shown in

the table below. Time, t (hours) No. of cars, n

1

2

3

4

5

30

75

180

330

500

a Plot these points on a graph. b Explain why the dots cannot be joined. 1 c Estimate the number of cars in the park 2 hours after the car park opens. 2

2 a Sketch the graph of the relation {(x, y): y = 1 − x2, x ∈ [−3, 3]}. b State the domain and range of this relation. 3 State the implied domains of the following functions. b y=

a y= x

5 x

c y = 12 − x

4 If g( x ) = x + 2, where x ≥ 0, then find: a g ( x2) b the domain and range of g ( x). 5 Determine which of the following relations are functions. a y = 2x2 − 1 b 3x + y = 2 d x2 + y2 = 10 e y3 = x

c x = y2 + 1 f y2 − x2 = 1

6 Express the following rules in full function notation.

1 b y = (2 − x ) x 7 Sketch the graph of the function described below. x ≤ −1 2 − x , − f ( x ) = 3, 1< x < 3 2 x − 5, x ≥3 8 Sketch the graph of each of the following, stating the domain and range. a y=

a y=

−

1− x2

b (x − 2)2 + (y + 1)2 = 9

9 a Sketch the graph of the relation x2 + y2 = 100. b From this relation form two one-to-one functions (with maximal domains) and state the domain

and range of each. 10 A chicken farmer delivers chicken manure according to the following fee schedule:

Less than half a truckload: $50 Half to a full truckload: $75 More than 1 but less than 2 truckloads: $100 Sketch a graph showing this informations. 2 11 Sketch a graph for the equation y = + 1. x+3 12 State the dilation factor, reflections and translations that have occurred to each of the following equations. −3 5 +7 a f (x) = − 2 x − 4 b f (x) = c f (x) = + 2 x−2 x2 M U lt ip l e C h oiCe

1 If A = {−2, −1, 0, 1, 2, 3} and B = {−2, 0, 2, 4, 6}, then A ∪ B is: a {−2, −1, 0, 1, 2, 3, 4, 6} C {−1, 1, 3, 4, 6} e ∅

b {−2, 0, 2} d {−1, 1, 3}

2 Which of the following statements is false? a Z⊂Q d {0, 1, 2, 3} ∈ N

202

Maths Quest 11 Mathematical Methods CAS

b 3.142 ∈ Q e (N ∪ Z) = Z

C π∈R

y 8

3 The rule describing the relation shown is: a y = 2x b y = 2x, x ∈ {1, 2, 3, 4} C y = 2x, x ∈ N

6

x 2 e y = 2x, x ∈ R+

4

d y=

2 0

1

2

3

x

4

4 Which one of the following graphed relations is continuous? y a y b

C

x

0

y

x

0

x

0 y

d

e

y

x

0

x

0 5 The interval shown below is: −5

−1 0 1

a [−5, −1] ∪ [0, 4] d (−5, −1) ∪ (1, 4)

4 x b [−5, −1) ∪ [0, 4] e [−5, −1) ∪ (1, 4]

C (−5, −1) ∪ (1, 4]

6 The set R+ \ {2} is correctly represented on which number line below? a

b

0

2

x

C

0

2

x

0

2

x

d

0

2

x

0

2

x

e

7 The domain of the relation shown below is: a b C d e

R \ {0, 1} R \ {1} R Z \ {1} R+ ∪ R−

y 4 2 0 1

x

8 A relation has the rule y = x + 3, where x ∈ R+. The range of this relation is: a R+ d R

b R+ \ {3} e (3, ∞)

9 The implied domain of the relation described by the rule y = a (5, ∞) d (0, 5)

b R+ e R−

10 The range of the function f ( x ) = 2 4 − x is: a R d [0, ∞)

b R+ e (2, ∞)

C [3, ∞)

1 is: x−5

C [5, ∞)

C R−

Chapter 4 • Relations, functions and transformations

203

y

11 The relation shown is:

one-to-one one-to-many many-to-many many-to-one none of the above 12 Which of the following is not a relation? a y = x2 b x2 + y2 = 3 e {1, 3, 5, 7, 9} d y=5−x a b C d e

0

C {(1, 1), (2, 1), (3, 2), (4, 3)}

13 Which one of the following graphed relations is not a function? y y a b x

0

x

C

y

x

0

0 d

e

y

y

x

0

x

0

x

14 Which of the following rules does not describe a function?

x b y = 2 − 7x C x=5 d y = 10x2 + 3 5 15 Which of the functions listed below is not one-to-one? a {(10, 10), (11, 12), (12, 13)} b {(5, 8), (6, 10), (7, 8), (8, 9)} d {(x, y): y = 5 − 2x} e f (x) = 2 − x3 16 Which of the graphs below represents a one-to-one function? a y=

a

0

d

b

y

x

0

0

e

y

C

y

x

e y = −8 C {(x, y): y = 4x}

y

0

x

y

x 0

x

17 The function with the domain −5 ≤ x ≤ 5, range −8 ≤ y ≤ 17 and rule y = x2 − 8 can be written in

function notation as:

a f : R → R where y = x2 − 8 C f : [−5, 5] → R where f (x) = x2 − 8 e f : [−8, 17] → R where f (x) = x2 − 8

b f : R → R where f (x) = x2 − 8 d f : R → [−8, 17] where f (x) = x2 − 8

18 The function f : {x: x = 0, 1, 2} → R, where f (x) = x − 4, may be expressed as: a {(0, −4), (1, −3), (2, −2)} d {(−1, −5), (1, −3), (2, −2)}

b {0, 1, 2} e {−4, −3, −2}

19 If g (x) = 6 − x + x2, then g (−2) is equal to: a 6 d 12

204

Maths Quest 11 Mathematical Methods CAS

b 8 e 5

C {(0, 4), (1, 3), (2, 2)}

C 0

20 If f (x) = 3x − 5, then f (2x + 1) is equal to: a 6x − 8

b 6x − 5

C 3x − 5

d 3x − 4

e 6x − 2

21 The graph at right is to be restricted to a one-to-one function. A possible restricted

y

domain could be: a [−3, 1) b (1, 3) C (0, ∞) d (3, ∞) e (−3, ∞)

9

−3 −1 0

x + 1, x < 0 0 ≤ x ≤ 2 is represented by which of 22 The hybrid function f ( x ) = x 2 , 2 − x , x > 2 the following graphs? a

b

y 4 1

1

−10 1 2 d

C

y 4

x e

1

x

y 4 1

−10

y 4

3

2

x

2

x

0

2

x

y 4 1

−1 0

2

x

−1 0

23 The inverse of the graph shown below is: y

x

a

b

y

x d

y

x e

x

C

y

y

x

y

x

Chapter 4 • Relations, functions and transformations

205

24 The graph of the circle relation (x − 2)2 + (y + 1)2 = 4 is: y a b

y

2

1

1 −1 0 −1

1

2

3

4

−1 0 −1

5 x

1

2

3

4 x

−2 −3

−2 −3 −4 y

C

d

4 3 2 1 x 0 −3−2−1 −1 1 2 3 4 5 6 7 −2 −3 −4 −5

y 6 5 4 3 2 1 0 x −7 −6−5−4−3−2−1 −1 1 2 3 −2 −3 −4

y 4

e

3 2 1 −5 −4 −3 −2 −1 0 −1

1 x

−2 25 The equation of the circle shown is: y 2 0 1 −2

a (x + 3)2 + y2 = 4 C (x + 3)2 + y2 = 2 e x2 + (y − 3)2 = 4

3

5 x

b (x − 3)2 + y2 = 2 d (x − 3)2 + y2 = 4

The circle with equation (x + 1)2 + (y − 4)2 = 9 applies to questions 26 and 27. 26 The domain is: a [−10, 8] C (−2, 4) e [−4, 2]

27 The range is a [−7, −1] C [1, 7] e [1, 7]

206

Maths Quest 11 Mathematical Methods CAS

b [−2, 4] d [−3, 3]

b [−5, 13] d [−3, 3]

28 A circle has its centre at (4, −2) and a radius of 5. The equation of the circle is: a b C d e

(x − 4)2 + (y + 2)2 = 25 (x − 4)2 + (y + 2)2 = 5 (x + 4)2 + (y − 2)2 = 5 (x + 4)2 + (y − 2)2 = 25 4x2 − 2y2 = 5

29 The graph that best represents the function f : [−2, 2] → R where f (x) = 4 − x 2 is: a b C y y y 2 2 2

−2

2 x

0

−2

e

y

2 x

0 −2

y 4

4

−4

−2

−2

−2 d

2 x

0

4 x

0

−4

4 x

0 −4

−4

30 The table of maths tutoring fees charged by a Year 11 student is as follows:

Hours (h)

Charge (C$)

0

50

2

80

4

100

Which of the following graphs best shows the information in the preceding table? a

c 100

b

50 0 d

C

50 2

4

6 h

0

c 100 50

2

4

6 h

2

4

6 h

0

2

4

6 h

c e 100

c 100

50

50 0

c 100

2

4

6 h

0

1 Consider the diagram shown at right. a Find an expression for the area, A, in terms of x and y. b Find an expression for the perimeter, P. c If the perimeter is 72 cm, express A as a function of x. d What is the domain of A(x)? e Sketch the graph of this function. f Hence find the maximum area.

ex ten d ed r eS p o n S e

10 m xm ym

xm

Chapter 4 • Relations, functions and transformations

207

2 For the graph at right: a state the domain b state the range c find the rule for x ∈ (−∞, −2) d find the rule for x ∈ (−2, 0] e find the rule for x ∈ [0, 3], given it is of the form y = ax2 f determine the rule when x ≥ 3 g describe the relation using hybrid function notation −2 … … of the form f ( x ) = … . …. … . … 3 A function f is defined as follows: f : [−2, a] → R, where f (x) = (x − 1)2 − 4. a Find f (−2), f (−1), f (0), f (1), f (3). b If f (a) = 12, find the value of a. c Sketch the function f, labelling the graph appropriately. d From the graph or otherwise, state the: i domain of f (x) ii range of f (x). 4 A new rectangular penguin enclosure is to have a maximum side length of 8 m. The width is to be twice the length (x). a Draw a diagram of the enclosure and label the sides. b Define a rule that gives the perimeter, P, of the new enclosure. c What is the largest value that x can be? d State the domain and range. e Write in function notation the rule for the perimeter. f Define the function for the area of the enclosure, A(x). g If the maximum area allowed is 18 m2, find the dimensions of the enclosure.

y 18

(3, 18)

4 0

4 x

5 Thomas is looking to connect to a mobile phone service. He has to decide on one of two plans from

Busytone Communications. The details are as follows.

diGital doC doc-9748 Test Yourself Chapter 4

208

Plan

Flag fall

Cost perminute (cents)

A

20

30

B

50

20

a Write a function A for the cost of making a call from plan A. b Write a function B for the cost of making a call from plan B. c What is the cost of a 2-minute call from: i plan A? ii plan B? d What is the length of one call costing $5 from: i plan A? ii plan B? e How long would a call be for the call costs to be the same for both plans? f If Thomas frequently makes calls lasting more than 4 minutes, which plan should he connect with?

Maths Quest 11 Mathematical Methods CAS

ICT activities Chapter opener diGital doC • 10 Quick Questions doc-9738: Warm up with ten quick questions on relations, functions and transformations (page 161)

4C

domain and range

tUtorial • We 7 eles-1461: Watch how to sketch relations and state their domain and range using interval notation (page 169) diGital doCS • SkillSHEET 4.1 doc-9739: Practise stating the domain and range of relations (page 171) • WorkSHEET 4.1 doc-9740: Use set notation and interval notation, recognise appropriate specific number fields, recognise whether graphs are discrete or continuous and state the domain and range of relations (page 172) • Investigation doc-9741: Investigate some interesting relations using a graphing program to sketch them (page 172)

4d

types of relations (including functions)

tUtorial • We 8 eles-1462: Watch a tutorial on identifying the type of relation given a graph (page 173)

4e power functions (hyperbola, truncus and square root function) interaCtiVitY • Domain and range int-0263: Consolidate your understanding of domain and range for relations and functions (page 175)

4F

Function notation

diGital doCS • SkillSHEET 4.2 doc-9742: Practise substituting values into functions (page 186) • SkillSHEET 4.3 doc-9743: Practise transposing equations (page 186) • doc-9744: Investigate the graph of a square root function (page 186)

tUtorial • We 16 eles-1463: Watch a tutorial on stating maximal domain and range for functions with defined rules (page 185)

4G Special types of function (including hybrid functions) eleSSon • Hybrid functions eles-0077: Watch an eLesson to learn more about hybrid functions (page 187) tUtorial • We 20 eles-1464: Watch a tutorial on sketching the graph of a hybrid function and stating its range (page 189) diGital doC • WorkSHEET 4.2 doc-9745: Recognise types of relations and functions, determine maximal domain and range, identify co-domains and sketch graphs of relations (page 191)

4i

Circles

diGital doCS • Investigation doc-9746: Investigate the graph of x2 + y2 = 25 by constructing a table of values and plotting the graph (page 193) • doc-9747: Investigate graphs of circle relations (page 195)

4J

Functions and modelling

tUtorial • We 24 eles-1465: Watch a tutorial on expressing the hire costs of a removal van as a hybrid function and sketching the graph of the function (page 197)

Chapter review diGital doC • Test Yourself doc-9748: Take the end-of-chapter test to test your progress (page 208)

To access eBookPLUS activities, log on to www.jacplus.com.au

Chapter 4 • Relations, functions and transformations

209

Answers CHAPTER 4 relationS, FUnCtionS and tranSForMationS

y 7 6 5 4 3 2 1 −1 0 −2

e

exercise 4a

1 a b c d e f g 2 a b c d e 3 a b c d e 4 E 5 a 6 a e i

Set notation ∅ {4, 6} {6] {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14} {4, 5, 6, 7, 8} {2, 8, 10, 12, 14} {4, 5} {2, 3} {−3, −2, −1} {−2, −1} {−3, −2, −1, 0, 1} {−3, −2, −1, 0, 1, 2, 3, 4} ∅ {b, c, d, f, g, h} {a, e, i} {b, c, d, f, g, h} {o, u}

C T F T

b b f j

exercise 4b

1 2 3 4 5

cT gT kF

bC f C

cC

2 −2 −1 0

x 200 190 180 170 160 150 140 130

80 70 60 50 40 30 20 10

d F h T l T

0 2 4 6 8

t (minutes)

Cost (¢)

M TWT F S S Day

exercise 4C

80 70 60 50 40 30 20 10

y 9

c

n

0

1 a b c d e f g h 2 a

1

2

−2 −1 0 −1

3

x

1

x

2

−3

y=x−2 2

4

5

x

6

0 2

b −9

−3

0

c 0 2 0

5

e 01

10

f 0 2 0 1 2 3 4 5 6 n

0

7

g −2 0 1 3

h −8

0 2

i 01

4

j −1 0 1 2 3 4 5 t (s)

b i Approx. 110 km/h ii Approx. 320 km/h

Maths Quest 11 Mathematical Methods CAS

∞) (−∞, 5) (−3, 4] (−8, 9) (−∞, −1] (1, ∞) (−5, −2] ∪ [3, ∞) (−3, 1) ∪ (2, 4]

d

350 300 250 200 150 100 50

−4

−2

3

c The variables are discrete. 9 a V (km/h)

−2

0

2

b P ($)

y

y

1

domain and range

[−2,

−6

P($) 300 340 380 420 460 500 540 550 500 450 400 350 300 250 200

0 5 10 15 20 25 n

c The variables are discrete.

b T (°C)

d D

C($) 140 146 152 158 164 170 176 182 188 194 200

b C ($)

7 a T (°C)

8 a

1 0

210

2

c Because the variables are continuous. d Approx. 11 minutes

4

d

1

0 2 4 6 8 10 t (minutes)

60

b

x

2

4

70

6 a

1

n 15 16 17 18 19 20 21 22 23 24 25

y 6

f

relations and graphs

B A E C a D e D

B F T F

−2

10 a

k 0 2

l −2 0 1

5

6

3 a [−4, 2)

b (−3, 1]

c (− 1, 3)

d

−

1 1 , 2 2 − ( ∞, −3] [0, ∞) (−∞, −2) ∪ (−2, ∞) (−∞, −2] ∪ [0, ∞) E B

e (3, ∞) f g (−∞, ∞) h i (−∞, 1) ∪ (1, ∞) j k (−∞, 2) ∪ (3, ∞) l 4 a D b 5C 6 7 a i {3, 4, 5, 6, 7} ii {8, 10, 12, 14, 16} b i {1.1, 1.3, 1.5, 1.7} ii {1.4, 1.6, 1.8, 2} c i {3, 4, 5, 6} ii {110, 130, 150, 170} d i {M, T, W, Th, F} ii {25, 30, 35} e i {3, 4, 5} ii {13, 18, 23} f i R ii [−1, ∞) 8 a R, R b R, (0, ∞) c [−2, 2], [0, 2] d [1, ∞), R e R, (0, 4] f R, (−∞, −3] g R\{0}, R\{0} h R, (−∞, 1] i R, R y

9 a

2 0

− 2

2

b

x

y = x3 + 1 x ∈ [−2, 2]

1 0

−2

2

x

−7

Domain = [−2, 2], range = [−7, 9] c

y y = x2 + 3x + 2

2

x

−2 −1 0

−

d

Domain = (−∞, ∞), range = [ 14 , ∞) y

−2 −1 0 −3

−2 −1 0

y=

x2

−1 0

1 2 3 4

x

2

y=5

−6

c [−4, 4] fR

b [0, ∞) e R\{0}

10 a R d R

exercise 4d types of relations (including functions) 1 a One-to-many b Many-to-one c Many-to-one d One-to-one e One-to-one f Many-to-one g Many-to-many h Many-to-one i One-to-one j Many-to-one k Many-to-many l Many-to-one 2 b, c, d, e, f, h, i, j, l 3C 4 a B b D c B 5 b {−3, −1, 0, 1, 2}, {−2, −1, 1, 3} c {3, 4, 5, 6}, {−1} e R, {2} g R, R

j [−1, ∞), [0, ∞) k R, R a, d, f, h, i and l are not functions.

power functions (hyperbola, truncus and square root function) 1 a Dilation factor of 3 away from the x-axis, horizontal translation of 5 units in the positive direction of the x-axis and a vertical translation of 1 unit up. b Dilation factor of 2 away from the x-axis and a vertical translation of 1 unit up. c Reflection in the y-axis, horizontal translation of 4 units in the positive direction of the x-axis and a vertical translation of 2 units down. d Dilation factor of 2 away from the x-axis and a horizontal translation of 4 units in the negative direction of the x-axis. e Horizontal translation of 3 units in the positive direction of the x-axis. f Reflection in the x-axis, dilation factor of 4 away from the x-axis and a vertical translation of 5 units down. 2 a

y

1

b

y=1 x

1 2 Asymptote x =0

2

x=1

exercise 4F

1 a b c d e 2 a c

−2 −1 −1

1 2 3 (3, −1)

x

i i i i i

1 2 3 9 12

Function notation ii 7 iv iii −5 ii 1 iii 3 iv ii 2 iii 6 iv ii 1 iii 16 iv ii 6 iv iii −4

16 0 9 a2 + 6a + 9 2 b −3 or 3 d 2 or 3

3 1 3 −

e 4 or 1 3 a 3 c e 4 a b c d e f g h 5 a

c

f

3−1

x

1 3 y= +3 ( x − 2)

e

x

y=2

1 2

d

y

x

3 5

x=1

y

d

b

−2 −1

Domain = [−1, 4), Range = [−7, 3)

0

(1, 4)

(−1, 4)

−5 −7

y

Domain = (−∞, ∞), Range = [−6 81, ∞)

− 4, x ∈ [−2, 1]

y = 2x − 5, x ∈ [−1, 4)

c

3

−4 y 3

1

x

1

Domain = [−2, 1], range = [−4, 0] e

y = 2x2 − x − 6

exercise 4e

Domain = (−∞, ∞), range = (−∞, 2] y 9

y

f

6 a b c

f −1 b 3

10 5 d 2 − x2 − 2x x x 10 10 − x +1 f − x−3 x −1 x+3 f : R → R, f ( x) = 4x + 1 f : [6, ∞) → R, f ( x) = x − 6 1 f : R\{1} → R, f ( x) = x −1 2 f : (−1, ∞) → R, f ( x) = x +1 f : (0, ∞) → R, f ( x) = (x + 2)2 f : [2, ∞) → R, f ( x) = x2 + 3x f : (−∞, 0] → R, f ( x) = 8 − x f : [0, ∞) → R, f ( x) = x2 + x i {0, 1, 2, 3} ii Z iii {−7, −4, −1, 2} i (0, 10] ii R 3

iii [ 10 , ∞) i {2, 4, 6, 8, 10} iii {1, 2, 3, 4, 5} i (−∞, 0) iii (0, ∞) i R+ iii (−2, ∞) i [−3, 3] iii [0, 3] i R i [0, ∞) i R

ii N ii R ii R ii R ii R ii [0, ∞) ii R

Chapter 4 • Relations, functions and transformations

211

d i R e i [4, ∞) f i (3, ∞)

(−∞, 5] [0, ∞) (0, ∞)

ii ii ii

exercise 4G Special types of function (including hybrid functions) 1 a, c, d, f, g 2 i a, b, c, d, f, h, i, j, k, l ii c, h, i, k 3 a (−∞, −1], [−1, ∞) b (−∞, 2), (2, ∞) c [−3, 0], [0, 3] d (−∞, 3], [3, ∞) e [−4, −2), (−2, 0] f (−∞, −1], [−1, 1] or [1, ∞) g (−∞, 0], [0, ∞) h [−2, 0], [0, 2] i (−∞, 0), (0, ∞) j (−∞, −3], [−3, ∞) 4 a D b E 5 a B b C

6 a

y 2 1

f(x)

b (−∞, 0) ∪ [1, ∞) 7 a y 3 2 1

y

1 −1 0

x

b g : (−∞, −1] → R, g ( x) = x2+ 2x + 1 or

x

g : [−1, ∞) → R, g ( x) = x2+ 2x + 1

exercise 4h inverse relations and functions 1 a {(7, 1), (5, 2), (3, 3)} y

d y

y=x

8 7 6 5 4 3 2 1

x 1 2 3 4 5 6 7 8 x

0

y

e

y

y=x

9 8 7 6 5 4 3 2 1

g(x)

x

0 1 2 3 4 5 6 7 8 9 x

−2 −1 0

1

b [1, ∞) c i 3 ii 1 y 8 a

2

x

y

iii 2

y=x

10 8 6 4 2

1 2 3 x

−3 −2 −1 0

y

f

c {(11, −2), (6, 0), (1, 2)}

5 4 3 2 1

0

−4 −5

x

2

4 6 8 10

x

3 a

(−∞,

b 0] ∪ (4, ∞) c i −5 ii 0 iii −3

y

y

2 a y iv 0

x + 2, x ≤ 0 9 f (x) = 2 x + 1, x > 0 10 f : [0, 1] → R, f ( x) = 1 − x 2 with range [0, 1] or f : [0, 1] → R, f ( x) = − 1 − x 2 with range [−1, 0]. 11 a

c g(x)

b {(3, 1), (6, 2), (9, 3)} x

01 2

−2 −1

y

12 a

y=x

v 7 x

x

b

b y

f(x)

y = 4x

y

y = x2 + 3

y=x

9

0

3

x

3

b f : (−∞, 3] → R, f ( x) = (x − 3)2 or

f : [3, ∞) → R, f ( x) = (x − 3)2

212

Maths Quest 11 Mathematical Methods CAS

x

0

3

x

c y

y

f

y = 1_2 x + 1

−1–2

0 −1–2

y

h

1– 2

3

x

1– 2

x

0

− 3

− 3

1

Not a function

0

x

1

d

y

4 a

y 9

y = x3 + 4 −9

0

9

x

exercise 4i

x

4

b d f

g y = 9 − x2 2 a Both [−3, 3] c Both [−5, 5]

h b d f

−

b

x2 + y2 = 1 x2 + y2 = 100 x2 + y2 = 8

y

−2

6, 6]

0

−2

2

y

−1

2

0

1

x

c

0

x

1– 3

[1, 7] and [−3, 3] d

Function

x

4

e

y

0

− 1–2

7

x

1– 2

x

d

6 x

2

−5

[−2, 6] and [−5, 3] e

Function f

−7

y 3 −2 −1 0

− 1–2 0

7 x

4

−3 0

− 1–3

y 7

−7

y

0 1

−4

c

x

2

3

1– 3

y 4

−4

0

y

d

−2

b

−2

[−2, 2] and [0, 4] Function

x

y 4

Function c

2

2

[−1, 1] and [−3, −1]

x

b

−

y = 16 − x 2 Both [−1, 1] Both [−10, 10] Both [− 2 2, 2 2] h [−4, 4], [−4, 0]

g [−3, 3], [0, 3] 3 a y

0

x

−1 −1 0 1

−3

Circles

1 a x2 + y2 = 9 c x2 + y2 = 25 e x2 + y2 = 6

−2

y

Not a function 0

e Both [

D B C E

7 a

−9 4

5 a b 6 a b

y 3

y −8

5

y

−3

2 x

0 −2

7 − 7

0

g

− 10

−2 3

−7

[−8, 2] and [−7, 3] f

2 3 x

0 − 10

Not a function

y 5

10

2 3 0

x

y

y

−2 3

5

Function

− 7

e

0

− 5

7 x

10 x

2 −1 0

3

6

x

[0, 6] and [−1, 5]

Chapter 4 • Relations, functions and transformations

213

y 10

g

4 a B=

n 12 c $90

b B (hours) 10

4 5 −11

[−11,

1] and

x

0 1 x −2

−5

[−2,

0

10]

0 1–

−1

2

2

60

120

b f : (− ∞, 2] → R, f ( x) =

n

17 550

x

−2 −1 0

− 3–2

8 a

4650.30

−3

[−1, 2] and [−3, 0] 8 y=

36 − domain 6] and range [0, 6] or y = 36 − x 2 ; domain [−6, 6] and range [−6, 0] 9 y = 2 + 9 − x 2 ; domain [−3, 3] and range [2, 5] or y = 2 − 9 − x 2 ; domain [−3, 3] and range [−1, 2] 10 a 2 cm, 13.8 cm b 3.9 cm/s exercise 4J

Functions and modelling

40, 70, 1 a C (t ) = 110, 160,

0 < t ≤1 1< t ≤ 2

c 6 a b 7 a b 8 a b 9 a c d

2

10 a b

b C ($) 110 70 40 1 2 3 4 5 6 t (hours)

Cost ($)

b

−1

0

Distance (km)

60t ,

0 ≤ t ≤ 1.5 1.5 ≤ t ≤ 2 80t − 70, 2 ≤ t ≤ 4

700

300

9 a

y

x2 + y2 = 100

10

10 x

0

(100 − x 2 ) with dom f = [−10, 10], ran f = [0, 10] and f 2 : [−10, 10] R, f ( x) = − (100 − x 2 ) with dom f = [−10, 10], ran f = [−10, 0]

10

200

100 75 50 25 0

100 1 2 3 4 5 t (hours)

1 −3−2−1 0 1 2 3

x

11

1 2 Number of truck loads y

2 1 −5 −4−3−2−1 −1

5 3

1 2 3

x

−3

3 a d (t ) = 90,

b Domain [0, 4]; range [0, 250] c i 60 km ii 170 km

x

Domain = [−1, 5]; range = [−4, 2]

Cost ($)

400

b The number of cars is a discrete variable. c 120 y 2 a 100 200

5

−4

Graph is not continuous as n ∈ N

0

0.80 0.60 0.40

0 (2, −1)

b f 1 : [−10, 10] → R, f ( x) =

500

2.00 1.70

y 2

1 a

100 < d ≤ 200 200 < d ≤ 700 d > 700

x

1

−10

Short anSWer

50 < d ≤ 100

0

Domain = [−1, 1]; range = [−1, 0]

−10

Chapter reVieW

0 < d ≤ 50

No. of cars (n)

0.40, 0.60, 2 a C (d ) = 0.80, 1.70, 2.00,

x

−1

0 1 2 3 4 5 6 7 8 9 101112 n

0

214

−1

6000 5000 4000 3000 2000 1000

160

b

Domain [37 001, 80 000]; range [4650.3, 17 550] $6150 P = 4x + 6 Domain (1, 6]; range (10, 30] A = x2 + 4x Domain (0, 8]; range (0, 96] P = 100 000(1.02)t $121 899 47 b 21 9 weeks 96 No, as t increases approaches zero, t+3 so N approaches 15. T = 6000 + 100n − 50n2 c $11 T

1 2 3 4 5 y

80 000 x ($)

37 001

[−6,

x2 ; −

2− x

y 5 4 3 2 1

7

5 a T = 0.3x − 6450 b T ($)

y

h

3 a Domain = [0, ∞) b Domain = R\{0} c Domain = (− ∞, 12] 4 a x + 2, x ≥ 0 b Domain = [0, ∞); range = [2, ∞) 5 a, b, e 1 6 a f : R \{0} → R, f ( x) =

12 a Dilation factor of 2 away from the −8

y = 1 − x2

b Domain = [−3, 3]; range = [−8, 1]

Maths Quest 11 Mathematical Methods CAS

x-axis, reflection in the y-axis and vertical translation of 4 units in the positive direction of the x-axis.

b Dilation factor of 5 away from the

x-axis and a horizontal translation of 2 units up. c Dilation factor of 3 away from the x-axis, reflection in the y-axis, vertical translation of 2 units in the positive direction of the x-axis and a horizontal translation of 7 units up.

1 4 7 10 13 16 19 22 25 28

A C B D B C D A D B

2 5 8 11 14 17 20 23 26 29

D E E D C C E E E C

extended reSponSe

1 a A = xy + 10y − x2 b P = 2x + 2y + 20 or

P = 2(x + y + 10)

c A = 260 + 16x − 2x2

3 6 9 12 15 18 21 24 27 30

B C A E B A D A C E

(5, 12)

292 260

(−2, 5)

130

(−1, 0) −2

0 MUltiple ChoiCe

f (x) 12

c

d (0, 13) e A (m2)

2 4 6 8 10 12 14 x (m) −

2 a x ∈ ( ∞, 4)\{ 2}

ii Range = [−4, 12]

2x

c y=4 d y = −2x

x

e y = 2x2 f y = −18x + 72

3 a 5, 0, −3, −4, 0 b a=5

(1, −4)

d i Domain = [−2, 5] 4 a

b [0, 18]

4, − 2 x , g f (x) = 2 2x , − 18 x + 72

5 x

1

(0, −3) −4

f 292 m2 −

0

x ∈(− ∞ , − 2) x ∈(− 2, 0] x ∈[0, 3] x ∈[3, 4]

b d e f g 5 a c d e

P(x) = 6x c 0<x≤4 Domain = (0, 4]; Range = (0, 24] P: (0, 4] → R where P(x) = 6x A: (0, 4] → R where A(x) = 2x2 Length = 3, Width = 6 A(x) = 0.2 + 0.3x b B(x) = 0.5 + 0.2x i 80 cents ii 90 cents i 16 min ii 22.5 min 3 min f Plan B

Chapter 4 • Relations, functions and transformations

215

ChaPTeR 5

Exponential and logarithmic functions DiGiTal DoC doc-9749 10 Quick Questions

ChaPTeR ConTenTS 5a 5B 5C 5D 5e 5F 5G 5h

Index laws Negative and rational powers Indicial equations Graphs of exponential functions Logarithms Solving logarithmic equations Logarithmic graphs Applications of exponential and logarithmic functions

introduction Functions in which the independent variable is an index number are called indicial or exponential functions. For example: f (x) = ax where a > 0 and a ≠ 1 is an exponential function. It can be shown that quantities that increase or decrease by a constant percentage in a particular time can be modelled by an exponential function. Exponential functions have applications in science and medicine (for example, decay of radioactive material, or growth of bacteria like those shown at right), and finance (for example, compound interest and reducing balance loans).

5a

index laws

Recall that a number, a, that is multiplied by itself n times can be represented in index notation: a× a × a … ×a= an n lots of a

Index (or power or exponent) Base

where a is the base number and n is the index (or power or exponent). The expression an is read as ‘a to the power of n’ or ‘a to the n’.

multiplication When multiplying two numbers in index form with the same base, add the indices. For example, 23 × 24 = 2 × 2 × 2 × 2 × 2 × 2 × 2 = 27

am × an = am + n

ChaPTeR 5 • Exponential and logarithmic functions

217

Division When dividing two numbers in index form with the same base, subtract the indices. am ÷ an = am − n 2×2×2×2×2×2 = 24 For example, 26 ÷ 22 = 2×2

Raising to a power To raise an indicial expression to a power, multiply the indices. (am)n = am × n = amn For example, (24)3 = 24 × 24 × 24 = 24 + 4 + 4 = 24 × 3 = 212

Raising to the power of zero Any number raised to the power of zero is equal to one. a0 = 1, a ≠ 0 3 3 3 − 3 0 For example 2 ÷2 =2 = 2 [1] or 23 ÷ 23 = (2 × 2 × 2) ÷ (2 × 2 × 2) =8÷8 =1 So 23 ÷ 23 = 1 [2] Using [1] and [2] we have 20 = 1.

Products and quotients Note the following. (ab)n = anbn For example, (2 × 3)4 = (2 × 3) × (2 × 3) × (2 × 3) × (2 × 3) n n a = a = 2 × 3 × 2 × 3 × 2 × 3 × 2 × 3 b bn = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 4 4 =2 ×3 Worked Example 1

Simplify. 6 2 3 5 a 2x3y2 × 4x2y b (2x2y3)2 × xy4 c (3a)5b6 ÷ 9a4b3 d 8 p m × (3 p) m 4 6p m Think

a

b

c

1

Collect ‘plain’ numbers (2 and 4) and terms with the same base.

2

Simplify by multiplying plain numbers and adding powers with the same base. (Note: y = y1.)

1

Remove the bracket by multiplying the powers. (The power of the 2 inside the bracket is 1.)

Write

a 2x3y2 × 4x2y

= 2 × 4 × x3 × x2 × y2 × y

= 8x5y3 b (2x2y3)2 × xy4

= 22 × x4 × y6 × xy4

2

Convert 22 to a plain number (4) first and collect terms with the same base.

= 4 × x4 × x × y6 × y4

3

Simplify by adding powers with the same base.

= 4x5y10

1

Write the quotient as a fraction.

2

Remove the bracket by multiplying the powers.

c (3a)5 b 6 ÷ 9a 4 b 3 =

(3a)5 b6 9a 4 b 3

=

243a5 b6 9a 4 b 3

=

27a5 b6 a 4 b3

3 4

Simplify by first cancelling plain numbers. Complete simplification by subtracting powers with the same base. (Note: a1 = a.)

218 Maths Quest 11 Mathematical Methods CAS

= 27ab3

d 1 Expand the brackets by raising each term to

the power of 3.

6 2 3 5 6 2 3 3 5 d 8 p m × (3 p) m = 8 p m × 3 p m 4 4

6p m

6p m

8 × 27 × p6 × p3 × m 2 × m 5 6 p4 m

2

Convert 33 to 27 and collect like variables.

=

3

Simplify by first reducing the plain numbers, and then reducing the variables by adding the indices for multiplication and subtracting the indices for division.

= 36p6 + 3 − 4m2 + 5 − 1

4

Simplify the indices of each base.

= 36p5m6

WoRkeD examPle 2 3

Simplify

6 a 4 b3 3 a 2 b ÷ . 16 a7 b6 2 a3 b2

Think

WRiTe

6a 4 b3 3a 2 b ÷ 16a 7 b6 2a3 b 2

3

1

Write the expression.

2

Change the division sign to multiplication and replace the second term with its reciprocal (turn the second term upside down).

=

6a 4 b 3 2a 3 b 2 × 16a 7 b6 3a 2 b

3

Remove the brackets by multiplying the powers.

=

6 a 4 b 3 23 a 9 b 6 × 16a 7 b6 33 a6 b3

4

Collect plain numbers and terms with the same base.

=

6 × 8a 4 + 9 − 7 − 6 b 3 + 6 − 6 − 3 16 × 27

5

Cancel plain numbers and apply index laws.

=

a0 b0 9

6

Simplify.

=

1 9

3

WoRkeD examPle 3

Write the following in simplest index notation and evaluate. 95 × 34 a 23 × 162 b 27 3 Think

WRiTe

a 1 Rewrite the bases in terms of their prime factors.

a 23 × 162 = 23 × (2 × 2 × 2 × 2)2

2

Simplify the brackets using index notation.

= 23 × (24)2

3

Remove the brackets by multiplying the powers.

= 23 × 28

4

Simplify by adding the powers.

= 211

5

Evaluate as a basic number.

= 2048 ChaPTeR 5 • Exponential and logarithmic functions

219

b 1 Rewrite the bases in terms of their prime factors.

b

95 × 34 (3 × 3)5 × 34 = 273 (3 × 3 × 3)3

2

Simplify the brackets using index notation.

=

(32 )5 × 34 (33 )3

3

Remove the brackets by multiplying the powers.

=

310 × 34 39

4

Write in simplest index form.

= 35

5

Evaluate as a basic number.

= 243

Complex expressions involving terms with different bases have to be simplified by replacing each base with its prime factors.

WoRkeD examPle 4

Simplify

3 4 n × 18 n + 1 . 63 n − 2

Think

WRiTe

34 n × 18n + 1 34 n × (3 × 3 × 2)n + 1 = 63 n − 2 (2 × 3)3n − 2

1

Rewrite the bases in terms of their prime factors.

2

Simplify the brackets using index notation.

=

34 n × (32 × 21 )n + 1 (2 × 3)3n − 2

3

Remove the brackets by multiplying powers.

=

34 n × 32 n + 2 × 2 n + 1 23n − 2 × 33n − 2

4

Collect terms with the same base by adding the powers in the products and subtracting the powers in the quotients.

= 34n + 2n + 2 − (3n − 2) × 2n + 1 − (3n − 2)

5

Simplify.

= 36n + 2 − 3n + 2 × 2n + 1 − 3n + 2 − = 33n + 4 × 2 2n + 3 = 33n + 4 × 23 − 2n

index laws

exercise 5a

Simplify each of the following.

1 We1a, b a

x2

×

2 We1c a

a7b8

x5

×

b 52 × 57 × (53)3

x3

c (xy)3 × x4y5

d 3m2p5 × (mp2)3 × 2m4p6

c (3x5)y11 ÷ 6x2y2

d p13q10 ÷ ( pq4)2

Simplify each of the following. ÷ a2b5

b 2a12b9 ÷ (2a)3b4

Simplify each of the following. 6 p8 m 4 × 2 p7 m 6 (3 x )2 y 2 × 5 x 6 y3 a b 9 p5 m 2 10 x 7 y

3 We1d

c

14u11v 9 × (3u 2 )3 v 21u6 v 5

b

5k 12 d 6 kd 4 ÷ 3 2 (2 k ) 25( k 2 d 3 )3

4 We2 Simplify each of the following.

15a8 b3 2a3 b ÷ a 9a 4 b5 3ab 2 220

2

Maths Quest 11 Mathematical Methods CAS

d

(5e3 )2 f 4 × 8e 4 f 3 20e f 5

3 p3 m 4 can be simplified to: p1m 2

5 mC a

a 3p2m2

b

6 5

B 3p4m6

C 3p3m8

D 3p3m2

C x30y16

D

C −3a6b

D

6x y x × can be simplified to: x 5 y3 (2 y)2 a 2x5y4

B 3x5y4 2

3ab3 a 2 b ÷ c − is equal to: ab a5 a −a18

B −3a6

3 x 9 y10 2

x n + 1 × y5 × z 4 − n x × y4 − n × z 3 − n n−2

a

×

42

× 8

c 53 × 152 × 32

34 × 272 f 64 × 35 Write the following in simplest index notation and evaluate.

45 27 (625)4 e (53 )5 We4

e

b 94 × 35 ÷ 27

a

9

f

(25)4 (125)3

8 × 52 23 × 10

c

(162 )3 (25 )4

d

272 (32 )3

g

411 ÷ 82 163

h

272 × 81 93 × 35

c

4 n × 7n − 3 × 493n + 1 14 n + 2

Simplify the following.

× 92 n + 1 12 x − 2 × 4 x b 6n − 2 6x − 2 n − 5n + 1 5n − 4 3 n 5 3 × 16 × 9 d e* 5n + 1 + 5n 4 n + 1 × 181 − n × 63 − 2 n *Hint: Factorise the numerator and denominator first. 362 n × 6n + 3 10 mC In simplest index notation, is equal to: 216n − 2 a 216n + 5 B 65n + 1 C 62n + 5 a

e 3a6

b6

( x n y m + 3 )2 x2 y × x n + 2 y3 − m x n − 5 × y5 − 3m

b 37 × 92 × 273 × 81

d 205 × 84 × 125 8

3x 5 2

Write the following in simplest index notation.

7 We3

24

b

e

− 3a15

6 Simplify each of the following. a

e 3

4

2n

5B negative and negative powers

D 69

e 62n + 9

rational powers

Wherever possible, negative index numbers should be expressed as positive index numbers using the simple rule: When an index number is moved from the numerator to denominator or vice versa, the sign of the power changes. 1 a −n = n , a ≠ 0 a This is easily verified as follows: 1 a0 = since a0 = 1 an an = a0 − n using the division rule for indices =a n simplifying the index. a n 1 1 an = n and − n = . In other words, 1 a a 1 A simple way to remember this rule is ‘change the level, change the sign’. −

−

ChaPTeR 5 • Exponential and logarithmic functions

221

WoRkeD examPle 5

Express each of the following with positive index numbers. 5 a 8 4 b x y

−4

−2

× ( x 4 y) − x 3 y3

TUToRial eles-1417 Worked example 5 −

5

Think

WRiTe

a 1 Remove the brackets by raising the denominator

and numerator to the power of −4.

a 5

−4

8

−

4

−

4

=

5 8

84 54

2

Interchange the numerator and denominator, changing the signs of the powers.

=

3

Simplify by expressing as a fraction to the power of 4.

8 = 5 4 b x y

b 1 Remove the brackets by multiplying powers.

−

× ( x 2 y) x 3 y3

2

Collect terms with the same base by adding the powers on the numerator and subtracting the powers on the denominator.

2

4

−

−

5

=

x4 y

−

=x

=

5

−

−

−

−

Rewrite the answer with positive powers.

−

x 6y 7 x 3 y3 6 − ( − 3)

= x 3y 3

−

−

−

=

× x 10 y x 3 y3

2

−

y

−

7−3

10

1 x 3 y10

Rational powers Until now, the indices have all been integers. In theory, an index can be any number. We will confine ourselves to the case of indices that are rational numbers (fractions). 1

a n , where n is a positive integer, is defined as the nth root of a: 1

an = n a For example, we know that a × a = a 1

1

1

a2 × a2 = a2 = a1 =a

but

+

1 2

1

a = a2

Therefore, 1

1

Similarly, 3 a = a 3 , 4 a = a 4 … etc. 1

a n is defined for all a ≥ 0 and n ≠ 0. In general, m an

222

Maths Quest 11 Mathematical Methods CAS

1 = an

m

=

( n a )m = n a m

WoRkeD examPle 6

Evaluate each of the following without a calculator. a

3 16 2

9 b 25

−3

2

Think

WRiTe 3

3

a 16 2 = (24 ) 2

a 1 Rewrite the base number in terms of its

prime factors. 2

Remove the brackets by multiplying the powers.

= 26

3

Evaluate as a basic number.

= 64

b 1 Rewrite the base numbers of the fraction in

terms of their prime factors.

b 9 25

−3

2

32 = 2 5 −

3

−

3

2

Remove the brackets by multiplying the powers.

=

3 5

3

Rewrite with positive powers by interchanging the numerator and denominator.

=

53 33

4

Evaluate the numerator and denominator as basic numbers.

=

125 27

−3

2

WoRkeD examPle 7

Simplify the following, expressing your answer with positive indices. a

7

128 × 4 64

b

3

x 2 y6 ÷ x 3 y 5

Think

a 1 Write the expression.

WRiTe

a

7

128 × 4 64 1

1

2

Write using fractional indices.

= 128 7 × 64 4

3

Write 128 and 64 in index form.

= (27 ) 7 × (26 ) 4

4

Multiply the powers.

= 21 × 2 4

5

Simplify the powers.

= 21 × 2 2

1

1

6

3

5

= 22 b 1 Write the expression.

b

3

x 2 y6 ÷ x 3 y5 1

1

2

Express the roots in index notation.

= ( x 2 y6 ) 3 ÷ ( x 3 y5 ) 2

3

Remove the brackets by multiplying the powers.

= x 3 y2 ÷ x 2 y 2

2

3

5

ChaPTeR 5 • Exponential and logarithmic functions

223

2

4

Collect terms with the same base by subtracting the powers.

= x3

5

Simplify the powers.

=x

6

Rewrite with positive powers.

=

−

3 5 2− 2y 2

−1 6y 2

−5

negative and rational powers

exercise 5B DiGiTal DoC doc-9750 SkillSHEET 5.1 negative and rational powers

Express each of the following with positive index numbers.

1 We5a a 6

−

3

−

4

−

−

−

−

2

−

3

−

−

−

(3 2 )2 × (2 5 ) 1 e (24 ) 2 × (34 ) 3 −

−

−

−

−

−

x 3 y 2 × ( xy 2 ) 3 (2 x 3 )2 × ( y 3 )2 −

f

−

−

−

Evaluate the following without a calculator.

3 We6 1 92 1 16 4 81 3

4 mC a

−

−

−

e

34 c 3 2

2

Simplify each of the following, expressing your answer with positive index numbers. ( x 2 )3 × ( y 4 ) 2 ( − m)2 × m 3 ×2 4 b c x 5 × ( y 2 )3 (p 2) 1 × p 4 2 3

( − 22))3

x5 (x 4 ) ÷ d x 3 (x 2 )

a

−

3 5

b

2 We5b a

1 5 1 x6 y2

1

2

3

b 27 3

c 83

d 814

f

−

3 25 2 16

g

3 81 4

h

− 2 8 3 27

25 × 125 simplifies to:

5 25 6

7

3

11

13

B 56

C 52

D 56

e 56

5 We7 Simplify each of the following, expressing your answer with positive indices.

d g

2

9 × 3 81

a

5 24

−1

×4

2

b

(64 m 6 ) 3 e 4m 2

3

−

( x + 1)2 x +1

5C

c

3

(xy3 ) ÷ ( x 2 y)

4

−2

×8

1

x3 × x6

f

1 x

−

4

h ( y − 4) y − 4

indicial equations 1

We can solve equations of the form x 3 = 2 as follows: 1

( x 3 ) 3 = 23 The left-hand side becomes x, so x = 8. However, when the unknown (or variable) is not a base number but is an index number, a different approach is required. Take the cube of both sides:

method 1: exact solutions without a calculator To attempt to solve index equations exactly, express both sides of the equation to the same base and equate the powers. If am = an, then m = n. 224

Maths Quest 11 Mathematical Methods CAS

Worked Example 8

Find the value of x in each of the following equations. a 3x = 81 b 4x − 1 = 256 c 63x − 1 = 362 x − 3 Think

a 1 Write the equation. 2

Express both sides to the same base.

3

Equate the powers.

b 1 Write the equation.

Write

a

3x = 81 3x = 34 ⇒x=4

b 4x − 1 = 256

2

Express both sides to the same base.

4x − 1 = 44

3

Equate the powers.

⇒x−1=4

4

Solve the linear equation for x by adding one to both sides.

x = 5

c 1 Write the equation.

c 63x − 1 = 362x − 3

2

Express both sides to the same base.

63x − 1 = (62)2x − 3

3

Remove the brackets by multiplying the powers.

63x − 1 = (6)4x − 6

4

Equate the powers.

⇒ 3x − 1 = 4x − 6

5

Subtract 3x from both sides to make x the subject.

−1

6

Add 6 to both sides to solve the equation.

x = 5

=x−6

More complicated equations can be solved using the same technique. Worked Example 9

Solve for n in the following equation. 23n × 16n + 1 = 32 Think

Write

1

Write the equation.

23n × 16n + 1 = 32

2

Express both sides using the same base, 2.

23n × (24)n + 1 = 25

3

Remove the brackets by multiplying the powers.

23n × 24n + 4 = 25

4

Multiply the terms on the left-hand side by adding the powers.

5

Equate the powers.

6

Solve the linear equation for n.

27n + 4 = 25 ⇒ 7n + 4 = 5 7n = 1 n=

1 7

In some cases indicial equations can be expressed in a quadratic form and solved using the Null Factor Law. Look for numbers in index form similar to a2x and ax appearing in different terms. Chapter 5 • Exponential and logarithmic functions 225

WoRkeD examPle 10

Solve for x if 52x − 4(5x) − 5 = 0. Think

WRiTe

5 − 2x

4(5x)

TUToRial eles-1418 Worked example 10

−5=0

1

Write the equation.

2

Rewrite the equation in quadratic form. Note that 52x = (5x)2.

(5x)2 − 4(5x) − 5 = 0

3

Substitute y for 5x.

Let y = 5x

4

Rewrite the equation in terms of y.

⇒ y2 − 4y − 5 = 0.

5

Factorise the left-hand side.

(y − 5)(y + 1) = 0

6

Solve for y using the Null Factor Law.

⇒ y = 5 or y = −1

7

Substitute 5x for y.

5x = 5 or 5x = −1

8

Equate the powers.

⇒ 5x = 51 and 5x = −1

9

State the solution(s).

⇒x=1

(5x = −1 has no solution.)

Note that in step 9, the possible solution 5x = −1 was rejected because there is no value of x for which it will be satisfied. Recall that exponential functions such as 5x are always positive.

method 2: Using a CaS calculator If answers are not exact, the CAS calculator can be used to solve indicial equations. WoRkeD examPle 11

Solve for x given 3

−

2x+1

=

1 . Write your answer correct to 2 decimal places. 45

Think 1

WRiTe

Use the solve feature of the CAS calculator: 1 solve 3 2 x + 1 = 5 , x 4

The result is x = 3.654 65.

Write the answer.

Solving 3

−

2

exercise 5C

1 for x gives x = 3.65, 45 correct to 2 decimal places. −

2x +1

=

indicial equations

1 We8a Find the value of x in each of the following equations. a 2x = 32 b 5x = 625 c 3x = 243 d 10

−

x

1 = 100

e 4

−

x

= 16

f

6x =

1 216

2 We8b Find the value of n in each of the following equations. 3n + 1 a 2 = 64 b 52n + 3 = 25 c 32 − n = 27 d 16n + 3 = 23 3 We8c

f

364n − 3 = 216

b 274 − x = 92x + 1

c 163x + 1 = 128x − 2

3x + 5 = d 9

c 34x × 27x + 3 = 81

d

1 243

Solve for x in each of the following equations.

a 2x × 83x − 1 = 64

226

1 7

Find x in each of the following.

a 42x = 8x − 1 4 We9

5 − 3n = e 49

Maths Quest 11 Mathematical Methods CAS

b 52x × 1253 − x = 25

812 − x = 92 x 27 x + 3

5 We10 Solve for x in each of the following. a 32x − 4(3x) + 3 = 0 b 22x − 6(2x) + 8 = 0 2x x c 3(4 ) = 15(4 ) − 12 d 25x − 30(5x) + 125 = 0 6 mC Consider the indicial equation 32x − 12(3x) + 27 = 0. The equation can be solved by making the

substitution: a y = 3x D y = 2x

B y = 2x e y = 3x

C y = 32x

7 mC The quadratic equation formed by the appropriate substitution in question 6 is: a y2 − 3y + 27 = 0 D y2 − 12y + 27 = 0

B y2 − 11y + 27 = 0 e y2 − 9y + 3 = 0

C y2 + 12y + 27 = 0

8 mC The solutions to the equation in question 7 are x equals: a 2 or 3 D 0 or 1

B 1 or 2 e 0 or 2

C 1 or 3

9 We11 Solve for x. Write your answer correct to 2 decimal places. a 2x = 3 b 3x = 8 c 10x = 45 d 4x = 10 10 mC The nearest solution to the equation 3x = 10 is: a x = 2.5 D x=2

5D

B x = 2.3 e x = 2.1

C x = 1.9

Graphs of exponential functions

Functions of the form f (x) = ax, where a is a positive real number other than 1 and x is a real number, are called exponential functions. In general, there are two basic shapes for exponential graphs: y = ax, a > 1 or y = ax, 0 < a < 1 y

y

y = ax, a > 1

y = ax, 0 < a < 1

1

1 Asymptote y=0

x

0

Asymptote y=0 Decreasing exponential x

0

Increasing exponential

However, in both cases: the y-intercept is (0, 1) the domain is R the asymptote is y = 0 (x-axis) the range is R+. x x Verify the shapes of these graphs by graphing, say y = 2x, y = 3x, y = 1 and y = 1 on a graphics 2 3 calculator. What is the effect of changing a on the steepness of the graph?

• • • •

Reflections of exponential functions The graph of y = a x is obtained by reflecting y = ax in the y-axis. −

y

The graph of y = −ax is obtained by reflecting y = ax in the x-axis. y

y=

ax,

a>1

y = ax, a > 1 1

y=

1 0

a−x,

a>1 Asymptote x y=0

−1

0

x y = −ax, a > 1

Asymptote y=0

ChaPTeR 5 • Exponential and logarithmic functions

227

Horizontal translations of exponential functions The graph of y = a x + b is obtained by translating y = ax: 1. b units to the right if b < 0 2. b units to the left if b > 0. For example, the graph of y = 2x − 3 is obtained by translating y = 2x to the right 3 units. Check this graph using a graphics calculator. Note also that

y y = 2x

3 units

2 1

()

2 x −3 = (2 x )(2 3 ) = 1 2 x , so the effect is identical to that of 8 multiplying by a constant. −

y = 2x − 3

Vertical translations of exponential functions

Asymptote y=0

x

−1 0 1 2 3 4

y

y = 10 x

10

The graph of y = ax + c is obtained by translating y = ax: 1. up by c units if c > 0 2. down by c units if c < 0. Furthermore the equation of the asymptote becomes y = c. For example, the graph of y = 10x − 5 is obtained by translating y = 10x down by 5 units. −5 units The equation of the asymptote is y = −5. −4. The y-intercept is Check this graph using a graphics calculator.

5 y = 10 x − 5 1 −1 −4 −5

1

x

Asymptote y = −5

Dilation from the x-axis The graph of y = Aax (for positive, real values of A) has a dilation factor of A. The graph is stretched along the y-axis, away from the x-axis (as each y-value is being multiplied by the constant A). Consider the graphs below. The y-intercept in each case is equal to A. Also, as A increases, the graph becomes steeper; as A decreases, the graph becomes less steep. The domain, range and asymptotes are the same as for f (x) = ax. y f(x) = 4 × 2x f(x) = 3 × 2x f(x) = 2x 4 3 (0, 3) 2 1 (0, 1) Asymptote x 0 y=0 −3 −2 −1 1 2 3

y

(0, 4)

Dilation from the y-axis

2 (0, 1) (0, 1–2) −1

(1, 2)

f(x) = 2x f(x) =

1– 2

f(x) =

× 2x 1– 4

× 2x

(1, 1) 1–

(0, 1–4) (1, 2 ) Asymptote x 0 y=0 1

1

The graph of y = akx (for k > 0) has a dilation factor of k from the y-axis. The graph is said to be stretched along the x-axis. Consider the graphs at below. The y-intercept is (0, 1) in each case. As k increases, the graph becomes steeper and closer to the y-axis. The domain, range and asymptotes are the same as for f (x) = ax. y 4

f(x) = 22x f(x) = 2x –x f(x) = 22

2 (0, 1) −3 −2 −1

228 Maths Quest 11 Mathematical Methods CAS

01 2 3

x

Asymptote y=0

Worked Example 12

Find the equation of the asymptote and the y-intercept for each of the following functions. Hence, sketch the graph of each function and state its domain and range. a f : R → R, f (x) = 2x + 3 − 5 x

b f : R → R, f ( x ) = 3 × 2 2 Think

a 1 Write the rule.

Write/Draw

a f (x) = 2x + 3 −5

2

The graph is the same as y = 2x translated 3 units left and 5 units down.

3

State the asymptote.

The asymptote is y = −5.

4

Evaluate y when x = 0 to find the y-intercept.

When x = 0, y = 23 − 5 = 3 Therefore, the y-intercept is (0, 3).

5

Locate the y-intercept and asymptote on a set of axes.

6

Sketch the graph of the exponential function using the y-intercept and asymptote as a guide.

y f(x) = 2x +3 − 5 3 x

0

Asymptote y = −5

−5 7

Use the graph to state the domain and range.

The domain is R and the range is (−5, ∞). x

b 1 Write the rule.

b f (x) = 3 × 2 2 0

2

Find the y-intercept by letting x = 0, or recall that the y-intercept is equal to A in f (x) = Aa x.

f (0) = 3 × 2 2 = 3 × 20 = 3×1 =3 The y-intercept is 3.

3

Locate the horizontal asymptote.

The horizontal asymptote is the x-axis.

4

Locate another point on the graph. This is necessary to be able to see the effect of the dilation. Locate the second point by substituting a value for x into the equation and evaluating a corresponding y-value.

f (2) = 3 × 2 2 = 3 × 21 = 3× 2 =6 Another point is (2, 6).

5

Sketch the graph. Note: The asymptote remains at y = 0 as there is no vertical translation.

y

2

–x

f(x) = 3 × 22 6

(2, 6)

(0, 3) 0 6

State the domain and the range.

1

2

Asymptote x y=0

The domain is R and the range is R+.

Chapter 5 • Exponential and logarithmic functions 229

WoRkeD examPle 13

Use a CAS calculator to solve 2x = 15 (correct to 2 decimal places) by finding the intersection of two graphs. Think

WRiTe

1

Enter the left-hand side as a function into the CAS calculator.

f1(x) = 2x

2

Enter the right-hand side as a function into the CAS calculator.

f2(x) = 15

3

Graph the two functions and use the intersection points feature of the CAS calculator. Write the answer.

The graphs intersect at (3.91, 15).

exercise 5D

Graphs of exponential functions

1 Sketch the graph of each of the following on separate axes. (Use a table of values or copy a DiGiTal DoC doc-9751 SkillSHEET 5.2 Substitution in exponential functions

CAS calculator screen). a y = 3x c y = 10x e y=4 x g y = −3 x

b y = 5x− x d y=2 − x f y= 3 h y = 0.5x

−

−

2 Sketch the following graphs, using a table of values or by copying a CAS calculator screen. State the

equation of the asymptote and the y-intercept for each. a y = 2(3x) c

b y = 0.5(4x)

1

y = 4 (2 x )

d

y = 4 1

x

3

Find the equation of the asymptote and the y-intercept for each of the following. Hence, sketch the graph of each and state its domain and range. a f : R → R, f (x) = 2x − 1 b f : R → R, f (x) = 3x + 2 c f : R → R, f (x) = 51 − x d f : R → R, f (x) = 2x + 3 e f : R → R, f (x) = 3x − 3 f f : R → R, f (x) = 2x + 3 − 1 x h f : R → R, f (x) = 102 − x + 5 6 g f : R → R, f (x) = +3

3 We12a

−

4 mC a The rule for the graph at right is: a B C D e

y

y = 3x − 2 y = 3x y = 2x − 3 y = 3x + 2 y = 3x − 1

3 2 1

b The rule for the graph at right is: a B C D e

Asymptote 0 1 2 3 x y=0 y

y = 2x − 3 y = 3x − 2 y = 2x + 1 − 3 y = 2x − 1 + 3 y = 2x − 1 − 3

x

0 −2 −3 −4

(1, −2)

y = −3

5 We12b Sketch the graph of each of the following, stating the domain and range. a f : R → R, f (x) = 22x b f : R → R, f (x) = 23x x c f : R → R, f (x) = 3 × 2 d f : R → R, f (x) = 5 × 32x x

e f : R → R, f ( x ) = 2 × 5 2 230

Maths Quest 11 Mathematical Methods CAS

x

f

f : R → R, f ( x ) = 2 × 2 3

6 Sketch a graph of f (x) = 4 × 23x + 3, stating the domain and range. Compare your answer to that found

using a CAS calculator. 7 We13 Use a CAS calculator to solve the following indicial equations using the intersection of two

graphs. Give answers rounded to 2 decimal places. a 2x = 10 b 10x = 20 c 3x = 12

5e

d 2x = x + 3

e 3x = x + 4

DiGiTal DoC doc-9752 WorkSHEET 5.1

logarithms

The index, power or exponent (x) in the indicial equation y = ax is also known as a logarithm. This means that y = ax can be written in an alternative form: loga (y) = x, which is read as ‘the logarithm of y to the base a is equal to x’. For example, 32 = 9 can be written as log3 (9) = 2. 105 = 100 000 can be written as log10 (100 000) = 5.

Logarithm y = ax Base numeral Base

ax = y is equivalent to x = loga (y).

In general, for a > 0 and a ≠ 1:

Using the indicial equivalent, it is possible to find the exact value of some logarithms. WoRkeD examPle 14

Evaluate the following without a calculator. 1 a log6 (216) b log 2 ( 8 ) Think

WRiTe

a Let x = log6 (216)

a 1 Let x equal the quantity we wish to find. 2

Express the logarithmic equation as an indicial equation.

6x = 216

3

Express both sides of the equation to the same base.

6x = 63

4

Equate the powers.

x=3

b 1 Write the logarithm as a logarithmic equation. 2

Express the logarithmic equation as an indicial equation.

8

1 b Let x = log2

2x =

1 8

= 1

3

2

= (2 1 )3 x 2 =2 3 −

3

Express both sides of the equation to the same base.

4

Equate the powers.

−

x = −3

logarithm laws The index laws can be used to establish corresponding rules for calculations involving logarithms. These rules are summarised in the following table. Name Logarithm of a product

Rule

Restrictions

loga (mn) = loga (m) + loga (n)

m, n > 0 a > 0, a ≠ 1

loga m = loga (m) − loga (n)

m, n > 0 a > 0 and a ≠ 1

Logarithm of a power

loga (m)n = n loga (m)

m>0 a > 0 and a ≠ 1

Logarithm of the base

loga (a) = 1

a > 0 and a ≠ 1

Logarithm of one

loga (1) = 0

a > 0 and a ≠ 1

Logarithm of a quotient

n

ChaPTeR 5 • Exponential and logarithmic functions

231

It is important to remember that each rule works only if the base, a, is the same for each term. The ‘logarithm of a product’ and ‘logarithm of a quotient’ rules formed the basis for the pre-1970s calculation device for multiplication and division — the slide rule. Worked Example 15

Simplify, and evaluate where possible, each of the following without a calculator. a log10 (5) + log10 (4) b log2 (12) + log2 (8) − log2 (3) Think

Write

a 1 Apply the ‘logarithm of a product’ rule. a log10 (5) + log10 (4) = log10 (5 × 4) 2

b 1

= log10 (20)

Simplify. Multiply the base numerals of the logs being added since their bases are the same.

b log2 (12) + log2 (8) − log2 (3) = log2 (12 × 8) − log2 (3)

= log2 (96) − log2 (3)

2

Apply the ‘logarithm of a quotient’ law.

= log2 (96 ÷ 3)

3

Simplify, noting that 32 is a power of 2.

= log2 (32) = log2 (2)5

4

Evaluate using the ‘logarithm of a power’ and ‘logarithm of the base’ laws.

= 5 log2 (2) =5

Worked Example 16

Simplify 3 log2 (5) − 2 log2 (10). Think

Write

3 log2 (5) − 2 log2 (10) = log2 (5)3 − log2 (10)2

1

Express both terms as logarithms of index numbers.

2

Simplify each logarithm.

= log2 (125) − log2 (100)

3

Apply the ‘logarithm of a quotient’ law.

= log2 (125 ÷ 100)

4

Simplify.

= log2

Worked Example 17

Simplify each of the following. log 8 (49) a b 2 log10 (x) + 1 c 5 log10 (x) − 2 log 8 (343) Think

a 1 Express each base numeral as powers

to the same base, 7.

Write

a

log8 (49) log8 (7)2 = log8 (343) log8 (7)3

2

Apply the ‘logarithm of a power’ law.

=

2 log8 (7) 3 log8 (7)

3

Simplify by cancelling out the common factor of log8 (7).

=

2 3

232 Maths Quest 11 Mathematical Methods CAS

5 4

or log2 (1.25)

b 1 Express 2 log10 (x) as log10 (x)2 and

1 as a logarithm to base 10 also.

2

b 2 log10 (x) + 1 = log10 (x)2 + log10 (10)

= log10 (10x2)

Simplify using the ‘logarithm of a product’ law.

c 1 Express 5 log10 (x) as log10 (x)5 and

2 as 2 log10 (10).

c 5 log10 (x) − 2

= log10 (x)5 − 2 log10 (10)

2

Express 2 log10 (10) as log10 (10)2.

= log10 (x)5 − log10 (10)2

3

Simplify using the ‘logarithm of a quotient’ law.

x5 = log10 2 10 x5 = log10 100

WoRkeD examPle 18

Evaluate each of the following expressions, correct to 3 decimal places. a log2 (5) b log7 (8) Think

WRiTe

a & b 1 On a Calculator page, complete the

entry lines as: log2 (5) log7 (8) 2

Write the answer.

a log2 (5) = 2.332 b log7 (8) = 1.069, correct to 3 decimal places.

logarithms

exercise 5e 1

Express the following indicial equations in logarithmic form. b 35 = 243 c 50 = 1 −2 − 1 d 0.01 = 10 e bn = a f 2 4= 16 a 23 = 8

Express the following logarithmic equations in indicial form.

2

= −1

b log10 (1000 000) = 6

c

log2

d log3 (27) = 3

e log5 (625) = 4

f

log2 (128) = 7

g

log3 1 9

−

= 2

h logb (a) = x

3 mC The value of log5 (25) is: a −2

B 5

C 1

D 2

= 512 is: B log3 (512) = 8 e log8 (3) = 512

4 mC When expressed in logarithmic form, a log3 (8) = 512 D log512 (3) = 8

5

1 2

a log4 (16) = 2

e

4

83

C log8 (512) = 3

When expressed in indicial form, log10 (10 000) = 4 is: a = 10 000 B 10 0004 = 10 C 10 00010 = 4 10 000 10 D 10 =4 e 4 = 10 000 mC

104

6 We14

Evaluate each of the following.

a log2 (16)

b log3 (81)

e log10 (1000)

f

log10 (0.000 01)

j

log2 1 64

i

log2 (32)

1 4 1 log3 243 logn (n5)

c log5 (125)

d log2

g log2 (0.25)

h

k log3

(−3)

l

ChaPTeR 5 • Exponential and logarithmic functions

233

7 We15

Simplify the following, and evaluate where possible.

a log2 (8) + log2 (10) d log6 (8) + log6 (7) g log2

1 3

+ log2 (9)

b log3 (7) + log3 (15) e log2 (20) − log2 (5)

c log10 (20) + log10 (5) f log3 (36) − log3 (12)

h log4 (25) + log4

i

1 5

log3 (8) − log3 (2) + log3 (5)

8 We16 Simplify each of the following. a 3 log10 (5) + log10 (2) b 2 log2 (8) + 3 log2 (3) c log5 (12) − 2 log5 (2) d 4 log10 (2) − 2 log10 (8) e g

1 3 1 2

1

log2 (27) − 2 log2 (36)

f

log2 (x − 4) + 3 log2 (x)

log3 (16) + 2 log3 (4)

h 2 log10 (x + 3) − log10 (x − 2)

9 We17a Simplify the following. a

log3 (25) log3 (125)

b

log2 (81) log2 (9)

c

2 log10 (8) log10 (16)

d

3log5 (27) 2 log5 (9)

e

log3 ( x 6 ) log3 ( x 2 )

f

log10 ( x 3 ) log10 ( x )

g

2 log2 ( x + 1)3 log2 ( x + 1) C

log10 ( x ) log10 ( y)

10 mC The expression log10 (xy) is equal to: a log10 (x) × log10 (y) D y log10 (x)

B log10 (x) − log10 (y) e log10 (x) + log10 (y)

11 mC The expression log5 (xy) is equal to: a x log5 (y)

B y log5 (x)

C 5 logx (y)

D log5 (x) + log5 (y)

e 5y

1

12 mC The expression 3 log2 (64) + log2 (5) can be simplified to: a log2 (40)

B 1

13 mC The expression a log4 (x3)

B

C

log2

64 15

320

D log2 (20)

e log2 3

D log4 (x5 − x2)

e log4 (x7)

log4 ( x 5 ) can be simplified to: log4 ( x 2 ) 5

log4 ( x 2 )

C

5 2

14 We17b Express each of the following in simplest form. a log3 (27) + 1 b log4 (16) + 3 d 2 + 3 log10 (x)

c 3 log5 (2) − 2

e 4 log3 (2) − 2 log3 (6) + 2

f

1 2

+ 3 log10 ( x 2 )

15 We18 Evaluate the following, correct to 3 decimal places where appropriate. a log2 (16) b log3 (81) c log10 (0.001) d log2 (9) e log3 (4) + log2 (7)

5F Solving logarithmic logarithms to the base 10

equations

Logarithms to the base 10 are called common logarithms and can be evaluated using the log function on a calculator. Note: The logarithm of a negative number or zero is not defined. Therefore: loga (x) is defined for x > 0, if a > 0. This can be seen more clearly using index notation as follows: • Let n = loga (x.) • Therefore, an = x (indicial equivalent of logarithmic expression). • However, an > 0 for all values of n if a > 0 (positive based exponentials are always positive). • Therefore, x > 0. 234

Maths Quest 11 Mathematical Methods CAS

WoRkeD examPle 19

Find x if log3 (9) = x − 2. Think 1

Write the equation.

2

Simplify the logarithm using the ‘logarithm of a power’ law and the fact that log3 (3) = 1.

3

Solve for x by adding 2 to both sides.

WRiTe

log3 (9) = x − 2 log3 (32) = x – 2 2 log3 (3) = x – 2 2=x–2 x=4

WoRkeD examPle 20

Solve for x if log6 (x) = −2. Think 1

Write the equation.

2

Express in index form.

3

Evaluate the index number.

WRiTe

log6 (x) = −2 Therefore, x = 6 2. −

x= =

1 62 1 36

WoRkeD examPle 21

Solve for x given that 2 log x (25) = 4, x > 0. Think

WRiTe

2 logx (25) = 4

1

Write the equation.

2

Divide both sides by 2.

3

Write as an index equation.

4

Express both sides of the equation to the same base, 5.

x2 = 52

5

Equate the bases. Note that x = −5 is rejected as a solution, because x > 0.

x=5

logx (25) = 2 Therefore, x2 = 25.

WoRkeD examPle 22

Solve for x correct to 3 decimal places, if 2x = 7. Think 1

Write the equation.

2

Take log10 of both sides.

3

Use the ‘logarithm of a power’ law to bring the power, x, to the front of the logarithmic equation.

4

Divide both sides by log10 (2) to get x by itself.

5

Evaluate the logarithms correct to 4 decimal places, at least one more than the answer requires.

6

Solve for x.

WRiTe

2x = 7 log10 (2x) = log10 (7) x log10 (2) = log10 (7) Therefore x =

logg10 ((7) logg10 ((2)

0.8451 0.3010 x = 2.808 x=

ChaPTeR 5 • Exponential and logarithmic functions

235

Therefore, we can state the following rule: If ax = b, then x =

logg a ( b) logg a ( a)

= logg a ( b) This rule applies to any base a, but 10 is the most commonly used base for this solution technique.

exercise 5F

Solving logarithmic equations

Find x in each of the following.

1 We19

1 10

a log2 (4) = x

b log9 (1) = x

e 2 log2 (8) = x

f

log3 (81) = 2x

b e h k

log3 (x) = 2 c log5 (x) = 4 log8 (x) = −1 f log3 (x) = −3 log2 (3x + 1) = 4 i log10 (2x) = 1 log3 (5) − log3 (4) = log3 (x) − log3 (8)

Solve for x. log2 (x) = 3 log10 (x) = 1 log3 (x − 3) = 3 2 log6 (3x) = 1

c log10

=x

d log3

1 9

=x

g log10 (1000) = 2x − 1

2 We20 a d g j

Solve for x given that:

3 We21

a logx (36) = 2 d

− 2 log

1 x 100

b logx (125) = 3

=4

e

g logx + 1 (27) = 3

h

1 log x 2 −

c 3 logx (16) = 6

(64) = 3

log3 x − 1

1 32

f

5 logx (625) = 10

= 5.

4 mC a The solution to the equation log7 (343) = x is: a x=2

B x=3

C x=1

D x=0

e x = −2

a 4096

B 512

C 64

D 2

e

D 1

e 9

b If log8 (x) = 4, then x is equal to: 1

1 2

c Given that log x (3) = 2 , x must be equal to: a 3

B 6

C 81

d The solution to the equation log3 (x) − 2 = log3 (x − 8) is: a x=8

B x=6

C x=9

D x = −4

e x=2

5 We22 Solve the following equations for x, correct to 3 decimal places. a 2x = 11 b 2x = 0.6 c 3x = 1.7 x x − 1 e 0.7 = 3 f 10 = 18 g 22x + 1 = 5 2 − x i 8 = 0.75

d 5x = 8 − h 10 2x = 7

6 mC The nearest solution to the equation 4x = 5 is: a x = 0.86 DiGiTal DoC doc-9753 WorkSHEET 5.2

236

C x = 1.25

7 mC The nearest solution to the equation a x = 0.18

5G DiGiTal DoCS doc-9821 Investigation investigating logarithmic graphs 1 doc-9822 Investigation investigating logarithmic graphs 2 inTeRaCTiViTY int-0264 logarithmic graphs

B x = 1.2 B x = 0.13

0.62x − 1

= 2 is:

C x = −0.18

D x=1

e x = 0.5

D x = −0.71

e x = −0.13

logarithmic graphs

The graphs of y = loga (x) and y = ax are reflections of each other across the line y = x. Functions such as these that are reflections of each other in the line y = x are called inverses of each other. Consider the logarithm loga (ax). This logarithm can be simplified using the log laws.

loga (ax) = x loga (a) =x×1 =x Notice how the logarithm with base a and the exponential with base a have a cancelling effect on one another, demonstrating that they are inverse operations. This is similar to the way that multiplication and division have a cancelling effect. Multiplication and division are also inverse operations of each other.

Maths Quest 11 Mathematical Methods CAS

Consider now the exponential aloga (x). As the logarithm with base a is the inverse operation to the exponential with base a, the expression aloga (x) simplifies to give x. That is, aloga (x) = x. The inverse properties of logarithms and exponentials can be used to plot the graphs of logarithmic functions. Alternatively, a table of values can be used. For example: y = log10 (x) −1

x y

0

undefined undefined

1

2

3

4

0

0.301

0.477

0.602

y Asymptote x=0 f(x) = log10 (x)

The graph of y = loga (x) does not exist for values of x ≤ 0. It is an increasing function. There is a vertical asymptote along the y-axis, and so there are no y-intercepts. The x-intercept for all values of a is always (1, 0). That’s because loga (1) = 0. Another point on the graph is (a, 1). That’s because loga (a) = 1. The domain of the function is R+ and the range is R.

0

x

1

WoRkeD examPle 23

Sketch the graph of f (x) = log2 (x). Think 1

2

WRiTe/DRaW

Realise that f (x) = log2 (x) is the inverse of f (x) = 2x, so these two graphs are reflections of each other across the line y = x. Alternatively, recall the basic shape of the logarithmic graph. Sketch the basic shape on a set of axes.

3

Mark the x-intercept (1, 0).

4

Mark a second point on the graph (a, 1), which in this case is (2, 1).

f(x) = 2x

y

y=x

f(x) = log2 (x)

1

(2, 1) 0 (1, 0)

x

WoRkeD examPle 24

Find the equation of the inverse of f (x) = 2x. Think

WRiTe

1

Write the function using (x, y) notation.

y = 2x

2

Interchange x and y.

x = 2y

3

Take the logarithm to base 2 of both sides.

log2 (x) = log2 (2y)

4

Simplify using loga (mn) = n loga (m).

log2 (x) = y log2 (2)

5

Simplify using loga (a) = 1.

log2 (x) = y

6

State the inverse of f.

exercise 5G

−

f 1(x) = log2 (x)

logarithmic graphs

1 Simplify the following expressions. log (b) a log2 (2x) b 10 10

c 3 × log5 (5y)

2 We23 Sketch the graphs of each of the following. a f (x) = log2 (x) b f (x) = log5 (x) d f (x) = log10 (x) e f (x) = log12 (x)

d 5 × 20 log20 (x) c f (x) = log8 (x) f f (x) = log15 (x) ChaPTeR 5 • Exponential and logarithmic functions

237

3 Compare the steepness of each of the graphs in question 2, and hence explain how changing the base, a,

affects the steepness of a logarithmic graph of the type f (x) = loga (x).

4 We24 Find the equation of the inverse of the following. a f (x) = 3x b f (x) = 4x c f (x) = log5 (x)

Further work on logarithmic graphs is available on your eBookPLUS.

DiGiTal DoCS doc-9754 logarithmic graphs doc-9755 Extension logarithmic graphs

applications of exponential and logarithmic functions 5h

Exponential and logarithmic functions can be used to model many practical situations in science, medicine, engineering and economics. WoRkeD examPle 25

A square sheet of paper that is 0.1 mm thick is repeatedly folded in half. a Find a rule that gives the thickness, T mm, as a function of the number of folds, n. b What is the thickness after 10 folds? c How many folds are required for the thickness to reach 6 cm? Think

a

b

c

238

1

T = 0.1 when n = 0 and doubles with each fold. This doubling implies that the base should be 2.

2

Complete a table of values showing the thickness, T, for values of n from 0 to 5.

3

Determine the rule for T(n). There is a doubling term (2n) and a multiplying constant for the starting thickness (0.1).

4

Compare the rule for T(n) against the table of values in step 2.

1

Substitute n = 10 into the formula for T.

WRiTe

a When n = 0, T = 0.1, and as n increases

by 1, T doubles. n T

0 0.1

1 0.2

2 0.4

T(n) = 0.1(2n)

b When n = 10,

T(10) = 0.1(210) T = 102.4 mm

2

Calculate T.

1

Change 6 cm to millimetres.

2

Substitute T = 60 into the formula.

3

Divide both sides by 0.1.

4

Take log10 of both sides.

log10 (600) = log10 (2n)

5

Use the ‘logarithm of a power’ law to bring the power n to the front of the logarithm.

log10 (600) = n log10 (2)

6

Divide both sides by log10 (2).

n=

7

Evaluate.

n ≈ 9.23

8

Round the answer up to the nearest whole number, as the number of folds are positive integers and if you round down the thickness will not have reached 60 mm.

Therefore, n = 10 folds.

Maths Quest 11 Mathematical Methods CAS

c 6 cm = 60 mm

When T = 60, 60 = 0.1 (2n) 600 = 2n

log10 (600) log10 (2)

3 0.8

4 1.6

5 3.2

WoRkeD examPle 26

The price of gold since 1980, P (dollars per ounce), can be modelled by the function P = 400 + 50 log10 (5t + 1), where t is the number of years since 1980. a Find the price of gold per ounce in 1980. b Find the price of gold in 2006. c In what year will the price pass $550 per ounce?

TUToRial eles-1419 Worked example 26

Think

a 1 State the modelling function.

WRiTe

a P = 400 + 50 log10 (5t + 1)

In 1980, when t = 0,

2

Determine the value of t represented by the year 1980.

3

Substitute t into the modelling function.

P = 400 + 50 log10 [5(0) + 1] = 400 + 50 log10 (1)

4

Evaluate P.

P = 400

b 1 Repeat part a by determining the value of

t represented by the year 2006. 2

b t = 2006 − 1980

= 26

When t = 26, P = 400 + 50 log10 [5(26) + 1] = 400 + 50 log10 (131) = 400 + 105.864 = $505.86

Substitute the value of t into the modelling function and evaluate P.

c 1 Since P = 550, substitute into the modelling

function and solve for t.

c

550 = 400 + 50 log10 (5t + 1)

2

Simplify by isolating the logarithm part of the equation.

150 = 50 log10 (5t + 1) 3 = log10 (5t + 1)

3

Express this equation in its equivalent indicial form.

103 = 5t + 1

4

Solve this equation for t.

5

Convert the result into years.

1000 = 5t + 1 999 = 5t 199.8 = t The price of gold will reach $550 in 1980 + 199.8 = 2180 (approximately).

applications of exponential and logarithmic functions exercise 5h

1 We25 Before a mice plague that lasts 6 months, the population of mice in a country region is

estimated to be 10 000. The mice population doubles every month during the plague. If P represents the mice population and t is the number of months after the plague starts: a express P as a function of t b find the population after: i 3 months ii 6 months c calculate how long it takes the population to reach 100 000 during the plague. ChaPTeR 5 • Exponential and logarithmic functions

239

2 WE26 The population of a town, N, is modelled by the function N = 15 000(20.01t), where t is the

number of years since 1980. Find the population in 1980. Find the population in: i 1985 ii 1990. What is the predicted population in 2015? In what year will the population reach 20 000?

a b c d

3 The weight of a baby, W kg, t weeks after birth can be modelled by W = 3 log10 (8t + 10). a Find the initial weight. b Find the weight after: i 1 week ii 5 weeks iii 10 weeks. c Sketch the graph. d When will the baby reach a weight of 7 kg? 4 If $A is the amount an investment of $P grows to after n years at 5% p.a. using compound interest: a write A as a function of P b use the function from a to find the value of $10 000 after 10 years c calculate how many years it will be until an investment of $10 000 reaches $26 500. 5 The value of a car, $V, decreases according to the function V = 25 000 2 0.1t, where t is the number of

years since the car was purchased. a Find the value of the car when new. b Find the value of the car after 6 years. c In how many years will the car be worth $10 000?

5

6 The temperature, T (°C), of a cooling cup of coffee in a room of

temperature 20 °C can be modelled by T = 90(3 0.05t ) , where t is the number of minutes after it is poured. a Find the initial temperature. b Find the temperature: i 3 minutes after pouring ii 6 minutes after pouring. c How long is it until the temperature reaches half its initial value? −

7 A number of deer, N, are introduced to a reserve. The deer population can be predicted by the model

N = 120(1.1t ), where t is the number of years since introduction. a Find the initial number of deer in the reserve. b Find the number of deer after: i 2 years ii 4 years iii 6 years. c How long does it take the population to treble? d Sketch the graph of N versus t. e Explain why the model is not reliable for an indefinite time period. 8 After a recycling program is introduced, the weight of rubbish disposed of by a household each week is given by W = 80(2 0.015t ), where W is the weight in kg and t is the number of weeks since recycling was introduced. a Find the weight of rubbish disposed of before recycling starts. b Find the weight of rubbish disposed of after recycling has been introduced for: i 10 weeks ii 40 weeks. c How long is it after recycling starts until the weight of rubbish disposed of is half its initial value? d i Will the model be realistic in 10 years time? ii Explain. 9 The number of hectares (N) of forest land destroyed by a fire t hours after it started is given by N = 40 log10 (500t + 1). a Find the amount of land destroyed after: i 1 hour ii 2 hours iii 10 hours. b How long does the fire take to burn out 155 hectares? −

240 Maths Quest 11 Mathematical Methods CAS

10 A discus thrower competes at several competitions during the year. The best distance, d metres,

that he achieves at each consecutive competition is modelled by d = 50 + log10 (15n), where n is the competition number. a Find the distance thrown at the: i 1st competition ii 3rd competition iii 6th competition iv 10th competition. b Sketch the graph of d versus n. c How many competitions does it take for the thrower to reach a distance of 53 metres?

11 The population, P, of a certain fish t months after being introduced to a reservoir is

P = 400(100.08t), 0 ≤ t ≤ 20. After 20 months, fishing is allowed and the population is then modelled by P = 15 000 + 924 log10 [10(t − 19)], t ≥ 20. a Find the initial population. b Find the population after: i 5 months ii 15 months iii 25 months iv 40 months. c How long does it take the population to pass 10 000? 7 12 A ball is dropped from a height of 5 metres and rebounds to 10 of its previous height. a Find the rule that describes the height of the ball (h metres) after n bounces. b Find the height after: i 4 bounces ii 8 bounces. c Sketch the graph of the height of the ball after n bounces. 13 A computer appreciates in value by 10% per year. If the computer costs $5000 when new, find: a the rule describing the value, V, of the computer at any time, t years, after purchase b the value of the computer after 6 years c the number of years it takes to reach double its original value. 14 From the start of 1996, a small mining town has seen a steady increase in population until 2000 as the

price of minerals improved and mining was extended. Year Population (P)

1996 700

1997 750

1998 804

1999 870

2000 925

Let 1996 be t = 0; then 1997 will be t = 1 and so on. Plot P against t. What does the shape of the curve look like? Calculate the ratio of the population in 1997 to the population in 1996. Calculate all the ratios in successive years, and hence estimate the percentage annual increase or growth. To obtain an accurate estimate of population growth, follow these steps. e On the table above, evaluate log10 (P). f Plot log10 (P) against t. Are the points approximately collinear? g Draw a line of best fit and find its gradient and the intercept on the y-axis. h Write the equation for the line. i Show that P = 700(1.07)t. Is this close to your estimate in d? j Use this formula to estimate the population in 2001 and 2002. k When might the population have reached 2000? l In fact there was a downturn in the population as the mine output decreased. From 2000 onwards there was an annual decline of 10% in population. During which year did the population reduce to below 600? a b c d

15 A used car dealership keeps data on the value of the Fraud Atlas (new at the start of 2005) over 5 years.

Year (t) Value (V) a b c d e

2005 45 000

2006 35 500

2007 28 000

2008 22 500

2009 18 000

Let t = 2005 be x = 0; then 2006 will be x = 1 and so on. Plot V against x. What does the shape of the curve look like? Calculate the ratios of values of the car in successive years, e.g. V2006 ÷ V2005 and so on. Estimate the annual rate of depreciation. For a more accurate result, evaluate log10 (V). Chapter 5 • Exponential and logarithmic functions 241

f g h i j k l

Plot log10 (V) against x. Are the points approximately collinear? Draw a line of best fit and find its gradient and the intercept on the y-axis. Write the equation for the line. Show that V = 45 000(0.79)x (or a formula close to it). What is the annual rate of depreciation? Use this formula to estimate the car’s values in 2010 and 2011. When will it reach a value of $7000?

16 Johannes Kepler was a German astronomer born in the 16th century. He used data collected by Tycho

Brahe to formulate an equation or law connecting the period of a planet’s revolution around the sun to the radius of its orbit. The following table contains the data Kepler used. The radius of the orbit is expressed as a proportion of Earth’s orbit (with the period given in days). Planet Mercury Venus Earth Mars Jupiter

Radius (R) of orbit 0.389 0.724 1.000 1.524 5.200

Period (T) 87.77 224.70 365.25 686.98 4332.62

a Plot T against R (using (0,0) too). What does the graph look like?

The graph has the form T = aRb, where a and b are constants. b To find them, find log10 (R) and log10 (T). c Plot log10 (T) against log10 (R) on graph paper. Are the points collinear? d Draw a line of best fit and find its gradient, correct to 2 decimal places. e Read off the intercept on the y-axis and write it as the equivalent logarithm. f Write an equation for the straight line. g By transposition, show that T = 365.25R1.50 (or a formula close to it). h If Saturn’s orbit has a radius of 9.510, find its period using the formula above. The actual period is 10 759.2 days. Why is there a difference in the results? i Kepler’s Law is T 2 = kR3. What is the value of k? 242 Maths Quest 11 Mathematical Methods CAS

Summary am × an = am + n am ÷ an = am − n (am)n = amn a0 = 1 (ab)n = anbn n n 6. a = a b bn • To simplify indicial expressions: – when dealing with questions in the form (expression 1) ÷ (expression 2), replace expression 2 with its reciprocal and change ÷ to × – remove brackets using laws 4, 5 and 6 – collect plain numbers and terms of the same base – simplify using laws 1, 2 and 3.

index laws

1. 2. 3. 4. 5.

negative and rational powers

•a

−n

=

1 ,a≠0 an

1

• an = n a m

1

n • a n = (a n ) m = ( n a ) m = a m

indicial equations

• If am = an, then m = n. • A graphics calculator may be used to solve indicial equations, using the solve function.

Graphs of exponential functions

• • • • •

f ( x) = ax, a > 1; f ( x) = ax, 0 < a < 1 The y-intercept is (0, 1). The asymptote is y = 0 (x-axis). The domain is R. The range is R+. y

y

1 0

f(x) = ax, a > 1 Asymptote x y=0

1

f(x) = ax, 0 < a < 1 Asymptote x y=0

0

y

• Reflections: f ( x) = ax, a > 1; f ( x) = a−x, a > 1

f(x) = ax, a > 1

1 0

f ( x) = ax, a > 1; f ( x) = −ax, a > 1

f(x) = a−x, a > 1 Asymptote x y=0

y f(x) = ax, a > 1 1 −1

0

x

Asymptote y=0

f(x) = −ax, a > 1

ChaPTeR 5 • Exponential and logarithmic functions

243

• Translations f ( x) = ax, a > 1; f ( x) = ax + b, a > 1, b > 0

−

x+b y f(x) = a , a > 1, b > 0

x b f(x) = a , a > 1

b 1 x

0

f ( x) = ax, a > 1; f ( x) = ax + c, a > 1, c > 0

Asymptote y=0

x y f(x) = a + C, a > 1, C > 0

f(x) = ax, a > 1

C

Asymptote y=C C

1 x

0

• Dilations f ( x) = ax, a > 1; f ( x) = Aax, A > 0, a > 1

y

f(x) = Aax, A > 1, a > 1 f(x) = ax, a > 1

A 1 0

f ( x) = ax, a > 1; f ( x) = akx, a > 1, k > 0

y 4

x

Asymptote y=0

f(x) = 22x f(x) = 2x –x f(x) = 22

2 (0, 1) −3 −2 −1

logarithms

244

Asymptote y=0

• If y = ax, then loga ( y) = x, where a = the base, x = the power, index or logarithm, and y = the base numeral. Log laws: • loga (m) + loga (n) = log a (mn) m, n > 0 m • loga (m) − loga (n) = loga n m, n > 0 • loga (mn) = n loga (m) • loga (a) = 1 • loga (1) = 0

Solving logarithmic equations

0 1 2 3 x

m>0

• Logarithmic equations are solved more easily by: 1. simplifying using log laws 2. expressing in index form 3. solving as required.

Maths Quest 11 Mathematical Methods CAS

• If ax = b, then x = • loga (ax) = x

log10 (b) = loga (b) log10 (a)

log ( x ) • a a =x

logarithmic graphs

• The logarithmic function f ( x) = loga (x) is the inverse function of the exponential function f ( x) = ax. y

f(x) = ax, a > 1

y=x f(x) = loga (x), a > 1

1 0 1

• • • •

x

The x-intercept is (1, 0). The asymptote is x = 0. The domain is R+. The range is R.

ChaPTeR 5 • Exponential and logarithmic functions

245

Chapter review S h oRT anS WeR

1 Simplify the following expression with positive indices. −

1

(16 x 6 y10 ) 2 ÷ 3 (27 x 3 y9 ) 2 Solve the following equations. a 4x3 = 500 3 Find the solutions to: a 9x − 1(3x) − 6 = 0 4 For the function with the rule f ( x) = 3x − 2 + 1: a find the y-intercept b state the equation of the asymptote c sketch the graph of f ( x) d state the domain and range.

b 8x + 1 × 22x = 43x − 1 b 2x + 2

−x + 1

= 3.

5 For the function with the rule f ( x) = 3 × 2x − 3: a find the y-intercept b state the equation of the horizontal asymptote c find a second point on the graph d sketch the graph of f ( x) e state the domain and the range of the function. 6 a Evaluate log3

1 . 27

b Express y in terms of x if log10 ( x) + log10 ( y) = 2 log10 (x + 1). 7 Simplify the following. a 3 log4 (5) − 2 log4 (6)

b

2 log5 ( x 2 ) 1 logg5 ( x ) 3

8 Solve each of the following. a log6 (x) = 3

b 2 logx (125) = 6

c log2 (3x + 6) − log2 (5) = 2

9 If y = loge (7x − 6) + 3, then what does x equal? 10 If f (x) = 3x: a sketch the graph of f (x) and label: i the y-intercept ii the equation of the asymptote b sketch the line y = x, use this line to sketch the inverse function g(x) = log3 (x), and label: i the x-intercept ii the equation of the asymptote. 11 The number of bacteria in a culture, N, is given by the exponential function N = 1500(20.4t ), where t is

the number of days. a Find the initial number of bacteria in the culture. b Find the number of bacteria (to the nearest 100) after: i 5 days ii 10 days. c How many days does it take for the number of bacteria to reach 12 000? mUlTiP l e Ch oiCe

1 When simplified, a

2

x 4 y7 7

5m 4 p2 (5m 2 p6 )3 may be simplified to: ÷ 2m 3 p 3m 7 p m2 3m10 3m 2 a 16 B C 47 p 50 p16 2 p32

3 The value of a

246

(2 xy3 )2 3 x 5 y 2 × is equal to: 7x3 4y 3x 4 y7 3 y7 B C 7 x2

1 20

−1 − 2 64 3 5 125

B 5

Maths Quest 11 Mathematical Methods CAS

D

3x 4 y6

e

x y

D

m15 p29

e

m 20 p24 25

is: 4

C 5

5

D 4

e −5

4 If 252 − x = 125, then x is equal to: 1

a 1

C −1

B 2

5 If 42x − 17(4x) + 16 = 0, then x is equal to: a 1 or 16

B 0 or 1

C 2 or 8

6 The rule for the graph below could be:

D 2

e 5

D 1 or 4

e 0 or 2

y

0

x −1 −2

a y = 3x − 2

B y = 3x + 2

D y = −2x

C y = 3x − 2

e y = 2x + 2

Questions 7 to 9 refer to the function defined by the rule y = 2x + 3 − 1. 7 The graph that best represents this function is: a B y y

C

y

7

−3 D

0

−3

x e

y

0 −1

3

8 The domain is: a (3, ∞)

9 The range is: a [−1, ∞)

−3

x

a 3

5

a 2

e R\{−3}

B R

C R+

D (1, ∞)

e (−1, ∞)

B log5 (x) = 250 e log250 (x) = 5

log3 25 is nearest to: log3 5 B 5

C log5 (250) = x

C 0

D 69

e 1

C 2

D 9

e 20

5 (x 4 )

log7 simplifies to: log7 ( x ) 3

5

B

log7 ( x 2 )

14 The solution to log5 (x) = 4 is: a 25

x

D R

B 7

13 The expression

0

C R+

11 The value of log7 (49) + 3 log2 (8) − 4 is: 12 The value of

B 125

C

log7 ( x 4 )

3

B 7

e

C 1

D 625

e 20

C 5

D 14

e

16 If log3 (2x − 1) + log3 (2) = 2, then x is equal to: a 2

B 1

5 8

D 4

15 The value of x if 2 logx (343) = 6 is: a 3

−1

B [−1, ∞)

a logx (5) = 250 D logx (250) = 5

x

y 1

10 When expressed in log form, 5x = 250 becomes:

a 3

x

0

0

C 3

5

D 2

e

7 11 4

ChaPTeR 5 • Exponential and logarithmic functions

247

17 The solution to the equation 43 − 2x = 12 is nearest to: a x = −1 D x = 0.2

B x = 0.35 e x=0

C x = 0.604

y

18 The equation of the graph shown at right is: a y = log16 (x)

(4, 2)

B y = log2 (x) C y = 2 log8 (x)

0

D y = log10 (x)

1

x

4

e y = 2 log10 (x)

19 The inverse of the graph below would be: y

(2, 1) 1 0 a

x B

y

y

x

0 −1

(1, −2) 1

C

D

y

e

x

1

x (−1, −2)

248

Maths Quest 11 Mathematical Methods CAS

1 0

y

−1

y

(−1, 2)

(2, 1) 0

x

(2, −1)

x

−x

1 a For the function f ( x ) = 5 : i find the y-intercept ii find the values f (1) and f (−1) iii find the equation of the asymptote iv sketch the graph of f (x) v state the domain and range. − b For the function g(x), where g(x) = f (x + 3) − 1 and f (x) = 5 x: i state the transformations to change f (x) to g(x) ii state the equation of the asymptote iii sketch the graph of g(x) iv state the domain and range of g(x).

e x Ten D eD ReS P o n S e

2 The number of lions, L, in a wildlife park is given by L = 20 (100.1t ), where t is the number of years

since counting started. At the same time the number of cheetahs, C, is given by C = 25(100.05t ). a Find the number of: i lions ii cheetahs when counting began. b Find the numbers of each after i 1 year ii 18 months. c Which of the animals is the first to reach a population of 40 and by how long? d After how many months are the populations equal, and what is this population? y 3 The graph of the function f : R → R, where f (x) = −A × 102x + 4, is shown at right. a Give the equation of the horizontal asymptote in the form y = c. c b The graph passes through the point (0, 3.8). Use this information to find the value of A. (0, 3.8) c Find the x-intercept, correct to 3 decimal places. d Find the values of: ii f (4). i f (−3) e Find the value of x if f (x) = −5.2. x f State the domain and range of f (x). 4 The temperature, T °C, of a coffee in a ceramic mug at time t minutes − after it is poured is given by T = 60(4 0.05t ) + 20. a Find the initial temperature of the coffee. b Find the temperature of the coffee, correct to one decimal place, at: i 2 minutes after it is poured ii 25 minutes after it is poured. c Sketch the graph of the equation for 0 ≤ t ≤ 50. If the coffee can be comfortably drunk when it is between temperatures of 30 °C and 45 °C, find: d the time available to drink the coffee e the final temperature the coffee will settle to. 5 The number of bacteria (N) in a culture is given by the exponential function N = 12 000(20.125t), where t is the number of days. a Find the initial number of bacteria in the culture. b Find the number of bacteria in the culture after: i 4 days ii 2 weeks. c Find the time taken for the bacteria to reach 32 000. When the bacteria reach a certain number, they are treated with an anti-bacterial serum. The serum − destroys bacteria according to the exponential function D = N0 × 3 0.789t, where D is the number of bacteria remaining after time t and N0 is the number of bacteria present at the time the serum is added. The culture is considered cured when the number of bacteria drops below 1000. d If the bacteria are treated with the serum when their numbers reach 32 000, find the number of days it takes for the culture to be classed as cured. e How much longer would it take the culture to be cured if the serum is applied after 6 weeks?

DiGiTal DoC doc-9756 Test Yourself Chapter 5

ChaPTeR 5 • Exponential and logarithmic functions

249

ICT activities Chapter opener DiGiTal DoC • 10 Quick Questions doc-9749: Warm up with ten quick questions on exponential and logarithmic functions (page 217)

5B

negative and rational powers

TUToRial • We5 eles-1417: Watch a tutorial on writing exponential expressions with positive indices (page 222) DiGiTal DoC • SkillSHEET 5.1 doc-9750: Practise working with negative and rational powers (page 224)

5C

indicial equations

TUToRial • We10 eles-1418: Watch a tutorial on solving an indicial equation by using substitution (page 226)

5D

Graphs of exponential functions

DiGiTal DoCS • SkillSHEET 5.2 doc-9751: Practise substituting values into exponential functions (page 230) • WorkSHEET 5.1 doc-9752: Write expressions with their simplest index notation, solve indicial equations and sketch graphs of exponentials (page 231)

5F

Solving logarithmic equations

DiGiTal DoC • WorkSHEET 5.2 doc-9753: Simplify logarithmic expressions and solve logarithmic equations (page 236)

250

Maths Quest 11 Mathematical Methods CAS

5G

logarithmic graphs

inTeRaCTiViTY • Logarithmic graphs int-0264: Consolidate your understanding of logarithmic graphs and their features (page 236) DiGiTal DoCS • Investigation doc-9821: Consolidate your understanding of logarithmic graphs and their features (page 236) • Investigation doc-9822: Investigate advanced logarithmic graphs and their features (page 236)

5h applications of exponential and logarithmic functions DiGiTal DoCS • doc-9754: Investigate graphs of exponential functions (page 238) • Extension doc-9755: Practise sketching logarithmic graphs and identifying rules of functions (page 238) TUToRial • We26 eles-1419: Watch a tutorial on calculating the price of gold, where the price is modelled by a logarithmic function (page 239)

Chapter review DiGiTal DoC • Test Yourself doc-9756: Take the end-of-chapter test to test your progress (page 249)

To access eBookPLUS activities, log on to www.jacplus.com.au

Answers CHAPTER 5 exPonenTial anD loGaRiThmiC FUnCTionS exercise 5a

index laws

1 a x10 c x7y8

b

x 3 y9 c 2 4 p10 m8 3a 3

c

a9b5 4

b

9 xy 4 2

125k 11d 6 24 c B b x5y6m − 1

5a A 6 a x3yn + 1z

bE

7 a 211

b 324

c 55 × 34

e

f 2

8a 8

3 24 b 59 049

d 1

e 5

g 16

h 3

e

−2

c 16 1

d x10 3 a 3

1 xy 2

e 213 × 38 b3

d 27

e

1

2 3

p2 m 1 f 4 x 6 y2 c 4 c

f

h 4

4E 5 a

d

7 33

b

1

5 x6

e 64m10

7 24

g (x

125 64

9

g 27

3 + 1) 2

7 D

h

c

exercise 5D

Graphs of exponential

1 a

y

1 a 5 c 5

e −2 2 a 5 d

3 −9 4

e

2 11 6

b

x

y=3

b

y

y = 0.5(4x)

2

0

x

1

y 5

y = 5x

1

y = 0, (0, 0.5) c

y

y = 1–4 (2x)

1 0 y 10

d

1

4– 3

0

1

1

y

x

−1 0 y

y = 4−x

4

0

f

y = 2x − 1

1– 2

4

x

b y = 0, (0, 9)

1 0

−1

x

y = 0, (0, 4) 1 3 a y = 0, (0, 2 ) dom = R, ran = (0, ∞)

2

e

x

y y = 4( 1–3 )x

x

1

1 4)

y = 0, (0,

y

y = 2−x

0

x

y = 10x

1 0

d

1

x

dom = R, ran = (0, ∞) y

y = 3x + 2

y 0

9

x

1

−3

0

x

c y = 0, (0, 5)

dom = R, ran = (0, ∞)

g

y −1 −1 0

c −1 9 8

x

1– 2 1– 4

−1

f

x

1

y = 0, (0, 2)

0.5

0

2 x3

indicial equations b4 d2 f −3 b

2

1

3 ( y − 4) 2

−1

6

3

1 y2

f x2

y = 2(3x)

0

y = −3x exercise 5C

y

8 B

x

1

2 a

b 1.89 d 1.66

negative and rational powers 212 52 b 2 c 316 3 b

0

b 1 or 2 d 1 or 2

functions

c

2 a −22

1 1– 2

−1 11

d

b 23x − 2 d 25n − 6 × 39n − 3

3

1 1 a 3 6

b 7

5

10 E exercise 5B

y = 0.5x

2

f 5

1

9 a 22 × 33n + 4 c 2n − 2 × 76n − 3

5 9 10 −5 7

y

10 E

b

d 222 × 58

− 18

6A 9 a 1.58 c 1.65

d 10e9f 2

15 4

b 7 −5 d

5 a 0 or 1 c 0 or 1

d p11q2

c 18u11v5 4a

4 a

b 518 d 6m9p17

2 a a5b3

c

h

10

3 a −3

−3 y = −3−x

y = 51 − x y x 5 0

x

ChaPTeR 5 • Exponential and logarithmic functions

251

c dom = R, ran = R+

d y = 3, (0, 4)

dom = R, ran = (3, ∞)

y

y y = 2x + 3

4

2 a 42 = 16

f(x) = 3 × 2x

e 5 = 625 −2

g 3

3

e y=

−2)

(0, dom = R, ran = (−3, ∞) y

0

d g j 7 a

x

1

c e g i 8 a

d dom = R, ran = R+

y = 3x − 3

y x

0

(1, 45)

45

f(x) = 5 × 32x

−2 −3

y 5

7

0

y = 2x + 3 − 1 0

e

x

−1

x

1

–x

−

6 log2 (80) log10 (100) = 2 log2 (4) = 2 log2 (3) log3 (20) log10 (250)

10

(2, 10)

4 3 x

y

dom = R, ran = (5, ∞)

f(x) = 2 ×

5

b E

y

6

3

f(x) = 4 × 23x + 3

(1, 4)

y 8

(1, 8)

f(x) = 23x

− 1–3

7 a 3.32 c 2.26 e 1.56,

0

x

x

d log10 (100x3) 6 f log10 ( 10 x )

−3.99

e log b (a) = n

1

g 30

i 5

k 10

j

6 2 B 3.459 1.292 0.661

b 5 f 25 b A

Maths Quest 11 Mathematical Methods CAS

( )

b

c 4 g 2 c E

b e h

−0.737 −

3.080

−0.423

h5

d 10 h1 dC c 0.483 f 2.255 i 2.138

7C

logarithmic graphs b b c 3y

y

1 0

d 2.44, −2.86

f log2 1 = − 4 16

c −3

6 3

b 1.30

13 C

b log4 (1024) = 5

f 27

2 a

d 4

Solving logarithmic equations b 0 c −1 d −2 f 2 g 2 b 9 c 625 d 10

1 a x

logarithms 1 a log 2 (8) = 3 b log 3 (243) = 5 c log 5 (1) = 0 d log 10 (0.01) = −2

1

12 D

b 4 e 4.070

exercise 5G y=3

exercise 5e 1

f 6 11 B

6 B (− 1–3, 5) 7 5 3

9

3

c 2 g 6

e 8

3 a e 4 a 5 a d g

x

y

b dom = R, ran = R+

( x + 3)2 x − 2

b 2

1 a 2 e 6 2 a 8

(3, 4)

0

x

x

h log10

1

4 2

1

g log3 (64)

exercise 5F –x 23

y = 102 − x + 5

f(x) = 22x

f log2 [x3(x − 4)]

15 a 4 d 3.170

2 x

f dom = R, ran = R+

h y = 5, (0, 105)

4 a A 5 a dom = R, ran = R+

b log2 (1728)

e log3 (4)

0

0 1

Undefined log3 (105) log6 (56) f log3 (3) = 1 h log4 (5)

e log2 = −1 2

8

2

f −5 i 5 l 5

−5

1

c log5 25

y=6 +3

0

h k b d

5A c 3

d log10 4

14 a log3 (81) = 4

f(x) = 2 × 52

−x

252

2

−2

10 E

y

0

−

2

y

dom = R, ran = (3, ∞)

0

4 C b 4 e 3

9 a 3 e 3

dom = R, ran = R+

g y = 3, (0, 4)

4 3 2 1

h bx = a

1

dom = R, ran = (−1, ∞)

y

=

c log5 (3)

f y = −1, (0, 7)

15

f 27 = 128

1 9

3D 6 a 4

3

x

0 −3,

d 33 = 27

4

(1, 6)

6

b 106 = 1 000 000

1 − c 2 1= 2

f(x) = log2(x)

(2, 1)

x

2

y

1 0

(5, 1) 5

f(x) = log5(x) x

d 5x

c

e The population will reach a limit at

y

some stage. 8 a 80 kg b i 72.1 kg c 67 weeks

(8, 1) f(x) = log8(x)

1 0

x

8

d i d

f

x

10

(12, 1) f(x) = log12(x) x

12

(15, 1) f(x) = log15(x) 15

rubbish will be disposed of in 10 years or so, which is unlikely. 9 a i 108 hectares ii 120 hectares iii 148 hectares b 15 h 10 a i 51.18 m ii 51.65 m iii 51.95 m iv 52.18 m b

d

0 2– 3

c 11 a b c 12 a b c

26 weeks A = P(1.05)n $25 000 90 °C i 76.3 °C 12 min 37 s 120 i 145

20

n

1 2 3 4

10

h 5 4 3 2 1

0

b $16 288.95 c 20 years b $14 427 c 10 years

f

n

1 2 3 4 5

2

N N = 120(1.1t) 120

t

x 0

1

2

3

4

log10(V) 4.5 4.4 4.3

0

g h i

1

2

3

4 t(years)

b The graph shows a slight upward curve

not starting from (0, 0); this suggests possible exponential growth. c 1.071 d 1.072, 1.082, 1.063. Estimated % annual growth is 7%. e

x(year)

4.6

700

iii 213

4

4.2

0

ii 176

3

V 4.653 4.550 4.447 4.352 4.255

800

ii 64.7 °C

2

down with a reducing slope — exponential decay. c 0.789, 0.789, 0.804, 0.800 d The ratio is about 0.795 or a 20.5% rate of depreciation. e

h = 5(0.7n), n ≥ 0, n ∈ J

1

b The curve starts from $45 000, curving

900

c 11 1 years

0

30

13 a V = 5000(1.1t) b $8857.81 c 8 years 14 a P(people)

t

0

⇒ P = 100.03t × 102.845 ⇒ P = 700 × 1.072t, i.e. 7.2% growth similar to d j P5 = 982 people; 2002 (t = 6), P6 = 1051 k During 2011 l During 2004

40

67 400 i 1005 ii 6340 iii 16 643 iv 17 146 17.48 months h = 5(0.7n) i 1.20 m ii 0.29 m

0 W = 3 log10(8t + 10)

t

4

g m ≈ 0.03; y-intercept = 2.845 h log10 (P) = 0.03 t + 2.845 i P = 10(0.03t + 2.845)

d = 50 + log10(15n)

W 3

3

15 a V($000)

52 51

3 The smaller a is, the steeper the graph. 4 a log3 (x) for x > 0 1 b 2 log2 (x) for x > 0 exercise 5h applications of exponential and logarithmic functions 1 a P = 10 000(2t) b i 80 000 ii 640 000 c 3.32 months 2 a 15 000 b i 15 528 ii 16 077 c 19 118 d 2022 3 a 3 kg b i 3.77 kg ii 5.1 kg iii 5.86 kg

2

50

x

c 5x

1

The points appear collinear.

ii No, the model suggests virtually no

1 0

−5– 4

0

t

0

y

c

W = 80(2−0.015t)

(10, 1) f(x) = log10(x)

1 0

d

2.9 2.8

y

e

7

3.0

y

1 0

4 5 6

3.1

ii 52.8 kg

W 80

d a a a b c a b

f log10(P)

t

0

P

2.845 2.875 2.905 2.940 2.966

1

2

3

4

j k l

1

2

3

4

x

The points are roughly collinear. m ≈ −0.099; y-intercept = 4.65 log10 (V ) = −0.099t + 4.65 log10 (V ) = −0.099t log10(10) + 4.65 log10(10) − = log10(10 0.099t) + log10(104.65) − ⇒ V = 10 0.099t × 104.65 − 4.65 = 10 × (10 0.099)t = 45 000(0.79)t The rate of depreciation is about 21%. In 2010 (x = 5), V = $13 847; in 2011, V = $10 939. During 2012

ChaPTeR 5 • Exponential and logarithmic functions

253

16 a

7 10 13 16 19

y y = 3x − 2+ 1

c

T 5000

—) (0,10 9

4000

(2, 2)

1

3000

x

0

1000 (0, 0)

1

2

3

4

5

R

The graph starts from (0, 0) and curves slightly up to the right like a positive exponential function. log10 R −

0.410

1.943

−0.140

2.352 2.563

0.183

2.837

0.716

3.637

log10 (T ) 4 3

d

y

i

ChaPTeR ReVieW ShoRT anSWeR

4 y2 3x 4 2 a x=5 3 a x=1

b x=5 b x = 0 or x = 1

4 a (0, 9 )

b y=1

1

10

254

1 −1 0

1

e The domain is R and the range is (−3, ∞). 6 a −3

b y=

(x + x

b 12

8 a x = 216

b x=5

14

y = 3x (1, 3)

1

y

x

y = 3x (1, 3)

3

y = log3(x)

(1, 1)

1

3

80 70 60 50 40 30 20 10

x

x=0

11 a 1500 b i 6000 ii 24 000 c 7.5 days

0

mUlTiPle ChoiCe

Maths Quest 11 Mathematical Methods CAS

2 C 5 E

3 A 6 C

−3

−1

1

x

iv Domain R, range ( −1, ∞) 2 a i 20 ii 25 b i L = 25, C = 28 ii L = 28, C = 30 c Lions by 1 year 1 month d 31 after 1 year 11 months 3 a y=4 b A = 0.2 c x = 0.651 d i 3.999 998 (≈ 4) ii −19 999 996 (≈ −2 × 107) e 0.831 f Domain R, range (−∞, 4) 4 a 80 °C b i 72.2 °C ii 30.6 °C c T

(3, 1) 01

y

c x= 3

0

1 B 4 B

left, vertical translation of 1 unit down

ii y = −1 iii y = 5−(x + 3) − 1

y = −1

3

y=x

v Domain R, range R+ b i Horizontal translation of 3 units to the

1)2

ey − 3 + 6 7 y

x

(−4, 4)

7 a log4 125 36

b

iii y = 0

x y = −3

y=0

1

E C B B

(−1, 5)

(1, 3)

0

−·4 −·3 −·2 −·1 0 ·1 ·2 ·3 ·4 ·5 ·6 ·7 ·8 log10 (R)

The points are very close to collinear. m ≈ 1.50 The y-intercept is 2.5614 or log10 (364.25) log10 (T ) = 1.50 log10 (R) + 2.5614 log10 (T ) = log10 (364.25 × R1.50) ⇒ T = 364.3 R1.50 T = 10 711.76 days. The difference may be due to rounding errors or ignoring the small effects of other planets and moons on Saturn’s orbit. k = 133 407.5625

9 12 15 18

y y = 5−x

y=0

10 a

1

h

D B D C

ii 5 , 5

f(x) = 3 × 2x − 3

3

9 x=

2

d e f g

1 a i 1 iv

d Domain = R, range = (1, ∞) 5 a y=0 b y = −3 c (1, 3) (for example)

log10 T

0

c

8 11 14 17

exTenDeD ReSPonSe

2000

b

A C A E C

d 5 a b c

T(x) = 60 (4−0.05t) + 20

(25, 30.6)

10 20 30 40 50 x

13.2 minutes e 20 °C 12 000 i 16 970 ii 40 363 11.3 days d 4 days e 4 more days

Chapter 6

Circular functions DiGital DoC doc-9757 10 Quick Questions

Chapter ContentS 6a 6B 6C 6D 6e 6F 6G 6h 6i

Trigonometric ratio revision The unit circle Radians Symmetry Identities Sine and cosine graphs Tangent graphs Solving trigonometric equations Applications

6a

trigonometric ratio revision

Recall that for a right-angled triangle: O = opposite A = adjacent H = hypotenuse. These ratios can be memorised using the term ‘SOH CAH TOA’.

sin (θ ) = H O θ

A

O H

cos (θ ) =

A H

tan (θ ) =

O A

WorkeD exaMple 1

Find the value of x in each of the following triangles. Express lengths correct to 2 decimal places and angles to 1 decimal place. a 9

b

64°

4.1

x x 27° c x

7.5

8.6

Chapter 6 • Circular functions

255

Write

think

a 1 Write the cosine ratio. 2

Replace A with x, H with 9 and θ with 64°.

3

Make x the subject of the equation.

4

Calculate x, correct to 2 decimal places.

cos (64 ) =

x 9

= 3.95 O b Use sin (θ ) = H

2

Replace O with 4.1, H with x and θ with 27°.

3

Make x the subject of the equation.

4

Calculate x, correct to 2 decimal places.

4.1 sin (27 ) = x x=

4.1 sin (27 )

= 9.03 c Use tan (θ ) =

c 1 Write the tangent ratio.

DiGital DoCS doc-9758 SkillSHEET 6.1 trigonometry review i

A H

x = 9 cos (64 °)

b 1 Write the sine ratio.

2

Replace O with 8.6, A with 7.5 and θ with x°.

3

Evaluate the right-hand side of the equation, keeping plenty of decimal places at this stage.

4

Make x the subject of the equation using inverse tan notation.

5

Calculate x, correct to 1 decimal place, using the inverse tan function of the calculator.

exercise 6a

a Use cos (θ ) =

tan ( x ) =

O A 8.6 7.5

= 1.146 67 x = tan 1 (1.14667) −

= 48.9°

trigonometric ratio revision

1 We1a Find the value of x in each of the following, correct to 2 decimal places. a b c 10 x 18°

doc-9759 SkillSHEET 6.2 trigonometry review ii

x

12 x

81°

2 We1b Find the value of y in each of the following. a b 8 54° y

36°

c

10.6 32° y

y 72°

6.8

2.4

3 We1c Find the value of a in each of the following. Give answers correct to the nearest tenth of a degree. b c a a 8 2.5 9 6.4 10.2 a a 3.7

256

Maths Quest 11 Mathematical Methods CAS

4 MC In the figure below: a

x

10 cm a x is nearest to: A 60 cm

21.98 cm 20°

y

B 22 cm

C 7.5 cm

D 8 cm

E 9 cm

B 64°

C 37°

D 26°

E 39°

B 6 cm

C 5 cm

D 10 cm

E 7 cm

b angle a is nearest to: A 53°

c y is nearest to: A 8 cm

5 MC If sin (a°) = 0.951, then tan a is equal to: A 0.000 29 D 1.32

B 0.0166 E 0.309

C 3.076

6 MC If 0° < a < 90° and cos a < 0.5, then which of the following is correct? A a < 30° D a < 45°

B a > 30° E a > 60°

C a < 60°

7 A tree 5 metres tall casts a shadow so that the angle of elevation from the end of the shadow to the top

of the tree is 35°. How long is the shadow?

5m 35° 8 A mathematically able tree removalist measures the angle of elevation of two points on a diseased

section of a large eucalypt at a distance of 40 m from the base of the tree. If the angles are 20° and 24° respectively, how tall is the diseased section of the tree? Diseased section

20° 24° 40 m 9 A 60-metre flying fox cable is set up to cross a river so that it will drop 10 metres vertically. What is the

angle of depression of the cable? 10 A soccer player has a shot for goal from the position shown in the figure below right. Find the ‘scoring

angle’, θ.

Goal 5m 7.3 m

12 m

θ

Chapter 6 • Circular functions 257

6B

The unit circle

So far in this chapter we have considered angles in a right-angled triangle. But trigonometric functions can be defined over a larger domain by considering their definition inside a unit circle. This is a circle with centre (0, 0) and a radius of 1 unit. A point P (x, y) is a point on the circle. Its location can also be defined as P (θ), where θ is the anticlockwise rotation from the positive x-axis. A negative θ value denotes a clockwise rotation. Using the SOH, CAH definitions in the diagram: x cosine (θ) = = x 1 y sine (θ) = = y 1 These may be abbreviated: x = cos (θ) y = sin (θ)

y 1 P(θ ) 1

y

θ 0 x

−1

1x

−1

The point P can also be denoted as (cos (θ), sin (θ)). Note: As θ increases, the x- and y-coordinates oscillate between −1 and +1, so −1 ≤ sin (θ) ≤ +1 and −1 ≤ cos (θ) ≤ +1 for all θ. The ‘tangent line’ on a unit circle is the line that is a tangent to the circle at the point (1, 0). Tangent (θ), or tan (θ) for short, is the height at which a line along an angle hits the tangent line. The diagrams below show tan θ for angles in the first and second quadrants. y

y

tan (θ ) θ

θ

x

x tan (θ)

The tangent line

sin (θ ) y . Note: Tangent (θ), or tan (θ), may also be defined as the ratio x . That is, tan (θ ) = cos (θ ) This formula and its derivation will be discussed in more detail later in this chapter.

Quadrants The coordinate axes divide the unit circle into four quadrants as shown in the diagram below left. The angle measurements, in degrees, between the quadrants are shown in the diagram below right. y

y 90°

2nd quadrant

1st quadrant θ°

0

3rd quadrant

x

180°

0

4th quadrant 270°

258 Maths Quest 11 Mathematical Methods CAS

x 0° or 360°

Therefore:

quadrant 1 is where 0° < θ < 90° quadrant 2 is where 90° < θ < 180° quadrant 3 is where 180° < θ < 270° quadrant 4 is where 270° < θ < 360°. y

Since y = sin (θ),

sine is positive in quadrants 1 and 2 and sine is negative in quadrants 3 and 4. Since x = cos (θ), cosine is positive in quadrants 1 and 4 and cosine is negative in quadrants 2 and 3. y Since tan (θ ) = , tangent is positive in quadrants 1 and 3 and x tangent is negative in quadrants 2 and 4. The diagram at right summarises which of the trigonometric functions is positive in each quadrant. The word ‘CAST’ will assist in recalling this. You could also use a short phrase, such as ‘All Stations To Croydon’.

Sine positive

All positive

Tangent positive

Cosine positive

x

WorkeD exaMple 2

If a is any angle, find the quadrant where: a sin (a°) > 0 and tan (a°) < 0 b sin (a°) < 0 and cos (a°) > 0. think

Write

a Sine is positive and tangent is negative in quadrant 2.

a Quadrant 2

b Sine is negative and cosine is positive in quadrant 4.

b Quadrant 4

exact values Calculated trigonometric values (sin, cos and tan) of most angles are rational approximations correct to several decimal places. However, for a few particular angles, exact trigonometric values can be determined. These include multiples of 90° and the ratios of isosceles and equilateral triangles. Exact values can be determined for 0° and any multiple of 90° using this diagram and the fact that y = sin (θ), x = cos (θ) and y tan (θ) . x The isosceles triangle with equal sides of one unit each gives the exact trigonometric values for 45°. The hypotenuse is calculated using Pythagoras’ theorem: hypotenuse =

12

0° or 360° x (1, 0)

0

(0, −1) 270°

45° 2 1 45°

SOH sin (45) =

180° (−1, 0)

(0, 1)

+ 12

= 2 So, using

y 90°

1 2

=

1 2 × 2 2

=

2 2

1

CAH

TOA

1 cos (45) = 2 =

2 2

tan (45) =

1 1

=1

Chapter 6 • Circular functions

259

The equilateral triangle with each side 2 units long, shown in the figure below, has been bisected.

Line of bisection

30° 2

60° 1

Using Pythagoras’ theorem: line of bisection = 22 − 12

= 3

So, using

SOH

CAH

TOA

1 3 1 3 × = 3 3 sin (30 ) =

1 2

3 2

cos (30) =

tan (30) =

3 3 3 tan (60 ) = 1

= sin (60 ) =

3 2

cos (60 ) =

1 2

= 3 These exact values are summarised in the table below. Angle (θ °) 0°

sin (θ) 0

cos (θ) 1

tan (θ) 0

30°

1 2

3 2

3 3

45°

2 2

2 2

1

60°

3 2

1 2

3

90°

1

0

Undefined

Worked Example 3

Without using a calculator, find: a sin (90 °) b cos (180 °) c tan (270 °). Think

Write

a 1 Sine corresponds to the y-coordinate on the unit circle. 2

The value of y at the point where θ = 90° is 1.

b 1 Cosine corresponds to the x-coordinate on the unit circle. 2

a

sin (90 °) = 1 b

The value of x at the point where θ = 180° is −1.

c 1 Tangent corresponds to the value of

y x on the unit circle.

2

The value of y is −1 and the value of x is 0 at the point where θ = 270°.

3

The value of

−1 0

is not defined.

260 Maths Quest 11 Mathematical Methods CAS

cos (180 °) = −1 c

tan (270 ) =

−1

0

tan (270 °) is undefined.

WorkeD exaMple 4

Without using a calculator, state the exact value of: a sin (45 °) b cos (60 °) c tan (30 °). think

Write

a Read the value from the table, or

determine using the special triangle, O sin (θ ) = . H

2

45° 1

45° 1

a sin (45) =

1 2

=

1 2

=

2 2

b cos (60) =

b Read the value from the table, or

determine using the special triangle, A cos (θ ) = . H

2

×

2 2

×

3

1 2

30° 3

60° 1 c Read the value from the table, or

determine using the special triangle, O tan (θ ) = . A

2

30°

1

2

3

1

=

1

=

3 3

3

60° 1

exercise 6B

c tan (30 ) =

3 3

3

the unit circle

a sin (40°) f sin (260°)

State whether the following values are positive (P ) or negative (N ). b sin (65°) c sin (110°) d sin (160°) g sin (215°) h sin (321°) i sin (425°) Compare your answers with those found using a calculator.

e sin (230°) j sin (−36°)

a cos (27°) f cos (295°)

State whether the following values are positive (P) or negative (N). b cos (68°) c cos (115°) d cos (200°) g cos (402°) h cos (−83°) i cos (−240°) Compare your answers with those found using a calculator.

e cos (250°) j cos (157°)

State whether the following values are positive (P ) or negative (N ). b tan (75°) c tan (118°) d tan (166°) g tan (308°) h tan (500°) i tan (−45°) Compare your answers with those found using a calculator.

e tan (199°) j tan (−137°)

a tan (12°) f tan (255°)

DiGital DoC doc-9760 the unit circle

4 We2 If a is any angle, find the quadrant where: a sin (a) < 0 and tan (a) > 0 b sin (a) > 0 and cos (a) > 0. 5 MC

a If 0° < a < 360°, sin (a) < 0 and cos (a) < 0, then which one of the following is true?

a 0° < a < 90° D 270° < a < 360°

B 90° < a < 180° e 0° < a < 180°

C 180° < a < 270°

a 0° < a < 90° D cos (a) > 0

B cos (a) < 0 e 270° < a < 360°

C 180° < a < 270°

a one solution D three solutions

B no solution e four solutions

C two solutions

a no solutions D one solution

B two solutions e four solutions

C three solutions

b If 0° < a < 360°, sin (a) > 0 and tan (a) < 0, then which one of the following is true?

c Given that 0° < a < 360°, then the equation cos (a) = −1 has: d If 0° < a < 360°, then the equation sin (a) = cos (a) has:

Chapter 6 • Circular functions

261

e If 0° ≤ a ≤ 180°, and sin (a) = cos (a), then a is equal to: a 60° D 0°

B 150° e 90°

6 We3 Find: a cos (180°) f cos (360°) 7 We4 DiGital DoC doc-9761 WorkSHEET 6.1

b sin (270°) g tan (270°)

c tan (360°) h tan (180°)

d sin (180°) i sin (630°)

e cos (270°) j cos (720°)

c tan (30°) h cos (0°)

d cos (60°) i sin (90°)

e sin (30°) j tan (90°)

State the exact value of:

a sin (60°) f tan (45°)

6C

C 45°

b cos (45°) g sin (45°)

radians

y

1

You are used to measuring angles in degrees (°), and will recall that there are 360° in a circle. An alternative unit for angle measurement is the radian (c). This is given on the calculator as (r). Radians are derived from the circumference of the unit circle, that is, a circle of radius 1 unit. In a unit circle, the circumference = 2πr = 2π(1) = 2π units. So, the angle swept in one revolution is equivalent to 2π c. Therefore 2π c = 360°, so: πc 180 1 = π c = 180° 1c = π 180

ius rad 1 radian

An arc length of one radian is 1c.

These formulas may be used to convert degrees to radians and vice versa. WorkeD exaMple 5

Convert the following angles into exact radians. a 120° b 270° using a CAS calculator think

a 1 Multiply 120 by

2

Write/DiSplaY

πc . 180

a 120 = 120 ×

=

2π c 3

=

3π 2

Simplify by dividing through by the highest common factor, 60.

b 1 Ensure your CAS calculator is in radians mode.

πc 180

b 270°

Enter the value of the angle in degrees. 2

Press ENTER to convert to radians.

3

Write the answer.

270 =

3π c 2

WorkeD exaMple 6

Convert the following angles to degrees. 3π c a using a CAS calculator 4 b 0.8π c think

a 1 Ensure your CAS calculator is in degrees mode.

Enter the value of the angle in radians. 262

Maths Quest 11 Mathematical Methods CAS

Write/DiSplaY

3π 4

a

r

x

2

Press ENTER to convert to degrees.

3

Write the answer.

b 1 Multiply 0.8π c by

= 135° 3π c = 135 4

180 . πc

b 0.8π c = 0.8π ×

180 π

2

Cancel out π c.

= 0.8 × 180°

3

Simplify.

= 144°

When working in degrees, all angles should be expressed to the nearest tenth of a degree unless otherwise stated.

Worked Example 7

Use a calculator to convert: a 57.2° to radians, correct to 3 decimal places b 2.75c to the nearest tenth of a degree. Think

Write

a 1 Multiply 57.2 by 2

a 57.2 = 57.2 ×

Evaluate and round off to 3 decimal places.

b 1 Multiply 2.75 by 2

πc . 180

180 . π

πc 180

= 0.998c b 2.75 = 2.75 ×

180 π

= 157.6°

Evaluate to 1 decimal place.

Exact values and radians Since π c = 180°,

πc πc πc πc = 90 = 30 = 45 and = 60 2 6 4 3 So the table of exact values can be written in terms of radian measurements: Angle (θ c)

sin (θ)

cos (θ)

tan (θ)

0

0

1

0

π 6

1 2

3 2

π 4

1 2

=

2 2

1 2

=

π 3

3 2

1 2

π 2

1

0

1 3

2 2

=

3 3

1 3 Undefined

Chapter 6 • Circular functions 263

Other important angles are shown on the circle below.

c 4–

π

45 °

5 —π c 6 15

35 °

0° 12

π

3 — c 4 1

60° –π c 3

c 2π— 3

90° –2

πc

y

c π– 6

0°

° 30

0° 0c

π c 180°

x 360° 2π c 330 1 1π c ° —

0° 21

6

7 —π c 4

c 5π— 3

πc 3— 270° 2

4π c — 3 24

0°

25 °

31 5°

0° 30

5 —π c 4 2

c 7π— 6

( π–6 sectors shaded) π – 6

Our special triangles can be updated to include radians as shown.

30° 2 2

Exercise 6C

π – 4

Radians

45°

1 π– 3

1

1 WE5 Convert the following angles into exact radians. a 30° b 45° d 20° e 50° g 270° h 360° j −225° k −420° 2 WE6 Convert the following angles to degrees. a

2π 3 −π f 6

π 5

b

e 3.6π

3

c f i l

2π 9 −π g 4 c

60°

1

60° 90° 150° 300° d

11π 6

h 0.375π

3 MC The smallest angle measurement listed below is: A 85°

B π

C 1.92c

D

2π 3

E

D

5π 4

E 3.84c

D

π 6

E 135°

4 MC The largest angle measurement listed below is: A −3π

4π 3

B

C 250°

5 MC If sin (a) = cos (a), then a could be equal to: A

π 3

B

π 4

C 50°

2π 5

6 WE7a Use a calculator to convert the following angles to radians, correct to 3 decimal places. a 49° b 78.2° c −125° d 191.4° e 234.6° f 327.5° g 170.25° h 216.8° 264 Maths Quest 11 Mathematical Methods CAS

7 We7b Use a calculator to convert the following radian measurements to degrees, to the nearest tenth

of a degree. a 0.35 e 15.6 8

b 1.47 f −2.71

c 3.142 g 4.06

Find the exact value of each of the following. π π a sin b cos 4 3 e tan

6D

π 4

f

π cos 6

d 0.958 h 8

c tan

π 3

d sin

π 6

g sin

π 2

h cos

π 4

Symmetry

The unit circle can be divided into symmetrical sections, as shown in the diagram below. Relationships between the circular functions — sine, cosine and tangent — can be established, based on these symmetrical properties. For simplicity, assume θ is an acute angle, although the following properties hold for any θ.

interaCtiVitY int-0265 Symmetry

c π – 2

P(π − θ )

1 y

P(θ ) = (cos (θ ), sin (θ )) = (x, y)

π c −x −1

θ

P( π + θ )

−y

0 or 2π c x 1 P(2π −θ )

−1 c 3— π 2

Quadrant 2 By symmetry: sin (π − θ ) = y = sin (θ ) cos (π − θ ) = − x = − cos (θ ) y tan (π − θ ) = − = − tan (θ ) x Quadrant 3 By symmetry: sin (π + θ ) = − y = − sin (θ ) cos (π + θ ) = − x = − cos (θ ) tan (π + θ ) =

− −

y = tan (θ ) x

Quadrant 1 As already seen: sin (θ ) = y cos (θ ) = x y tan (θ ) = x Quadrant 4 By symmetry: sin (2π − θ ) = − y = − sin (θ ) cos (2π − θ ) = x = cos (θ ) −y = − tan (θ ) tan (2π − θ ) = x

Note: An angle measurement is assumed to be in radians unless the degree symbolis given. Provided an angle is expressed as π ± θ or 2π ± θ, the trigonometric function remains the same, only the sign (+ or −) may change. Use this diagram to determine the sign.

S

A

T

C

Chapter 6 • Circular functions

265

WorkeD exaMple 8

a b c d

If sin (θ) = 0.93, find sin (180° + θ). If cos (θ) = 0.44, find cos (360° − θ). If tan (θ) = 1.72, find tan (π − θ). If cos (θ) = 0.83, find cos (π + θ).

tUtorial eles-1420 Worked example 8

think

a 1 Sketch the angle on a unit circle and relate it

Write/DraW

y

a

to the first quadrant. 180° sin (180° + θ ) + θ) (180°

2

By symmetry sin (180 ° + θ ) = −sin (θ ).

3

Replace sin (θ ) with 0.93.

b 1 Sketch the angle on a unit circle and relate

S θ

A θ

T

C

sin (θ ) x

sin (180 ° + θ) = −sin (θ) = −0.93 y

b

it to the first quadrant. S

A θ θ C

T

2

By symmetry cos (360 ° − θ ) = cos (θ ).

3

Replace cos (θ ) with 0.44.

c 1 Sketch the angle on a unit circle and relate

cos (θ ) x cos (360° − θ ) (36 0° −θ )

cos (360 ° − θ) = cos (θ) = 0.44 y

c

it to the first quadrant. π− π

θ

S

A

T

2

By symmetry tan (π − θ ) = −tan (θ ).

3

Replace tan (θ ) with 1.72.

d 1 Sketch the angle on a unit circle and

tan (θ )

θ

θ

x tan (π − θ )

C

tan (π − θ) = −tan (θ) = −1.72 y

d

relate it to the first quadrant. cos (π + θ ) π +θ

266

2

By symmetry cos (π + θ ) = −cos (θ ).

3

Replace cos (θ ) with 0.83.

Maths Quest 11 Mathematical Methods CAS

S

A θ

θ

T

cos (θ ) C

cos (π + θ) = −cos (θ) = −0.83

x

WorkeD exaMple 9

Find the exact value of each of the following. a tan (150 °) b sin (330 °)

tUtorial eles-1421 Worked example 9

7π d tan 6

3π 4

c cos

think

Write/DraW

a 1 Express tan (150 °) as tan (180 − 30 °). 2

a tan (150 °) = tan (180 − 30 °)

Sketch the angle on a unit circle and relate it to the first quadrant.

y

(18

0° −

30

°)

T

3

By symmetry tan (180° − 30 °) = −tan (30 °).

4

Replace tan (30 °) with its exact value,

1

or

3

3 . 3

=

− 3 3

A 30°

T

3

By symmetry sin (360° − 30 °) = −sin (30 °).

4

Replace sin (30 °) with its exact value, 2 .

C

sin (30°) x sin (360° − 30°)

(36

0°

−3

0°)

sin (330 °) = −sin (30 °)

1

=

π 3π c 1 Express cos . as cos π − 2

or

3

y

S

− 1

b sin (330 °) = sin (360 ° − 30 °)

Sketch the angle on a unit circle and relate it to the first quadrant.

4

C

tan (150 °) = −tan (30 °)

b 1 Express sin (330 °) as sin (360° − 30 °). 2

tan (30°) = tan (180° − 30°) x

A 30°

S

−1 2

π 3π = cos π − 4 4

c cos

4

y

π

Sketch the angle on a unit circle and relate it to the first quadrant.

π– 4

−

S

A π – 4

cos ( π–4 ) C

T

x

cos (π − π–4 )

3

π π By symmetry cos π − = − cos . 4 4

4

Replace cos π with its exact value of 4

3π π cos = − cos 4 4 1 2

or

2 . 2

=

− 1 2

or

− 2 2

Chapter 6 • Circular functions

267

7π as tan π + π . d 1 Express tan 6

2

7π = tan π + π 6 6

d tan

6

Sketch the angle on a unit circle and relate it to the first quadrant.

y

π+

3

π π By symmetry tan π + = tan . 6 6

4

π Replace tan with its exact value 6

π– 6

S

A

T

C

π– 6

tan (π–6 ) = tan (π + π–6 ) x

7π π tan = tan 6 6 1 3

=

3 . 3

or

1 3

3 3

or

WorkeD exaMple 10

3π If sin = 0.924 , evaluate each of the following. 8 11π . 8

5π 8

a sin

b sin

think

Write

3π 8π − 3π 5π a 1 Express sin = sin π − . as sin 8

8

8

8

2

3π 3π Using symmetry express sin π − as sin . 8 8

3π = sin 8

3

3π Replace sin with 0.924. 8

= 0.924

3π 11π as sin 8π + 3π = b 1 Express sin sin π + . 8

8

8

268

3

8

11π = sin 8

b sin

π + 3π 8

3π 3π By symmetry express sin π + as − sin . 8 8

3π = − sin 8

4

3π Replace sin with 0.924. 8

= −0.924

Symmetry

1 We 8 If sin (θ °) = 0.63, find: a sin (180 ° − θ ) b sin (180 ° + θ ) 2

2

exercise 6D

DiGital DoC doc-9760 the unit circle

3π 5π = sin π − 8 8

a sin

If cos (θ) = 0.25, find:

a cos (π − θ )

If tan (θ) = 2.1, find:

a tan (2π − θ )

Maths Quest 11 Mathematical Methods CAS

c sin (−θ )

d sin (360 ° − θ ).

b cos (π + θ )

c cos (2π − θ )

d cos (−θ ).

b tan (−θ)

c tan (π + θ )

d tan (π − θ ).

4

Given that sin (a) = 0.3, cos (b) = 0.7 and tan (c) = 0.9, write down the value of each of the following. a sin (180 ° + a) b cos (180 ° − b) c tan (360 ° − c) d sin (−a °) e sin (180 ° − a) f cos (−b °) g cos (360 ° − b) h tan (180 ° − c) i tan (180 ° + c)

5 We9a, b Find the exact value of each of the following. a sin (150 °) b cos (135 °) c tan (240 °) e sin (240 °) f cos (210 °) g tan (120 °) i cos (−60 °) j sin (−135 °) k tan (180 °) 6 We9c, d

5π a cos 4

Find the exact value of each of the following. 5π 7π b sin c tan 4 6

2π 3

d sin

7π 3

f

−π tan 6

g sin

h cos

11π tan 6

j

− 5π sin 4

k cos (π)

l

e cos i

d tan (330 °) h sin (300 °) l sin (270 °)

4π 3

7π 6

−π sin 2

π π π If sin = 0.383, cos = 0.924 and tan = 0.414, evaluate each of the following. 8 8 8 9π 17π 7π a sin b cos c tan 8 8 8

7 We 10

d cos

−π 8

e sin

7π 8

f

15π tan 8

Given that sin (75°) = 0.966, cos (75°) = 0.259 and tan (75°) = 3.732, find the value of each of the following. a sin (105 °) b cos (255 °) c tan (285 °) d sin (255 °) e cos (435 °) f tan (−75 °) 9 If sin (0.7) = 0.644, cos (0.7) = 0.765 and tan (0.7) = 0.842, find the value of each of the following. (Hint: π = 3.142, approximately.) a sin (2.442) b cos (3.842) c tan (5.584) d sin (−0.7) 8

6e

identities

y

An identity is a relationship that holds true for all legitimate values of a variable or variables. For example, a simple identity is x + x = 2x.

P(θ ) 1

the pythagorean identity Consider the right-angled triangle in the unit circle shown. Applying Pythagoras’ theorem to this triangle gives the identity: sin2 (θ ) + cos2 (θ ) = 1

sin (θ)

θ

0

cos (θ )

x

D

the tangent Consider the unit circle on the right. A tangent is drawn at A and extended to the point C, so that OC is an extension of OP. This tangent is called tangent (θ ), which is abbreviated to tan (θ ). Triangles ODP and OAC are similar, because they have their three corresponding angles equal. tan (θ ) sin (θ ) It follows that: = (corresponding sides) 1 cos (θ ) or tan (θ ) =

sin (θ ) cos (θ )

y 1

B P(θ ) C 1

0

sin (θ )

θ

cos (θ )

D 1A

tan (θ ) x

(as mentioned in an earlier section).

Chapter 6 • Circular functions

269

Another relationship between sine and cosine — complementary functions Consider the unit circle shown on the right. The triangles OAB and ODC are congruent because they have all corresponding angles equal and their hypotenuses equal (radius = 1). Therefore all corresponding sides are equal and it follows that: sin (90 ° − θ ) = cos (θ °) = x and cos (90 ° − θ ) = sin (θ °) = y or −1 π sin − θ = cos (θ ) 2

y 1 D x

y

θ

0

π and cos − θ = sin (θ ) 2

C(90° − θ ) 1

θ

1

x

B(θ ) y A1

x

−1

We say that sine and cosine are complementary functions. Though not required for this course, you may like to try to find the complementary function for tangent, that is, tan (90 ° − θ ) = ? Worked Example 11

If sin (θ) = 0.4 and 0° < θ < 90°, find, correct to 3 decimal places: a cos (θ) b tan (θ). Think

Write

a 1 Use the identity sin2 (θ ) + cos2 (θ ) = 1. 2

Substitute 0.4 for sin (θ ).

3

Solve the equation for cos (θ) correct to 3 decimal places.

4

Retain the positive answer only as cosine is positive in the first quadrant.

b 1 Use the identity tan (θ ) =

a sin2 (θ ) + cos2 (θ ) = 1

(0.4)2 + cos2 (θ ) = 1

sin (θ ) . cos (θ )

cos2 (θ ) = 1 − 0.16 = 0.84 cos (θ ) = ± 0.84 = 0.917 or −0.917 For 0° < θ < 90°, cos is positive so cos (θ ) = 0.917. b tan (θ ) =

2

Substitute 0.4 for sin (θ ) and 0.917 for cos (θ ).

3

Calculate the solution correct to 3 decimal places.

sin (θ ) cos (θ )

0.4 0.917 = 0.436 =

Worked Example 12

Find all possible values of sin (θ) if cos (θ) = 0.75. Think 1

Use the identity sin2 (θ ) + cos2 (θ ) = 1.

2

Substitute 0.75 for cos (θ ).

3

Solve the equation for sin (θ ) correct to 3 decimal places.

4

Retain both the positive and negative solutions, since the angle could be in either the first or fourth quadrants.

270 Maths Quest 11 Mathematical Methods CAS

Write

sin2 (θ ) + cos2 (θ ) = 1 sin2 (θ ) + (0.75)2 = 1 sin2 (θ ) = 1 − 0.5625 = 0.4375 sin (θ ) = ± 0.4375 = 0.661 or −0.661

WorkeD exaMple 13

Find a° if 0° < a° < 90° and

a sin (a°) = cos (42°)

think

Write

a sin (a °) = cos (42 °)

a 1 Write the equation. 2

Replace cos (42 °) with sin (90 ° − 42 °) (complementary functions).

sin (a °) = sin (90 ° − 42 °) sin (a °) = sin (48 °) a° = 48° b cos (a °) = sin (73 °)

b 1 Write the equation. 2

b cos (a°) = sin (73°).

Replace sin (73 °) with cos (90 ° − 73 °).

cos (a °) = cos (90 ° − 73 °) cos (a °) = cos (17 °) a° = 17°

WorkeD exaMple 14 2

If 0° < a° < 90° and cos (a°) = 3 , find the exact values of: a sin (a°) b tan (a°) c cos (90 − a°) d sin (180 + a)°. think

tUtorial eles-1422 Worked example 14

Write/DraW

1

Draw a right-angled triangle.

2

Mark in angle a°, its adjacent side (A = 2) and the hypotenuse (H = 3).

a° H=3

A=2

O= 5 3

Use Pythagoras’ theorem to calculate the opposite side (O) to a°.

a 1 Use the right-angled triangle to find

2

Substitute O = 5 and H = 3.

b 1 Use the right-angled triangle to find

2

O . H

O . A

Substitute O = 5 and A = 2.

c 1 Use the identity cos (90 ° − a °) = sin (a °). 2

Substitute sin (a ) =

5 . 3

d 1 Use the symmetry property

sin (180 ° + a °) =

2

−sin

Substitute sin (a) =

(a °).

5 . 3

O2 = 32 − 22 =5 O= 5 a sin (a) =

O H

=

5 3

b tan (a) =

O A

=

5 2

c cos (90 ° − a °) = sin (a °)

5 3

=

d sin (180 ° + a °) = −sin (a °)

=

−

5 3

(Note: The above results could have been obtained using the identities directly.)

Chapter 6 • Circular functions

271

exercise 6e

identities

1 Copy and complete the table below, correct to 3 decimal places:

θ°

30°

81°

129°

193°

260°

−47°

350°

sin2 (θ ) cos2 (θ ) sin2 (θ ) + cos2 (θ ) 2 We11a If sin (θ ) = 0.8 and 0° < θ < 90°, find, correct to 3 decimal places: a cos (θ ) b tan (θ ). 3 We11b If cos (θ ) = 0.3 and 0° < θ < 90°, find, correct to 3 decimal places: a sin (θ ) b tan (θ ). 4 We12 Find all possible values of the following, correct to 3 decimal places. a cos (x °) if sin (x °) = 0.4 b cos (x °) if sin (x °) = −0.7 c sin (x °) if cos (x °) = 0.24 d sin (x °) if cos (x °) = −0.9 5

6

7

Use the diagram at right to find the exact values of:

b sin (x °)

a c c cos (x °).

3

Use the diagram at right to find the exact values of: a b b cos (x) c tan (x).

x

Find the exact values of: 12 a cos (x) if sin (x) = 13 and 90° < x < 180° b sin (x) if cos (x) = c cos (x) if sin (x) =

−3

2 7 x

and x is in the third quadrant

5 −7 25

9 We13 Find a° if 0° ≤ a° ≤ 90° and: a sin (a °) = cos (20 °) b sin (a °) = cos (58 °) d cos (a °) = sin (82 °) e sin (8 °) = cos (a °) g sin (89 °) = cos (a °) h cos (17 °) = sin (a °).

a

36°

c 54° b

c cos (a °) = sin (39 °) f cos (44 °) = sin (a °)

Copy and complete the following table. sin (θ )

0.8

0.28

cos (θ )

0.6

0.96

tan (θ )

272

bc

8

and x is in the fourth quadrant 3π 3 < x < 2π . d sin (x) if cos (x) = 2 and 2 8 MC Examine the diagram at right and answer the following questions. a sin (54 °) is equal to: a cos (54 °) B cos (36 °) C tan (36 °) D sin (36 °) e tan (54 °) b cos (54 °) is equal to: a tan (36 °) B cos (36 °) C tan (54 °) D sin (36 °) e sin (54 °) c tan (36 °) is equal to: cos (36 ) sin (36 ) a B sin (36 °) cos (36 °) C sin (36 ) cos (36 ) D sin (54 °) cos (54 °) e sin (36 °) + cos (36 °) d tan (54 °) is equal to: cos (36 ) cos (54 ) a B sin (54°) − cos (54 °) C sin (36 ) sin (54 ) D sin (54 °) cos (54 °) e sin (36 °) cos (36 °)

10

5

Maths Quest 11 Mathematical Methods CAS

0.77 0.3 3.18

0.573 0.447

1.207

2

0.7

11 We14 a d g j

If 0° < a°, b°, c° < 90° and sin (a ) =

sin (b °) tan (a °) sin (90 ° − a °) sin (180 ° − a °)

b e h k

2 , 5

cos (b ) = 5 , tan (c ) =

tan (b °) sin (c °) cos (90 ° − b °) cos (180 ° + b °)

6F Sine and cosine the graph of y = sin (x)

11 , 5

3

c f i l

find:

cos (a °) cos (c °) sin (90 ° − c °) tan (180 ° + c °).

graphs

To get an idea of the shape of the y = sin (x) graph, we can construct a table of values, plot the points and join them with a smooth line. The values for the table can be generated from the unit circle. You may recall that continuing on the unit circle past 2π, the sine values begin repeating themselves. For example, 9π π π π sin = sin 2π + = sin . We have gone exactly once around the circle from . 4 4 4 4 x y = sin (x)

−π

0

4

−π 2

−π 4

0

π 4

π 2

3π 4

π

5π 4

3π 2

7π 4

2π

−0.71

−1

−0.71

0

0.71

1

0.71

0

−0.71

−1

−0.71

0

−3 π

interaCtiVitY int-0251 Sine and cosine graphs

y ( –2π , 1)

1 0.71 (−π, 0) −π

π ( –4 ,

(0, 0) 0

− π–2 − π–4

(− 3—4π , − 0.71) ( − π–2 , − 1)

–π 4

0.71) –π 2

y = sin (x)

π ( 3— , 0.71) 4

(2π, 0)

(π , 0) π

π ( 5— , −0.71) 4

−0.71

−1

2π

3π — 2

x

( 7—4π , −0.71) ( 3—2π , −1)

You can verify the shape of this curve using a CAS calculator. It can be observed that the curve repeats itself in cycles after an interval of 2π units. Due to this repetition it is called a periodic function; the period is the interval between repetitions. The y period of y = sin (x) is 2π radians (or 360°). When dealing with graphs of circular functions Period such as sin (x), unless otherwise stated, we assume that the units for x are radians. Amplitude The mean position of the graph of y = sin (x) is y = 0 and the maximum and minimum values are x 1 and −1 respectively. The distance from the mean Mean position to the maximum (or minimum) position position is called the amplitude of the periodic function. Period The amplitude of y = sin (x) is 1 unit.

the graph of y = cos (x) As for the graph of y = sin (x), the graph of y = cos (x) can be established by first completing a table of values. x

−π

−3π 4

−π 2

−π 4

0

π 4

π 2

3π 4

π

5π 4

3π 2

7π 4

2π

y = cos (x)

−1

−0.71

0

0.71

1

0.71

0

−0.71

−1

−0.71

0

0.71

0

Chapter 6 • Circular functions

273

When these points are plotted on a set of axes and joined with a smooth curve, the graph looks like the diagram at right. It can be seen that the basic shape of the curve y = cos (x) is the same as that of y = sin (x). It is also a periodic function with: period = 2π amplitude = 1.

y y = cos (x) (−π–4 , 0.71)

1 (0, 1) (π–4 , 0.71)

( π–2 , 0)

(− π–2 , 0) −π (− 3—4π, −0.71)

(−π , −1)

− π–2 − π–4

0

π –

π –

4

2

−0.71 3π −1 ( —4 , −0.71)

(2π , 1)

(7—4π , 0.71)

0.71)

( 3—2π , 0) 3π — 2

π

2π

x

( 5—4π , −0.71)

(π , −1)

π Note: The graph of y = cos (x) is exactly the same as that of y = sin (x) translated units or 90° to 2 the left. In general, the graphs of y = a sin (bx) + c and y = a cos (bx) + c (for a, b, c ∈R) have the following properties: 1. Amplitude = | a | 2π 360 2. Period = (or ) b b 3. Mean position at y = c 4. Maximum = c + a and minimum = c − a π 1 5. When c = 0, x-intercepts for cosine are at x = ± and every 2 period to the left and right of these; 2 b π 1 x-intercepts for sine are at x = ± and every 2 period to the left and right of these. b Note: | a | means the size or magnitude of a, expressed as a positive number, as the amplitude must always be positive. If a < 0, then the resulting graph is a reflection in the x-axis of the graph for which a > 0. For example, the graph of y = −2 sin (x) is the reflection of y = 2 sin (x) in the x-axis (think of the x-axis as a plane or flat mirror). Worked Example 15

State i the period and ii the amplitude of each of the following functions. a y b y = −1.5 sin (4 x) 4

0

3— π 2

3π

x

−4 Think

a 1 The cycle repeats after 3π units. 2

The distance from the mean position to the maximum position is 4 units.

b 1 Write the formula for the period of y = a sin (bx).

Write

a i Period = 3π ii Amplitude = 4

b i Period =

2π where b = 4 b

2

Substitute b = 4.

=

2π 4

3

Simplify.

=

π 2

4

By rule, the amplitude is | a | or | −1.5 |.

274 Maths Quest 11 Mathematical Methods CAS

ii Amplitude = | −1.5 |

= 1.5

WorkeD exaMple 16

Sketch the graphs of the following functions and state i the period and ii the amplitude of each. 2 x x ∈ [0, 12π] a y = 4 cos b y = sin (4 x) x ∈ [0, 2π] 3 3 think

tUtorial eles-1607 Worked example 16

Write/DraW

a 1 Write the formula for the period.

a i Period =

1

2π b 2π

2

Substitute b = 3 .

=

3

Simplify the value of the period.

= 6π

4

The amplitude is the value in front of cos, written as a positive value.

5

Draw a set of axes.

6

Because the amplitude is 4, mark or imagine horizontal guidelines at y = −4 and 4.

7

Sketch one cycle of the graph every period (every 6π) along the x-axis, for x ∈ [0, 12π], showing key x-values.

1 3

ii Amplitude = 4

y 4 0

3— π 2

9— π 2

3π

6π 15—2π 9π

π 21 — 2

12π x

5— π 4

3— π 2

−4

b 1 Write the formula for the period.

b i Period =

2π b

2

Substitute b = 4.

=

2π 4

3

Simplify the value of the period.

=

π 2

4

The amplitude is the value in front of sin, written as a positive value.

5

Draw a set of axes.

6

Because the amplitude is 3 , mark or imagine horizontal guidelines at −2 2 y = 3 and 3 .

2

ii Amplitude =

y 2– 3

0 − 2–3

7

Sketch one cycle of the graph every period (every π2 ) along the x-axis, for x ∈ [0, 2π].

8

Determine the x-intercepts and mark these on the graph.

2 3

–π 4

–π 2

3— π 4

π

7π — 4

2π x

Chapter 6 • Circular functions

275

WorkeD exaMple 17

Sketch the graph of y = cos (2x) − 3, x ∈ [0, 2π], and state: a the period b the amplitude c the maximum and minimum values. think

Write/DiSplaY

a 1 Write the formula for the period.

a Period =

2π b 2π 2

2

Substitute b = 2.

=

3

Simplify the value of the period.

=π

b 1 The amplitude is the value in front of cos,

written as a positive value. 2

3

4

Sketch the graph of y = cos (2x) first. Since the amplitude is 1, mark or imagine horizontal guidelines at y = +1 and y = −1. Sketch one cycle of the graph every period (every π) along the x-axis, for x ∈[0, 2π].

y 1 0

–π 2

−3.

y= The maximum and minimum are at c + a and c − a.

3— π 2

2π x

−1

Because the mean position is now at y = −3, translate y = cos (2x) vertically to this position.

c Maximum = −3 + 1

= −2 Minimum = −3 − 1 = −4 y 1 0 −1 −2 −3 −4 −5

π — 2

WorkeD exaMple 18

Sketch the graph of the following function. π f: [−2, 4] → R, f (x) = 2 cos x 2 think

276

π

Determine the x-intercepts and mark these on the graph.

c 1 The mean position of y = cos (2x) − 3 is at

2

b Amplitude = 1

Write/DraW

2π π , where b = . 2 b

1

The period of the function is

2

Simplify the value of the period.

3

The amplitude is 2.

4

Draw a set of axes.

5

Using an interval of 4 (the period), mark the x-axis from −2 to 4 (the specified domain).

Maths Quest 11 Mathematical Methods CAS

Period =

2π π 2

=4 Amplitude = 2

π

3— π 2

2π x

6

7

8

9

10

Show −2 and 2 on the y-axis, since the amplitude is 2.

πx y = 2 cos — 2 (4, 2)

( )

y 2

Visualise or check the general shape of the graph of y = a cos (bx).

−2

Starting from the point (0, 2), complete one cycle of the cosine function forward to (4, 2).

−1

0

1

2

3

x

4

−2

(−2, −2)

Complete half of a cycle back from (0, 2) to (−2, 2). The x-intercepts are halfway between the maximum and minimum points.

Sine and cosine graphs

exercise 6F

State i the period and ii the amplitude for each of the following functions.

1 We15a a y

b

1

2 0

2π

4π

x

x

3— π 2

−1.5

x

π

y 3 − –3π

0

−2 h

y 2.5 0

− –2π

π

−2.5

DiGital DoCS doc-9762 Sine graphs doc-9763 Cosine graphs

3π x

0

f

0

−π

−4 g

x

y 2

e

0

5— π 4

–π 4

−1

y 4

y 1.5

0

−2 d

c

y

π

x

π

2π

−3

y 0.5

y

i

8 6

x

0

–π 6

–π 2

x

4 2

−0.5 −2π

−π

0

x

2 We15b State i the period and ii the amplitude of each of the following functions. a y = sin (x) b y = 3 sin (x) c y = 2 sin (2x) d y = −4 sin (3x) g y = 0.4 cos

j

x +1 3

y = −sin (6x) + 4

e y=

1 2

x sin 3

h y = −3 cos (5x)

k y=

1 5

sin (π x )

f i

l

y = 2 cos (4x)

πx y = 2.5 cos 4 πx cos 2 y= 4

Sketch the graph of the following functions, showing one complete cycle, and state i the period and ii the amplitude of each. x a y = sin (2x) b y = 2 cos (x) c y = 3 sin 2

3 We16

1

d y = 4 cos (2x)

e y = 2 sin (3x)

x g y = 5 sin

h y = − 4 cos

3 j y = −3 sin (2x)

x 2

2 3

x cos 2

f

y=

i

y = 2 cos (4x)

Chapter 6 • Circular functions

277

4 MC Parts a to c refer to the graph shown below. y 3

0

x

–π 2

−3 a The amplitude of the function is:

π 2

a

B −3

C π

D 3

e 6

D 6

e 3

b The period of the function is:

π C π 2 c The equation of the function could be: a y = 3 sin (2x) B y = 3 sin (x) x D y = −3 sin (2x) e y = 3 sin 2 a 2π

B

C y = 3 cos (2x)

y

5 MC The equation of the curve at right could be:

x 2 x D y = cos 2

a y = 2 cos (x)

B y = 2 sin

C y = −2 cos (x) e

x y = − 2 cos 2

2

0

π

2π

−2

State the equation of each of the functions graphed below.

6

y

a

y

b

1.5

2 3π x

0 −1.5

0

−π–4

π – 4

x

3π — 4

−2 y

c

d

0

−2π

4π x

2π

0

y

−1

x

y

f

1 0

3π — 2

−4

−5 e

y 4

5

3 2π — 3

4π — 3

x − π– 6

0

−3

278

3π

Maths Quest 11 Mathematical Methods CAS

π– 6

x

x

7 We17 Sketch the graph of each of the following functions. b f (x) = 3 cos x for x ∈ [−4π, 4π]

a f (x) = cos (2x) for x ∈ [−2π, 2π] c f (x) = 2 sin (3x) for x ∈ [0, 2π]

d f (x) =

e f (x) = 1.5 sin x for x ∈ [−6π, 6π] 3

f

−3

2 cos (2x) for x ∈ [0, 2π]

f (x) = −4 sin (4x) for x ∈ [0, π]

8 We17 Sketch the graph of the following functions and state i the period, ii the amplitude, and iii the

maximum and minimum values of each. a y = cos (2x) + 1 x ∈ [0, 2π] x c y = 2 cos + 3 x ∈ [0, 2π] 2

b y = 2 sin (x) − 2

x ∈ [−2π, 2π]

x d y = sin – 1

x ∈ [0, 6π]

3

9 We18 Sketch the graphs of each of the following functions. a f : [−π, 2π] → R, f (x) =

3 2

2x sin 3

b f : [−6, 6] → R, f (x) = 1.8 cos

πx 3

3π x d f : [0, 8] → R, f (x) = −3 cos

c f : [0, 6] → R, f (x) = −1.4 sin π x

4

2

10 For each of the functions graphed below, state the rule using full function notation. a

b

y 3

y f(x)

5– 2

f(x) 0

8— π 6

0

x

−3

c

5π x

− 5– 2

d

y

y

2

1.8 f(x) 0

−1

f(x)

x

1

−1

−2

e

f

y

y

3

f(x)

f(x) x

− 1–3 0

y 0

−3

1

x

−2.4

−3

−2

x

2.4

0

−1

3

−1.8

3

g

0

y

h — 2

π

3— 2

2π

x

1 −π

−—2 −10 −2 −3 −4

— 2

π

2π x DiGital DoC doc-9764 WorkSHEET 6.2

Chapter 6 • Circular functions

279

6G Tangent graphs The graph of y = tan (x) To manually plot the graph of y = tan (x), the following table of values may be used. x y = tan (x)

−π

0

−3π

−π 2

−π 4

0

π 4

π 2

3π 4

π

5π 4

3π 2

7π 4

2π

Undefined

−1

0

1

Undefined

−1

0

1

Undefined

−1

0

4

1 π

Note: Multiples of 8 could be used to give more points to plot and a clearer indication of the shape of the graph. sin ( x ) Note the presence of some undefined y-values. This is because y = and cos (x) = 0 at these cos (x) values. These undefined values are shown as vertical asymptotes at the given value of x for which they occur. (An asymptote is a line that a graph approaches, but never quite reaches. In the case of y = tan (x), y approaches (but never actually reaches) −∞ and +∞ for particular x-values.) The graph of y = tan (x) is shown in the figure below. Vertical asymptotes y y = tan (x) 5π 3π (− — , 1) ( π–4 , 1) ( —4, 1) 4 1 (π , 0) (−π, 0) (0, 0) 0 π π π π −π π – – − –2 − –4 4 2 −1 ( 3—4π, −1)

(2 π , 0) 2π −1)

π 3— 2

x

( 7—4π,

The features of the graph of y = tan (x) are as follows: − π π 3π 1. It has vertical asymptotes through x = … , , ,… 2 2 2 2. It has no amplitude.

3. It has a period of π. 4. It has a range of R (the set of all real numbers). In general, the graph of y = a tan (bx) + c has the following properties. y

3π −— 2b

− π–b

π −— 2b

0

π — 2b

π– b

π 3— 2b

x

1. No amplitude π 2. Period = . b π 3. When c = 0, the graph has x-intercepts at x = ± and every period to the left and right of these. b −π π 4. It has asymptotes at x = and x = and every period to the left and right of these. 2b 2b (2n + 1)π 5. The following formula for asymptotes applies: xasymptote = ± , where n = 0, 1, 2, . . . 2b 6. Mean position at y = c. The graph is translated vertically to c. 280 Maths Quest 11 Mathematical Methods CAS

WorkeD exaMple 19

For each function below, state i the period and ii the equation of the two asymptotes closest to the y-axis. 1 x + 3 a y = 2 tan (4 x) b y = 3 tan 2 think

Write

a 1 Write the formula for the period.

a i Period =

π b

=

π 4

2

Substitute b = 4.

3

Write the formula to find the two closest asymptotes to the y-axis.

4

Substitute b = 4.

x=±

π 2×4

5

Simplify.

x=±

π 8

b 1 Write the formula for the period.

ii Asymptotes: x = ±

b i Period =

1 . 2

2

Substitute b =

=

3

Simplify.

4

Write the formula to find the two closest asymptotes to the y-axis.

5

Substitute b =

6

Simplify.

π 2b

π b π 1 2

= 2π

1 . 2

ii Asymptotes: x = ±

x=±

π 2b

π 1 2× 2

x=±π

WorkeD exaMple 20

π x over [−2, 4] without a Sketch the graph of the function y = 3 tan 2 CAS calculator. think

tUtorial eles-1423 Worked example 20

Write

π . 2

π π 2 =2

1

Find the period using b =

2

Find the two asymptotes closest to the y-axis π π by substituting b = into x = ± . 2 2b

Asymptotes: x = ±

State the other asymptotes by adding/subtracting the period, 2 units, to/from the first asymptotes, x = ± 1.

Other asymptotes are: x = −1 − 2, x = 1 + 2, x = 1 + 2 + 2 etc.

3

Period =

= ±1

π 2×

π 2

Chapter 6 • Circular functions

281

4

State all asymptotes in the domain [−2, 4].

5

Evaluate y when x = 2 and x = 2 to establish two definite points on the graph.

6

Draw a set of axes using x ∈ [−2, 4].

7

8

9

1

π 1 When x = 2 , y = 3 tan 4 =3×1 =3 −π −1 When x = 2 , y = 3 tan 4 = 3 × −1 = −3

−1

πx y = 3 tan — 2

( )

y

Mark in the vertical asymptotes at x = −1, x = 1, x = 3.

3

Using the asymptotes as a guide, sketch the standard tan curve.

0

1

−2 −1 − –2

1– 2

1

2

3

4

x

−3

Verify that this graph is correct by using a CAS calculator.

tangent graphs

exercise 6G DiGital DoC doc-9765 tangent graphs

All asymptotes in the domain [−2, 4] are: x = −1, x = 1, x = 3.

State i the period and ii the equation of the two asymptotes closest to the y-axis for each of the following. x x a y = tan (2x) b y = tan (3x) c y = tan d y = tan 3 2 x x e y = 3 tan (x) f y = 2 tan g y = − tan h y = −2 tan (2x) 4 2 i y = tan (x) − 1 j y = tan (2x) − 3

1 We19

2 Sketch the graph of each function in question 1, showing the first two cycles. 3 We20

The function y = 3 tan (4x) has a period equal to: π π B C π D 4 2

MC

a 4π

4 MC Use the graph to answer questions a and b. a The period of the function is equal to:

π 4 π C 2 e 3 b The equation of the function is: a

a y = 1.5 tan (x) C y = 1.5 tan (2x) e y = 3 tan (x)

B π D 2π

y

1.5 0

x 2 D y = 1.5 tan (3x)

e 2π

π – 2

π

2π

B y = 1.5 tan

5 We20 Sketch the graphs of each of the following functions over the given domain. a y = 3 tan DiGital DoC doc-9766 Investigation tangent graphs

282

x , [0, 2π] 2 2π x , [0, 5] 5

d y = 1.8 tan

Maths Quest 11 Mathematical Methods CAS

b y = tan

π x , [0, 3] 3

e y = −2 tan (π x), [−1, 1]

c y = 2 tan

x , [0, 2π] 3

3π

x

6H

Solving trigonometric equations

To find the solution to the equation sin ( x ) =

1

2

where x ∈ [0, 4π], we can consider the graph of y = sin (x).

y 1 1– 2

0

π – 2

π 3— 2

π

2π

π 5— 2

3π

π 7— 2

4π

x

y = 8 sin (θ )

−1 1

π

By drawing a horizontal line through y = , it can be seen that there are four solutions in the domain 2 [0, 4π]. π π The solution for 0 < x < , that is, in the first quadrant, is (from our knowledge of exact values). 2 4 Note: For inexact solutions in the first quadrant, use a calculator. The sine function is also positive in the second quadrant. y π 3π Using sine symmetry, the next solution is π − = 4 4 (different symmetry properties are used for cosine and sin ( π–4 ) sin (π − π–4 ) S A tangent). Since the graph is periodic, any further solutions are π – 4 found by adding (or subtracting) the period (in this case 2π) to x (or from) each of the first two solutions. T C For example, two further solutions are: 3π π + 2π and + 2π 4 4 9π 11 π = and 4 4 π 3π 9π 11π Therefore, four solutions in the specified domain are , , and . 4 4 4 4 However, if a domain is not specified, there are an infinite number of solutions, as multiples of 2π π 3π can be added (or subtracted) indefinitely to (or from) and . In this situation a general solution is 4 4 obtained where the solutions are in terms of a parameter, n, where n is an integer, i.e. n ∈ Z. π π The general solution for the first quadrant solution x = becomes x = 2nπ + where n ∈ Z. 4 4 3π 3π becomes x = 2nπ + The general solution for the second quadrant solution x = where n ∈ Z. 4 4 π π 3π Note that the general solution x = 2nπ + can be expressed as x = 2nπ + π − = (2n + 1)π − . 4 4 4 Substituting different integer values of n will give specific solutions as shown in the table below. π– 4

−

n

x = 2nπ +

π , n ∈Z 4

x = 2nπ +

3π , n ∈Z 4

−1

x = − 2π +

π − 7π = 4 4

x = − 2π +

3π − 5π = 4 4

0

x = 0+

1

x = 2π +

2

x = 4π +

π π = 4 4 π 9π = 4 4

π 17π = and so on 4 4

x = 0+

3π 3π = 4 4

x = 2π +

3 π 11π = 4 4

x = 4π +

3 π 19π = and so on 4 4

Chapter 6 • Circular functions 283

In general, if: − − • sin (x) = a, then x = 2nπ + sin 1 (a) and x = (2n + 1)π − sin 1 (a) where a ∈ [−1, 1] and n ∈ Z. −1 − • cos (x) = a, then x = 2nπ ± cos (a) where a ∈ [ 1, 1] and n ∈ Z − • tan (x) = a, then x = nπ + tan 1 (a) where a ∈ R and n ∈ Z. When solving trigonometric equations, the following need to be determined: 1. The first quadrant angle, irrespective of the sign 2. The two quadrants in which the given function is positive or negative 3. Two solutions between x = 0 and x = 2π (use the appropriate sine, cosine or tangent symmetry property). If more solutions are required: 1. Repeatedly add (or subtract) the period to the two solutions as many times as required, noting solutions after each addition or subtraction. 2. Stop when all solutions within the specified domain are found. 3. If no domain is given, a general solution is required. WorkeD exaMple 21

Find θ to the nearest tenth of a degree if cos (θ °) = −0.58, given that θ ∈ [0°, 360°]. think

Write/DraW

Write the equation.

cos (θ °) = −0.58

2

Find the first quadrant angle by solving cos (θ °) = 0.58 to the nearest tenth of a degree.

First quadrant angle = cos−1 (0.58) = 54.5°

3

Identify where the cosine function is negative.

Cosine is negative in quadrants 2 and 3.

4

Use cosine symmetry to find the solutions.

(1

1

80

y

− )°

.5

54 54.5° −0.58 + 0.58

180°

x

(1

80

+

54

.5

)°

(−0.58)

5

Simplify the solutions.

6

Since the period is 360°, no further solutions are required for θ ° ∈ [0°, 360°].

θ = (180 − 54.5)° or (180 + 54.5)° = 125.5° or 234.5°

WorkeD exaMple 22

Find the value of x in each of the following equations if x ∈ [0, 4π]. Give answers correct to 3 decimal places, unless exact answers may be found. a sin (x) = 0.3 −1 b cos ( x) = 2 think

a 1 Write the equation.

284

Write

a sin (x) = 0.3

x ∈ [0, 4π]

2

Use a CAS calculator to find the first quadrant angle (since no exact answer exists). Ensure your CAS calculator is set to radians.

First quadrant angle = sin−1 (0.3) = 0.305

3

Identify where sine is positive.

Sine is positive in quadrants 1 and 2.

Maths Quest 11 Mathematical Methods CAS

4

y

Use sine symmetry to find the two solutions for x ∈ [0, 2π]. 2.83

c

.305 0.3 x

7c

0.3

For x ∈ [0, 2π] x = 0.305 or (π − 0.305) = 0.305 or 2.837 5

Convert the specified domain to a decimal.

[0, 4π] = [0, 12.566]

6

The period of y = sin (x) is 2π, so add 2π (= 6.283) to each of the solutions above.

For x ∈ [0, 4π] x = 0.305, 2.837, (0.305 + 6.283), (2.837 + 6.283)

7

Adding 2π (= 6.283) to the last two solutions would give solutions beyond the specified domain, so stop here.

x = 0.305, 2.837, 6.588, 9.120

b 1 Write the equation.

b cos( x ) =

−1 2

x ∈ [0, 4π] −1 1 2

Recall a special triangle to find the first quadrant angle. Note: This is not a solution in this case.

First quadratic angle = cos

3

Identify where cosine is negative.

Cosine is negative in quadrants 2 and 3.

4

Use cosine symmetry to find the two solutions for x ∈ [0, 2π].

2

=

π 3

y –π 3

2— π 3 1 −— 2

1 — 2

x

4— π 3

For x ∈ [0, 2π] π π x = π − or π + 3 3 2π 4π = or 3 3 5

The period of y = cos (x) is 2π, so add 2π to each of the solutions above for x ∈ [0, 4π].

6

Adding 2π (=

6π ) to the last two solutions 3 would give solutions beyond the specified 12π domain (4π or ), so stop here. 3

For x ∈ [0, 4π] 2π 4π 2π 4π x= + 2π , + 2π , , 3 3 3 3 2π 4π 2π 6π 4π 6π = + + , , , 3 3 3 3 3 3 x=

2π 4π 8π 10π , , , 3 3 3 3

Chapter 6 • Circular functions 285

WorkeD exaMple 23

Find solutions to 2 sin (x) = 0.984 over the domain [0, 2π]. think

Write/DraW

x ∈ [0, 2π]

2 sin (x) = 0.984

1

Write the equation.

2

Divide both sides by 2 to get sin (x) by itself.

3

Determine the first quadrant angle in radians, correct to 3 decimal places.

First quadrant angle = sin−1 (0.492) = 0.514

4

Identify where the sine is positive.

Sine is positive in quadrants 1 and 2.

5

Use sine symmetry to find the solutions.

sin (x) = 0.492

y

(π −

0.51

4) c 0.492 π

0.492 x 0.514c

x = 0.514 or (π − 0.514) 6

Simplify the solutions.

7

Since the period is 2π, no further solutions are required over the domain [0, 2π].

= 0.514 or 2.628

Note: Any equation not in the form sin (A) = B (or cos or tan) should be transposed before the solutions are found.

WorkeD exaMple 24

Find the general solution of the following equations. 1 a cos ( x) = b 2 sin (x) − 1 = 0 c tan ( x) = 3 2 Hence, find all the solutions for −2π ≤ x ≤ 2π for each equation. think

a 1 Write the general solution for

cos (x) = a.

286

tUtorial eles-1424 Worked example 24

Write −1

a x = 2nπ ± cos

(a)

2

Substitute a =

1 into the general equation 2 − 1 , recognising that it and evaluate cos 1 2 is an exact angle.

− 1 x = 2nπ ± cos 1 2 π x = 2nπ ± 4

3

Write the two separate solutions and specify n ∈ Z.

x = 2nπ +

4

Substitute n = −1, n = 0 and n = 1 into each of the general solutions.

n = − 1: x =

Maths Quest 11 Mathematical Methods CAS

π π and x = 2nπ − , n ∈ Z 4 4 − 9π

and x =

−

7π 4

4 −π π n = 0: x = and x = 4 4 π 7π π 9π n = 1: x = 2π − = and x = 2π + = 4 4 4 4

5

Write the solutions for −2π ≤ x ≤ 2π.

b 1 Rearrange the equation and substitute

1 a = 2 into the general equations. Evaluate − 1 sin 1 ( 2 ), recognising that it is an exact angle.

x=

−

7π − π π 7π , , , 4 4 4 4 1 2

b sin ( x ) =

−1

x = 2nπ + sin x = 2nπ

Specify n ∈ Z.

(a) and x = (2n + 1)π − sin

− 1 + sin 1 ( 2 )

and x = (2n + 1)π

−1

(a)

− 1 − sin 1 ( 2 )

π π and x = (2n + 1)π − or 6 6 5π x = 2nπ + , where n ∈ Z 6 x = 2nπ +

2

Substitute n = −1 and n = 0 into each of the general solutions.

n = − 1: x = − 2π +

π − 11π = and 6 6

π − 7π = 6 6 π π 5π n = 0: x = and x = π − = 6 6 6 x = −π −

3

Write the solutions for −2π ≤ x ≤ 2π.

c 1 Write the general solution for tan x = a.

x=

−

11π − 7π π 5π , , , 6 6 6 6

c x = nπ + tan

−1

(a)

−

x = nπ + tan = ( 3) π x = nπ + , n ∈ Z 3 1

2

n = − 2 : x = − 2π +

Substitute n = −2, n = −1, n = 0 and n = 1 into each of the general solutions.

−5

π 3

=

n = − 1: x = − π + = n = 0: x =

−

π 3

3

Write the solutions for −2π ≤ x ≤ 2π.

− 5π

3

,

−

π 3

2π π 4π , , 3 3 3

Solving trigonometric equations

exercise 6h 1

x=

4π 3

π 3

2π 3

n = 1: x = π + =

π 3

Find the exact value of x in terms of π in each of the following equations, given that x is in the first quadrant. a cos ( x ) =

1 2

b sin ( x ) =

c tan ( x ) =

3

d cos ( x ) =

3 2

sin ( x ) =

3 2

e tan (x) = 1

f

1 2

DiGital DoC doc-9767 trigonometric equations

Chapter 6 • Circular functions

287

2 We21 Find the value of θ to the nearest tenth of a degree in each of the following equations, given

that θ ∈ [0°, 360°].

b cos (θ) = −0.25 e cos (θ) = 0.195 h cos (θ) = 0.757

a sin (θ) = 0.6 d sin (θ) = −0.85 g sin (θ) = −0.333

c tan (θ) = 5.72 f tan (θ) = −0.837

3 We22 Find the value of x in each of the following equations if x ∈ [0, 4π]. Give answers correct to

3 decimal places, unless exact answers are found. a sin (x) = 0.8 b cos (x) = −0.5 d tan ( x ) =

−1

e sin (x) =

3

c tan (x) = 1.5

−0.39

cos ( x ) =

f

− 2 2

2 2

and 0° < x < 360°, then x is equal to: a 150° or 210° B 135° or 225° C 225° D 135° or 315° e 120° π b If 3 tan (x) = 1 and 0 < x < , then x is equal to: 2 π π π π π a B C D e 6 2 3 4 5 5 We23 Find solutions, exact where possible, to the following over the domain [0, 2π]. a 2 sin (x) = 0.586 b 2 cos (x) = 1 c 2 tan ( x ) = 2 3 d 2 sin (x) + 0.893 = 0 3 tan ( x ) + 1 = 0 e 1.8 cos (x) + 1.236 = 0 f 4 MC a If sin ( x ) = cos ( x ) =

The solution to the equation 2 cos (x) + 1 = 0 over the domain [0, 2π] is: 5π 7π 2π 4 π π 2π π 5π a B C D , , , , 6 6 3 3 3 3 6 6

6 MC

e

4π 5π , 3 3

7 Solve each of the following, to the nearest tenth of a degree, over the domain [0°, 360°]. a 4 sin (x) = 1 b 3 cos (x) = −2 c 2 tan (x) − 7 = 0 d 4 + sin (x) = 3 e 1 + 2 cos (x) =

2

f

3 tan (x) + 9 = 0

8 Solve the following equations over [0°, 360°]. a sin (x + 30)° = 1 c tan (x + 45)° = 1

b cos (x − 60)° = 0 d 2 sin (x − 60)° = 1

2 cos (x + 90)° + 1 = 0 x =1 g 2 cos 2

f

e

i

3 tan (2 x ) = 1

k tan (2 ( x − 30) ) = 9 We24

3

sin (2x) = 0 x h 2 sin = 2 2 2 cos (3 x ) = − 1 j l

sin

(

1 2

)

( x − 40) = 0.6

Find the general solution of the following equations.

a 2 cos ( x ) − 3 = 0

1 3 2 sin ( x ) − 1 = 0 .

b tan ( x ) = c

Hence, find all solutions for −2π ≤ x ≤ 2π for each equation. 10 Find the general solution of the equation 2 sin (2x) − 1 = 0. Hence, find all solutions for −π ≤ x ≤ π. 11 Find the general solution of the equation 2 cos (3x) − 1 = 0. Hence, find all solutions for −π ≤ x ≤ π.

6i

applications

Many situations arise in science and nature where relationships between two variables exhibit periodic behaviour. Tide heights, sound waves, biorhythms and ovulation cycles are examples. In these situations trigonometric functions can be used to model the behaviour of the variables. The independent variable (x) is often a measurement such as time. When modelling with trigonometric functions you should work in radians unless otherwise instructed. 288

Maths Quest 11 Mathematical Methods CAS

WorkeD exaMple 25

E. coli is a type of bacterium. Its concentration, P parts per million (ppm), at a particular beach over a 12-hour period t hours after 6 am is described by the function πt P = 0.05 sin + 0.1. 12 a Find i the maximum and ii the minimum E. coli levels at this beach. b What is the level at 3 pm? c How long is the level above 0.125 ppm during the first 12 hours after 6 pm?

tUtorial eles-1425 Worked example 25

think

a Write the function.

Write/DraW

a P = 0.05sin

i 1 The maximum value of the sine function

π t + 0.1 12

i The maximum P occurs when sin

is 1. 2

3

πt Substitute sin = 1 into the equation 12 for P and evaluate.

Max. P = 0.05(1) + 0.1 = 0.15

State the solution.

The maximum E. coli level is 0.15 ppm.

ii 1 The minimum value of the sine function

ii The minimum P occurs when sin

is −1.

b

π t = 1. 12

π t = − 1. 12

2

πt Substitute sin = − 1 into the equation 12 for P and evaluate.

Min. P = 0.05(−1) + 0.1 = 0.05

3

State the solution.

The minimum E. coli level is 0.05 ppm.

1

At 3 pm it is 9 hours since 6 am.

2

Substitute t = 9 into the equation for P and evaluate.

b

At 3 pm, t = 9. 9π When t = 9, P = 0.05 sin + 0.1 12 3π = 0.05 sin + 0.1 4 1 + 0.1 2 = 0.035 + 0.1 = 0.135 = 0.05

c

3

State the solution.

The E. coli level at 3 pm is approximately 0.135 ppm.

1

A sketch graph will give a better understanding of this question.

2

State the amplitude.

Amplitude = 0.05

3

Calculate the period.

Period = 2π π 12 Period = 24

c

Chapter 6 • Circular functions

289

πt The basic graph is P = 0.05 sin . 12

4

Identify the basic graph.

5

State the translations needed.

6

Sketch the graph of P.

7

Draw a horizontal line through P = 0.125.

No horizontal translation is needed; the vertical translation is 0.1 units up. P (parts per million) 0.150 0.125 0.100 0.050 0

2

4

6

8

12 t (hours)

8

Identify where P > 0.125 from the graph.

The graph shows that P > 0.125 between the first two points where P = 0.125.

9

Solve the equation P = 0.125 to find the first two values of t.

When P = 0.125, πt 0.05 sin + 0.1 = 0.125 12 πt 0.05 sin = 0.025 12 πt sin = 0.5 12 πt π π = or π − 12 6 6 π 5π = or 6 6 t 1 5 = or 12 6 6 t = 2 or 10

10

Find the difference between the solutions t = 2 and t = 10.

P > 0.125, for 10 − 2 = 8 hours

11

State the solution.

The E. coli level is above 0.125 parts per million for 8 hours.

exercise 6i

applications

1 The weight of a rabbit over a period of time is modelled by the graph shown below. W (kg) 4 3 2 1 0

a State i the amplitude and ii the period. b Express W as a function of t. 290

10

Maths Quest 11 Mathematical Methods CAS

3

6

t (days)

2 The graph below shows the heart rate of an athlete during a particular hour of a workout. H (beats/min) 160 110 60 0

15

30

45

60

t (min)

a Find the initial heart rate. b State i the amplitude and ii the period. c Express H as a function of t. 3 WE25 The height above the ground, h metres, of a child on a swing at any time, t seconds, after being

released is:

πt h = 1 + 0.6 cos 2

Find: a the maximum height of the swing b the height after 4 i 3 seconds and ii seconds 3 c the length of time that the swing is below 1.5 metres, travelling from one side to the other.

4 The temperature, T (°C), inside a building on a given day is given by the function:

πt T = 8 sin + 18 12 where t is the number of hours after 8 am. a What is the maximum temperature in the building and the time at which it first occurs? b Find the temperature at i 8 pm, ii 6 pm and iii 12 am (midnight). 5 The displacement, x (in mm), of a harp string t seconds after it is initially plucked is modelled by the

function: x (t) = 12 sin (20π t) a b c d

What is i the amplitude and ii the period of this function? How many vibrations (that is, cycles) will it complete in one second? Find the displacement after 0.08 seconds. At what time will its displacement first be 6 mm? Chapter 6 • Circular functions 291

6 The height of a bungee jumper, h metres, above a pool of water at any time, t seconds, after jumping is

described by the function: h (t) = 20 cos (0.8t) + 20 a What is the initial height of the bungee jumper? b When, if at all, does the bungee jumper first touch the water? c Assuming the cord is perfectly elastic, how long is it until the bungee jumper returns to the lowest

position?

7 A cyclist rides one lap of a circular track at a constant speed so that her distance, d metres, from her

starting point at any time, t seconds, after starting is:

πt d = 50 − 50 cos 30 Find: a the time taken to complete one lap b the radius of the track c the maximum distance from the start d the length of the track e the distance from her starting point after i 15 seconds and ii 40 seconds f the times at which she is 93.3 metres from her starting points. 8 The depth of water, d metres, at a port entrance is given by the function:

πt d (t ) = 4.5 + 1.5 sin 12 where t is in hours. a Find i the maximum and ii the minimum depth at the port entrance. b A certain ship needs the depth at the port entrance to be more than 5 metres. The ship can be loaded and unloaded, and in and out of the port, in 9 hours. Assuming that the ship enters the port just as the depth at the entrance passes 5 metres, will the ship be able to exit 9 hours later? How long will it have to spare, or by how many minutes will it miss out? 9 The Australian dollar’s value (a) in US dollars was observed to follow the equation

a = 0.9 + 0.01t + 0.02 sin (0.5π t) over a period of 8 days (t represents the number of days). Using a CAS calculator, sketch a graph of a for 0 ≤ t ≤ 8. When will the Australian dollar first reach US$0.95? At what other times will it be worth US$0.95? Find all of the maximum turning points of the graph (to 3 decimal places). What is the highest value reached?

a b c d e

292 Maths Quest 11 Mathematical Methods CAS

Summary trigonometric ratio revision

• SOH CAH TOA O A cos (θ ) = • sin (θ ) = H H

tan (θ ) =

O A

H

θ

• Angles of elevation and depression:

O

A

n

tio leva of e

gle Horizontal

An

gle

An the unit circle

n

sio

res

ep of d

–π 2

• sin (θ) = y • cos (θ) = x

90° y 1st quadrant P(θ )

2nd quadrant S

A

T

θ cos (θ ) C

π , 180°

sin (θ ) x 0, 360°, 2π

3rd quadrant

4th quadrant 270° 3π — 2

• tan (θ) = the height of the point where the angle line meets the tangent line (at the RHS of the unit circle)

tan (θ )

y

θ

• tan (θ ) = radians

x

y x

• π c = 180° 180 • 1c = π c π • 1 = 180

Chapter 6 • Circular functions

293

π– 6

• For exact values use special triangles:

30° 2 2 π– 4

Angle (θ) 0 π or 30° 6 π or 45° 4 π or 60° 3 π or 90° 2

45°

cos (θ) 1

1 2

3 2

1 2

=

2 2

1 2

=

3 2

1 2

1

0

60°

π– 3

1

sin (θ) 0

3

1 1

tan (θ) 0 1 3

2 2

=

3 3

1 3 Undefined

Symmetry

• Relating a given angle back to the first quadrant of the unit circle is often helpful. Provided an angle is expressed as π ± θ or 2π ± θ, the trigonometric function (sin or cos) remains the same, 3π 4π 6π 8π only the sign (+ or −) may change. So π may be written as etc. , , , 3 4 6 8 • sin (π + θ) = −sin (θ) • sin (2π − θ) = −sin (θ) • sin (π − θ) = sin (θ) − − • cos (π − θ) = cos (θ) • cos (π + θ) = cos (θ) • cos (2π − θ) = cos (θ) • tan (π − θ) = −tan (θ) • tan (π + θ) = tan (θ) • tan (2π − θ) = −tan (θ)

identities

• sin2 (θ) + cos2 (θ) = 1 sin (θ ) • tan (θ ) = cos (θ ) π • sin (90 ° − θ °) = cos (θ °) or sin − θ = cos (θ °) 2

π • cos (90 ° − θ °) = sin (θ °) or cos − θ = sin (θ °) 2 Sine and cosine graphs

• Basic graph types: 1. y = a sin (bx) + c y = a cos (bx) + c 2π 2. Period = , amplitude = | a | b y a 0 −a

294

Maths Quest 11 Mathematical Methods CAS

y y = a sin (bx)

–π b

y = a cos (bx)

a

2— π b

x = period

0 −a

π — 2b

–π b

3— π 2b

2— π b

x

• When sketching: 1. Recall the basic graph type. 2. Find the period and amplitude. 3. Sketch in sections of one period. 4. Find x-intercepts between other known intercepts or minimum and maximum points. 5. Translate the graph vertically by c. tangent graphs

• y = a tan (bx) + c π • Period = , no amplitude b π • Asymptotes at x = ± and every period to the 2b left and right of these • The formula for asymptotes applies: (2n + 1)π xasymptote = ± , 2b where n = 0, 1, 2, . . .

y

3π −— 2b

− π–b

π −— 2b

0

π — 2b

π– b

x

π 3— 2b

Solving trigonometric equations

• When solving trigonometric equations, the following need to be determined: 1. The first quadrant angle, irrespective of the sign 2. The two quadrants in which the given function is positive or negative 3. Two solutions between x = 0 and x = 2π (use the appropriate sine, cosine or tangent symmetry property). • If more solutions are required: 1. Repeatedly add (or subtract) the period to the two solutions as many times as required, noting solutions after each addition or subtraction. 2. Stop when all solutions within the specified domain are found. 3. If a general solution is required, the answer will be in tems of n, n ∈ Z. In general, if: − − • sin (x) = a, then x = 2nπ + sin 1 (a) and x = (2n + 1)π − sin 1 (a) where a ∈ [−1, 1] and n ∈ Z. −1 • cos (x) = a, then x = 2nπ ± cos (a) where a ∈ [−1, 1] and n ∈ Z − • tan (x) = a, then x = nπ + tan 1 (a) where a ∈ R and n ∈ Z.

applications

• • • •

To find the maximum value of a function, replace sin x or cos x with +1. To find the minimum value of a function, replace sin x or cos x with −1. Initial values occur at t = 0. A sketch graph may provide greater understanding.

Chapter 6 • Circular functions

295

Chapter review S h ort anS Wer

1 The angle of elevation from an observer to an aircraft when it is 4.5 km away is 30°. How high is the

aircraft above the ground if the observer’s eye level is 1.75 m above the ground?

4.5 km

30° 1.75 m 2 Find the exact values of: a tan (120 °) 3 4

3 If cos (θ ) = a sin (θ)

5π . 6

b cos (150 °)

c sin

π

and 0 < θ < 2 , find the exact values of: b tan (θ).

4 Convert the following angles to exact radians. a 140° b 310°

c 1°

5 Convert the following radian measures to degrees. a

2π 5

b 1

c

6 If cos (θ) = 0.69, find: a sin (180 ° − θ °)

7π 2

b cos (2π − θ) c tan (π + θ). π 7 If cos (x) = and < x < π, evaluate: 2 a sin (x) b tan (x). 8 Sketch the graphs of the following functions. x a y = −4 sin over [−2π, 4π] b y = 1.5 cos (2x) over [−π, π] 2 9 Sketch the graph of: −1 x a y = 2 tan (4x) for x ∈ [−π, π] b y= tan for x ∈ [0, 12π]. 6 2 1 10 Find all of the solutions to the equation sin ( x ) = over the domain [0, 3π]. 2 11 Solve the equation 3 + 2 cos (x) = 0 over the domain [0°, 360°]. 3 12 Find the general solution of the equation sin ( x ) = . Hence find all solutions for −2π ≤ x ≤ 2π. 2 13 If cos (θ) = 0.40 and sin (θ) = 0.92, find: a tan (θ) b cos (180° − θ ) c sin (−θ) d cos (90° − θ ). 14 With the aid of a diagram if necessary, find exact values for: π 7π a sin b cos 6 6 −0.9

5π 3

c tan

d cos (π).

15 Solve the following.

3 2 1 c tan ( x ) = 3 e tan (x °) = −1 a sin ( x ) =

296

Maths Quest 11 Mathematical Methods CAS

b cos ( x ) =

−1

d cos (x °) = 1

2

16 The sound level of a siren follows the rule L (t) = 6 sin (π t) + 80, where L is the sound level measured

in decibels (dB) and t is the time in seconds. a What is the amplitude and period of L(t)? b What are the minimum and maximum sound levels of the siren? c How long does it take for the sound to reach its maximum level the first time? d Find the first time the sound level reaches 83 dB. e Sketch the graph of L(t) for 0 ≤ t ≤ 2. M U ltip l e C ho iC e

1 The relationship between a, x and y in the triangle shown is:

x a y a tan (a) =

y x

B

sin (a) =

x y

C tan (a) =

x y

D cos (a) =

y x

e sin (a) = xy

2 In the triangle shown, the value of x is closest to:

4.2

x 56°

a 2.35

B 2.83

C 5.07

D 7.51

e 3.48

3 The angle that the 3-metre ladder makes with the wall in this diagram is closest to:

3.0 m

1.2 m a 23.6°

B 21.8°

4 In which quadrants is tan (x) positive? a 1 and 2

B 1 and 4

C 66.4°

D 18°

e 76°

C 2 and 4

D 2 and 3

e 1 and 3

5 The value of cos (320°) is: a positive, as 320° is in the 1st quadrant C negative, as 320° is in the 3rd quadrant e positive, as 320° is in the 4th quadrant

B negative, as 320° is in the 2nd quadrant D negative, as 320° is in the 4th quadrant

6 If tan (a°) < 0, sin (a°) < 0 and 0° < a° < 360°, then which one of the following is correct? a 0° < a° < 90° D 0° < a° < 180°

B 180° < a° < 270° e 270° < a° < 360°

C 90° < a° < 180°

Chapter 6 • Circular functions

297

7 If tan (θ ) = a

3, then sin (θ) could be equal to:

1 2

3 2

B

C −1

D

5π 9

D

8 The angle 150° is equivalent to: a

2π 3

7π 6

B

9 The angle that is equivalent to a 405°

B 315° 4π is: 10 The value of cos 3 a

3 2

1

B 2

C

9π is: 4

C 540°

−

C

1

e 0

2

5π 6

D 270°

3 2

D

4π 3

e

e 300°

1

−1

e

2

π Use the following information to answer questions 11 to 13: sin (a) = 0.6 and 0 < a < . 2 11 sin (2π − a) is equal to: a 0.36 B 0.8 C −0.6 D −0.8 12 cos (a) is equal to: a 0.6 B 0.8 C −0.6 D −0.8 13 tan (π − a) is equal to: a 0.75 C 1.33 B −1.33 D −0.75 14 The one value that is equal to sin (53°) is: a tan (37 °) B cos (37 °) C cos (53 °) D sin (37 °) Questions 15 to 17 refer to the function f (x) = −2 sin (3x). 15 The amplitude of f (x) is equal to: a 3 D 2 B −2 C −3 16 The period of f (x) is equal to: 2π a 6π B 4π C D 2π 3 17 The range of f (x) is: a [0, 2] B [0, 4] C [−2, 0] D [−1, 1] 18 The rule for this graph is: y x a y = 3 cos 2 3

2

e 1 e 0 e 1.4 e cos (127 °)

e 1

e 3 e [−2, 2]

B y = 3 sin (2x)

x 3 D y = 3 cos (2x) e y = −3 cos (2x) C y = 2 cos

0

π

2π

3π

4π x

−3

19 The function y = 5 tan (4x) has a period and asymptote respectively of:

π 2 π π D 4π and x = 8π e and x = 4 2 20 If sin (3x) = 0.966, then x could be equal to: a 65° B 75° D 25° e 20° 21 If tan (2x) = 0.839 has one solution of x = 20°, then another solution could be: a 160° B 110° D 380° e 70° a 4π and x = 2π

298

Maths Quest 11 Mathematical Methods CAS

B π and x =

C

π π and x = 4 8

C 105°

C 200°

x 22 The maximum value of y = 3 sin − 1 is: a −1

B 0

2

C 2

23 The minimum value of h = cos (3π t) + 5 is: a 6

B 5

C 8

D 3

e −4

D 3

e 4

1 The temperature in an office is controlled by a thermostat. The preferred temperature, P, can be set to

values between 18 °C and 25 °C. The temperature, T (°C) in the office at time t hours after 9 am is given by the rule T = P + 2.4 sin (π t). If the preferred temperature on the thermostat has been set to 23 °C:

ex ten D eD r eS p o n S e

a find the maximum and minimum temperatures b find the temperature at i noon and ii 3.30 pm c sketch the graph of the function between 9 am and 5 pm.

Freddy feels thirsty if the temperature is above 24.2 °C. d Find the amount of time between 9 am and 5 pm that Freddy feels thirsty. 2 A ‘standing wave’ on a guitar string may be

y π x , 20 L where x cm and y cm are defined on the diagram x at right. a Find the period of the standing wave. b If the frets coincide with the mean positions of the wave, find the value of L. c If the frets were to be spaced at 16 cm, what would be the equation of a similar standing wave of amplitude 0.3 cm such that a fret is at each mean position?

approximated by the function y = 0.3 sin

3 The number of rabbits in a national park is observed for one year.

At any time t months after observation begins, the number is modelled by the function:

πt P = 2 − 0.8 sin 6 where P is in thousands. a Find: i the maximum number of rabbits ii the minimum number of rabbits iii the median number of rabbits. b Find i the period and ii the amplitude

of the function. Chapter 6 • Circular functions

299

c d e f

Sketch the graph of the function. Find the population after 5 months. How long is the population below 1600? How long is the population above 2100?

4 The height (in cm) that a clock’s pendulum swings above its base can be approximated by the function

7π H = 14 + 5.9 cos t at any time t seconds after being released. Give all answers correct to 4 3 decimal places. a Find i the maximum and ii the minimum heights that the pendulum reaches. b Find the height after i 1.5 seconds and ii 1 minute. c Sketch the graph of the function for the first 2 seconds. d On the same set of axes, sketch the median position. e Find the length of time that the pendulum is below 14 cm travelling from one side to the other. f Find the number of times the pendulum swings in 1 minute. The pendulum is found to be losing time and needs its swing adjusted to 75 swings per minute. g Find the new function H(t) that approximates the height of the pendulum.

5 The depth, h(t), of water in metres at a point on the coast at a time t hours after noon on a certain day is

2π (t + 2) given by h(t ) = 2.5 + 0.5 cos . 11

DiGital DoC doc-9768 Test Yourself Chapter 6

300

Use a CAS calculator to answer the following. a What is the depth of the water at noon (correct to 2 decimal places)? b What is the period of h(t)? c What is the depth of the water (and what time does each occur) at: i high tide? ii low tide? d Sketch the graph of h(t) for 0 ≤ t ≤ 12. e The local people wish to build a bonfire for New Year’s celebrations on a rock shelf near that point. They estimate that they can pass the point safely and not get splashed by waves if the depth of water is less than 2.25 m. Between what times can they work? f How long do they have?

Maths Quest 11 Mathematical Methods CAS

ICT activities Chapter opener DiGital DoC • 10 Quick Questions doc-9757: Warm up with ten quick questions on circular functions (page 255)

6a

trigonometric ratio revision

DiGital DoCS • SkillSHEET 6.1 doc-9758: Practise identifying the hypotenuse, adjacent and opposite sides of a right-angled triangle with respect to a given angle (page 256) • SkillSHEET 6.2 doc-9759: Practise calculating trigonometric ratios (page 256)

6B

the unit circle

DiGital DoCS • doc-9760: Investigate the unit circle (page 261) • WorkSHEET 6.1 doc-9761: Locating exact values in the unit circle, finding unknown angles and sides in a right-angled triangle and converting radians to degrees and degrees to radians (page 262)

6D

Symmetry

interaCtiVitY • Symmetry int-0265: Use the interactivity to consolidate your understanding of symmetry in the unit circle (page 265) tUtorialS • We 8 eles-1420: Using symmetry find the value of trigonometric functions of angles given in radians and degrees (page 266) • We 9 eles-1421: Find the value of trigonometric functions in exact values (page 267) DiGital DoC • doc-9760: Investigate the unit circle using a spreadsheet (page 268)

6e

identities

tUtorial • We 14 eles-1422: Watch how to find the value of trigonometric functions in exact values (page 271)

6F

Sine and cosine graphs

interaCtiVitY • Sine and cosine graphs int-0251: Use the interactivity to investigate the features of sine and cosine functions (page 273)

tUtorial • We 16 int-0298: Watch how to sketch the graphs of a sine and cosine function over a set domain, stating the amplitude and period of each (page 275) DiGital DoCS • doc-9762: Investigate sine graphs (page 277) • doc-9763: Investigate cosine graphs (page 277) • Worksheet 6.2 doc-9764: Sketching trigonometric graphs, solving trigonometric equations and recognising exact values (page 279)

6G

tangent graphs

tUtorial • We 20 eles-1423: Watch a tutorial on how to sketch the graph of a tangent function over a given domain (page 281) DiGital DoCS • doc-9765: Investigate tangent graphs (page 282) • Investigation doc-9766: Investigate tangent graphs (page 282)

6h

Solving trigonometric equations

tUtorial • We 24 eles-1424: Watch a tutorial on finding a general solution to three trigonometric equations using a CAS calculator (page 286) DiGital DoC • doc-9767: Investigate solutions to trigonometric equations (page 287)

6i

applications

tUtorial • We 25 eles-1425: Watch a tutorial on applying understanding of trigonometric graphs and equations to a real world model (page 289)

Chapter review DiGital DoC • Test Yourself doc-9768: take the end-of-chapter test to test your progress (page 300)

To access eBookPLUS activities, log on to www.jacplus.com.au

Chapter 6 • Circular functions

301

Answers CHAPTER 6 CirCUlar FUnCtionS exercise 6a

1 2 3 4 5 8

a a a a

3.09 8.41 18.2° D

C 3.25 m

trigonometric ratio revision b 1.88 c 4.00 b 2.97 c 12.50 b 24.3° c 51.1° bA c B 6E 7 7.14 m 9 9.6° 10 23.1°

4 a c e g i

−0.3

5 a

1 2 − 3 2

1 a e i 2 a e i 3 a e i 4 a b 5 a d 6 a d g j 7 a e

the unit circle P bP cP N f N gN P j N P bP cN N f P gP N j N P bP cN P f P gN N j P Quadrant 3 Quadrant 1 C bB B e C −1 b −1 0 e 0 Undefined h 0 1 3 2 1 2

b

i 1

e i

radians b f j

2 a b e f 3 E 4 C 5 B 6 a 0.855 d 3.341 g 2.971 7 a 20.1° d 54.9° g 232.6° 8 a d

=

2 2

1 2

g 1

302

π 4 π 2 − 5π 4 120° −30°

0.63 −0.63 −0.25 0.25 −2.1 2.1

c g k c g

1 3 2 2

1 2 −

i d P h N d N h P

e

d N h N

7 a d 8 a d 9 a c

c A c 0 f 1

=

d

i

d

j 2 2

b f

3 3

j

−0.383

h 2π

5π 3 d 330° h 67.5° l

c −2.182 f 5.716

b e h b e h

1.365 4.095 3.784 84.2° 893.8° 458.4°

b

1 2

c

e 1

f

c 180° f −155.3°

1 2 = 2 2

3 2

1 a i 4π

3 −

g

d

3

h

c −1 3 3

−

g

2 2

−0.63

i i 2π

2 a i 2π

3 2

h

−0.924

0.383 − 0.259 0.259 − 0.765 −0.644

c i π

− 3 2

l −1 c f c f

0.414 −0.414 − 3.732 −3.732

g i 6π

ii 0.4

i i 8

ii 2.5

k i 2

1 5

0

identities

sin2 θ + cos2 θ

30°

0.25

0.75

1

81°

0.976

0.024

1

129°

0.604

0.396

1

193°

0.051

0.949

1

260°

0.970

0.030

1

350°

0.030

0.970

1

−47°

0.535

0.465

1

2 a 0.6 3 a 0.954 4 a −0.917, 0.917 c ±0.971 5 a 2

b b

−5 13

b

8 a B 9 a 70° e 82°

b b b d

c

7 4

c c

5

bD b 32° f 46°

b

−2

c

24 25

0

d

3 7 7

d

c C c 51° g 1°

−1

d A d 8° h 73°

2

e

cos θ

0.6

0.96

d 2

0.25 −2.1 −2.1

g

1 5

j

2 5

Maths Quest 11 Mathematical Methods CAS

e

=

5 5

=

2 5 5

f

0.3 0.638 0.447 0.819

h k

c 11 6

4 5 −3 5

ii 2

i 4π

ii 3

i π

ii 4

f

5 6

i

5 6

l

=

3— π 4

i

π – 3

5 5

ii

1 2

i 4π ii

2 3

2—π x 3

y

0

− 2–3

g

π

2π

3π 4π x

i 6π

y 5 0

11 5

2π 3

2– 3

0.7

1 5

x

y

0

0.28 0.954 0.77 0.894 0.573

2

π

–π 2

–π 4

1– 2

0.8

4 3

i 2π

4π x

2π

y 4

−4

sin θ

b

ii 1

−3

10

4 5

i π

x

y

− 1–2

11 a

ii

3

0

=

3— π 2π 2

–π 2

−5

3π

6π

x

3 0.5 3 4

ii 1

x

y

0

2 3 3 7

π

–π 2

−1

4

ii 3

2

1.333 3.180 −0.714, 0.714 ±0.436

5 3

−4

y

1

ii 2

f i

1

cos2 θ

7 a

ii

ii

π 2 2π h i 5 π j i 3 l i 4

1 2

e i 6π

3 a

sin2 θ

−0.63 −0.25

g i π

3 2

d

k −1 b e b e b d

e i 2π

− 3 3 − 3 2

l −1

k 0

2

Sine and cosine graphs ii 2 ii b i π 3π ii 1.5 d i ii 2 4π ii 2 f i ii 3 2π ii 2.5 h i ii 3 ii 4 ii 1 b i 2π ii 2π ii 2 d i ii 3

c i 3π

tan θ 1.333 0.292 3.180 1.207

Symmetry b d b d b d

3

exercise 6F

0.3 0.7 −0.9

−

c

− 3 2 − 2 1 2 −

−0.7

θ

1 2

π 9

− 2 2

0.924 0.966 −0.966 0.644 −0.842

1

h 1

π 3 3π 2 − 7π 3 40° −45°

1 2 −

f

exercise 6e

i −1

3 3

6 a

b

6 a 6

h

exercise 6D

1 a c 2 a c 3 a c

g

j Undefined

π 6 5π 18 5π 6 36° 648°

1 2

c

f 1

exercise 6C

1 a

2 2

0.9 0.3 0.7 0.9

e

exercise 6B

b d f h

−

ii 5

1 4

i 4π

y

h

4 0

ii 4

f

y

π 2π 3π 4 π x π x

0

π i 2

y 2

π – 2

ii 2

–π 8

−2

j

–π 2 3— π 8

–π 4

−3

10 a f : 0,

−4

8 a i π 0

ii 1

iii 2, 0

y

x

1

i π

3

2π x

π

b i 2π

d f : [− 1, 3] → R, f ( x ) = 1.8cos

iii 0, −4

ii 2

4 a D 5E

bC

2x 3

d y = 4 cos

3x e y = − sin

f y = −3 cos ( 3x)

c i 4π

y

− 7—π − 5—π 4

4

−2π

− 3—π 2

− π–

−π

2

π– 4

0

π 3— 4

0 −1

π– 2

−1

π 7— 4

π 5— 4

π

π 3— 2

2π

x

d i 6π

2π

iii 0, −2

ii 1

y

y

π

2π

3π

4π

−1

3

−2 −4 π

0

4π

x

c

9 a

π 2— 3

0 − 2– 3

b

f i 4π g i 2π

ii ii

h i

ii

π 2 i i π

ii

π 2

ii

2π x

π

1 0 π π – –

3 3 − —— 4

8 4

π – 2

3— π 4

π – 3

π – 2

π

x

y 1.8

π

2π

0

−6

x

6

x

b

y

−1.8

c

y

y 1.4

1 0 π π — –

1.5 −6π

ii

2 a y

y

−π

2π x

π 4— 3

−3

e

e i π

2

y 3 0

x

ii

− 3–

−2

d

6π

ii

ii

j i

y 2 0

5π

x

π 4 π x=± 6 x = ±π 3π ± 2 π x=± 2 x = ±2π x = ±π π x=± 4 π x=± 2 π x=± 4

ii x = ±

d i 3π

−3

−3

tangent graphs

π 1 a i 2 π b i 3 c i 2π

π

0

b

exercise 6G

iii 5, 1

5 4 3 2 1

1 − 3—π − π– 4 4

ii 2

y

2x 3

2

g f : [0, 2π] → R, f (x) = −sin (x) − 1 h f : [−π, 2π] → R, f (x) = 2 sin (4x) – 1

−5

b y = 2 cos (2x)

x 2

−1 3π x − f f : [ 3 ,1] → R, f ( x ) = 2.4 cos 2

−3 −4

c y = 5 sin

7 a

−1 −2

c A

6 a y = 1.5 sin

2 x

0

−2

πx 2

3

x

−3

2x 5

2π x e f : [0, 3] → R, f ( x ) = − 3sin

y –π π 2

5 cos 2

c f : [−1, 1] → R, f (x) = 2 sin (π x)

0 −1

ii 3

3x 8π → R, f ( x ) = 3sin 2 6

b f : [0, 5π ] → R, f ( x ) =

2

y

0

8 x

0

−4

i

y 3

d

4

−1.5

12 6

0

6π x

0

2

4

6

π 2— 3

x

x

−1.4

Chapter 6 • Circular functions

303

c

y

y

j

x = π–

2 1 1 0 π π –

2π

2

d

0 −1 −2 −3 −4 −5 −6

3π 4π x

y

1 0 3π 3π 3π 9π — — — 4

e

2

2

y

π – 4

3— π 4

π – 2

b

0

π 3—π

2π

2

1 a

2 a b c d e f g h 3 a

3

4 2

π x

d

3

0 –π π–

equations

4

3 B 4 a D b B 5 a y

6π x

exercise 6h

x = 3—π

4

–π π 2

x

2π

c

x

d e

f

b

y

f

y

4 a 5 a

2

b

1

0 π 2π 4π 6π 8π x

0 3 –

3

3– 2

4

x

c d

g

e y

0

c

f

2

π – 2

3π 4π x

π 2π

0

−1

h

y

−2 3

y

d

3— π 3— π 2π 4 2

x

y

1.8 0 ππ – – 8 4

3— π 4

π – 2

π

x

0 5 5 – –

5– 2

8 4

15 — 4

5

x

−2

i

y

x = –π 2

304

1– 4 –π 2

π

3—π 2

2π x

−1 − 1– 0 2 −2

Maths Quest 11 Mathematical Methods CAS

π π π b c 4 6 3 π π π e f 6 4 3 36.9° and 143.1° 104.5° and 255.5° 80.1° and 260.1° 238.2° and 301.8° 78.8° and 281.2° 140.1° and 320.1° 199.5° and 340.5° 40.8° and 319.2° 0.927, 2.214, 7.210, 8.497 2π 4π 8π 10π , , , 3 3 3 3 0.983, 4.124, 7.266, 10.408 5π 11π 17π 23π , , , 6 6 6 6 3.542, 5.882, 9.826, 12.165 π 7π 9π 15π , , , 4 4 4 4 C b A 0.297, 2.844 π 5π , 3 3 π 4π , 3 3 5.820, 3.604 2.328, 3.955 5π 11π , 6 6

6 D 7 a 14.5°, 165.5° b 131.8°, 228.2° c 74.1°, 254.1° d 270° e 78°, 282° f 108.4°, 288.4° 8 a 60° b 150°, 330° c 0°, 180°, 360° d 90°, 210° e 45°, 135° f 0°, 90°, 180°, 270°, 360° g 120° h 90°, 270° i 15°, 105°, 195°, 285° j 45°, 75°, 165°, 195°, 285°, 315° k 60°, 150°, 240°, 330° l 113.74°, 326.26°

π , n ∈Z 6 − − 11π π π 11π For −2π ≤ x ≤ 2π : x = , , , 6 6 6 6 π b General solution: x = n π + , n ∈ Z 6 − 11π − 5π π 7π For −2π ≤ x ≤ 2π : x = , , , 6 6 6 6

9 a General solution: x = 2nπ −

y

2

3 2 1 0 −1 −2 −3

e

x = 3—π

Solving trigonometric

1– 2

1 x

c General solution: x = (2n + 1)π −

π or 4

3π π and 2n π + , n ∈ Z 4 4 − 7π − 5π π 3π For −2π ≤ x ≤ 2π : x = , , , 4 4 4 4 10 General solution: (12n + 5)π (12n + 1)π x= and , n ∈Z 12 12 − − 11π 7π π 5π For −π ≤ x ≤ π : x = , , , 12 12 12 12 (6n ± 1)π 11 General solution: x = , n ∈Z 9 −π ≤ x ≤ π : For − 7π − 5π − π π 5π 7π x= , , , , , 9 9 9 9 9 9 x = 2n π +

exercise 6i

applications

1 a i 1 kg ii 6 days b 2 a b c 3 a b c 4 a b

5 a

πt W = cos + 3 3 110 beats/min i 50 ii 60 min πt H = 50sin + 110 30 1.6 m i 1m ii 0.7 m 3.254 s 26 °C at 2 pm i 18 °C ii 22 °C iii Approx. 11.1 °C i 12 mm

c t = 6.582; t = 6.987 d (1.206, 0.931) and (5.266, 0.971) e $0.98 when t = 8

Chapter reVieW Short anSWer

1 2251.75 m 2 a 3 a 4 a 5 a b c 6 a b c 7 a b

−

b

3

7 4

b

7π 9 72° 57.3° 630° 0.724 0.690 1.049 0.436 − 0.484

b

−

3 2

c

1 2

c

π 180

7 3

31π 18

13 a

14 a c

c e 16 a b c

π 2π 3π 4π x

3 2 d −1

3

s

80 74

−π − –π 0 2

0

π x

–π 2

40 m 3.9 s 7.8 s 60 s 50 m 100 m 314.16 m i 50 m ii 75 m f 25 s and 35 s 8 a i 6 m ii 3 m b Yes, by approx. 24 minutes 6 a b c 7 a b c d e

1

t

2

MUltiple ChoiCe

1 6 11 16 21

−1.5

9 a

y 2

C E C C B

2 7 12 17 22

D B B E C

3 8 13 18 23

A D D A E

4 9 14 19

E A B C

5 10 15 20

E E D D

extenDeD reSponSe

1 S 120

1 a 25.4 °C, 20.6 °C b i 23 °C ii 25.4 °C c π − 7— 8

−π

π − 5— 8 π − 3— 4

π − 3— 8

− –2π

− π–8

0 − –4π

–π 8

3— π 8 –π 4

5—π 8 –π 2

7— π 8 π 3— 4

T (°C) π

x

27 26 25 24

25.4

23 −2

b y

22 21 20 19

20.6

18 (8, 0.98)

0

0.96 0.93 0

0.90 0

−

86

positive to the right, then the string is 11.41 mm to the left (or vice versa).

0.99

d 0.92

L (t)

1.5

1 S 10

9 a a ($A)

b

e

y

b

1 2 −

1 6

d

−4

b −0.40

π 2π 3π 5π b , , 3 3 4 4 π 7π d 0°, 360° , 6 6 135°, 315° Amplitude = 6, period = 2 Minimum = 74, maximum = 86 1 s 2

15 a

−2π −π 0

23 10 −

0.92

c

y 4

8 a

ii b 10 c −11.41 mm; if the displacement is

d

π 3π 9π 11π , , , 4 4 4 4 11 150° and 210° π 12 General solution: x = 2n π + and 3 π x = (2n + 1) π − , n ∈ Z 3 − 5π − 4π π 2π , , , For 2π ≤ x ≤ 2π : x = 3 3 3 3 10

1 2 3 4 5 6 7 8 t (days)

b t = 4.253 (3 d.p.), ∴ on the 5th day

3π 6π

9π 12π x

1

2

3

5 4 t (hours)

6

7

8 t

d 2 hours, 40 minutes 2 a 40 cm

π c 0.3sin x 16

b 80 cm

Chapter 6 • Circular functions

305

3 a i 2800 ii 1200 b i 12 months ii 0.8 c P (thousands)

iii 2000

306

5 a 2.71 m b 11 hours c i 3 m at 9 pm ii 2 m at 3:30 pm d h(t) (m)

2.0 0

ii 8.1 cm ii 8.1 cm

3.0 (0, 2.71) 2.5

8.1

3 6 9 12 t (months)

f Approx. 5 months 4 a i 19.9 cm b i 11.742 cm

g H = 14 + 5.9cos

14

d 1600 e 4 months 1 2

8π t 5

H 19.9

2.8 2 1.2 0

c

d See c. e 0.571 s f 52.5 times

Maths Quest 11 Mathematical Methods CAS

1

2

t

1.5 0

(9, 3) (12, 2.25)

(3.5, 2) 3

6

e 1:40 pm and 5:20 pm f 3 h 40 min

9

12 t (hours)

Exam practice 2 CHAPTERS 1–6 1 Determine the centre and radius of the circle described by x2 + 6x + y2 + 8y – 24 = 0.

4 marks

2 a Sketch the function f defined by:

20 minutes

2x + 2 f (x) = x + 2 − 2x + 8

−

3< x <0 0≤x<2 2<x ≤5

4 marks

b State the range of f.

1 mark

3 Solve for x in each of the following. a 23x = 5 b log2(x + 1) − log2(3x) = 2

2 marks 3 marks

M U ltip l e C ho iC e

1 For the function f (x) = (x − 2)2 + 3, what are the coordinates of the turning points for f (x − 1)? a (–3, 3)

B (–2, 3)

C (1, 3)

D (2, 3)

e (3, 3)

1] → R, where f (x) = 6 − 5x − is: C [6, ∞) D [–6, 1] 3 Which one of the following represents a many-to-one relationship? 2 The range for the function f: a [–5, 1]

[–5,

B

y

y

y

x

x

e

each question is worth one mark.

e [0, 12.25]

C

y

x

D

12 minutes

x2,

B [0, 6]

a

S ho rt a n S W er

y

x

x

4 62x × 3x × 2x + 1 simplifies to: 2 a 2 2 x + 2 x × 32 x

B 364x + 1

D 23x + 1 × 33x

2 e 62 x +3 x

3 2 C 62 x + 2 x

π x + 1 = 0 between the interval [−π, 2π] are: 3 −π 7π 5 7π 7 11 a B C and and and 6 6 2 2 2 2 − 1 7 11 11π D e and and 2 2 2 2 6 A possible equation for the rule of the function whose graph is shown below is: y a y = −2 sin (x) B y = −2 sin (3x) π x C y = − 2 sin D y = −2 cos (x) 3 x 5 The two x-values that satisfy the equation 2 sin

πx e y = − 2 cos

−3

0

3

3

Exam practice 2

307

7 Consider the function f (x) = 3 − −

2 . The equations of the asymptotes would be: (2 x + 1)

1 f (x) = 3 2 B x = −1 f (x) = −3 C x=0 f (x) = 0 1 D x= f (x) = 3 2 e x=2 f (x) = 1 8 For the function g(x) = 4 − 2 − x , the implied domain and range would be respectively: a (−∞, 2) and (4, ∞) B R /{2} and R + C [2, ∞) and [4, ∞) − − D ( ∞, 2] and ( ∞, 4] e R ∪ {0} and R+ a x=

e x t enDeD r e S ponS e 35 minutes

DiGital DoC doc-10162 Solutions exam practice 2

308

1 Rock pools in a tropical lagoon form when the depth of water in the lagoon falls below 6 metres. The

π (t − 3) depth of water in a lagoon can be modelled using d (t ) = 1.5 sin + 5.5, where d is the depth 6 in metres and t is the time in hours after 7 am in the morning. a Determine the maximum and minimum water depth. 2 marks b Determine the time, in hours, taken for the water depth to increase from the minimum depth to the maximum depth. 1 mark c Sketch the graph of d(t) in the time interval 0 ≤ t ≤ 24. 3 marks d At what time will the depth of water first be 6 metres deep? 1 mark e Billy and Tommy swim only in the safety of the rock pools. They arrive at the lagoon at 9:30 am and leave at 4:15 pm. i Determine how long they will be able to swim in the rock pools. Write your answer correct to 2 decimal places. 1 mark ii Determine the total time Billy and Tommy would be able to swim in the rock pools. Write your answer to the nearest whole minute. 2 marks kt 2 Over the summer months, the number of mosquito larvae in a dam was modelled using M = M0 e , where M0 is the initial population, t is the time in days and k is a positive constant. The number of mosquito larvae was recorded each morning at 8 am. The first day of recording was taken as t = 0. The recording lasted only until the end of summer (90 days). a On the first day of recording, there were 150 mosquito larvae. Determine the value of M0. 1 mark b At 8 am on the fifth day there were 235 mosquito larvae. i Write an equation that determines the number of larvae on the fifth day. 1 mark ii Show that the value of k is 0.11. 2 marks c If the mosquito larvae continue to increase according to the model, determine the expected number of larvae present in the dam at the end of summer (90 days). 2 marks d Frogs were introduced to the dam during the twentieth day. Mosquito larvae are part of the frogs’ diet. Since the frogs were introduced, the number of mosquito larvae has decreased. The decrease in mosquito larvae can be modelled using D = 200 e(20 – t) + 300, where D is the decrease in mosquito larvae after the twentieth day. The recording of the mosquito larvae has remains the same as before. i State the domain of the function D. 1 mark ii Determine the number of mosquito larvae expected in the dam over the long term. 1 mark iii By how much had the mosquito larvae decreased in the first twenty-four hours of recording after the frogs were introduced? 1 mark

Maths Quest 11 Mathematical Methods CAS

ChApTer 7

Matrices diGiTAL doC doc-9769 10 Quick Questions

ChApTer ConTenTS 7A 7B 7C 7d

Addition and subtraction of matrices Multiplying matrices Solving matrix equations Matrices and transformations

7A Addition introduction

and subtraction of matrices

In all walks of life we deal with the mathematical phenomenon called the matrix. A train timetable, a football scorecard and examination results are all examples of matrices. Anytime there is a rectangular grouping of numbers, there is a matrix. A summary of cricket scores after three matches is shown in the table below. Overs

Maidens

Wickets

Runs

Smith

10

3

2

35

Brown

14

2

3

29

Nguyen

9

2

3

39

Removing the headings from the table leaves only an array of numbers: 10 3 2 35 14 2 3 29 9 2 3 39 This is called a matrix. The plural of matrix is matrices. The matrix above is a (3 × 4) matrix as it has 3 rows and 4 columns. This number of rows and columns gives the order, or the dimensions, of the matrix.

6 3 5 −4

is a (4 × 1) matrix since it has 4 rows and 1 column.

A matrix with m rows and n columns is called an (m × n) matrix. ChApTer 7 • Matrices

309

Each individual entry in a matrix is called an element. In general, the elements of a matrix are identified by the row and column that pinpoint their position in the matrix. Thus if A is a (3 × 2) matrix, we write: a11 a12 A = a21 a22 a31 a32

The elements of the matrix A are referred to as aij, where i refers to the position of the element in the row and j refers to the position of the element in the column. So, a21 refers to the element that is in row 2, column 1.

Addition and subtraction of matrices Imagine, for the three players in the cricket team, that a fourth match gave the following results: Total

Overs

Maidens

Wickets

Runs

Smith

3

1

0

10

Brown

10

1

2

34

Nguyen

15

2

4

54

Clearly we can add these results to the previous results, giving the total figures for four matches: Total

Overs

Maidens

Wickets

Runs

Smith

13

4

2

45

Brown

24

3

5

63

Nguyen

24

4

7

93

In matrix form we write this as: 10 3 2 35 3 1 0 10 13 4 2 45 14 2 3 29 + 10 1 2 34 = 24 3 5 63 9 2 3 39 15 2 4 54 24 4 7 93 Matrices are added by adding corresponding elements. It follows that: 13 4 2 45 3 1 0 10 24 3 5 63 − 10 1 2 34 24 4 7 93 15 2 4 54

10 3 2 35 14 2 3 29 = 9 2 3 39

Matrices are subtracted by subtracting corresponding elements. The matrices we have added or subtracted have the same order, i.e. the same number of rows (three) and the same number of columns (four). Addition and subtraction of matrices can be performed only if the matrices are of the same order.

The commutative and associative laws for addition of matrices The Commutative Law Simply stated, if this law holds, matrices can be added in reverse position. Consequently for matrix addition, A + B = B + A, provided A and B are of the same order. And clearly with subtraction, A − B = A + (−B) = (−B) + A, if A and B are the same order. The Commutative Law holds for matrix addition for matrices of the same order. 310 Maths Quest 11 Mathematical Methods CAS

The Associative Law Simply stated, if this law holds, any number of matrices (of the same order) can be added together, adding any pair of matrices first. Thus: A + B + C + D = (A + B) + C + D = (B + C) + A + D = (B + D) + A + C provided the matrices are the same order. Thus for subtraction, A − B − C = [A + (−B) + (−C)] = [A + (−B)] + (−C) or A + [(−B) + (−C)] and so on. The Associative Law holds for matrix addition for matrices of the same order. Worked exAmpLe 1

Show that the Commutative Law holds for matrices A and B given below. 1 A= 0 2

−

2 1 0

1 0 4 , B = 5 3 0 0 3

−5

1 2

−4

Think 1

2

3

4

TUToriAL eles-1426 Worked example 1

WriTe

For the Commutative Law to apply, A + B = B + A. Both matrices have the same order so it is possible to add them. Find A + B by adding the corresponding elements in each matrix.

Find B + A by adding the corresponding elements in each matrix.

Write the answer.

Both matrices are (3 × 3) so it is possible for addition to occur. 1 A+ B = 0 2

2 −1 0

1 0 4 + 5 3 3 0 0

−5

2 2 = 5 2 2 0

−9

1 0 B+ A= 5 3 0 0

−4

2 2 = 5 2 2 0

−9

1 2

−4

5 5 1 1 + 0 2 2

2 −1

0

4 3

−5

5 5

The Commutative Law holds for matrices A and B as A + B = B + A.

Worked exAmpLe 2

1 A= 0 2

3 1 − 4

0 5 5 4 − , B = 1 0 and C = 7 0 2 3 3 1

4 2

−2

Find, if possible: A+B A−B A−C the answers to parts a, b and c using a calculator.

a b c d

ChApTer 7 • Matrices

311

Think

WriTe

1 3 0 5 a A and B have the same order, so it is a A+ B = 0 1 + −1 0 possible to add them. Add the corresponding 2 −4 2 3 elements of each matrix. 1 8 = −1 1 4 −1 1

b A and B have the same order, so it is possible b A − B = 0

to subtract them. Subtract the corresponding elements of each matrix.

c Subtraction cannot be performed since the

order of A is (3 × 2) and the order of C is (3 × 3).

d 1 Define the matrices using the CAS

calculator.

2

1 = 1 0

3 1 − 4

0 5 − − 1 0 2 3

1 −7 −2

c A – C cannot be calculated because A and C are of

different order. 1

d 0

2

3 1 −4

→a

0 5 − 1 0 →b 2 3 5 4 7 0 3 1

2

Use the CAS calculator to perform the matrix addition A + B.

3

Use the CAS calculator to perform the matrix subtraction A − B.

4

Use the CAS calculator to perform the matrix subtraction A − C.

4 →c 2

−2

1 a + b = −1 4 1 a−b= 1 0

8 1 −1

1 − 7 −2

ERROR (The two matrices are of different orders.)

Special types of matrices In life it is convenient to associate shapes with names. Thus we all know what a round shape looks like, what a pear-shaped person looks like and so on. A similar principle can be applied to describing the shapes of matrices. 312

Maths Quest 11 Mathematical Methods CAS

Type Column matrix

Shape Consists of a single column

Row matrix

There can be any number (1 × 2) of columns, but only one (1 × 10) row. (1 × 56) (1 × n) Consists of an Any number of rows. (1 × 1) equal number of Must have the same (2 × 2) rows and columns number of columns. (6 × 6) (20 × 20) A diagonal Must be a square matrix (1 × 1) matrix is a square (3 × 3) matrix with every (6 × 6) element, not on the (25 × 25) leading diagonal, equal to 0. A unit matrix Must be a square matrix (1 × 1) is a diagonal (2 × 2) matrix with all the (6 × 6) elements on the (20 × 20) leading diagonal equal to 1. Any A zero matrix can Any shape be any shape, but order all the elements are 0. Must Any shape If matrices have are equal, then both same their shape and all corresponding elements order must be identical.

Square matrix

Diagonal matrix

Unit matrix

Zero matrix

Equal matrices

Description Order There can be any number (2 × 1) of rows, but only one (5 × 1) column. (27 × 1) (m × 1)

Consists of a single row

Example

Order of example (4 × 1)

−1 3 2 0 [5

2 4

−1

3

0

(1 × 5)

4]

(2 × 2)

0

−5

2 0 0 0 0 0

0 0 −3

(3 × 3)

(3 × 3)

0 0 0 0 0 0

(2 × 3)

x y −1 3 2 p 3q = 4 − 9 1 4 1 4 x = –1, y = 3, p = 2, q = –3

Both (3 × 2)

1 0 0 0 1 0 0 0 1

Worked exAmpLe 3

2x + y Solve for x and y in the following equal matrices. 3 x − 2 y Think

2

Corresponding elements are equal, so write these as equations. Solve the simultaneous equations.

3

Write the answer.

1

4 = 13

WriTe

TUToriAL eles-1427 Worked example 3

2x + y = 4 3x – 2y = 13 2x + y = 4 [1] 3x – 2y = 13 [2] Multiply equation [1] by 2. 4x + 2y = 8 [3] Add equations [2] and [3]. 7x = 21 x=3 Substitute x = 3 into [1]. y = −2 x = 3 and y = −2

ChApTer 7 • Matrices

313

multiplication of a matrix by a scalar As in ordinary arithmetic, multiplication is simply a shorthand method of addition. Rather than 4 + 4 + 4 + 4 + 4 = 20, we say 5 × 4 = 20. Similarly, this method also applies to matrices. 2 1

2 + 0 1

−5

4 = 0 2

0

− 10

−5

is the same as: 2 2× 1

4 = 0 2

−5

0

− 10

Thus if A = [a b c], then 3A = [3a 3b 3c]. A scalar is a single (real) number. When a matrix is multiplied by a scalar, each element of the matrix is multiplied by the scalar.

Worked exAmpLe 4

If A = a 3A

1 and B = 7 b 4B c 3A + 4B

3

−2

4 5

−

0 2

, find : d 2(A + B)

Think

a Multiply each element of A by 3.

e 5(B − A) using a calculator.

WriTe

3

−2

a 3A = 3

4 5

9 12 = − 6 15 b Multiply each element of B by 4.

c Add the two matrices found in parts a and b.

1 7

0 −2

4 = 28

0 −8

b 4B = 4

9 12 4 + − 6 15 28

c 3A + 4B =

13 12 = 22 7 d 1 Find A + B by adding the

corresponding elements of A and B.

3 4 1 + − 2 5 7

b A+ B =

4 4 = 5 3 2

Multiply each element of the resulting matrix by 2.

4 4 2( A + B) = 2 5 3 8 8 = 10 6

314

Maths Quest 11 Mathematical Methods CAS

0 −2

0 −8

e 1 Define the matrices using your CAS

e

calculator.

3 −2

1 7 2

Perform the matrix arithmetic 5(b − a) and write the result.

4 5

→a

0 →b

−2

− 10 5(b − a) = 45

− 35

− 20

If A and B are matrices of the same order and a and b are real numbers, then: 1. aA + bA = (a + b)A 2. aA + aB = a(A + B) 3. (ab)A = a(bA). If aA = 0, then a = 0, or A is a zero matrix.

Addition and subtraction of matrices

exercise 7A

Show that the Commutative Law holds for matrices A and B given below.

1 We1

A=

3 1 6 and = B −4 2 2

1

−5

5 2 3 0 2 We2 , C = 1 − 1 2 1 4 find each of the following, if they exist. b B–A a A+B c B+D d A+C e C – D. 3 Find [2 −5 3 0 4] – [−3 4 1 6 −3]. 4 We3 Solve for x and y in the following equal matrices. 1 3 If A = 2 1

x + 2y 2 x − 3 y 5

=

5 −3

, B =

−

−

3 0 − 2

0 and D = 2 1

4 , −1

−3

5 − 11

State the order of each of the following matrices. 2 4 5 a 0 1 −2 b 4 −3 6 c [−2x + y]

d [1

6 Find the matrix P such that P – [−1

2

−5]

1 3 −4 0

4

5

3]

= [−3 4 2].

p 8 7 If 3 q = − 9 , find p, q and r. r 10 − 8 What does [ 5

2 4] − equal? 1 ChApTer 7 • Matrices

315

3[2 3 4] – 5[−1 A [11 −1 27] C [11 19 27] e [1 −1 27]

9 mC

−2

−3]

equals: B [0 d [1

19 19

−3]

27]

x

3 2 y = : − 1 2m 2 p 12

10 mC If 2

A x = 1, y = 3, p = −1, m = 3 C x = 1, y = 3, p = −1, m = 6 e x = 2, y = 6, p = −1, m = 12

11 We4

mC

B x = 1, y = 6, p = −1, m = 3 d x = 2, y = 3, p = −1, m = 3

If A = [0 4] and B = [−3 0], what is −4A − 3B?

−16]

A [9 C [9 −19] e [−13 −16]

B [5 d [5

−19] −16]

12 George, Linda and Mary sat two examinations in each of English, Mathematics and Science. In the first

examination they scored 40%, 50% and 60% respectively for English; 62%, 87% and 28% respectively for Mathematics; and 100%, 14% and 73% respectively for Science. a Write this information as a (3 × 3) matrix, labelling names on columns and subjects on rows. In the second examination their results were 50%, 65% and 80% respectively for English; 87%, 76% and 45% respectively for Mathematics; and 95%, 37% and 89% respectively for Science. b Write this information as a (3 × 3) matrix, labelling names on columns and subjects on rows. c Add the matrices together. d Hence calculate their average mark in each subject. e Write Mary’s average as a (1 × 3) matrix. 13 At the Queen Victoria market, 5 people each bought 3 apples and 2 oranges, 7 people each bought

1 apple and 6 oranges, and 8 people each bought 4 apples and 2 oranges. Write this information in matrix form, and use matrix addition to calculate the total number of apples and oranges bought by the group.

7B

multiplying matrices

Multiplying matrices is a procedure that takes place in two distinct stages: Stage 1: Identification of the correct conditions for multiplication to occur Stage 2: The multiplication itself. Stage 1: We need to establish whether multiplication is possible. This can best be thought of as two tiles, side by side, in the game of dominoes.

Here we can join the number three in each domino. A similar principle applies to the multiplication of matrices, if we think of each domino as row– column. Specifically, only those matrices that have the same number of columns in the first matrix as the number of rows in the second matrix can be multiplied. Further, if we can join the dominoes (matrices) as shown, that is by placing a three against a three, not only can we multiply the matrices, but the order of the resulting matrix is given by the two outside numbers, in this case a (2 × 4) matrix. In general we can multiply an (m × n) matrix by an (n × p) matrix, because the inside numbers are the same (both n). These inside numbers are called the inner product. The outer numbers (outer product) give the order of the resultant matrix (m × p). 316

Maths Quest 11 Mathematical Methods CAS

Stage 2: We know from our dominoes that we can multiply because the inner product is equal (3 × 3). The resulting answer matrix is of order (2 × 4). For example: 4 2 and B = − 1 7 3 0 2

2 5 A= − 1 2

−1

a A × B = AB = e

0 2 − 2

1 2 − 3

c d g h

b f

(2 × 3) × (3 × 4) = (2 × 4) In the answer matrix, the letter a marks the intersection of the first row and the first column. To calculate the value of a, we multiply the first row of the first matrix by the first column of the second matrix. (In matrix multiplication we always multiply rows of the first matrix by columns of the second matrix). −1

× –4 × −1 2 5 –1 × –0 a = 2 × 4 + 5 × −1 = 8 + −5 + 0 =3 2

5

−1

+

×0

To calculate the value of f, at the intersection of the second row and second column of the answer matrix, we multiply the second row of the first matrix by the second column of the second matrix. −1

2

3

f = −1 × 2 + f = −2 + 14 + 6 f = 18

2 × 7 2 2×7

+

3×2

This process is repeated for each element in the answer matrix, 3 37 so A × B = − 6 18 b11 b12 a11 a12 In general, if A = and B = b21 b22 a21 a22

12 −2

15 . −6

,

a11 × b11 + a12 × b21 a11 × b12 + a12 × b22 then A × B = a21 × b11 + a22 × b21 a21 × b12 + a22 × b22

.

Worked exAmpLe 5

4 1 −3 1 0 P= , Q = 2 5 7 3 2 a Write the order of the two matrices. c Does QP exist? Think

a 1 Matrix P has two rows and two columns. 2

Matrix Q has two rows and three columns.

b PQ is the product of a (2 × 2) and a

(2 × 3) matrix, so it exists.

TUToriAL eles-1428 Worked example 5

b Does PQ exist? d Calculate the product(s) that do exist. WriTe

a P is a (2 × 2) matrix.

Q is a (2 × 3) matrix. b PQ exists.

ChApTer 7 • Matrices

317

c QP is the product of a (2 × 3) and a

c QP does not exist.

d 1 Multiply the rows of matrix P

d PQ = 1

(2 × 2) matrix, so it does not exist.

0 4 1 3 2 2 5

with the columns of matrix Q.

7

−3

1× 4 + 0 × 2 1×1+ 0 × 5 1× −3 + 0 × 7 = − 3 × 4 + 2 × 2 3 × 1 + 2 × 5 3 × 3 + 2 × 7 2

4 1 = 16 13

Simplify PQ.

5

−3

Note: In worked example 5, PQ exists but QP does not exist, indicating that matrix multiplication is not commutative.

exercise 7B

multiplying matrices

1 3 A= 2 1

1 We5

, B =

5 −3

the following, if they exist. a AB b BA

diGiTAL doC doc-9770 WorkSHEET 7.1

Evaluate 2

3

1 3 ,Q = If P = 2 1

5 0 5 3 0 R= 1 4 − 1 4 1 0 −1 b R(P + S) c R(P + SQ).

a QR

−

0 5 and N = − − 0 3 4 a Show that MN ≠ NM. b Evaluate M2. 2 3 0 M+ c Find 2 N × 0 0 3 1 2

c BD

−

3 0 − 2

0 and D = 2 1

d AC

−1

0 and S = − 1

.

0 2

.

5 a Use multiplication to simplify the matrix equation below.

1 5 x 11 − = 3 4 y 5 b Write the answer to 5a in a form that does not contain matrices. c Solve for x and y. 6 A is a (2 × 2) square matrix. 2 −1 0 3 4 0 . 0 4

a Find A2 − 3 A − b Find A3. 318

4 . Find each of −1

−3

e CB

1 3 − × 5 0 3 − . 2 0

2

4 M=

5 2 3 0 , C = 1 − 1 2 1 4

−

Maths Quest 11 Mathematical Methods CAS

4 , evaluate:

−3

7 Write the simultaneous equations

2x + 5y = 11 + 2y = 12

−3x

as a matrix equation. −2 8 E = 2 3 4 and F = − 3 . −4 a Find EF. b Find FE. 1 9 If A = 3

c Does the Commutative Law hold?

1 0 4 2 and I = , find A + 5A − 3I. −2 0 1

3 4 3. , find P 1 2

10 a If P =

1

0 . 9 12

b Find the (2 × 2) matrix Q such that Q × P =

3 P= 1

11 mC

A

−5

3

−

3 9

2 P= 3

12 mC

1 9

−2

A

−3

0 and Q = −2

B

−3

3

−2

8 9

1 . PQ equals: 5

C

−5

3

3 − 11

d

d

−3

3

−

3 9

e

e

7 2 6 3

−3 −

5

−

8 9

. P2 equals: 0

−1

7 6

−

−1

B

3

−2 −

1 6

C

−2 −3

7 2 9 3

The order of the answer matrix when we multiply a matrix of order (2 × 3) by a matrix of order (1 × 2) is: A (1 × 2) B (1 × 3) C (2 × 1) d (3 × 2) e (3 × 1)

13 mC

14 The three matrices A, B and C have orders (a × b), (a × c) and (b × a) respectively. a For which pairs of matrices does the product exist? State the order of the answer(s). b Is it possible to multiply all three matrices together? If so, state the order of the answer. c For which pairs of matrices does the sum exist? State the order of the answer(s).

1 0 2 2 and I = , evaluate M and solve for a and b such that aM − bM − I = 0. 0 1 16 Fast-food chain McDonuts has outlets in the centre of the city. An overall view of sales is maintained to monitor demand. Hamburgers, drinks and chips were surveyed. In January, Store A sold 1200 hamburgers, 2367 drinks and 4219 serves of chips. a Write this as a (1 × 3) matrix. b The selling price of each product is $4.50, $1.95 and $2.45 respectively. Write this as a (3 × 1) matrix. c Multiply these matrices to calculate the total income from these three items. 17 Mayfleet, the international hotel chain, is building a new hotel in Arcadia Waters South. The guest accommodation comprises 80 standard rooms, 45 suites and 3 presidential suites. Each standard room has 1 king-size bed, 1 table, 2 chairs and 1 TV. Each suite has 1 king-size bed, 2 single beds, 2 tables, 5 chairs and 3 TVs. The presidential suites have 2 king-size beds, 6 single beds, 7 tables, 18 chairs and 6 TVs. a Set this information out in the form of two matrices. b Multiply these matrices to calculate the total number of pieces of furniture that Mayfleet must purchase to fit out the guest rooms.

15 If M =

3

−2

5 0

ChApTer 7 • Matrices

319

7C Solving matrix equations The multiplicative identity matrix The unit matrix, previously defined as a significant square matrix, does not alter any other matrix by which it is multiplied. It is denoted by I and is known as the multiplicative identity matrix. 1 0 0 I = 0 1 0 is an example of a multiplicative identity matrix. 0 0 1 2 3 If A = 0 2

−1 −4

4 3

1 0 and I = , then: 0 1 1 0 2 3 IA = 0 1 0 2

2 3 Notice that 0 2

−1 −

4 4 3

2 3 = 0 2

4 . 4 3

−1 −

1 4 1 0 is not possible. − 4 3 0 1 −

Thus, it follows that AI = IA = A, if A is a square matrix.

The inverse matrix When any square matrix is multiplied by its inverse we obtain the identity, I. The inverse of matrix A is written as A 1. By definition: −

A × A 1 = A 1 × A = I. −

−

1 0 Thus, if A is a (2 × 2) matrix, then A × A 1 = . 0 1 Similarly, if A is a (3 × 3) matrix, then: −

A× A

−1

1 0 0 = 0 1 0 0 0 1

.

Calculation of the inverse of a (2 × 2) matrix a b −1 Only square matrices have inverses. If A = , then its inverse A equals c d 1 d −b . The number (ad – bc) is called the determinant of the matrix A and is ad − bc − c a written as det (A) or | A |. The determinant may be positive, negative or zero. Notice that if det (A) = 0, the inverse does not exist, as 1 is undefined. In this case, matrix A is called a 0 singular matrix. Worked exAmpLe 6

If F =

1 −2

5 3

, find F 1 . −

Think

320

WriTe

1

Find det (F ).

det (F ) = (1 × 3) − (−2 × 5) = 13

2

Swap the elements in the main diagonal.

3 1

Maths Quest 11 Mathematical Methods CAS

3

Multiply the elements on the other diagonal by −1.

2

4

Write the inverse of F.

F

5

Alternatively, using your CAS calculator, define the matrix f.

1 5 − → f 2 3

6

Use the inverse feature of the CAS calculator to − determine f 1. Note that this is the same result as in step 4.

−1

− f 1

−5

=

1 3 13 2

=

3 13

−5 13

2 13

1 13

1

−5

Use of matrices to solve simultaneous linear equations Matrices may be used to solve simultaneous equations of the form: 2x + 5y = 11 −3x + 2y = 12 These simultaneous equations may be expressed as the matrix equation: 2 5 x 11 − = 3 2 y 12 2 5 x 11 If A = − , X = y and B = , 3 2 12 then AX = B, − − A 1AX = A 1B − IX = A 1B − X = A 1B det (A) = 2 × 2 − (−5 × 3) = 19 1 2 −5 − A 1= 19 3 2 x 1 2 y = 19 3 =

−

5 11 2 12

1 − 38 19 57

−2 = 3 x = –2, y = 3 Worked exAmpLe 7

Solve the simultaneous equations using matrix methods. 3x – y = −7 −2x + 5y = 9 Think 1

Write the simultaneous equations as a matrix equation.

WriTe

3 − 2

−

1 5

TUToriAL eles-1429 Worked example 7

x −7 y = 9 ChApTer 7 • Matrices

321

2

3

Calculate the inverse of the coefficient matrix.

Multiply both sides of the equation by this inverse matrix.

det

3 − 2

= 15 − 2 = 13 5

−1

The inverse is

1 5 1 . 13 2 3

1 5 1 13 2 3

3 − 2

1 x 1 5 1 = 5 y 13 2 3

−

−

7 9

x 1 − 26 I = y 13 13 4

Simplify.

5

Write the answers.

x −2 y = 1 x = −2 and y = 1.

Worked exAmpLe 8

Solve the simultaneous equations using matrix methods. Interpret your findings. 2x − 4y = 10 3x − 6y = 5 Think 1

WriTe

2 3

Write the simultaneous equations as a matrix equation.

2

Calculate the inverse of the coefficient matrix.

3

Interpret the result.

−4 −6

x 10 y = 5

2 −4 − det = 12 + 12 = 0 −6 3 The inverse does not exist as this is a singular matrix. There is no point of intersection between the lines. It follows that the lines are parallel.

Worked exAmpLe 9

Solve the following in terms of r.

rx – y = 1 x – ry = 2

Think

TUToriAL eles-1430 Worked example 9

WriTe

1

Write the simultaneous equations as a matrix equation.

r 1

x 1 y = −r 2

2

Calculate the inverse of the coefficient matrix.

r det 1

−1

−1 −r

= (r × − r ) − (1 × −1) = − r 2 + 1

The inverse is

322

Maths Quest 11 Mathematical Methods CAS

1 −r2

+ 1

−r −1

1 r

.

3

4

Multiply both sides of the equation by this inverse matrix.

x 1 y = − 2 r + 1

Write the answers.

Thus x= or x=

−r −1

1 1 r 2

−r

− +2 1 + 2r ,y= − 2 r +1 r +1

− 2

r−2 1 − 2r . ,y= 1− r2 1− r2

This method can be used to solve a range of problems efficiently. Worked exAmpLe 10

A cubic equation is of the form y = ax3 + bx2 + cx + d. The points (1, 1), (0, 3), (−1, 3), (2, −1) lie on the curve. Use matrices to find a, b, c and d. Think

WriTe

1

Substitute the coordinates given to form equations involving a, b, c and d.

y = ax3 + bx2 + cx + d 1 = a(1)3 + b(1)2 + c(1) + d 3 = a(0)3 + b(0)2 + c(0) + d 3 = a(−1)3 + b(−1)2 + c(−1) + d −1 = a(2)3 + b(2)2 + c(2) + d

2

Simplify.

1=a+b+c+d 3=0+0+0+d 3 = −a + b − c + d −1 = 8a + 4b + 2c + d

3

Set up a matrix equation, AX = B.

1 1 0 0 −1 1 8 4

1 0 −1 2

1 1 1 1

4

Using your CAS calculator, define matrix a.

1 1 0 0 −1 1 8 4

1 0 −1 2

1 1 1 1

→a

5

Using your CAS calculator, define matrix b.

6

Using the CAS calculator’s matrix multiplication − and matrix inverse features, calculate X = A 1B.

7

Write the answer.

1 3 3 −1

=

1 3 3 −1

→b

−1 A B= 1

a b c d

− 1 −4 3 3 1 3

a = 3 , b = − 1, c =

−4 3

,d =3

ChApTer 7 • Matrices

323

Solving matrix equations

exercise 7C

Find the determinant of each of the following.

1

a 1

2 3 4

diGiTAL doC doc-9771 WorkSHEET 7.2

3 − 6 5

c

x 2 3 y

f

b

d 0

0 2 4

−2

e

1 −4

3 −3

x2 x

−x −

6

x−2 9 − and det (A) = 3. Evaluate x. x + 1 3 2 −5 3 We6 a Find the inverse of matrix A = − 1 3 2 A=

. 1 1 b Show that A × A = A × A = I, where I is the multiplicative identity matrix. 4 If it exists, find the inverse of each of the following matrices. − −5 0 a 2 b 1 1 c − 1 − 2 2 0 15 − − 3 8 d 3 e 6 − − 1 2 1 3 −

−

x +1 2 If A = , for what values of x is A singular? x − 1 4 6 We7, 8 Solve each of the pairs of simultaneous equations by a matrix method. a 2x – 5y = 6 b −x – 3y = 7 c 2x + 4y = 1 −5x + y = − 7 x+y=2 2x + 4y = 0

−

3 9

5

Solve for x and y in terms of p.

7 We9

d x – y = −1

2x – y = 6

2px – y = 3 3x + py = 2

8 Solve for x, y and z, given 3x – 2y + z = 1, −x – y – 2z = 13 and x + 3y – 2z = 9. 9 mC A 30 d 16

10 mC

A

d

1 For what value of p does |D| = 0, where D = p B 20 e None of these The inverse of 1 3 is: −1 4

1 −1 7 1

−

−1 − 1

7 1

3 4 −

−

2 ? 3

B 4

1 3 1

3 4

1 4 7 1

e

C 13

C

1 4 7 1

1

−3 −

1

−3

x

11 mC If −x + y = 4 and 4x + 6y = 5, then equals: y

A

1 10 −

d

324

1 10

−

6 4 6 − 4

− −

1 4 1 5 −

1 − 1

4 5

Maths Quest 11 Mathematical Methods CAS

B

1 10 1 10

−

e

−

1

−4

1 −4

1 4 6 5

−

−

1 4 5

−6

C

−1

10

−

6 4

1 4 1 5

−

1 4

12 If A =

−2

−

12

1 1

12 , find A3 and A−1. − 2 1 0

13 P = and Q = 1 0 . Show that P and Q do not commute when multiplied. 0 1

cos (θ )

14 Find the multiplicative inverse of the matrix

sin (θ )

(θ ) . cos (θ )

− sin

0 1 1 0 and J = − , find the condition for pI + qJ to have a multiplicative inverse. 1 0 0 1

15 If I =

2 −1 1 0 1 1 and N = − . If MX = XN, show that X = k . 0 2 1 2 0 0 The point of intersection of the lines 2x + 3y = a and x – 2y = 5 is (7, 1). Use a matrix method to find a. We10 The points (1, 3), (2, 5), (−1, −1) and (3, 15) lie on the curve y = ax3 + bx2 + cx + d. Establish the equation of the curve. Solve for x, y and z in terms of r. rx + y + z = 1, 2x – ry – z = 3, 3x + 4y – rz = 5 Solve for a, b, c and d. a+b+c+d=0 a – 2b + 3c + d = 7 2a + b – 2c – d = −1 3a + 2b + c – 4d = 11 Show that the following equations have a unique solution only if b ≠ 1, −2. x+y+z=3 x + by + z = 3b −2x + 4y + bz = 2 Tercell, the clothing shop in Church Street, sells 15 suits, 45 shirts and 43 ties per week. It has another branch in High Street that sells 24 suits, 23 shirts and 37 ties per week. If the profit from each sale is $100 per suit, $18 per shirt and $14 per tie, use a matrix method to calculate the total weekly profit from the two shops. Slapstick, Brush and Company manufactures four main brands of paint, which they mix up using four basic ingredients — urg, blog, slug and nog. The four main brands of paint are Quix, Runns, Splash and Thix. Each litre of Quix requires 0.2 kg of urg, 0.5 kg of blog, 0.3 kg of slug and 1 kg of nog. Each litre of Runns requires 0.1 kg of urg, 0.4 kg of blog, 0.7 kg of slug and 0.8 kg of nog. Each litre of Splash requires 0.2 kg of urg, 0.7 kg of blog, 0.1 kg of slug and 1 kg of nog. Each litre of Thix requires 1.0 kg of urg, 0.5 kg of blog, 0.3 kg of slug and 0.0 kg of nog. Each week the company uses 2000 kg of urg, 1945 kg of blog, 1800 kg of slug and 3230 kg of nog. Use a matrix method to find out how many litres of each paint the company produces per week.

16 M = 17 18 19 20

21

22

23

7d

matrices and transformations

Matrices can be used to describe transformations (translation, reflection, rotation and dilation) of points or curves in the x–y plane. The point (x′, y′) is the image of the point (x, y) following a sequence of transformations.

Translation

inTerACTiViTY int-0266 matrices and transformations

A point P (x, y) is moved to the point P′ (x′, y′) by translating a units in the positive direction of the x-axis (right) and b units in the positive direction of the y-axis (up). A horizontal translation of a units and a vertical translation of b units can be represented by: x x' x a (x′, y′) = (x + a, y + b) or T = = y + y y' b

. ChApTer 7 • Matrices

325

Worked Example 11

− The point (2, −3) is translated by the matrix 4 . Find the new coordinates of the point. 2 Think

Write

1

x x' x a Use the formula T = = y + . y y' b

x' 2 y' = − + 3

2

Add the matrices to find the image point (x, y).

x' − 2 y' = − 1 (x′, y′) = (−2, −1)

2

−4

Other transformations by matrix multiplication It is useful to summarise the effects of various transformations as set out in the table below. Here we deal with (2 × 2) matrices that cause a change in the position of a point. These can be done on calculators or by hand, using the usual matrix multiplication techniques. Matrix

What it does

Calculation

Change

1 0 0 1

Nothing

x x' 1 0 x x T = = y = y y y' 0 1

0 1 1 0

Reflection in the line y=x

x x' 0 1 x y T = = y = y y' 1 0 x

0 − 1

−1

Reflection in the line y = −x

x x' 0 T = = − y y' 1

1 0

0 −1

Reflection in the x-axis

x x' 1 T = = y y' 0

−1 0 0 1

Reflection in the y-axis

x x' − 1 0 x T = y = = y y' 0 1

k 0 0 1

Dilation by factor k from the y-axis

x x' k 0 x kx T = = y = y y y' 0 1

(x′, y′) = (kx, y)

1 0 0 k

Dilation by factor k from the x-axis

x x' 1 0 x x T = = y = ky y y' 0 k

(x′, y′) = (x, ky)

Anticlockwise rotation of A degrees about the origin

x x' cos ( A) T = = y y' sin ( A)

0

cos ( A) sin ( A)

( A) cos ( A)

− sin

326 Maths Quest 11 Mathematical Methods CAS

x = 0 y

−1

0 x y = −1

(x′, y′) = (y, x)

− x

(x′, y′) = (−y, −x)

x

(x′, y′) = (x, −y)

y

(x′, y′) = (−x, y)

−y

−y

−x

( A) x cos ( A) y

− sin

(x′, y′) = (x, y)

x′ = x cos (A) − y sin (A) y′ = x sin (A) + y cos (A)

Worked exAmpLe 12

The point (−5, 1) is translated by the matrix y = x. Find its new coordinates.

and then reflected in the line 5

−3

Think 1

WriTe

Write the point (−5, 1) in matrix form.

TUToriAL eles-1431 Worked example 12

−5 (− 5, 1) = 1

x x' x a Use the formula T = = y + y y' b to translate the point.

x' − 5 − 3 + y' = 1 5

3

Add the matrices to find the image point (x′, y′).

x' − 8 y' = 6 (x′, y′) = (−8, 6)

4

0 1 The reflection matrix in y = x is from the 1 0 table.

x' 0 1 − 8 6 = y' = 1 0 6 − 8

2

x x' 0 1 x Use the formula T = = y y y' 1 0 to reflect the point. 5

The new point is (6, −8).

Write the answer.

Worked exAmpLe 13

Find the image of the point (3, 8) under 30° anticlockwise rotation about the origin. Think

WriTe

1

Write the point (3, 8) in matrix form.

2

Use the rotation formula x x' cos ( A) T = = y y' sin ( A) where A is 30°.

3

Evaluate using exact values.

4

Multiply the matrices.

5

Write the answer.

( A) x , cos ( A) y

− sin

3 (3, 8) = 8 x' cos (30 ) y' = sin (30 ) 3 2 = 1 2

−

1 2 3 2

3 3 −4 = 2 3 4 3+ 2

(30 ) 3 cos (30 ) 8

− sin

3 8

3 3 3 − 4, 4 3 + . As a The new point is 2 2 decimal approximation, the point is (−1.40, 8.43).

ChApTer 7 • Matrices

327

exercise 7d

matrices and transformations Find the new coordinates of the point P (3, −2) after the following transformations.

1 We11,12

− − 0 1 c 1 0 b 1 0 followed by 3 1 0 0 2 2 0 1 d Reflection in the y-axis e Translation of 6 units to the left and 4 units upwards, followed by reflection in the x-axis a

2

Find the new coordinates of Q (m, −n) after the following transformations.

0 1 followed by reflection in the line y = x 1 0 2n b Reflection in the line y = −x followed by a translation of − 3m 3 Two transformations as shown are applied to the point M (x, y). a Reflection by the matrix

3 0 0 1 x 0 1 1 0 y a Find the single (2 × 2) matrix that can be used to describe the combined transformations. b If M is the point (−1, 0), find its new coordinates after the transformations. 4 Write the matrices for the following rotations about O. a 90° clockwise b 180° clockwise d 360° clockwise e 90° anticlockwise g 120° clockwise h 150° anticlockwise j 120° anticlockwise

c 270° clockwise f 45° anticlockwise i 300° clockwise

5 mC Consider the point (−3, −5). a The image of this point after a reflection through the line y = x followed by a rotation of 180°

anticlockwise is: d (5, 3) A (−5, −3) B (−3, 5) C (−5, 3) b The image of the point (−3, −5) after a rotation of 270° clockwise followed by 4 a translation of − is: 2 A (1, −3)

B (9, −5)

c The image of the point

C (7, 1)

d (−1, 1)

e (3, 5)

e (6, 4)

(−3, −5)

after a reflection in the y-axis, followed by a reflection in the x-axis and then an anticlockwise rotation of 45° is: A (

−

2, − 2 2)

d ( −4 2,

2)

B (

−

2, −4 2)

− C (2 2, 2 2)

− 2, 4 2) e (

6 mC The point (x, y) is dilated by a factor of 3 from the x-axis, then reflected in the line y = x and then

translated 2 units to the left and 1 unit down. The coordinates of the final point are: B (3x + 2, y – 1) C (y – 2, 3x + 1) − − d ( 2 + 3y, 1 + x) e (y + 2, 3x + 1) A (3x – 2, y – 1)

7 a Combine a dilation of factor 2 from the x-axis and factor 3 from the y-axis to find the new

coordinates of the square (0, 0), (1, 0), (1, 1), (0, 1).

b Write down the single (2 × 2) matrix for the combined dilations. c Write down the single matrix that would represent dilations of factor n from both the x-axis and

the y-axis. 8 The point P (a, b) is subjected to a sequence of transformations as follows:

• a dilation of 2 away from the y-axis • a translation of 5 to the left and 3 downwards • a reflection in the x-axis • a reflection in the line y = x. Find the new coordinates of P after all of the above transformations have been completed in the given order, using matrix methods. 328

Maths Quest 11 Mathematical Methods CAS

9 WE13 Find the images of each of the following points under each of the anticlockwise rotations, i–v. a A (4, 0) i 30°

b B (5, 4) ii 45°

c C (−3, 6) iii 90°

d D (0, 9) iv 180°

v

−60°

10 Find the image of the point (7, −2) after a rotation of 270° anticlockwise followed by a translation

2 of . 4 11 The triangle ABC, with vertices A (2, −1), B (−4, 0) and C (5, 2), is rotated by a clockwise rotation of 45° (i.e. an anticlockwise rotation of −45°). a Find the new coordinates (A′, B′ and C′) of the rotated triangle. b Compare the area of triangle A′B′C′ to that of triangle ABC. 12 Consider the point A (a, b). a Is an anticlockwise rotation of 30° equivalent to a reflection in the line y = x followed by a clockwise rotation of 60°? Give reasons for your answer. b Is a reflection in the x-axis followed by an anticlockwise rotation of 45° equivalent to a transformation by the identity matrix? Give reasons for your answer. 13 The point P (3, 4) is transformed to become the point P′ ( − 2 3, 13). a Prove that this transformation represents an anticlockwise rotation of θ ° about the origin. b Calculate the value of θ in degrees.

Chapter 7 • Matrices 329

Summary Addition and subtraction of matrices

Matrix definitions: • A matrix is a rectangular array of numbers arranged in rows and columns. • An m × n matrix has m rows and n columns. • The individual entries in the matrix are called the elements of the matrix. Operations with matrices: • Addition and subtraction of matrices are performed by adding or subtracting corresponding elements. • Addition and subtraction can be performed only if the matrices are of the same order. • Scalar multiplication is performed by multiplying every element of the matrix by the scalar. x y ax ay a = p q ap aq • The Commutative Law holds: A + B = B + A • The Associative Law holds: A + B + C = (A + B) + C = A + (B + C )

multiplying matrices

• Matrices are multiplied in the following way. b11 b12 a11 a12 In general, if A = , and B = b21 b22 a21 a22 a11 × b11 + a12 × b21 a11 × b12 + a12 × b22 then A × B = . a21 × b11 + a22 × b21 a21 × b12 + a22 × b22 • The orders are (m × n) × (n × p) = (m × p). • Matrix multiplication is not usually commutative. That is, AB ≠ BA.

Solving matrix equations

• I is the multiplicative identity matrix, where AI = IA = A and A is a square matrix. • The multiplicative inverse of matrix A is A 1, and A × A 1 = A 1 × A = I. −

−

−

a b 1 d −b • If A = , then its inverse is A 1 = . ad − bc − c a c d The number (ad – bc) is called the determinant of the matrix A and is written as det (A) or |A|. If det (A) = 0, then A 1 does not exist and A is said to be singular. −

−

matrices and transformations

• Matrices can be used to represent transformations of points in the x–y plane. • A horizontal translation of a units and a vertical translation of b units can be represented by (x′, y′) = (x + a, y + b) x x' x a or T = = y + . y y' b • Pre-multiplying by a (2 × 2) matrix causes a change of the position of a point. For example, 0 1 pre-multiplying by causes a reflection in the line y = x. 1 0 x x' 0 1 x y T = = y = y y' 1 0 x • An anticlockwise rotation about the origin can be represented by x x' cos ( A) T = = y y' sin ( A)

330

Maths Quest 11 Mathematical Methods CAS

( A) x cos ( A) y

− sin

, where A is the angle of rotation.

Chapter review 1 State the order of:

a

1 0 8 4

b

0 − 1 2

2 Find the matrix D such that

1 2 4 0 4

2 5 0 −5 −1

S ho rT A n S W er

3 1 − D = 5 0 3

0 . 2

−2

−4

3 x = . Find x and y. − 1 y

3

4 A=

1 −2 − − and B = [1 0 4 1]. Evaluate: 0 3

2

5 P= , Q = [0 1

3] and R = [0 2

a PQ

−1].

a AB

Evaluate:

b QP

2 − 5

6 Find the inverse of

b BA

c PR

d P(Q + R)

. 6

−4

x y , show that AA−1 = I. 2 3 8 Solve for x, y, and z. 7 For A =

2x – 3y + z = 3 + y – 2z = 1 x–y–z=6

−3x

9 Find the new coordinates of the point R(−p, q) after reflection in the x-axis followed by a translation

of a . b 10 The point (3, 2) is dilated from the y-axis. Its new coordinates are (6, 2). Write down the (2 × 2) dilation matrix. 11 The point A (1, 2) is rotated clockwise about the origin through an angle of 60°. Find the new

coordinates of A in exact form. 2 1 0 1 − 12 M = . Find MN(NM) 1. , N = −1 4 3 5 1 4 0 13 P = − 2 0 , Q = − 2 −1 3

−

0 4 . Find (QP)−1. 5 1

1 3 1 3 0 , C = − . , B = −2 0 2 4 1 a Find |A|. b Find the inverse of C if it exists.

14 A =

c Calculate BC. ChApTer 7 • Matrices

331

3 2

0 , find P2 – 2P. Hence, express the inverse of P in terms of P.

15 If P =

−1

3 0

m U LT ip L e C h oiCe

5 c , N = . M and N commute under multiplication. Evaluate c. 0 3

5 −3

16 M =

−5 − 2 − 3 0 2

1

−1

equals:

x

−4

z A 2, 3, 0

3 2[−5 A

3

−6]

[−4

= 6 0

−y

2 If 3

– 3 [−2

−9]

2 − 1

4 If A =

d

5 −2 −6

e

−1 2

−4

9 , x, y and z are respectively equal to:

−4

B 2, −3, 0

−1]

equals:

B [−16

−15]

C 6, −9, −4

d 6, −9, 0

e 2, 3, −4

C [−4

d [−16

e [4

−15]

−9]

9]

and B = [3 0], AB equals:

6 0

0 0

B 4 − 3

A

4 C 5 −2

4 B −1 − 2

−6 A 5 2

C [6]

d [5]

e

6 0 − 3 0

1 2 2 , M – 2M equals: 4 0

5 If M =

−

7

A

− 12

−

d 3 − 4

8

−2

0

−6

8

B 11 − 12

−6

7 − 12

−6

e

C 7 4

8

−6

8

1

3 A (1 × 2)

B (2 × 3)

C (3 × 2)

d (2 × 1)

e (3 × 1)

B −7

C −5

d −6

e −4

C 4, −1

d −4, 1

e 1, −2

1 6 P = 2 , Q = [1 2] and R = 3

7

−1

−2

−3

1

A 7

2 4

. The order of PQR is:

equals:

x +1

2 is singular if x equals: −3 x − 4 B 1, 2 A −1, −2

8

332

Maths Quest 11 Mathematical Methods CAS

−1 2 is: 1 4

9 The inverse of

A

d

1 6

−4

−1

2

1

2 1

4 −1

−

B

2

e

−1

1 4 6 − 1 −1

6

−1

−2

−

4 −1

C

−1

1 2 − 1

−

2

−4

2

−1

. The coordinates of the new point are: 1 A (0, 0) C (0, 2) B (−4, 0) d (−4, −2) e (0, −2) −1 11 The point (a, b) is reflected in the line y = x and then translated by the matrix . Its new coordinates are: 3 A (a − 1, b + 3) B (a + 1, b − 3) C (b + 1, a − 3) d (b − 1, a + 3) e (−a + 1, −b + 3) 10 The point (−2, −1) is translated by the matrix

−2

12 Reflection in the line y = −x followed by a dilation of factor 2 from the y-axis is equivalent to pre-

multiplying by the single matrix: 1 2 0

A 0

B 0 − 2

0

−1

C

−1 −

0

0 2

d

1 0 0 2

e

0 − 1

0

−2

13 The point (2, −1) is reflected in the y-axis followed by an anticlockwise rotation of 90° about the origin.

The coordinates of the image point are: B (−1, −2)

A (1, −2)

C (−2, 1)

e (2, −1)

d (1, 2)

1 0

14 The coordinates of the image of (3, −4) under the transformation given by A (3, 4)

B (3, −4)

C (−3, 4)

0 are: −1

d (−3, −4)

e (4, 3) ex Ten d ed r eS p o n S e

1 A and B are both square matrices. a Show that A2 – B2 = (A – B)(A + B) only if A and B are commutative for multiplication.

1 9

b Use part a above to evaluate A2 – B2, where A =

7 B and = −3 6

−2

. −3

−2

2 A and B are both (2 × 2) matrices. a Show that (A + B)2 = A2 + 2AB + B2 only if A and B are commutative for multiplication. b Use the matrices in question 1b to prove the result in question 2a numerically. 3 Potatoes cost $p per kg, carrots cost $q per kg, onions cost $r per kg and eggs cost $s per dozen.

George buys 14 kg of potatoes, 10 kg of carrots, 8 kg of onions and 2 dozen eggs for a total cost of $60. Mary buys 10 kg of potatoes, 11 kg of carrots, 3 kg of onions and 5 dozen eggs for a total cost of $52. Cindy buys 4 kg of potatoes, 21 kg of carrots, 12 kg of onions and 1 dozen eggs for a total cost of $56. Amy buys 15 kg of potatoes, 5 kg of carrots, 11 kg of onions and 4 dozen eggs for a total cost of $69. Establish a matrix equation and hence calculate the values of p, q, r and s. a 4 a The point P (1, −3) is translated by the matrix . Find its new coordinates. b b The new point is reflected in the line y = −x. Find the new coordinates. c This point is now rotated counterclockwise through 90° before being reflected in the x-axis. Find

the new coordinates.

d Find the transformation(s) needed to return the point found in part b to the point (1, −3). ChApTer 7 • Matrices

333

4 0 . 0 1 a Find the coordinates of the new vertices, calling them O′, A′ and B′. b Has the area altered? If so, how? c The new triangle is now reflected in the line y = −x. Find the coordinates of the new vertices, calling them O″, A″ and B″. −3 d Finally, O″A″ B″ is translated by . Find the final coordinates of the vertices. 1 3i 0 , where i2 = −1. 6 A= −i 2 3 a Find A – 2A2. – b Find A 1. 1 0 2 3 4 c Show that A2 – A – 2I = (A – 2I)(A + I), where I = . Note: I = I = I = I etc. 0 1 7 Use a matrix method to solve the following simultaneous equations. mx − 3 y = 6 5 The triangle O (0, 0), A (2, −3), B (−5, −4) is transformed by the matrix

2x + m2 y = 3 8 Use a matrix method to solve the following simultaneous equations. a+b+c+d = 5 −a + b − c + d

= −5

8a + 4 b + 2c + d = 19 −

8a + 4 b − 2c + d = − 25

9 Tickets for a one-way trip on a Melbourne to Sydney passenger train can be purchased as either adult,

child (under 15 years old) or pensioner. The table below shows the number of passengers and the total takings for three trips. Number of adult passengers 145 130 142

Number of child passengers 103 110 115

Number of pensioner passengers 121 90 80

Total takings ($) 20 260 18 400 19 200

a Let x equal the cost of an adult’s ticket. Let y equal the cost of a child’s ticket. Let z equal the cost

of a pensioner’s ticket. Construct three equations in terms of x, y and z.

b Using matrices, express the equations in the form AX = B. − c Use your CAS calculator to find A 1. d Use your CAS calculator to determine the costs of a train ticket for an adult, a child and a

pensioner. 10 Describe the transformation given by the following matrices. It may be helpful to use a graph and consider the effect of the matrix on the points (a, b). b

0 1 1 0

f

e

diGiTAL doC doc-9772 Test Yourself Chapter 7

334

1 0 1

a 0

i

−3

0

−

0 3

Maths Quest 11 Mathematical Methods CAS

−1

0

0 − 1

0 1 − 1 0 0 1

j

0

−1

1 0

c

0 − 1

g

0 −1 0

−1

0 0 1

d

−1

3 0 0 3

h

ICT activities Chapter opener diGiTAL doC • 10 Quick Questions doc-9769: Warm up with ten quick questions on matrices (page 309)

7A

Addition and subtraction of matrices

TUToriALS • We1 eles-1426: Watch a tutorial on showing that the Commutative Law holds for two matrices (page 311) • We3 eles-1427: Watch a tutorial on solving for variables in two equivalent matrices (page 313)

7B

multiplying matrices

TUToriAL • We5 eles-1428: Watch a tutorial on recognising the order of matrices, and the product of two matrices if they exist (page 317) diGiTAL doC • WorkSHEET 7.1 doc-9770: Determine the order of a matrix, add, subtract and multiply by a scalar and practise matrix multiplication (page 318)

7C

Solving matrix equations

• We9 eles-1430: Watch a tutorial on using matrices to solve simultaneous literal equations (page 322) diGiTAL doC • WorkSHEET 7.2 doc-9771: Calculate determinants and inverses of matrices and solve matrix equations (page 324)

7d

matrices and transformations

inTerACTiViTY • Matrices and transformations int-0266: Use the interactivity to consolidate your understanding of matrices and transformations (page 325) TUToriAL • We12 eles-1431: Watch a tutorial on finding the new coordinates of a point after translating then reflecting it in the line y = x (page 327)

Chapter review diGiTAL doC • Test Yourself doc-9772: Take the end-of-chapter test to test your progress (page 334)

To access eBookPLUS activities, log on to www.jacplus.com.au

TUToriALS • We7 eles-1429: Watch a tutorial on using matrices to solve simultaneous equations (page 321)

ChApTer 7 • Matrices

335

Answers CHAPTER 7 mATriCeS exercise 7A

matrices

1 A+ B =

4 1 = B+ A − 2 3

− 5 2 a 1 6 − 2 1 3 c Does not exist. 5 −7 − 1 3 3 −1 [5 −9 2 −6 x = −1, y = 3 a 3×3 c 1×1 [−4 6 −3]

−

−

e 3 4 5 6

8

10 0 8 , NM = − 10 4 0 10

7] b 4×1 d 1×4

6 a

8 a [−29]

−4 − b 6 − 8 c No

11 12 13 14

57.5 70 81.5 36.5 97.5 25.5 81 45

42 + 32 16

−7

multiplying matrices b Does not exist.

9 − 2 3 − 6 8

2 [−13]

336

0 − 2 −3

27

− 19

−

6

−9 − 12

a AC, CA, CB c None

0

− 13

7 x=

1 c e 2

a

−2

1 3

b

0

−2

b Yes. ACB

1 0 1 2 1 45 3 1 2 2 5 3 2 6 7 18 6 Solving matrix equations b 8 d 0 f 5x2

9 xy − 6 −4, 3 −1 5 3 a 1 13 − 3 2

Maths Quest 11 Mathematical Methods CAS

−

1 4 0.5 1.5

16

−2

29

− 9

b x = 14, y = −7 d 7, 8

3p + 2 4p− 9 ,y= 2 2 p2 + 3 2p + 3

− 5 − 19 3

8 3, 3,

b [131 108 191 439 233] exercise 7C

0.5

− 0.5

d

c x = 22 , y = 22

8 − 12 − 16

0 1

b

3 3 6 5 x = ±3

−

−1

17 a 80

0 − 1 2

−2

1 e 1 3

4

59 92 23 36

6 a x= 7 ,y= 7

15 ; a = − 0.1, b = − 0.3 − 10 4.50 16 a [1200 2367 4219] b 1.95 2.45 c $20 352.20

1 a Does not exist.

28 d − 1

D C B

= 54 68

e

8 0

b

−1 4

c None

15 − 6

36.5 81

− 11

4

−2 −

10 a

d 74.5

5 2 −5 2 3 − 1 13 0 1 =I = 13 0 13 4 a

15 16 12 3

90 115 140 c 149 163 73 195 51 162

− 17

−5

−1 1 − 13 3

60 − 12

60

c − 108

9

50 65 80 b 87 76 45 95 37 89

6 5

0

40 50 60 12 a 62 87 28 100 14 73

c

1 2

− −

2 x + 5 y 11 7 − 3 x + 2 y = 12

9 C 10 B 11 A

exercise 7B

−6

2 4 ] − does not exist. 1

13 15 10 + 7

−

1 2

b

1 −1 5 − − 1 13 3 2 13 0 =I 0 13

2 3

1 = 13

4 a MN =

10

e 70

12 − 1 15

− 22

x + 5 y 11 = 5 a − 3 x + 4 y 5 b x + 5y = 11, −3x + 4y = 5 c x = 1, y = 2

7 p = 3 , q = −3, r = 3 8 [ −5

3 a

4

3 0 5 − 3 1 4 d Does not exist.

b

b

5 35 − b 5 1 −1 2 c R ( P + SQ ) does not exist.

Addition and subtraction of

9 E 10 E 11 D

12 I;

−1 2

3 2

− 3 2

−1 2

2 0 1 1 ≠ 1 0 1 1

13

sin(θ ) − sin(θ ) cos(θ ) p2 + q2 ≠ 0 Check with your teacher. a = 17 y = x3 – 2x2 + x + 3 − r 2 − 7r + 2) r 2 + 8r + 11 (3 x= 3 , ,y= r + 9r + 5 r 3 + 9r + 5 − 2 5r + 9r − 7 z= 3 r + 9r + 5 a = 1, b = −1, c = 2, d = −2 − 10 10 x= , y = 3, z = ; unique solution b+2 b+2 cos(θ )

14 15 16 17 18 19

20 21

if b ≠ 1 or −2. If b = −2, division by 0 If b = 1, equations 1 and 2 become identical. 22 $2912 + $3332 = $6244 23 2834, 120, 1120, 1037

iii (−9, 0) iv (0, −9)

exercise 7d

matrices and transformations b (−6, 0) c (3, −4) 1 a ( 2, 3) d (−3, −2) e (−3, −2) 2 a (m, −n) b (3n, −4m) −

0 3 1 0

3 a

1 0

−

3 −1

1 3

i

f

1 2

1 1

h

1 2

−

1

j

1 2

3 1 b B

−

1 2

1 1 − −

1 −1

−

3 3 5 a D c E 6D 7 a (0, 0), (3, 0), (3, 2), (0, 2) 3 0 b 0 2 n 0 0 n 8 (3 − b, 2a − 5) c

ii (2 2, 2 2) iii (0, 4)

b i

(

− 2,

1 2

9 2

ii

(

iii

(−4,

,

5 2

)

c

i ii

(

5 2

(

(

− 9 2

−5 3 2

− 3, 3

,

2

ChApTer reVieW

−3 2

)

)

d i ii

( (

(

−9 2

9 3 2

,

−9 2

,

9 2

)

)

0

8

0

0

0

− 12

1 2 0 −3

3 3 2

)

+3

1 5 9 13

5

d Impossible 6

y = − x 2 3x − 2 y 0 = 0 3x − 2 y

− 12

− 37

8 x = 11 , y = 11 , z = 11 9 (a – p, b – q)

2 0 10 0 1 11

b [−2]

6 4 5 2

(

1 2

+ 3,1 −

1 2

)

3

3

−

2 6 10 14

B C B A

3 A 7 B 11 D

4 E 8 B 12 E

1 a (A − B)(A + B) = A2 + AB − BA − B2 =

b [3]

C B E A

exTended reSponSe

3 4

−

8

17

+ 3 3,

2

4

1 0 0 1 =I

iv (3, −6) v

−

0

−1

1 3 x − 2 y

iii (−6, −3) −3

−3

y 1 × 2 3 3 x − 2 y

+3 3

)

mULTipLe ChoiCe

2

1 2 2 1 3 x = 3 and y = −1

7 x

+2

5

1 a 4×1 b 5×2

2 −6

− 1 27

16 c = 3

ShorT AnSWer

6

+ 2 3, 2

3 −1

)

5)

−3 3

1 3

+2 3

iv (−5, −4) v

) )

5 a

3)

5 3 2

,

0 6 0 3 0 4 −2 c − 0 2 1

iv (−4, 0) v (2,

2 −3 2

− b moves. Only if tan 1 = 22.5° would a this be true. 13 a OP = 5 = OP′. The transformation could be a rotation (in this case anticlockwise) about the origin. (It could also be a translation, but that is not the question.) b 80.72°

1 −2 4 a 0 3

9 a i (2 3, 2)

−2

2

7 2

b Areas are the same. 12 a No. Single matrix ≠ combined matrices. b No. Under the identity matrix nothing

−

3

)

−3)

( C' = (

d

−

1 2 −

0 − 1

9 3 9 , 2 2

− 3 , B' = (− 2 2, 2 2) 11 a A' = 1 ,

1 0 0 1

−1

0 e 1 g

−1 b 0

0

0 1

c

10 (0,

b (0, −1)

0 1 4 a − 1 0

(

v

1 5 12 62 − 19 13 64 − 257 67 7 4 14 a | A | = 6 b C is not a square matrix. No inverse. 3 c BC = 2 15 P(P − 2) = 3I P−2 1 1 0 = The inverse of P is . 3 3 2 − 3

12

A2 − B2 only if AB – BA, i.e., if A and B are commutative for multiplication. − 54 12 b − − 6 42 a (A + B)(A + B) = A2 + AB + BA + B2 = A2 + 2AB + B2 only if AB + BA = 2AB, i.e., if A and B are commutative for multiplication. b Not equal p = 2, q = 1, r = 2 and s = 3 a (a + 1, b − 3) b (3 − b, −a − 1) c (1 + a, −3 + b) d Translation by (−a, −b) a O′ (0, 0), A′ (8, −3) and B′ (−20, −4) b Yes, by a factor of four. The new triangle is now reflected in the line y = −x. c O″ (0, 0), A″ (3, −8) and B″ (4, 20). d The final coordinates of the vertices are (−3, 1), (0, −7) and (1, 21). − 27i + 18 0 a A3 − 2A2 = − − 14 − 8i 14 − 8i − i 0 − 1 b A 1= 3 − 2 3 i c A2 − A − 2I − 3i 0 2 0 = 9 0 − − − 2 −i 0 2 4i 1 =

0 − 3i 4i − 2 i − 3

− 11

(A − 2I )(A + I ) 3i − 2 = 2 =

−

0 3i + 1 i−2 2

−

0 i +1

0 − 3i 4i − 2 i − 3

− 11

ChApTer 7 • Matrices

337

x 1 6m 2 + 9 = 3 y m + 6 3m − 12

7

2 −1 = 3 1 9 a 145x + 103y + 121z = 20 260 130x + 110y + 90z = 18 400 142x + 115y + 80z = 19 200 145 103 121 x 20 260 b 130 110 90 y = 18 400 19 200 142 115 80 z 8

338

a b c d

−1

145 103 121 c 130 110 90 142 115 80 155 − 1135 101 6068 12 136 1517 − − = 119 2791 67 3034 30 340 1517 67 2049 − 64 6068 60 680 1517 d $70 for an adult’s ticket, $50 for a pensioner’s and $35 for a child’s. 10 a None b Rotation through 180° c Reflection in the x-axis

Maths Quest 11 Mathematical Methods CAS

Reflection in the y-axis Reflection in the line y = x Clockwise rotation of 90° Reflection in the line y = −x Dilation by a factor 3 from both the x-axis and the y-axis i Dilation by a factor 3 from both the x-axis and the y-axis and rotation through 180° j Anticlockwise rotation of 90° d e f g h

Exam practice 3 CHAPTERS 1–7 S ho rT A n S W er

3 1 1 Let A = . − 2 2 a Determine det (A) − b Hence, find A 1.

20 minutes

1 mark 2 marks

3 2 and A = 1 2

2 Let B = a B+A b AB c

1 2

. Determine the following.

−1

4 −2 0

2 marks 3 marks

A − 2B

3 marks

3 The point P (4, −2) has been dilated by a factor of k from the y-axis to form a new

coordinate P′. The new coordinate of P′ is (4, −4).

a Show that k = 1.

2 marks

b P′ is now translated by the matrix

2 −

3

to form a new coordinate R. The coordinates

of R are (2a, b − 1). Determine the value of a and b. 1 1 Let C = 3

−

a 1

A −3 d

2 marks

m U LTip L e C ho iC e

. If C is a singular matrix, then the value of a would be: − 1 B C 0 3

1 3

10 minutes

each question is worth one mark.

e 3

2 The point (1, 4) has been translated by the matrix

−

a . If the coordinates of the new point are b

(−1, 2), then the values of a and b would be: A a = − 2, b = 1 B a = − 2, b = − 2 2 − d a = − 1, b = 1

C a = − 2, b = 2

e a = − 1, b = − 2

2

3 A is a matrix of order m × n, B is a matrix of order p × n and C is a matrix of order n × p. Which one of

the following is correct? (AB)C would result in a matrix of order m × n. A + C would result in a matrix of order (m + n) × (n + p). BC would result in a matrix of order n × p. (AC)B would result in a matrix of order m × n. (BA)C would result in a matrix of order p × (nm) × p.

A B C d e

4 Which one of the following matrix equations would provide a solution to the simultaneous equations

x – 2y = 3, 3y – 2z = 1 and 3x + 4z = 2? x 3 1 A y = 1 = 3 z 2 3 x 1 d y = 3 z 3

2 − 2 4

−

−1

x 1 −2 B y = 0 4 z 3 −2

3 e 1 2

0 3 −2 0 4

−2

x 3 1 y = 1 0 z 2 3

−2

−1

3 1 2

0 − 3 2 0 4

x 1 C y = 3 z 3

−

2 − 2 4

−1

3 1 2

−1

Exam practice 3

339

e x T ended r e SponS e 35 minutes

1 A triangle ABC has been transformed by a factor of 3 from the x-axis and a reflection in the line y = −x.

The transformed triangle has vertices A′, B′ and C′. a i Write down the (2 × 2) matrix that can be used to describe a dilation by a factor of 3 from the x-axis. ii Write down the (2 × 2) matrix that can be used to describe a reflection in the line y = −x. iii Hence, write down the (2 × 2) matrix that can be used to describe the combined transformations. 1+1+1 = 3 marks b A′ has coordinates (−2, 3). Using your matrix from part a iii, determine the coordinates of A. 2 marks c The triangle A′B′C′ is rotated anticlockwise by 90° to form three new vertices, A″, B″ and C″. i Write down the (2 × 2) matrix to describe this transformation. ii Hence, determine the coordinates of B″ if B′ is (−5, −3). 1+1 = 2 marks 2 Jim owns a garden centre. He sells four varieties of fruit trees: apple, plum, nectarine and cherry. He

recorded the number of each tree he sold over a four-week period during spring. In the first week, he sold 4 apple trees, 2 plum trees, 3 nectarine trees and no cherry trees. In the second week, he sold 2 apple trees, 1 plum tree, 3 cherry trees and no nectarine trees. In the third week, he did not sell any plum or apple trees but he did sell 2 nectarine trees and 1 cherry tree. During the fourth week, Jim’s assistant did not record the number of plum trees sold. He did record that he sold 2 apple trees and 3 cherry trees but did not sell any nectarine trees. Let the unit cost, in dollars, of each type of tree sold be represented by the following variables: e = cost of an apple tree f = cost of a plum tree g = cost of a nectarine tree h = cost of a cherry tree. a Write a (1 × 4) matrix to record the number of apple, plum, nectarine and cherry trees sold in the first week. 1 mark b The (4 × 1) matrix C below shows the unit price, in dollars, of each type of fruit tree.

diGiTAL doC doc-10163 Solutions exam practice 3

340

e f C= g h 1 mark Using the matrix above, write the cost of 3 cherry trees. c The matrices W3 and C show the number of trees sold in the third week and the unit cost of each tree. e f W3 C = 0 0 2 1 g h i Write the order of the product of W3C. 1 mark ii In the context of this problem, what would the matrix W3C determine? 1 mark d The number of trees sold over weeks 2 to 4 can be represented in the following matrix, T. 2 1 0 3 T = 0 0 2 1 2 p 0 3 where p is the number of plum trees sold during the fourth week. By adding in your matrix from part a, write down the matrix that represents the total number of fruit trees sold over the four-week period. 1 mark The unit cost of a plum tree is $16.50. The amount of money Jim earns from selling the fruit trees for each of the first three weeks is $195.30, $175.35, $74.65 and ($158.85 + the money from the selling of p plum trees) respectively. e Write down a matrix, A, that shows the amount Jim earned over the four-week period. 1 mark f Set up a matrix equation that would determine the unknown values of e, g, h and p. 2 marks g Using an appropriate method, solve the matrix equation to find the unit cost of an apple tree, a nectarine tree and a cherry tree. 3 marks h The amount of money collected from selling the fruit trees during the fourth week was $208.35. Determine the number of plum trees that were sold during the fourth week. 1 mark

Maths Quest 11 Mathematical Methods CAS

ChapTer 8

Rates of change diGiTal doC doc-9773 10 Quick Questions

ChapTer ConTenTS 8a 8B 8C 8d 8e 8F 8G 8h 8i

Identifying rates Constant rates Variable rates Average rates of change Instantaneous rates Motion graphs (kinematics) Relating the gradient function to the original function Relating velocity–time graphs to position–time graphs Rates of change of polynomials

8a

identifying rates

In any subject dealing with things that move, grow or change it is often important to know the rate (‘speed’) with which quantities change. An example is the amount of water a city consumes per year. Knowing this rate is important so that measures can be taken to ensure there are adequate supplies of water stored in dams or reservoirs. Rates describe how much one quantity changes with respect to another. Example 1. The speed of a car is 70 km/h. This rate is read as 70 kilometres per hour, meaning that the car will travel 70 kilometres in an hour. Example 2. The council rates for a household are $560 per year. As with many rates, examples 1 and 2 involve time as the second variable. But this is not always the case. Example 3. The price of petrol is $1.129 per litre. This rate means that the petrol will cost 112.9 cents for every litre purchased. Rates usually have units associated with them but not always. 1 Example 4. The gradient of a road is 25 . This rate is read as 1 in 25 and means the road rises one unit vertically for every 25 units in length (horizontally). This is often expressed as a percentage, i.e. a 4% grade in this case. Road

1

25

In general, for two quantities A and B: Rate of change of A with respect to B = amount of A per B change in A . = change in B ChapTer 8 • Rates of change

341

Worked example 1

Which of the following represent a rate? a 20 m/s b 75 cents per packet

c $13

Think

WriTe

a 20 m/s is a rate as it compares distance and time. A

a 20 m/s is a rate.

distance of 20 metres is travelled each second. b 75 cents per packet is a rate as it compares cost and

b 75 cents per packet is a rate.

quantity of items. There is a cost of 75 cents for every packet. c $13 is not a rate as there is only one quantity.

e $13 is not a rate.

Worked example 2

A car travels 259 kilometres using 35 litres of petrol. Express this rate in km/L. Think

WriTe

distance travelled amount of petrol used

Rate =

259 km 35 L

Divide 259 by 35 to compare the number of kilometres travelled for each 1 litre of petrol.

=

7.4 km 1L

Express as a simplified rate.

= 7.4 km/L

1

Rate =

2

3

Note: Petrol consumption is usually expressed in L/100 km. 35 In this case, that would be × 100 = 13.5 L/100 km. 259

exercise 8a

identifying rates

1 We1 Which of the following represent a rate? a 100 km/h b $140 per week d 80 beats per minute e $2.75 g 25 kg h 10 000 litres j 18 degrees k 1.4% per quarter m 5 per person n A gradient of 0.8 p Cars per hour 2 a B C d e F G h i

3

342

c f i l o

64% $35.90 per m2 $11.00 per dozen 33 3.4

Match each of the rates you identified in question 1 to the most appropriate description below. The rental fee for a house The slope of a building’s roof The speed of a car The number of cards dealt in a game of poker The cost of oysters at a restaurant The pulse rate of a person walking The number of cars entering a car park The price of carpet The inflation rate one year ago

A horseriding school wishes to organise four-day riding treks for a group of twelve people. The following arrangements are made. a Each day the group will ride off at 9 am. b They will travel 20 km each day. c There will be four rest stops per day. d Each person must bring sleeping gear. e They will sleep two people per tent. f Meals will be supplied. g The cost will be $200 per person. h Three spare horses will be taken. Which of these arrangements represent a rate?

Maths Quest 11 Mathematical Methods CAS

4 Find the cost of the following items. a 5 dozen eggs at $2.12 per dozen c 27 hours of work at $11.25 per hour e 3.8 m3 of sand at $19.75 per m3

b 42.5 litres of petrol at 73.9 cents per litre d 4.9 kg of oranges at $1.14 per kg f 55 427 people at $11 per person

5 mC Francis delivers 440 pamphlets in 2 hours and earns $5 per 100 pamphlets delivered. a The rate of delivery of pamphlets per hour is: a 220 B 20 C 880 d 440 e 2.5 b How long will it take Francis to earn $55? a 4 hours B 4 hours, 20 mins C 5 hours d 6 hours e 3 hours 6 We2 Express the following as the rate indicated in the brackets. a Running 100 m in 12 seconds (m/s) b A rise of 21 °C in 6 hours (°C/h) c A meal costing $250 for 10 people ($/person) d 2.34 kg of washing powder added to 9 litres of water (g/L) e 32 m2 of lawn costing $149.50 ($/m2) f Losing 12.76 kg in 11 weeks (kg/week)

diGiTal doC doc-9774 SkillSHEET 8.1 introduction to rates of change

7 A 100-litre tank filled with water is leaking water at the rate of 2.5 litres per hour. a How long will it take for the tank to be half full?

Once it is half full, water is added to the tank at the rate of 3 litres per hour. b How much longer will it take until the tank is full again? c mC Which of the graphs below correctly shows the volume of water in the tank, v litres, as a function of time, t hours? a

v

100

100

50

50

50

120 t

v

20

120 t

20

120 t

20

120 t

v

e

100

100

50

50 100120 t

8B

v

C

100

20 d

v

B

Constant rates

When the rate of change of one quantity with respect to another does not alter, the rate is constant. For example, if petrol is $1.60 per litre, then every litre of petrol purchased at this rate always costs $1.60. This means 10 litres of petrol would cost $16.00 and 100 litres of petrol would cost $160.00. Calculating the gradient from the graph: rise run $(160 − 16) = (100 − 10) litres $144 = 90 litres

Gradient =

=

$1.60 1litre

Price ($)

160

16 10

100

= $1.60/litre Number of litres (L) = rate That is, the gradient of this graph represents a rate. A straight line graph shows that the rate of change between two quantities is constant.

ChapTer 8 • Rates of change

343

Worked Example 3

The graph shown represents the height, H, in metres (m) of a hot air balloon and time, T, in hours (h) after it is launched from a mountain top. Use the graph to find: a the gradient b the rate of change c the rule for height as a function of time.

Height (m)

1100

500

Time (h) Think

a Use the gradient of a straight line, m =

5

Write

rise . run

b Rate of change is the same as the gradient. Include

appropriate units. c As the graph is a straight line, express the rule in the

form y = mx + c. Remember that m is the gradient and c is the y-intercept.

1100 − 500 5− 0 600 = = 120 5

a Gradient =

b Rate of change = 120 m/h c The rule is of the form H = mT + c where

m = 20 and c = 500. H = 120T + 500

Worked Example 4

The table below shows the number of people, N, left inside an office building T minutes after the alarm goes during a fire drill. a Show that the relationship is linear. T 0 1 2 3 4 b Find the rate of change. N 140 105 70 35 0 c Express N as a function of T. Think

a 1 Plot the graph of the points given.

Write/draw

a

N 140 105 70 35 1

2

Check whether the points are collinear, that is, whether they lie on a straight line.

b 1 Calculate the gradient, m.

2

Use the gradient to state the rate of change. Include the appropriate units.

c Express the rule in the form y = mx + c where m = –35

and c = 140. Use N and T instead of y and x.

344 Maths Quest 11 Mathematical Methods CAS

2

3

4

T

A straight-line graph was produced, so the relationship is linear. 0 − 140 4−0 − 140 = = −35 4 Rate of change = −35 people/min The number of people left inside the building is decreasing at a constant rate.

bm=

c The rule is N = −35T + 140.

exercise 8B 1 a b c d e f g h i j k 2

a

Which of the following rates are constant? A person’s pulse rate when running 3 km The rate of growth of Australia’s population A person’s pulse rate when lying down The daily hire rate of a certain car The rate of growth of a baby The rate of temperature change during the day The commission rate of pay of a salesperson The rate at which the Earth spins on its axis The rate at which students arrive at school in the morning The rate at which water runs into a bath when the tap is left on The number of hours of daylight per day Which of the following graphs show a constant rate of change?

t

t

t

t

d

d

0 g

d

0 j

c

d

0 f

d

0 3

t

d

0 i

b

d

0 e

Constant rates

t

t

0 h

d

0

d

d

t

t

0

d

0

t

For the constant rates of change found in question 2, which of these rates are: i positive? ii negative? iii zero?

The distance travelled by a car is shown in this graph. The graph shows that the car is: a slowing down B speeding up C travelling uphill d travelling at constant speed e stationary 5 mC The volume of water, v litres, in a container is shown as a function of time, t minutes, in the graph. The rate of change of water in the container is: a 2 L/min B 8 L/min C −2 L/min d −8 L/min e 0 L/min

Distance (metres)

4 mC

0

Time (hours)

v (litres) 18 10

0

4

t (min)

ChapTer 8 • Rates of change

345

6 We3 For each of the functions graphed below, state: i the gradient ii the rate of change iii the rule. a d (m) b v (litres) 260

200

0 c

50 4

t (h)

0 d

w (kg) 100

30

t (min)

h (m) 75

60

0 e

40

0

t (weeks) f

w (g) 16

t (years)

10

w (g) 30

10

0

8

5 0

t (min)

120 v (litres)

20

7 mC Water is being pumped out of a swimming pool holding 30 000 litres at a constant rate. It takes

25 minutes to remove 1000 litres of water. a The rate of change of the volume of water in the pool is: a 400 L/min B 1200 L/min e 40 L/min d −40 L/min b The time taken to empty the pool is: a 12 hours, 5 minutes B 12 hours, 30 minutes d 2 days e 18 hours 8

C

−1200

C 6 hours, 30 minutes

The number of crates of fruit picked by a fruit-picker over the course of a day is shown in the graph.

Number of crates

13

7 6

5 pm

11 am 12 pm 1 pm

8 am

0 Time

If the fruit-picker is paid $12 per crate, answer the following. a What is the rate of pay per hour in the first 3 hours? b Explain what probably happened between 12 pm and 1 pm. c What is the rate of pay per hour in the last 4 hours? d Suggest two possible reasons why the line is not as steep in the afternoon. e How much is earned for the day? 346

L/min

Maths Quest 11 Mathematical Methods CAS

9 An athlete runs at a constant speed of 6 m/s for the first 420 m of a race. The next 400 m are run at a

constant speed of 5 m/s and the final 180 m at a constant speed of 6 m/s. a How far is the race? b How long does it take to run the first 420 m? c How long does it take to run the next 400 m? d How long does it take to run the final 180 m? e Sketch the graph of distance, D metres, versus time, T seconds. 10 We4 A spring is stretched a distance x cm beyond its natural length when a weight of W kg is attached to one end as shown in the following table. x

0

5

10

15

20

25

W

0

1

2

3

4

5

diGiTal doC doc-9775 plotting relations

a Show that there is a linear relationship between the variables. b What is the rate of change of W with respect to x? c Express W as a function of x.

A bucket contains 40 litres of water when it starts leaking through a hole in the bottom, such that the volume of water, V litres, in the bucket at any time, t hours, is given by the rule V = 40 − 6t. a At what constant rate, in litres/hour, is the water leaking? b How long will it take for the bucket to be empty? 12 Water is poured at a constant rate into the vessels a, b, c and d shown below. 11

a

b

c

d

Match the appropriate graph below, representing height of water, h, at any time, t, with each vessel above. i

ii

h

0

8C

t

iii h

h

0

t

0

Variable rates

If a rate is not constant (is changing), then it must be a variable rate. For example, the amount of electricity used per hour by a household is most likely to be a variable rate as the need for electricity will change throughout the day. Graphically, a variable rate will not be linear, because the gradient (or rate of change) is not constant. The distance–time graph at right shows a variable rate between t = 0 and t = 3, because the gradient is changing over this interval. However, it shows a constant rate between t = 0 and t = 1; t = 1 and t = 2; and t = 2 and t = 3, because the gradient is not changing over these intervals. Relationships that give curves when graphed show that the rate of change is continually changing; that is, a variable rate exists.

iv h

t

t

0

d

0

1

2

3

t

ChapTer 8 • Rates of change

347

Worked example 5

Consider the graph shown. For each interval, state whether: a the rate of change is constant or variable b the rate of change is positive, negative or zero.

B

C TUTorial eles-1432 Worked example 5

A Think

D

WriTe

a Consider whether the graph is linear for each

a Interval AB: variable rate of change

interval. The rate of change is constant if the graph is linear; otherwise it is variable.

Interval BC: constant rate of change Interval CD: constant rate of change

b Since the gradient of the graph gives a measure

of the rate of change, consider whether the gradient is positive, negative or zero in each interval.

exercise 8C 1 a b c d e f g h 2

a

b Interval AB: positive rate of change

Interval BC: zero rate of change Interval CD: negative rate of change

Variable rates

List which of the following rates are variable. The pulse rate of a person watching a horror movie The cost per kilogram of tomatoes at the local greengrocer today The speed of a car travelling through the city The cost per kilogram of tomatoes at the local greengrocer during the year The number of hours of sunlight per day The hourly rate of a receptionist The hourly rate of a private taxi driver The speed of a marble rolling down an incline Which of the following graphs show variable rates of change between points A and B? b

B

B A A

c

d

B

B A

A

f

e

A

B

B

A

348

Maths Quest 11 Mathematical Methods CAS

The graph at right shows how the distance of a runner from the starting line changes over the duration of a race. 100 For the graph shown, state whether: a the rate of change is constant or variable 50 b the rate of change is positive or negative. A c mC The rate of change of distance with respect to time is greatest at point: a A B B C C d D e E d The runner is travelling slowest (but still moving) at point: a A B B C C d D e E e At point E, the runner is: a moving very fast B speeding up d not moving e just beginning to move

3 We5

Distance (m)

4

E

D C B 6 Time (s)

12

C slowing down

The following table shows how weight, W (kg) varies with time, t (min), for a particular box as it is filled. t (min) W (kg) a b c d

5

0 0

1 3

2 6

3 9

4 12

Plot a graph of W versus t. What do you notice about the plotted points? Is the rate of change variable or constant? What is the rate of change in kg/min? The following table also shows weight, W (kg), varying with time, t (min). t (min) W (kg)

0 2

1 4

2 7

3 11

4 18

Plot these points on a graph. Do these points form a straight line? Is the rate of change variable or constant? What is the rate of change during: i the first minute? ii the second minute? iii the fourth minute? 6 The following tables of values show distance travelled, d km, at various times, t hours. Decide whether the rate of change of distance with respect to time appears constant or variable. a b c d

a

b

8d

t d

0 0

1 5

2 10

3 15

4 20

t d

0 0

1 15

2 20

3 45

4 80

diGiTal doC doc-9775 plotting relations

average rates of change

If a rate is variable, it is sometimes useful to know the average rate of change over a specified interval. For example, a tree grew from 6.2 m this time last year to 6.75 m now. The tree’s average rate of growth =

change in height change in time

(6.75 − 6.2) m 1 year = 0.55 m/year =

This means that the tree grew by 0.55 metres over the past year but not necessarily constantly at that rate during the year. ChapTer 8 • Rates of change

349

Worked example 6

Calculate the average rate of change between points P and Q marked on the graph. Q

Distance (m)

48 P

30

0

2

Think

Time (s)

7

WriTe

48 − 30 18 = = 3.6 5 7−2

1

Find the gradient of the chord PQ.

Gradient of PQ =

2

Use the gradient to state the average rate of change. Include the appropriate units.

Average rate of change = 3.6 m/s

Worked example 7

Find the average rate of change of height between t = 1 and t = 3 from the table below. t (min)

0

1

2

3

4

5

h (m)

20

60

90

130

140

145

Think

Calculate the average rate of change of height with respect to time by considering the change in each quantity. When the time changes from t = 1 min to t = 3 min, the height changes from 60 m to 130 m.

WriTe

Average rate of change of height change in height = change in time (130 − 60) m = (3 − 1) min 70 m = 2 min = 35 m/min

Worked example 8

Over a period of 6 hours, the temperature of a room is described by the function T(h) = h2 − 4h + 22, where T is the temperature in degrees Celsius after h hours. a What is the initial temperature of the room? b Sketch the graph of the function over the given time interval. c Draw a chord between the points where h = 1 and h = 5. d What is the gradient of this chord? e What is the average rate of change of temperature between h = 1 and h = 5? Think

a The initial temperature is the temperature at the

start of the time period. Substitute h = 0 into the function to find T(0).

350

Maths Quest 11 Mathematical Methods CAS

WriTe/draW

a When h = 0, T(0) = 0 − 0 + 22

= 22 The initial temperature is 22 °C.

TUTorial eles-1433 Worked example 8

b 1 To find the turning point, let h =

2

−

b . 2a

T (2) = 4 − 8 + 22 = 18 The turning point of the parabola is (2, 18).

Evaluate T (2).

c 1 Find the required points:

c T (°C) 27

T(1) = 12 − 4(1) + 22 = 19 T(5) = 52 − 4(5) + 22 = 27.

2

Indicate the points (1, 19) and (5, 27) on the graph and join with a straight line. rise and the points (1, 19) Use gradient = run and (5, 27).

d

Use the gradient to state the average rate of change. Include appropriate units.

e

exercise 8d

18

(2, 18) 1

2

3

5 H (hours)

4

27 − 19 5−1 8 = 4 =2

d Gradient =

e The average rate of change is 2 °C/h.

average rates of change b

y 8

y 10

P

0 −2

2

diGiTal doC doc-9776 SkillSHEET 8.2 Gradient of a straight line

Q

P

6

0 2

19

(5, 27)

22 (1, 19)

Calculate the average rate of change between the points P and Q marked on each graph.

1 We6 a

−

b 2a 4 = 2 =2

bh=

5

x

Q

5

x

Which of the following graphs have a straight line drawn that can aid in finding the average rate of change over a specified interval? a

y

b

x

0 c

y

0

0 d

x

y

x

y

0

x

ChapTer 8 • Rates of change

351

3

For each of the distance–time graphs below:

i draw a chord to the graph for the interval t = 1 to t = 3 ii find the gradient of this chord iii hence, find the average speed from t = 1 to t = 3. a d (km) b d (km) 70 70 50

10 0 c

1

0

t (h)

3

d

d (km) 70

1

3

t (h)

1

3

t (h)

1

3

t (h)

d (km) 70 60

40

0 e

1

0

t (h)

3

f

d (km) 70

d (km) 90 70

0 4

1

3

0

t (h)

The graph at right represents the weight of a non-uniform beam, W (kg), versus its length, L (m). a Find the gradient of the chord to the graph for the interval L = 5 and L = 20. b Find the gradient of the chord from L = 15 and L = 20. c What is the average rate of change between: i L = 5 and L = 20? ii L = 15 and L = 20? d Are these rates the same? Explain your answer.

W (kg) 32.5 30

10 0

Consider the graph at right. The chord with the highest gradient is the one from point 0 to point number: a 4 B 3 C 2 d 1 e 5

5 10 15 20 L (m)

5 mC

2

3

4

5

1

0 6

The total number of people at the zoo at various times of the day is shown in the table below. am T (time of day) N (number of people at the zoo)

10.00 0

pm 11.00 200

a Plot the graph of N versus T. b Draw chords to the graph for the interval: i 10.00 am to 1.00 pm ii 1.00 pm to 3.00 pm iii 3.00 pm to 5.00 pm. 352

Maths Quest 11 Mathematical Methods CAS

12.00 360

1.00 510

2.00 540

3.00 550

4.00 550

5.00 550

c Find the gradient of each of these chords. d What is the average rate of change from: i 10.00 am to 1.00 pm? ii 1.00 pm to 3.00 pm? iii 3.00 pm to 5.00 pm? e Briefly describe what these rates suggest about the number of people attending the zoo during the

course of the day. The height, h metres, reached by a balloon released from ground level after t minutes is shown in the table below.

7 We7

t (mins) h (m)

0 0

2 220

4 360

6 450

8 480

10 490

a Without drawing the graph, find the average rate of change of height with respect to time

between: i t = 0 and t = 2 ii t = 2 and t = 4 iii t = 4 and t = 6 iv t = 6 and t = 8 v t = 8 and t = 10. b Is the average rate of change for each 2-minute interval increasing or decreasing? 8 We8 The temperature of an iron rod placed in a furnace is described by the function T(t) = t2 + 20

between t = 0 and t = 10. T represents the temperature of the rod in degrees Celsius and t is the time in seconds. a What is the initial temperature of the iron rod? b Sketch the graph of the function over the given time interval. c Draw a chord between the points where t = 2 and t = 8. d What is the gradient of this chord? e What is the average rate of change of temperature between t = 2 and t = 8?

9 The weight of a rabbit in its first eight weeks is described by the function W(t) = t3 − 3t2 + 200, where

W represents the weight of the rabbit in grams and t is the time in weeks. a What is the weight of the rabbit at birth? b What is the average rate of change of weight of the rabbit between: i weeks 1 and 3? ii weeks 1 and 2? iii weeks 2 and 4? iv weeks 2 and 6?

8e

diGiTal doCS doc-9777 Gradient between two points doc-9778 WorkSHEET 8.1

instantaneous rates

If a rate is variable, it is often useful to know the rate of change at any given time or point, that is, the instantaneous rate of change. For example, a police radar gun is designed to give an instantaneous reading of a vehicle’s speed. This enables the police to make an immediate decision as to whether a car is breaking the speed limit or not. y

P 0

Tangent x

Instantaneous rates can be found from a curved graph by: 1. drawing a tangent to the curve at the point in question 2. calculating the gradient of the tangent over an appropriate interval (that is, between two points whose coordinates are easily identified). Note: The gradient of the curve at a point, P, is defined as the gradient of the tangent at that point. ChapTer 8 • Rates of change

353

Worked example 9

a Use the following graph to find the gradient of the tangent

W (kg)

at the point where L = 10. b Hence, find the instantaneous rate of change of weight, W, with respect to length, L, when L = 10.

10 Think

WriTe/draW

a 1 Draw in the required tangent. Do this ‘by eye’

a

(estimation). 2

W (kg) 12

To find the gradient of the tangent, choose a convenient interval (between the points where L = 5 and L = 15).

8 4 0

3

Use gradient =

appropriate units.

10 15 20 L (m)

5

4−8 15 − 5 −4 = 10 = − 0.4

rise . run

b Use the gradient to state the rate of change. Include

Gradient =

b Rate = −0.4 kg/m

The weight is decreasing with respect to length at a rate of 0.4 kg/m.

Worked example 10

a By plotting the graph of y = x2 − 2, find the instantaneous rate of change

where x = 1.

b Find the instantaneous rate of change of y = x2 − 1 at x = 2 using a calculator. Think

a 1 Plot the graph. 2

Draw a tangent to the curve at the point where x = 1. Do this by eye (estimation).

WriTe/draW

a

y 2 1 −2 −1 0 −1

1

2 x

−2 −3

3

354

L (m)

Find the gradient of the tangent. (Use points (1.5 O) and (0, −3) on the tangent line.)

Maths Quest 11 Mathematical Methods CAS

0 − (− 3) 1.5 − 0 3 = 1.5 =2

Gradient =

4

Use the gradient to state the rate of change. Note: There are no units to include as the required rate of change is the gradient.

b 1 Define the function using the CAS calculator.

Hence, the rate of change is 2.

b f1(x) = x2 − 1

2

Use the draw tangent feature of the CAS calculator to draw a tangent to f1(x) at (2, 3).

f1(2) = 22 − 1 =3

3

Use the coordinates and equations feature of the CAS calculator to determine the equation of the tangent line. Point the cursor to (2, 3) on the graph.

ytangent = 4x – 5

4

Answer the question.

The equation of the tangent is ytangent = 4x – 5. The gradient is 4. Hence, the instantaneous rate of change at (2, 3) is 4.

instantaneous rates

exercise 8e

Describe each of the rates below as either an instantaneous rate (I) or an average rate (A).

1

a Bill walks 12 kilometres in 2 hours. b An aircraft leaves the runway at 270 km/h. c A household uses 560 litres of water in one

day. d The pulse rate of a runner as he crosses the

finish line is 141 beats per minute. e A gas heater raises the temperature of a

room by 10 °C in half an hour. f A baby puts on 300 g in one week. g A road drops 20 m over a distance of 100 m. h Halfway along a flying fox, Jill is travelling

at 40 km/h. 2

a

Which of the following graphs have a tangent drawn at the point where x = 4? y

0 c

4

x

y

0 e

b

x

y

0

0 d

4

x

4

x

y

0 f

4

y

4

x

y

0

4

x

ChapTer 8 • Rates of change

355

g

y

h

0

x

4

y

0

x

4

In each of the following graphs, find an approximate value for the gradient of the tangent at the point indicated.

3 We9a a

y

b

x=1

3 2

y 20

x=2

10

1 0 c

1

x

2

0

y

d

5

2

y x=4

22 20

x = 1.7

x

4

2 0

1 1.7

4

x

0

4

x

8

Find the approximate value for the instantaneous rate of change of weight with respect to time at the time indicated on the graph.

4 We9b a

b

w (g)

t = 20

130

160

70

t = 10

80 0

c

w (g)

0

10 20 30 t (min)

w (g)

d

140

5 10 15 t (min)

w (g) 110 t=1

t=4 50

40 0

2

4

0 1

6 t (min)

5

t (min)

5 The weight of a person over a 40-week period is illustrated in the graph below. diGiTal doC doc-9779 Gradient at a point

Weight (kg)

90 80 70 60 50 40

0

10

20 30 Time (weeks)

40

a Estimate the maximum weight and the time at which it occurs. b Estimate the average rate of change of weight between week 10 and week 20. c Estimate the rate of change of weight at 30 weeks. 356

Maths Quest 11 Mathematical Methods CAS

6 The change of temperature in a restaurant kitchen over a 12-hour period is illustrated in the graph below.

Temperature (°C)

35 30 25 20 15 10 5 0

0 1 2 3 4 5 6 7 8 9 10111213

Time (hours) a When is the rate of temperature rise greatest? b Estimate the rate of temperature rise at 6 hours. 7 a Plot the graph of y = x2 from x = 0 to x = 4. b Draw tangents at the points where x = 1, x = 2 and x = 3. c mC The gradient of the tangent at x = 1 is closest to: a 0 C 2 B −1 d mC The gradient of the tangent at x = 2 is closest to: a 8 B 6 C 4 e mC The gradient of the tangent at x = 3 is closest to: a 6 B 9 C 3

d 3

e 4

d 2

e 0

d 7

e

−3

8 We10a a Plot the graph of y = 9 − x2 from x = −4 to x = 4. b Draw tangents at the points x = −2, x = 0 and x = 2. c Hence, estimate the instantaneous rate of change at the points: i x = −2 ii x = 0 iii x = 2. 9 We10b A tank is being filled at a variable rate. The depth of the water, H cm, at any time, t minutes, is

described by the rule H = t2 + 2t. At what rate is the depth of water changing after 2 minutes?

8F

motion graphs (kinematics)

In this section we will consider only objects moving in straight lines, either right and left, or up and down. Definitions 1. Position, x, describes the location of an object. 2. Distance, d, is how far an object has travelled. 3. Displacement, s, describes the change in an object’s position; that is, displacement = change in position = final position − initial position, or s = xfinal − xinitial. distance d 4. Speed = or speed = . time taken t 5. Velocity is the rate of change of position with respect to time, so change in position displacement s velocity = = , that is, v = . change in time t time taken Consider an object that begins at the –3-metre mark on a number line, moves 5 metres to the right, and then moves 5 metres to the left, taking 2 seconds to do so. For this example we have: x 1. Position (initially and finally) = −3 m −4−3−2−1 0 1 2 3 4 2. Distance = 10 m 3. Displacement = 0 m 10 m 4. Speed = = 5 m/s 2s 0m 5. Velocity = = 0 m/s 2s Distance and speed are (technically) always positive. Displacement and velocity can be either positive or negative, depending on the direction of motion.

inTeraCTiViTY int-0267 motion graphs (kinematics)

ChapTer 8 • Rates of change

357

Note: Some texts use displacement and position interchangeably (perhaps assuming displacement from a fixed origin). Worked example 11

Consider a lift starting from the ground floor, moving up to the top floor, stopping and then coming down to the fifth floor at the times shown on the diagram. Find: Top a the total distance travelled by the lift t = 14 s t = 20 s b the displacement of the lift after 25 s c the average speed of the lift d the average velocity of the lift.

TUTorial eles-1434 Worked example 11

Positive 100 m

150 m t = 25 s (5th floor)

t=0 Think

WriTe

a Add the distance travelled up (150 m) to the

distance travelled down (100 m). b At t = 0 s, the position of the lift is 0 m.

At t = 25 s, the position is +50 m.

c Average speed =

Ground floor

total distance travelled time taken

d Average velocity =

change in position change in time

a Total distance = 150 m + 100 m

= 250 m

b Displacement = change in position

= +50 – 0 = +50 m

250 m 25 s = 10 m/s

c Average speed =

(+ 50 − 0) m 25 s = +2 m/s

d Average velocity =

Worked example 12

Use the position–time graph at right, which shows the position of a particle, x, travelling in a horizontal straight line at any time, t, to find: a where the journey started b in which direction the particle moved initially c when and where the particle changed direction d when and where the particle finished its journey. Think

a Find x when t = 0.

x (m) 20 10 5 0

2

4 t (s)

WriTe

a When t = 0, x = 10 m. The journey started

10 metres from the origin or reference point. b Compare the position at t = 0 to the position a

short time after this. Notice that the graph has a negative gradient during this time.

358

Maths Quest 11 Mathematical Methods CAS

b As the position changes from 10 m at t = 0 s to

5 m at t = 2 s, the particle was moving towards the origin or reference point in the negative x-direction.

c Find x and t where the graph turns or changes

c After 2 s at a position of 5 m, the particle changed

direction. The particle changes direction at t = 2. When t = 2, x = 5.

direction.

d Find x and t where the curve stops.

d The particle finished its journey at a position of

20 m after 4 s. Worked example 13

Using the graph in worked example 12, consider the interval between t = 2 and t = 4 to decide whether: a the velocity of the particle is positive or negative b the particle is speeding up or slowing down. Think

WriTe

a 1 For this graph,

as

rise (or gradient) is the same run

a

change in position (or velocity). change in time So the gradient of a position–time graph gives the velocity. 2

Consider the gradient between t = 2 and t = 4.

From t = 2 to t = 4 the gradient is positive, so the velocity of the particle is positive.

b Consider how the gradient between t = 2 and

b The gradient is increasing, so the velocity is

t = 4 is changing.

increasing. This means the particle is speeding up.

Similarly, for the interval between t = 0 and t = 2 in worked example 12: x (m) 20

Positive gradient getting steeper Negative gradient getting less steep

10 5 0

2

4

t (s)

1. the velocity is negative because the gradient is negative 2. the magnitude of the velocity is decreasing because the magnitude of the gradient is decreasing. The negative gradient indicates the direction in which the particle is moving. 3. the speed is decreasing between t = 0 and t = 2 because the magnitude of the gradient is decreasing.

exercise 8F 1

motion graphs (kinematics)

Match the correct description (a, B, C or d) to each of the quantities (a, b, c or d) below. Quantity a Distance b Displacement c Speed d Velocity

a B C d

Description Rate of change of displacement with respect to time Change in position Length travelled Distance travelled with respect to time

2 State whether each of the following are true or false. a Speed can be positive or negative. b Displacement can be positive or negative. c Distance travelled is positive only. d Velocity is positive only. ChapTer 8 • Rates of change

359

A parachute ride takes people in a basket vertically up in the air from a platform 2 metres above the ground, then drops them back to the ground. Use the diagram below showing the position of the parachute basket at various times to find: a the total distance travelled by the parachute basket t = 55 s t = 60 s during a ride b the displacement of the parachute basket after 80 s c the average speed of the parachute basket during the ride d the average velocity of the parachute basket during the ride.

3 We11 diGiTal doC doc-9780 SkillSHEET 8.3 displacement and distance

19 m

t=0 2 m Platform 4

Consider the position and direction, at various times, of a particle travelling in a straight line as indicated below. t=5

a b c d e

t=2

t=0

−3 −2 −1

0

1

2

3

4

5

6

7

x

Where does the particle start? Where does the particle finish? In which direction does the particle move initially? When does the particle change direction? mC Which position–time graph below most accurately represents the particle’s motion? a

x (m) 6 5 4 3 2 1 0 −1 −2 −3

d x (m) 6 5 4 3 2 1 0 −1 −2 −3

360

t = 80 s Ground

B

2

5 t (s)

e

2

Maths Quest 11 Mathematical Methods CAS

5 t (s)

x (m) 6 5 4 3 2 1 0 −1 −2 −3

C

2

5 t (s)

2

5 t (s)

x (m) 6 5 4 3 2 1 0

x (m) 6 5 4 3 2 1 0 −1 −2 −3

2

5 t (s)

The total distance travelled in the first 5 seconds is: a 4m B 13 m d 14 m e 11 m g The displacement of the particle after 5 seconds is: B 14 m a −3 m e 0m d −4 m h The average speed in the first 2 seconds is: a 3 m/s B −2.5 m/s d 2.5 m/s e 1 m/s i The average velocity between t = 2 and t = 5 is: a 3 m/s B −2 m/s d 2 m/s e 1 m/s j The instantaneous speed when t = 2 is: a 2.5 m/s B 0 m/s d 2.8 m/s e 1 m/s f

C 9m

C 4m

C 6 m/s

C

−3

m/s

C 3 m/s

The following position–time graphs show the journey of a particle travelling in a straight line. For each graph, find: i where the journey started ii in which direction the particle moved initially iii when and where the particle changed direction iv when and where the particle finished its journey.

5 We12

a x (m) 8

0

b x (m) 12

c x (m) 12

10

10

4

3

1 2 3 4 5 t (s)

−3

0

d x (m) 18

e x (m)

0

1 2 3 4 5 6 t (s) f

5

1 2 3 4 5 6 7 8 t (s)

x (m) 5 4 2

0

1 2 3 4 5 t (s)

0

1 2 3 4 5 6 t (s)

−3 0

1 2 3 t (s)

−5

−5 6 For each position function of a particle given below (a–f ), sketch the position–time graph. In each case

explain: i where the particle started its journey ii in which direction it moved initially iii whether the particle changed its direction and, if so, when and where that happened iv where the particle finished its journey. a x(t) = 2t, t ∈ [0, 5] b x(t) = 3t − 2, t ∈ [0, 6] c x(t) = t2 − 2t, t ∈ [0, 5] d x(t) = 2t − t2, t ∈ [0, 4] e x(t) = t2 − 4t + 4, t ∈ [0, 5] f x(t) = t2 + t − 12, t ∈ [0, 5]

diGiTal doC doc-9781 SkillSHEET 8.4 interval notation

ChapTer 8 • Rates of change

361

7 a Plot the position–time graph for x(t) = 4t − t2. b Find the gradient at: i t=0 ii t = 1 iii t = 2 iv t = 3 v t=4 c Hence, give the instantaneous rate of change of position with respect to time (that is,

velocity) at: i t=0 ii t = 1 iv t = 3 v t=4 d Sketch the velocity–time graph from t = 0 to t = 5.

iii t = 2

8 We13 For each of the following position–time graphs, determine: i whether the velocity is positive or negative ii whether the particle is speeding up or slowing down. x b x c x a

0

d

e

x

0 9

t

0

t

t

x

0

t

Match the following position–time graphs (a–f ) with the corresponding velocity–time graphs (a–F ) on the opposite page. Position–time graphs a

c

x (m)

x (m)

3

3

3

2

2

2

1

1

1

0

d

b

x (m)

−1

1

2

3

0

4 t (s)

−1

1

2

3

0

4 t (s)

−1

−2

−2

−2

−3

−3

−3

e

x (m)

x (m)

f

3

3

2

2

2

1

1

1

−1

1

2

3

4 t (s)

0 −1

−2

−2

−3

−3

Maths Quest 11 Mathematical Methods CAS

1

2

3

4 t (s)

1

2

3

4 t (s)

1

2

3

4 t (s)

x (m) 4

3

0

362

0

t

0 −1 −2 −3 −4

Velocity–time graphs B v (m/s)

a v (m/s)

C v (m/s) 1

0

1

2

3

4 t (s)

1

2

3

1

2

3

4 t (s)

0

0

4 t (s)

e v (m/s)

d v (m/s)

0

0

1

2

3

4 t (s)

1

2

3

4 t (s)

F v (m/s)

1

2

3

0

4 t (s)

−1

10 A ball is projected vertically upwards from the top of a building 25 m high. Its position relative to the

11

The diagram at right shows the distance–time graph of two swimmers in a 200-metre race. a Who won the race? b Who was in front after 150 metres? c What was the approximate winning margin (in metres)? d What was the average speed of the winner? e For approximately how many seconds did Bill lead the race? f How many times did the lead change?

Distance (metres)

ground is given by the equation x = 25 + 20t − 5t2, where t is the time in seconds. Sketch a position–time graph for the ball and hence find: a the greatest height reached b when the ball reaches the ground c when the velocity of the ball is zero d an estimate for the velocity at which the ball is initially projected.

200 150

Alan

100 50 0

Bill 20 40 60 80 100 120 Time (seconds)

diGiTal doC doc-9782 WorkSHEET 8.2

ChapTer 8 • Rates of change

363

relating the gradient function to the original function 8G

Earlier we considered gradients calculated at a single point on a graph. However, if the gradients are calculated at every point within the domain of the function, the results can be graphed. This is the graph of the gradient function. The gradient of the function below has been calculated at several points as shown, and a table of gradient values has been constructed. Graph of original function y y = x3 − 2x2 + 2

6 4 Gradient = 7 2

0 1 −2 −1 −2

x Gradient

2

x

−1.0

−0.5

0

0.5

1.0

1.3

1.5

2.0

7

2.75

0

−1.25

−1

0

0.75

4

The table of values above may by be used to sketch a gradient function such as that at right. 4 Note: The gradient function is zero at x = 0 and x = 3 , which corresponds to the zero gradient (turning points) of f (x). The gradient function is negative in 4 the interval 0 < x < 3 , as is the gradient of f (x). The gradient function is positive elsewhere, which corresponds to the positive gradient of f (x).

Graph of gradient function y 8 6 4 2 −2 −1−2 0 1

2

relating the gradient function to the original function exercise 8G

1 Select several points (at least 6) on each of the graphs below and construct a table of gradient values. diGiTal doCS doc-9779 Gradient at a point doc-9775 plotting relations

Use each table to plot the gradient function in each case. You may use a CAS calculator to work through these problems. a

y y = x2 − 4 4

b

4

2

2

x −2−1 01 2 −2

e

y

2

x

−3−2−1 01 2 x −2 y = x3 + 3x2 + 2

2 1 y = ex

−3−2−1 0 1 2 3

f

y

4 2

4

−4

6

πx y = sin —– 4

0 1 2 3 4 5 6 7 8x

x

−1

y 6

y = 2x + 3

−3−2−1 0 1 2 3 −2

−4

d

c

y

y 40 32 24 16 8

y = x4

x −2−1 01 2

2 What do you notice about the gradient function of the quadratic function in question 1a? 3 Investigate other quadratic functions and their related gradient functions. What do you notice? 364

Maths Quest 11 Mathematical Methods CAS

x

4 Investigate the gradient functions of other cubics besides that shown in question 1c. What types of

functions are the gradient functions of cubics? 5 Which functions have a gradient function similar to the original function? 6 What do you notice about the original functions at x-values where the gradient function intersects the

x-axis? 7 When the gradient function is negative, what do you notice about the original function?

Relating velocity–time graphs to position–time graphs 8H

The gradient of a position–time graph gives the velocity, because velocity is the rate of change of position with respect to time. Therefore, by measuring the gradient of a position–time graph at various points, a velocity–time graph can be derived. Worked Example 14

The position–time graph for a particle moving in a straight line is shown below. x 3 2 Gradient = −2

1 Gradient = 2 0 −1

1

2

3

−2 Gradient = −1

4

5

t

Gradient = 1 Gradient = 0

The gradient of the curve at various times is indicated on the graph. Use this information to draw a velocity–time graph for the particle. Think 1

2

Set up a table of corresponding velocity and time values from the graph. Use the table of values to plot the velocity– time graph.

Write/draw

t v

0

1

−2

−1

2 0

3 1

4 2

v 3 2 1 −3 −2 −1 0 −1

1

2

3

4

5

t

−2

Similarly, a position–time graph can be derived from a velocity–time graph. The value of the velocity at a given time is equal to the gradient of the position–time graph at that time. By obtaining several of these values and knowing the initial position of a particle, the general shape of the position–time graph can be produced. The following example demonstrates the steps involved. Chapter 8 • Rates of change 365

Worked example 15

v 4 3 2 1

The velocity–time graph for a particle moving in a straight line and starting at the origin is shown in the diagram. Sketch the corresponding position–time graph.

0 −1 −2 −3 −4 Think 1

TUTorial eles-1435 Worked example 15

1 2 3 4 5

t

WriTe/draW

Set up a table of corresponding velocity and time values. (These velocity values represent the gradient of the position–time curve at the given times.)

t

0

v

1

4

2

2

3

4

0

−2

−4

x

2

Set up the axes for the position–time graph.

3

Draw in a curve starting at (0, 0) with a gradient of 4 decreasing to a gradient of 0 at t = 2 (the turning point). From t = 2 to t = 4, the gradient changes from 0 to −4. This means the curve will become steeper but with a negative slope.

0

1

2

3

4

5

t

relating velocity–time graphs to position–time graphs exercise 8h

For each position–time graph for a particle moving in a straight line, draw a velocity–time graph. The gradient of the curve at various times is indicated on each of the graphs.

1 We14

b

x

a

x Gradient = 0

Gradient = 3

Gradient = −3 0

1

2

3

4

5

Gradient = −2

t

6

Gradient = 2 Gradient = −2 Gradient = −1 2 mC

0

Gradient = 1 Gradient = 0

Gradients are shown at various points on a position–time graph. m=0

x

m = −6 m=6 m = −12

m = 12 0

366

Maths Quest 11 Mathematical Methods CAS

1

2

3

4

t

1 2 Gradient = 2

t

The velocity–time graph obtained from this position–time graph is: v 12

a

B

0

1

2

3

v 12

0

t

4

v 12

C

1

2

3

4

0

t

−12

2

3

t

4

−12

v 12

d

1

e

0

1

2

3

v 12

t

4

0

1

2

3

4

t

−12

Sketch the velocity–time graph for each of the following position–time graphs.

3 a

x 5

b

x 4

c

x 1

d

4

x 8 7

3

2

0

2 1

0

0 −1

1

2

3

t

1

2

3

4

t

1

6

t

2

5

−1

4 3

−2

2

−2

1 −4

−3

−1

−4 −5

0

1

2

3

4

t

−2

−6

−3 −4

−8

−5 −6 −7 −8

For each velocity–time graph shown below, sketch a position–time graph, given that the particle starts at the origin.

4 We15 a

b

v

c

v

−3

3

0

v

4

t

0

4

t

0 −1

1

4

t

−2

ChapTer 8 • Rates of change

367

v

d

v 8

e

v 3

f

3

0 −1

3

4

0

t

2

0

t

4

1

2

3

t

4

−3

−8

A particle starting at the origin moves with velocity v(t) = 2t − 4.

5

a Complete the table below.

t

0

1

2

3

4

v

−4

−2

0

2

4

Gradient of x–t graph b Sketch the position–time graph. 6

The velocities for a particle starting at the origin are given as a function of time. Sketch a position–time graph for each using t ∈ [0, 4]. a v=t+2 b v=2−t c v = 3t d v = −t

v 3

0

x

B

0

d

2

x

C

0

4 t

x

e

3

4

t

x

0

4 t

2

2

−3

A position–time graph is obtained from the velocity–time graph shown at right. For a particle starting at the origin, this position–time graph will look like:

7 mC

a

1

4 t

2

x

2 0

2

0

4 t

2

4 t

−1

8

Briefly describe the motion of a car during the first 30 seconds for each of the velocity–time graphs below. a v (m/s)

b v (m/s)

20

18

0

368

Maths Quest 11 Mathematical Methods CAS

30 t (s)

0

c v (m/s) 25

15

30 t (s)

0

10

25 30 t (s)

Sketch a velocity–time graph for each car described below for its first 20 seconds of motion.

9

a A car, initially at rest, increases its speed at a constant rate until it reaches a speed of 35 m/s after

10 s. It then maintains this speed. b A car is travelling at a constant speed 30 m/s. After 15 s the brakes are applied, decreasing its

speed at a constant rate until it stops after another 5 s. 10 A parachutist jumps from an aircraft and freefalls for 6 seconds. a If a parachutist falls y metres in t seconds where y = 5t2, find the average speed of the parachutist between: i t = 0 and t = 3 ii t = 3 and t = 6. b What is the speed of the parachutist after 6 seconds of freefall? c When the parachute is released (after 6 seconds of freefall), the speed of the parachutist is reduced by 2 m/s every second until a speed of 4 m/s is reached. How long after jumping from the aircraft does it take the parachutist to reach a speed of 4 m/s?

rates of change of polynomials

8i

We have seen that instantaneous rates of change can be found from a graph by finding the gradient of the tangent drawn through the point in question. The following method uses a series of approximations to find the gradient. Worked example 16

a Plot the graph of y = 2x2, x ∈ [0, 4]. b Draw a line through the following points on the curve and find the gradient of each line. i x = 2 and x = 3 ii x = 2 and x = 2.1 iii x = 2 and x = 2.01 iv x = 2 and x = 2.001 c Hence, predict the gradient of the tangent at the point on the curve where x = 2. Think

a

1

Use a table of values or a CAS calculator.

2

Plot the points using graph paper.

WriTe/draW

a

x y

0 0

1 2

2 8

3 18

4 32

y 32 i ii iii

18

iv

8 0 b i 1 Mark the points on the curve where x = 2 and

x = 3 and rule a line through them.

2

Select suitable values to calculate the rise . gradient using run When x = 2, y = 8; when x = 3, y = 18.

1

2

3

4

x

b i Line i

18 − 8 3− 2 10 = 1 = 10

Gradient =

ChapTer 8 • Rates of change

369

ii 1 Again, mark the points and rule a line

ii Line ii

through them. 2

When x = 2.1, y = 2(2.1)2 = 8.82

Select suitable points to calculate the gradient. Calculate the y-value for x = 2.1 using the rule y = 2x2.

8.82 − 8 2.1 − 2 0.82 = 0.1 = 8.2

Gradient =

iii 1 Again, mark the points and rule a line

iii Line iii

through them. 2

Select suitable points to calculate the gradient. This may be difficult, so select an x-value and calculate the y-value using the rule y = 2x2.

iv 1 Again, mark the points and rule a line

When x = 2.01, y = 2(2.01)2 = 8.0802 8.0802 − 8 Gradient = 2.01 − 2 0.0802 = 0.01 = 8.02 iv Line iv

through them. 2

When x = 2.001, y = 2(2.001)2 = 8.008 002 8.008 002 − 8 Gradient = 2.001 − 2 0.008 002 = 0.001 = 8.002

Select suitable points to calculate the gradient. Calculate the y-value for x = 2.001 using the rule y = 2x2.

c Continue to bring the second x-value closer and

c The gradient of the tangent at the point on

closer to 2. As the second x-value becomes closer to 2, the gradient of the line is becoming closer to 8. Notice that the slope of each line has progressively become closer to the slope of the tangent at x = 2.

the curve where x = 2 appears to be 8.

Worked example 17

Consider the function f (x) = 2 x2. a Using a CAS calculator, calculate the gradients of the chords connecting two points on the curve with the following coordinates: i x = 3 and x = 4 ii x = 3 and x = 3.1 iii x = 3 and x = 3.01 iv x = 3 and x = 3.001. b Hence, predict the gradient of the curve at x = 3. Think

a 1 Using the average rate of change

function of the CAS calculator, complete the function entry lines as: avgRC(2x2, x, 1) | x = 3 avgRC(2x2, x, 0.1) | x = 3 avgRC(2x2, x, 0.01) | x = 3 avgRC(2x2, x, 0.001) | x = 3 370

Maths Quest 11 Mathematical Methods CAS

WriTe

a i ii iii iv

avgRC(2x2, x, 1) | x = 3 avgRC(2x2, x, 0.1) | x = 3 avgRC(2x2, x, 0.01) | x = 3 avgRC(2x2, x, 0.001) | x = 3

2

Record the results.

b As the interval becomes smaller and

the two points come closer together, the chord approximates the tangent. The gradient of this tangent is the gradient of the curve at x = 3.

i Between x = 3 and x = 4, the gradient is 14. ii Between x = 3 and x = 3.1, the gradient is 12.2. iii Between x = 3 and x = 3.01, the gradient is 12.01. iv Between x = 3 and x = 3.001, the gradient is 12.002. b The gradient of f (x) = 2x2 approaches 12 as the interval

decreases. So, the gradient of f (x) = 2x2 at x = 3 is 12.

The previous example demonstrates that the gradient of a tangent to a function at a given point, x = a, can be approximated by finding the gradient of a line joining the point x = a and another point as close as possible to x = a.

Hybrid functions A hybrid function is one that has different rules over different sections of its domain. For example: x + 2, x ≤ 0 f (x) = 2 x>0 y x , The graph of f (x) is shown at right. This is a discontinuous function because there is a break (or jump) in 2 y = x2 the graph at x = 0. There is also no defined gradient at the point x = 0 on y=x+2 this graph, because at x = 0 two different values of the gradient can be obtained. x 0 −2 2 If the rate of change of y at x = 0 is calculated using points just to the left of x = 0, and then calculated using points just to the right of x = 0, different values are obtained: rise 2 = =1 Straight line: m = run 2 y Parabola: At the turning point (x = 0), gradient = 0. Not all hybrid functions have a break in them. For example: x , x ≤ 0 f (x) = 2 x , x > 0

Gradient of straight line is 1 at x = 0. 2

−2

0

y = x2

x

Gradient of parabola The graph of f (x) is shown at right. y=x is 0 at x = 0. However, even for this graph, there is no defined gradient at x = 0. The gradient of the tangent approaching x = 0 from the left is equal to the gradient of the straight line, that is, 1. The gradient of the tangent to the parabola approaching x = 0 from the right, is equal to zero. Therefore the gradient (or instantaneous rate of change) does not exist at x = 0. If the hybrid function is ‘smooth’ at the point where the combined functions meet, then the gradient does exist. A function is smooth at a given point if the left-hand gradient equals the right-hand gradient at that point. For example: y x ≤1 2 x , f (x) = 2 x + 1, x > 1 y = x2 + 1 The left-hand gradient of f (x) at x = 1 is the gradient of the line y = 2x, which equals 2. The right-hand gradient of f (x) at x = 1 is the gradient of the Both gradients 2 are 2 at x = 1. parabola y = x2 + 1, which also equals 2. So the left-hand gradient equals the right-hand gradient. Therefore, this hybrid function x 0 1 is smooth at x = 1, and the gradient (or the instantaneous rate of y = 2x change) at x = 1 is equal to 2. Chapter 8 • Rates of change 371

In summary, for a hybrid function to have a gradient (or instantaneous rate of change) at the point in the domain where the rule changes: 1. the function must be continuous at that point 2. the left-hand gradient must equal the right-hand gradient at that point. This makes the curve smooth. Worked example 18

Verify that the rate of change of the hybrid function 2 x, x≤2 f ( x) = 2 x>2 x , does not exist at x = 2, using a CAS calculator. Think

WriTe

1

Define a hybrid function using your CAS calculator.

2 x x ≤ 2 Define f ( x ) = 2 x x > 2

2

Set f 1(x) = f (x) for graphing.

f 1(x) = f (x)

3

Using the CAS calculator’s draw tangents feature, draw the tangent to y = 2x at x = 2.

Gradient = 2

4

Draw the tangent to y = x2 at x = 2.

Gradient = 4

5

Since the left-hand gradient and right-hand gradient are not equal, the rate of change does not exist at x = 2. Write the answer.

The left-hand gradient is 2 and the right-hand gradient is approximately 4 around the point x = 2. As the two results are not the same, the rate of change of the hybrid function 2 x , x≤2 f (x) = 2 x>2 x , does not exist at x = 2.

Worked example 19

4 − 2 t, t ∈ [0, 1] For the displacement function x( t ) = find: 2 3 − t , t ∈(1, 6] a the starting position b the finishing position c the displacement between t = 0 and t = 2 seconds d the instantaneous velocity at t = 1, if it exists. Think

372

WriTe

a Evaluate x(0).

a x(t) = 4 – 2t

b Evaluate x(6).

b x(t) = 3 – t2

c Evaluate x(2) − x(0). Note that x(0) = 4 from part a; x(2) = 3 − t2 at t = 2.

c x(2) – x(0) = (3 – 22) – [4 – 2(0)]

Maths Quest 11 Mathematical Methods CAS

x(0) = 4 – 2 × 0 =4 The starting position is 4 units. x(6) = 3 − 62 = 3 – 36 = −33 The finishing position is −33 units. =3–4–4 = −5 The displacement is −5 units.

d 1 Velocity is the rate of change of displacement,

d

therefore find the instantaneous rate of change of x at t = 1, if it exists. 2

To evaluate the left-hand gradient at t = 1, use the line x = 4 – 2t. Note that m = −2 in this rule.

Left-hand gradient = −2

3

To evaluate the right-hand gradient at t = 1, use the line x = 3 – t2. Calculate values for x when t = 1 and t = 1.001.

x(t) = 3 – t2 x(1) = 3 – 12 =2 x(t) = 3 – t2 x(1.001) = 3 – (1.001)2 = 1.998

4

Evaluate the right-hand gradient at t = 1.

5

The left-hand gradient equals the right-hand gradient, so the gradient exists and equals −2.

1.998 − 2 1.001 − 1 − 0.002 − Right-hand gradient ≈ = 2.001 0.001 Right-hand gradient ≈ −2 Right-hand gradient ≈

The instantaneous velocity at t = 1 is −2 units per second.

rates of change of polynomials

exercise 8i

1 We16 a Plot the graph of y = x2, x ∈ [0, 4]. b Draw a line through the following points on the curve and find the gradient of each line. i x = 2 and x = 4 ii x = 2 and x = 3 iii x = 2 and x = 2.5 iv x = 2 and x = 2.1 c Hence, predict the gradient of the tangent at the point on the curve where x = 2. d Use a CAS calculator to find the average rates of change for the intervals in part b. Does this

diGiTal doC doc-9779 Gradient at a point

change your prediction for the gradient? 2a Plot the graph of y = x3, x ∈ [0, 2]. b Draw a line joining the following points on the curve and find the gradient of each line. i x = 1 and x = 2 ii x = 1 and x = 1.5 iii x = 1 and x = 1.1 iv x = 1 and x = 1.01 c Hence, predict the gradient of the tangent to the curve where x = 1. d Use a CAS calculator to find the average rates of change for the intervals in part b.

Does this change your prediction for the gradient? 3 We17 a If y = x2 − x + 1, copy and complete the following table of values.

x

3

2

1.5

1.1

1.01

1

y b Hence, find the gradient of a line joining the points on the curve where x = 1 and: i x=3 ii x = 2 iii x = 1.5 iv x = 1.1 v x = 1.01. c Predict the gradient of the tangent at the point on the curve where x = 1.

The gradient of a tangent to a curve at the point where x = 3 is best approximated by the gradient of the line joining the points on the curve where: a x = 3 and x = 4 B x = 2 and x = 4 C x = 3 and x = 3.2 d x = 3 and x = 3.001 e x = 3 and x = 3.01

4 mC

ChapTer 8 • Rates of change

373

The rate of change at a point P on a curve is best approximated by finding the gradient of a line joining the point P and another point that is: a furthest from it B closest to it C anywhere on the curve d the origin e on the x-axis Use a CAS calculator to find the gradient of the following functions at the point indicated. a f (x) = x2 at x = 1 b f (x) = 3x2 at x = 2 c f (x) = x2 + 2x at x = 2 d f (x) = x3 + x at x = 1 e f (x) = 2 − 3x − x2 at x = 2 f f (x) = x3 − 2x + 3 at x = 1.5 g f (x) = x3 + x2 − 2x at x = 3 h f (x) = 12 x2 − 3x + 5 at x = 0.5 Hailstones fall x metres in t seconds where x = 4.9t2. Estimate the speed of a hailstone after: a 2 seconds b 3 seconds. h (m) The height, h metres, of a roller-coaster is given by − 1 3 3 2 50 h= d + d 80 8 where d is the horizontal distance from 0, in metres. Estimate the gradient of the curve describing the path of the rollercoaster, accurate to 2 decimal places, at the point where d is: 0 20 d (m) a 10 metres b 15 metres c 20 metres d 0 metres. We18 Verify that the rate of change of the function 2 x , x ≤1 f (x) = 2 x >1 2 x , does not exist at x = 1. We18 Verify that the rate of change of the function − 2 x , x<2 f (x) = − 2 x≥2 x , does not exist at x = 2. 5 − 4t , t ∈[0, 1] We19 For the displacement function x (t ) = , find: 2 t ∈(1, 6] 3 − 2t , a the starting position b the finishing position c the displacement between t = 0 and t = 2 seconds d the instantaneous velocity at t = 1, if it exists. 4t , t ∈[0, 2] For the displacement function x (t ) = , find: 2 t ∈(2, 6] 4+t , a the starting position b the finishing position c the displacement between t = 0 and t = 2 seconds d the instantaneous velocity at i t = 2 and ii t = 3, if it exists.

5 mC

6

7

8

9

10

11

12

374

Maths Quest 11 Mathematical Methods CAS

Summary identifying rates

• A rate describes how much one quantity changes with respect to another. • For two quantities A and B: change in A The rate of change of A with respect to B = amount of A per B = . change in B

Constant rates

• • • •

Variable rates

The gradient of a graph represents the rate of change. A positive gradient shows a positive rate of change or increasing function. A negative gradient shows a negative rate of change or decreasing function. A linear or straight line graph shows that the rate of change is constant.

v

t

• A variable rate exists when the rate of change is not constant. • On a graph, a variable rate of change can be seen when the points do not lie on a straight line. d

0

average rates of change

1

2

3 t

• Over a specified interval: change in A 1. Average rate of change of A with respect to B = . change in B 2. Average rate of change of a function f (t) over the time interval t = a to t = b is: change in f (t ) f (b) − f (a) = change in t b−a

y Rise Run

x

0

rise • When using a graph, find the gradient of the chord formed between the end points of the run interval. • When using a table of values or rules, find the change in each quantity over the specified interval. instantaneous rates

• The instantaneous rate of change is the rate of change at a particular instant rather than over an interval. • The instantaneous rate of change can be found by calculating the gradient of the tangent drawn to the curve at the required point.

y P Tangent 0

motion graphs (kinematics)

• Position, x, describes where an object is or was. • Distance, d, is how far an object has travelled. • Displacement, s, describes the change in an object’s position; that is, s = xfinal – xinitial. d distance • Speed = , or speed = . t time taken s • Velocity = displacement , or v = . t time taken

x

x

0

t

• The gradient of a position–time graph for an object gives the velocity of that object. • Distance and speed are always positive. Displacement and velocity can be either positive or negative depending on the direction of motion.

ChapTer 8 • Rates of change

375

relating velocity–time graphs (the gradient function) to position– time graphs (the original function)

• To draw a velocity–time graph (or gradient function) from a position–time graph (or original function): 1. Calculate the gradient of the graph at given time values. For an x–t graph this gives the velocity. 2. Construct a table of values showing the velocity (or gradient) at given time values (or x-values). 3. Use this table to plot the velocity–time graph (or gradient function). • To draw a position–time graph (or original function) from a velocity–time graph (or gradient function): 1. Set up a table of corresponding velocity (gradient) and time values (x-values) from the graph. 2. On a set of axes, plot a point for the initial position. Starting from this point, use the values in the table to estimate the appropriate relative slope at each given time (or x-value) to produce a curve.

rates of change of polynomials

• The gradient of a tangent to a function at a given point, x = a, can be approximated by finding the gradient of a line joining the point x = a and a point as close as possible to x = a. The closer the two points are, the more accurate the estimate of the gradient. • For a hybrid function to have a gradient (or instantaneous rate of change) at the point in the domain where the rule changes: 1. the function must be continuous at that point 2. the left-hand gradient must equal the right-hand gradient at that point.

376

Maths Quest 11 Mathematical Methods CAS

Chapter review 1 A car manufacturer claims a vehicle uses 12 litres of fuel per 100 km. a Express this rate in km/litre. b How far could the car travel using 35 litres? c How many litres are required to travel 765 km?

S ho rT a n S W er

2 a Determine the rate at which water is leaking from a tank, using the graph below. w (m3) 840 600

0

8

t (hours)

b How long does it take the tank to empty? 3 a Use the table below to plot the graph of M versus t.

t M

0

1

2

−9

−7

−1

3 9

4 23

b Is the rate of change of M constant or variable? Explain your answer. c Estimate the gradient when t = 2. 4 The amount of substance, A kg, in a container at any time, t hours, is A = t2 − 3t + 4, t ∈ [0, 5]. a Find the average rate of change during the first 4 hours. b At what rate is the amount changing after 4 hours? 5 Calculate the average rate of change of y = tan (x) between x =

π 4

π

and x = 3 . 6 A particle is moving in a straight line according to the position–time function x(t) = t2 − 2t + 5, where t ∈ [0, 5] and t is given in seconds. a Sketch the position–time graph. b Where did the journey start? c When and where did the particle change direction? x (m) d What was the distance travelled in the first 4 seconds? 7 From the position–time graph at right, find: 200 a the initial position b how long the object was not moving c the velocity during the first 5 seconds 30 d the velocity during the last 4 seconds e the total distance travelled 0 5 8 12 t (s) f the average speed for the journey. 8 For the velocity–time graph below, sketch a position–time graph, given that the object starts at the origin. v 2

0

3

6

t

−2 9 The height, h metres, of a projectile above the ground at any time, t seconds, is given by the function

h(t) = 5 + 12t − t2. a Find the rate of change of height when t equals: i 4 seconds ii 6 seconds iii 10 seconds. b Briefly describe the change of height of the projectile within the first 12 seconds. 2 x , x ≤ 2 10 Verify that the rate of change of the function f ( x ) = 2 does not exist at x = 2. x , x > 2

ChapTer 8 • Rates of change

377

2t ,

t ∈[0, 1]

11 For the displacement function x (t ) = 2 1 + t , a the starting position b the finishing position c the displacement between t = 0 and t = 2 d the instantaneous velocity at i t = 1 and ii t = 2, if it exists. m U lT ip l e C h oiCe

t ∈(1, 6]

, find:

1 Which one of the following is not a rate? a 35 km B 18 breaths per minute d 50 km/h e −8 °C/hour

C $2.50 per kg

2 Which one of the following is a constant rate? a The number of people entering the zoo per hour B The number of days it rains in Sydney per year C The hourly rate of pay of a tutor d The number of crates of fruit picked per hour by a fruit-picker e The number of patients visiting a doctor per day 3 The graph below shows the weight, w kg, of a tank of water t minutes after water is poured into it. w (kg) 70

20 0

t (min)

40

The rate of change of w in kg/min is: a 1.75 B 0.8 C 2.25 4 The graph below has a negative rate of change in the interval: a OP B PQ C QR

d 1.25

e 2

d RS

e OQ

d 6

e 2

y S

P R Q

x

O 5 The gradient of the chord PQ below is: a 3

B 12

C y

1 3

Q

21 P

9 0

5

9

x

6 From the table below, the average rate of change of H between t = 2 and t = 5 is: a 20 m/h B 38 m/h C 30 m/h d 190 m/h

t(h) H(m)

378

0 0

1 20

Maths Quest 11 Mathematical Methods CAS

2 40

3 70

4 120

5 190

6 280

e 50 m/h

7 The instantaneous rate of change of v at t = 4 on the graph below is: a 13.75 L/min B 12.5 L/min C 6.25 L/min d 55 L/min

e 8 L/min

v (L) 80 55 30 0

4

8 t (min)

8 Which one of the following describes an instantaneous rate? a The temperature dropped 15°C in 2 hours. B Harry was caught doing 85 km/h. C Jill ran 5 km in 28 minutes. d Jenny earned $80 in 7 12 hours. e A balloon rose 600 metres in 90 minutes. 9 The following diagram shows the position of a particle (x metres) travelling in a straight line at various

times (t seconds). t=5

t=2

t=0

−3 −2 −1

0

1

2

3

4

x

5

a The displacement after 5 seconds is: B 11 m C 7m a −2 m b The average speed in the first 5 seconds is: − a 11 m/s B 75 m/s C 11 m/s 5 5 c The average velocity in the first 5 seconds is: − a 11 m/s B 35 m/s C 35 m/s 5

d

−3

m

e 5m

d

−

7 5

m/s

e 1 m/s

d

−

11 5

m/s

e 1 m/s

10 The graph of a gradient function is shown below. y 2

Gradient function x

0

The graph of the original function could be: a

y 4

y

B

C

0 0 d

x

2

−2 e

y

4

y

x 0

x

y

2 1 0

1

x

0

1

x

ChapTer 8 • Rates of change

379

11 A position–time graph is shown below. x

0

t

2

Which one of the following diagrams represents its velocity–time graph? a

0

d

B

v

t

2

v

0

0

0

t

2

v

t

2

v

e

t

2

C

v

0

t

2

12 The velocity–time graph of a particle travelling in a straight line and starting at the origin is shown

below. v 3

0 −1

3

t

4

Which one of the following represents its position–time graph? a

x

0

d

B

3

4

0

e

3

4

Maths Quest 11 Mathematical Methods CAS

3

4

t

4

t

x

t 0

380

C

x

t

x

0

x

0

3

4

t

13 The rate of change of a polynomial f (x) when x = 3 is closest to: a

f (3.1) − f (3) 3.1 − 3

B

d

f (3) − f (0) 3− 0

e

f (3.001) − f (3) 3.001 − 3 f (3.00 001) − f (3) 3.00 001 − 3

14 The rate of change of y = x2 + x when x = 2 is: B 3 a 2 e 6 d 5

C

f (3.5) − f (3) 3.5 − 3

C 4

1 A soft-drink vending machine at a railway station dispenses cans of soft drink. It has a capacity of

600 cans when full. On a particular day: • the machine is half full at 6.00 am • in the next four hours drinks are dispensed at a constant rate of 15 per hour • at 10.00 am the machine is filled • in the next four hours the machine dispenses an average of 60 cans per hour at a steadily increasing rate • finally, from 2.00 pm to 6.00 pm an average of 30 cans per hour are dispensed at a steadily decreasing rate. a Sketch a graph showing the number of cans in the machine over the period from 6.00 am to 6.00 pm. Use the graph to: b find the number of cans in the machine at 6.00 pm c estimate at what time in the afternoon the machine is half full d estimate the rate at which cans are dispensed at 1.00 pm.

ex Ten d ed r eS p o n S e

2 The concentration (x mg/L) of a tranquilliser in a patient’s bloodstream at any time, t hours, after it is

administered is given by the following rule: 2t ,t≥0 t2 + 1 The tranquilliser is only effective if the concentration is at least 0.5 mg/L. a Determine the concentration at: i t=0 ii t = 1 iii t = 2 iv t = 4. b With the aid of a CAS calculator, sketch the graph for the concentration function. c Find the exact length of time that the tranquilliser is effective. d Determine the average rate at which the tranquilliser is absorbed into the bloodstream from t = 1 to t = 3 hours. Explain why the rate is negative. e Determine the instantaneous rate at which the tranquilliser is absorbed into the bloodstream at t = 0.5 and t = 2 hours. (Give answers to 2 decimal places.) f Verify that the rate at which the tranquilliser is absorbed into the bloodstream is 0 at t = 1. Hence, give the time and concentration when the concentration is at a maximum. x=

3 The position (x centimetres from the origin, 0) of an object travelling in a straight line at any time,

t seconds, is given by:

2t + 1, x= 2 t − 2t + 5,

t ∈[0, 2] t ∈(2, 5]

Determine the initial and final position of the object. Sketch the position–time graph. Verify that the velocity exists at t = 2 and determine this velocity. Find the displacement from t = 1 to t = 4. The average velocity from t = 1 to t = b is 2.9 cm/s. Find the value of b. Determine the time, t = c, for which the displacement from t = 0.5 to t = 2 is equal to the displacement from t = 2 to t = c. g Sketch the velocity–time graph for this object. a b c d e f

ChapTer 8 • Rates of change

381

4 A skateboarder starts on a 1 m long flat section before coming down on an inclined ramp. When

diGiTal doC doc-9783 Test Yourself Chapter 8

382

he reaches the top of the ramp a stopwatch is started. The distance he travels is known to follow an exponential curve d = at for the first 2 seconds of motion, where d is in metres and t in seconds. If his speed initially (at t = 0) is 1 m/s, find his equation of motion; that is, find a. Use the following steps to help. a Using a CAS calculator, find the gradient of d = 2t at t = 0. b Using a CAS calculator, find the gradient of d = 3t at t = 0. Clearly the value of a must be between 2 and 3. c Using trial and error, find the value of a correct to 2 decimal places. d Show that the accurate value is a = 2.718 281 8 (correct to 7 decimal places). e Find the value of 2.718 281 82 (correct to 5 decimal places). f Find the gradient of d = 2.718 281 8t at t = 2 (correct to 5 decimal places). g What do you notice? Note: 2.718 281 8 . . . is a very important transcendental number in mathematics and has the symbol e. You will find it on your calculator.

Maths Quest 11 Mathematical Methods CAS

ICT activities Chapter opener diGiTal doC • 10 Quick Questions doc-9773: Warm-up with ten quick questions on rates of change (page 341)

8a

identifying rates

diGiTal doC • SkillSHEET 8.1 doc-9774: Practise determining rates of change and using them in calculations (page 343)

8B

Constant rates

diGiTal doC • doc-9775: Investigate plotting relations using a spreadsheet (page 347)

8C

Variable rates

TUTorial • We 5 eles-1432: Watch a tutorial on identifying constant or variable rates of change (page 348) diGiTal doC • doc-9775: Investigate plotting relations using a spreadsheet (page 349)

8d

average rates of change

TUTorial • We 8 eles-1433: Watch a tutorial on calculating the average rate of change of the temperature of a room over a set period of time (page 350) diGiTal doCS • SkillSHEET 8.2 doc-9776: Practise finding the gradient of a straight line (page 351) • doc-9777: Investigate the gradient between two points using a spreadsheet (page 353) • WorkSHEET 8.1 doc-9778: Identify rates, form equations and make calculations involving rates (page 353)

8e

instantaneous rates

TUTorial • We 11 eles-1434: Watch a tutorial on calculating the average speed and velocity of a lift (page 358) diGiTal doCS • SkillSHEET 8.3 doc-9780: Practise calculating displacement and total distance travelled (page 360) • SkillSHEET 8.4 doc-9781: Practise describing intervals using interval notation (page 361) • WorkSHEET 8.2 doc-9782: Calculate displacement, instantaneous rates of change and starting and finishing positions, given position functions (page 363)

8G relating the gradient function to the original function diGiTal doCS • doc-9779: Investigate the gradient at a point using a spreadsheet (page 364) • doc-9775: Investigate plotting relations using a spreadsheet (page 364)

8h relating velocity–time graphs to position–time graphs TUTorial • We 15 eles-1435: Watch a tutorial on sketching a position–time graph of a particle moving in a straight line (page 366)

8i

rates of change of polynomials

diGiTal doC • doc-9779: Investigate the gradient at a point using a spreadsheet (page 373)

Chapter review diGiTal doC • Test Yourself doc-9783: Take the end-of-chapter test to test your progress (page 382)

To access eBookPLUS activities, log on to www.jacplus.com.au

diGiTal doC • doc-9779: Investigate the gradient at a point using a spreadsheet (page 356)

8F

motion graphs (kinematics)

inTeraCTiViTY • Motion graphs (kinematics) int-0267: Learn how distance-time, velocity-time and acceleration-time graphs are related graphically (page 357)

ChapTer 8 • Rates of change

383

Answers CHAPTER 8 raTeS oF ChanGe

exercise 8C

exercise 8a

1 2

3 4

5 6

7

identifying rates a, b, d, f, i, k, m, n, p a C bA d F f H i E k I mD n B p G b, c, e, g a $10.60 b $31.41 c $303.75 d $5.59 e $75.05 f $609 697 a A bC a 8.33 m/s b 3.5 °C/h c $25 per person d 260 g/L e $4.67/m2 f 1.16 kg/week a 20 h b 100 h c C

exercise 8B

e 9 a c e

b

throughout the time interval, except at the region E. E shows a constant rate of change. The rate of change is positive throughout the time interval, except at region E. E has a zero rate of change. D A D

c d e 4 a W (kg)

scarcer. $156 1000 m 80 s

b 70 s d 30 s

0

5 4 3 2 1 0

exercise 8d

average rates of change

2 5

70

Q

384

P 1

iii 0 km/h P Q

0

1

t (h)

3

iii −10 km/h b 0.5 ii 0.5 kg/m

70 50

N 600 500 400 300 200 100 0

0 10 am 11 am 12 pm 1 pm

b, c

T (°C) 120 (8, 84)

20 1

t (h)

3

c i d (km) 70

Q P

ii 15

Maths Quest 11 Mathematical Methods CAS

d e 9 a b

1

3

t (h)

iii 15 km/h

(2, 24) 2 4 6 8 10 t (s)

10 10 °C/s 200 g i 1 g/week iii 10 g/week

exercise 8e

b 6 h 0

T

c i 170 ii 20 iii 0 d i 170 people/h ii 20 people/h iii 0 people/h e Most people arrive in the morning, few

P

iii 10 km/h

40

2 pm 3 pm 4 pm 5 pm

Q

ii 10

b i d iii

t (h)

3

ii −10

iii 30 km/h

b i d (km)

2 3

t (h)

3

in the middle of the day and none later in the afternoon. 7 a i 110 m/min ii 70 m/min iii 45 m/min iv 15 m/min v 5 m/min b Decreasing 8 a 20 °C

b −4 2 a, c 3 a i d (km)

0

5 10 15 20 25 x (cm)

Q

1

0

b 0.2 kg/cm c W = 0.2x 11 a 6 L/h 12 a iv c ii

0

1 2 3 4 t (min)

b No c Variable d i 2 kg/min ii 3 kg/min iii 7 kg/min 6 a Constant b Variable

10 a w (kg)

P

70

change.

ii 30 150180 t (s)

t (h)

3

iii −5 km/h

e i d (km)

5 D 6 a, b

10 0

70

1

4 a 1.5 c i 1.5 kg/m d No, because it is a variable rate of

1000

0

0

1 2 3 4 t (min)

W (kg)

820

Q

ii −5

The points form a straight line. A constant 3 kg/min 20 16 12 8 4

1 a

P

70 60

90 70

0

b c d 5 a

d (metres)

420

d i d (km)

ii 0 f i d (km)

12 9 6 3

Constant rates

1 c, d, g, h, j 2 b, d, e, g, h, j 3 i b, e, h ii g, j iii d 4 D 5 C 6 a i 50 ii 50 m/h iii d = 50t b i 7 ii 7 L/min iii v = 7t + 50 c i −1 ii −1 kg/week iii w = 100 − t d i 0 ii 0 m/year iii h = 75 e i 0.75 ii 0.75 g/min iii w = 0.75t + 10 f i −0.25 ii −0.25 g/L iii w = −0.25v + 35 7 a D b B 8 a $24/h b Rest c $18/h d The picker is tiring a little or fruit is

Variable rates

1 a, c, d, e, g, h 2 b, d, e, f 3 a The rate of change is variable

1 a A e A 2 a, c, d, h

ii −2 g/week iv 28 g/week

instantaneous rates b I c A d I f A g A h I

3 a 1 c −1 4 a 2.5 g/min b 16 g/min c −25 g/min d 12 g/min 5 a 85 kg after 35 weeks b Approx. −1 kg/week c Approx. 2.0 kg/week 6 a Approx. 6 hours b Approx. 3.5 °C/hour y 7 a, b

b 2.5 d 0.25

x

6 a

4 0

0

i x=0 iii No b

ii Right iv x = 10

x 16

0 −2

4 1

2

3

x

4

d C

e A

c

−2

ii Right iv x = 16

x 15

5

−4 −3 −2 −1 0 −2

1

2

3

4

x

1 a c 2 a c 3 a b c d 4 a c e g i 5 a b c

d e f

ii 0

0 −1

1 2 3 4 5

t

i x=0 iii Yes, t = 1, x = −1

−7

exercise 8F

4 m/s m/s

−2

2 −4 2 m/s −4 m/s

iii 0 iii 0 m/s

x

0 −2 −4 −6

t

6

i x= iii No

y 9

9 6 cm/min

ii v ii v

−2

4 2

8

c i 4

4

b i iv c i iv d

12

c C 8 a, b

t

1 2 3 4 5

−5

t

1 2 3 4 5

16

0

x

7 a

10 8 6 4 2

iii −4

x

d

8 a i Positive ii Speeding up b i Negative ii Slowing down c i Negative ii Speeding up d i Positive ii Slowing down e i Positive ii Neither — velocity is constant. 9 a C b E c B d F e A f D 10 x (m) 45

1 0

motion graphs (kinematics) b B d A b True d False

C D False True 40 m −2 m (or 2 m below the platform) 0.5 m/s −0.025 m/s (or 0.025 m/s downwards) x=1 b x = −3 Right d t=2 C f D D h D C j B i x=0 ii Right iii t = 2, x = 8 iv t = 5, x = −3 i x=4 ii Right iii t = 4, x = 12 iv t = 6, x = 10 i x=0 ii Right iii t = 3, x = 12 and t = 6, x = 3 iv t = 8, x = 10 i x=0 ii Left iii t = 1, x = −5 iv t = 3, x = 18 i x = −3 ii Left 1 iii t = 1 , x = −6 iv t = 5, x = 5 2 i x=2 ii Left iii t = 3, x = −5 and t = 5, x = 5 iv t = 6, x = 4

ii Left iv x = 15

25

t

1 2 3 4 5

0

−8

i x=0 iii Yes, t = 1, x = 1 e

ii Right iv x = −8

x

9

a b c d 11 a c e

1

2

3

5 t (s)

4

45 m t=5 t=2 20 m/s Alan 12 metres 40 s

b Bill d 2.0 m/s f Twice

exercise 8G relating the gradient function to the original function

4

1 a

0

1 2 3 4 5

t

4

ii Left iv x = 9

−4 −2 0 2 −4

x 18

4

x

Gradient function

b 0

y 12 8

i x=4 iii Yes, t = 2, x = 0 f

t

1 2 3 4 5

1 2 3 4 5

y 12 8

t

4 −4 −2 0 2 −4

−12

i x = −12 iii No

Gradient function 4

x

ii Right iv x = 18

ChapTer 8 • Rates of change

385

c

v

3 a

y 12

5 a t

5

8

Gradient function

4

0 2

−4 −2 −4

x

4

0

1

2

t

3

0 −4

0 1

π – 4

− π–

c

x

2

1

2

t

4

3

0 2

4

6

x

8

d

0

2

4

2

3

t

4

6 a x

b

x

c

x

t

4

0 3 0

2

4 a

1 −3 −2 −1

−1 −2

1

2

t

1 2 3 4

3 x 0 0

b

−5

the gradient function of e is

t

7A 8 a It travels at a constant velocity of

c x

ex.

20 m/s.

b It starts at rest, increasing its velocity

at a constant rate for 15 seconds when it reaches 18 m/s. It then maintains this velocity. c It starts from rest, increasing its velocity at a constant rate for 10 seconds when it reaches 25 m/s. It maintains this velocity for a further 15 seconds, then decreases its velocity at a constant rate for 5 seconds, bringing it back to rest.

or inverted.

0

exercise 8h

relating velocity–time graphs to position–time graphs

1 2 3 4 5 t

d x

v 3

0

6 t

1

2

3

4 t

9 a v (m/s)

e x

35

−3 v 2

0

f 2

4

t

4

6 There are stationary (turning) points there. 7 The original function will also be negative

3

x

−8

4 Quadratic functions 5 The gradient function of sin (x) is cos (x); x

d

t

4

0

0

functions are always linear.

t

4

x

2 It is a straight line. 3 The gradient functions of quadratic

1

t

4

x

−3

0

2

12

−4

1

2

3

4

0

t

10

20 t (s)

b v (m/s)

x

t

30 0

2 C

386

−2

v

4 3

−2

4

12 y 5

0

1

0

2 t

1

−2

f

b

4

2

v

0

4

1 a

3

0

2

Gradient function

y

2

−8

−2 −1

e

−4

0

4

Gradient 4 (and original) function 2

Gradient of x–t graph

1 2

−

4

b x

8 y 8 6

−

v

b d

0

V

Maths Quest 11 Mathematical Methods CAS

1

2

3

4

t 0

15 20

t (s)

10 a i 15 m/s b 60 m/s c 34 seconds

ChapTer reVieW

ii 45 m/s

y

i (4, 16)

16

ii

0

(1, 1) 2

3

x

4

ii 5

7

ii

180

iii 3.31

iv 3.0301

ii 4.75

iii 3.31

iv 3.03, no

3

2

1.5

y

7

3

1.75 1.11 1.0101

1.1

ii 2 v 1.01

7 a 19.6 m/s b 29.4 m/s 8 a 3.75 b 2.81 c 0 d 0 9 Teacher to check. 10 Teacher to check. 11 a 5 c −10 12 a 0 b 40 c 8 d i 4

1.01

iii 1.5

1 1

1 0.5

0

1

2

12 4 4.75 −2.5

2

3

4

5

e 0.96 mg/L at t = 0.5, −0.24 mg/L at t = 2 f The maximum concentration is 1 mg/L

t

after 1 hour.

3 a Initial position is 1 cm, final position is

20 cm.

b

x (cm)

x = t2 − 2t + 5

20

(5, 20)

15 (4, 13) 10

3

x = 2t + 1

during first 6 seconds then decreases after that. 10 Teacher to check. 11 a 0 b 37 c 5 d i 2 ii 4 mUlTiple ChoiCe

ii 6

1 4 7 9 10 13

A B C

a D

D E

2 5 8 b 11 14

C A B C A D

3 D 6 E c B 12 C

(3, 8) (2, 5)

1 0

t

6

9 a i 4 m/s ii 0 m/s iii −8 m/s b The height of the projectile increases

b −69 d −4 m/s

t (hours)

concentration is decreasing.

1

5

b d f h

4

d −0.2 mg/L; it is negative because the

b x=5m c t = 1 s and x = 4 m d 10 m 7 a x = 30 m b 3s c 34 m/s d −50 m/s e 370 m f 30.83 m/s 8 x

0

3

c 2 3 hours x (t)

0

ii 4.75

4D 5B 6 a 2 c 6 e −7 g 31

b 240 c Approx. 3 pm d Approx. 84 cans/hour 2 a i 0 ii 1 iii 0.8 iv 0.471 mg/L b x (mg/L)

x 20

x

2

8.00 am 10.00 am12.00 pm 2.00 pm 4.00 pm 6.00 pm

Time

t

5

π

5 4

x

b i 3 iv 1.1 c 1

4

10

2 (1.1, 1.331) (1, 1) (1.01, 1.030301) 1

b i 7 c 3 d i 7 3 a

3

15

iii iv

1

2

line.

6 a

5

0

240

c Approx. 8.5 4 a 1 kg/h b 5 kg/h 12( 3 − 1) 5 Average gradient =

6

(1.5, 3.375)

1

b Variable, as the graph is not a straight

iv 4.1

i

(2, 8)

3

300

0

−9

iii 4.5

8

4

360

60

Tangent at t = 2

(2.1, 4.41)

(2, 4)

420

120

9

(2.5, 6.25)

b i 6 c 4 2 a y

480

M

8

1

540

iv

(3, 9)

0

600

23

iii

12

8.33 km/L 291.6 km 91.8 L 30 m3/h 28 hours

Number of cans

1 a b c 2 a b 3 a

rates of change of polynomials

4

1 a

ShorT anSWer

exercise 8i

1 a

exTended reSponSe

1

2

3

c 2 cm/s e b = 3.5 g v (cm/s)

4

5 t (s)

d 10 cm f c=3

8 7 6 5 4 3 2 1 1

2

3

4

5 t (s)

loge (2) b loge (3) a = 2.72, correct to 2 decimal places loge (2.7182818) = 1 e 7.389 06 7.389 06; 2.718 281 82 d = (2.718 281 8 x ) x = 2 dt g The gradient is the same as 2.71828182.

4 a c d f

ChapTer 8 • Rates of change

387

Chapter 9

Differentiation DiGitaL DoC doc-9784 10 Quick Questions

Chapter ContentS 9a 9B 9C 9D 9e 9F 9G

Introduction to limits Limits of discontinuous, rational and hybrid functions Differentiation using first principles Finding derivatives by rule Rates of change Sketching graphs containing stationary points Solving maximum and minimum problems

9a

introduction to limits

In mathematics it is important to understand the concept of a limit. This concept is especially important in the study of calculus. In everyday life we use the term limit to describe a restriction put on a quantity. For example, the legal blood alcohol concentration limit for a driver is normally 0.05 g/100 mL. As the number of standard alcoholic drinks consumed in 1 hour approaches 2, the average adult’s blood alcohol concentration approaches 0.05. Likewise, some time after a celebration, a person who has been drinking heavily at an earlier time may have a blood alcohol concentration that is approaching the legal limit of 0.05 from a higher level, as the number of drinks not yet metabolised by their body approaches 2. We could say that as the number of standard drinks remaining in the body approaches 2, the blood alcohol concentration approaches 0.05. In essence the blood alcohol concentration is a function, say f (x), of the number of drinks, x, remaining in the body.

Chapter 9 • Differentiation

389

WorkeD exampLe 1 1 1 Add the following series of numbers and state what value it is approaching. 12 + 14 + 18 + 16 + 32 +...

think

Write

1

Add the first 2 terms.

The sum of the first 2 terms is

3 4

(= 0.750).

2

Add the first 3 terms.

The sum of the first 3 terms is

7 8

(= 0.875).

3

Add the first 4 terms.

15 (≈ 0.938). The sum of the first 4 terms is 16

4

Add the first 5 terms.

The sum of the first 5 terms is

31 32

(≈ 0.969).

5

Add the first 6 terms.

The sum of the first 6 terms is

63 64

(≈ 0.984).

6

Give the upper limit.

The sum is approaching 1.

Expressing limits in mathematical language, we say that a limit can be used to describe the behaviour of a function, f (x), as the independent variable, x, approaches a certain value, say a. In some cases the function will not be defined at a. Using the correct notation for the example on blood alcohol concentration, we would write: lim f ( x ) = 0.05 x→2

This is read as ‘the limit of f (x) as x approaches 2 is equal to 0.05’. WorkeD exampLe 2

By investigating the behaviour of the function f (x) = x + 3 in the vicinity of x = 2, show that a by hand b using a CAS calculator. lim f ( x ) = 5: x→ 2

think

a 1 Create a table of values for x and f (x) in

the vicinity of x = 2.

2

Consider the values taken by f (x) as x approaches 2.

Write/DraW

a

x f(x)

1.95 1.99 1.995 2 2.005 2.01 2.05 4.95 4.99 4.995 5 5.005 5.01 5.05

As x approaches 2 from the left and the right, f (x) approaches a value of 5. So lim f ( x ) = 5.

y

3

x→2

−3 b 1 Using the Lists and Spreadsheets

390

f(x)

5

0

b 1.95

feature of the CAS calculator, enter the x-values into column A (a[ ]).

1.99 1.995 2 2.005 2.01 2.05

2

Enter a second column as a[ ] + 3.

3

Find the left- and right-hand limits as x approaches 2.

1.95 4.95 1.99 4.99 1.995 4.995 2 5 2.005 5.005 2.01 5.01 2.05 5.05 The left- and right-hand limits are equal, so when f (x) = x + 3, lim f ( x ) = 5 .

Maths Quest 11 Mathematical Methods CAS

x→2

2

x

Limits of continuous functions A continuous function has a graph that forms a continuous line; that is, it has no breaks. If a function is continuous at the point where a limit is being found, then the limit always exists and can easily be found by direct substitution. WorkeD exampLe 3

Find lim( x 2 + 2):

a by hand

x→ 2

b using a CAS calculator.

think

Write/DraW

a 1 To consider whether the function is continuous,

y

a

sketch the graph of y = x 2 + 2 in the vicinity of x = 2.

f(x)

6 2 0 2

2

As the graph is continuous, substitute x = 2 (from lim ) into x2 + 2 to evaluate the limit.

The function is continuous at x = 2. lim( x 2 + 2) = 22 + 2 x→2 =6

x→2

b lim( x 2 + 2)

b 1 Using the limit feature of the CAS calculator,

x→2

determine the required limit. 2

Record the result.

3

Write the answer.

x

6 lim( x 2 + 2) = 6

x→2

In general if a function, f (x), is continuous when x = a, then lim f ( x ) = f ( a). x→ a

theorems on limits theorem 1 For the constant function f where f (x) = c, then lim f ( x ) = c . x→a

If lim f ( x ) = A and lim g( x ) = B , then the following theorems apply. x→a

x→a

theorem 2 For example:

lim[ f ( x ) + g( x )] = lim f ( x ) + lim g( x ) = A + B

x→a

x→a

x→a

lim[ f ( x ) − g( x )] = lim f ( x ) − lim g( x ) = A − B

x→a

theorem 3 For example:

x→a

lim( x 2 x →1

+ 2x) =

For example:

x→a

+ lim 2 x = 1 + 2 = 3 x →1

lim[ f ( x ) × g( x )] = lim f ( x ) × lim g( x ) = A × B

x→a

theorem 4

lim x 2 x →1 x→a

x→a

lim [ x 2 (3 x )] = lim ( x 2 ) × lim (3 x ) = 1 × −3 = −3

x→− 1

x → −1

x → −1

f (x) A f ( x ) xlim lim = →a = if B ≠ 0 x → a g( x ) lim g( x ) B x→a

x 2 + 4) x + 4 xlim( →5 = lim x →5 x + 1 lim( x + 1) 2

x →5

29 = 6 Chapter 9 • Differentiation

391

These theorems on limits can be expressed in word form: 1. The limit of a sum equals the sum of the limits. 2. The limit of a difference equals the difference of the limits. 3. The limit of a product equals the product of the limits. 4. The limit of a quotient equals the quotient of the limits.

exercise 9a

introduction to limits

Add the following series of numbers and state what value it is approaching.

1 We1

4 + 2 + 1 + 12 + 14 + . . . 2 The diagram at right shows regular polygons with 3, 4 and 5 sides. As the number of sides gets very large (→ ∞), what shape emerges?

n=3

n=4

n=5

If n represents the number of sides of a regular polygon, then which of the following correctly describes the situation given in question 2 above? B lim C lim a n

3 mC

n→∞

D n→0

n→ 0

e n→5

1 as n gets infinitely large. n b Write this using limit notation. 1 + … and n represents the number of terms to be summed in the series, 5 a If S = 1 + 12 + 14 + 81 + 16 copy and complete the following table: a Find the value of

4

b

n

1

2

S

1

1 12

3

4

5

6

10

Which of the following is equal to lim S?

mC

n→∞

a 1.75

B 1.95

D 1

e 0

C 2

y

6 We2 By investigating the behaviour of the function f (x) = x + 5 in the vicinity of

3 2 1

x = 3, show that lim f ( x ) = 8. x →3

From the graph at right it can be seen that the lim f ( x ) is equal to:

7 mC DiGitaL DoC doc-9785 SkillSHEET 9.1 Substituting into a function

a 2

B −1

D 1

e 0

x→2

C 3

a

y 30

b

f(x)

4

c x

4

f(x)

y 2

f(x)

−3 0

d

x→4

y 0

14

0

x

4

x e

y 21

y

f

f(x)

4

y

f(x)

15 10

f(x)

5 0

392

0

−1

For each of the functions graphed below, find lim f ( x ).

8

4

Maths Quest 11 Mathematical Methods CAS

x

1 0

4

x

0

f(x)

4

x

2

x

9 We3a Evaluate the limits below. a lim(2 x + 5) b lim(10 − 2 k ) x→2

e lim(h 2 + 3h)

f

h→ 5

10 mC

c lim(49a + 1)

k →4

We3b

lim (3 p2 − 2 p + 3)

p→ − 2

x→0

B 3

a 4

g

lim(8 − 5 x 2 )

x→2

d lim( x 2 + 2) x→0

h lim( x 3 + 2 x ) x→0

If f (x) = 3x − 2, then the value of lim f ( x ) is equal to:

a 0

11 mC The

a→1

lim ( x 2 x→ − 3

C 2

D −2

e 1

C 9

D −4

e 0

− 5) is equal to: B −5

12 Find the values of the following limits.

x +3 x→2 x

a lim

2 x − 3x x→0 x +1

2

b lim

2

c

lim

x→ − 2

6x − 2 x

x + 3x − 2 x →3 x−2 2

d lim

DiGitaL DoC doc-9786 Investigation Sneaking up on a limit

Limits of discontinuous, rational and hybrid functions 9B

Limits of discontinuous functions

If a function is discontinuous at the point where the limit is being investigated, then the limit will exist only if the function is approaching the same value from the left as from the right. Consider the discontinuous functions graphed below. 1. From the left, lim f ( x ) = 1. (The symbol x → 1− indicates that we are x →1−

y

1

letting x approach 1 from the left side.) From the right, lim f ( x ) = 1.

0

x →1+

Left limit = right limit. Therefore, lim f ( x ) = 1.

x

1

x →1

y

2. From the left, lim f ( x ) = 2. x →1

−

f(x)

4

From the right, lim f ( x ) = 4. x →1+

2

Left limit ≠ right limit. Therefore, lim f ( x ) does not exist. x →1

0 1

Limits of rational functions

2

x

Finding the limit of a rational function involves simplifying the function before direct substitution can take place and recognising any values of x for which it is discontinuous.

WorkeD exampLe 4

x2 − x : x→1 x − 1 i by hand ii using a CAS calculator. x2 − x b Sketch the graph of f ( x ) = , stating the value of x for which it is discontinuous. x−1 a Find lim

think

a i 1 We cannot substitute x = 1 directly as we will get 00 ,

which is undefined, so we factorise the numerator.

Write/DraW

x2 − x x ( x − 1) = lim x →1 x − 1 x →1 x − 1

a i lim

Chapter 9 • Differentiation

393

2

Cancel (x – 1) from the numerator and denominator.

x, x ≠ 1 = lim x →1

3

Now substitute x = 1 (from lim) and evaluate.

=1

ii 1 Using the limit feature of the CAS calculator, enter

ii lim

x →1

x 2 − x =1 x→1 x − 1

the function and the value of x for the limit. 2

Record the result.

1

3

Write the answer.

x 2 − x lim =1 x→1 x − 1

x2 − x is the same as the graph x −1 of f (x) = x, except where the point (1, 1) does not exist.

b The graph of f ( x ) =

y

b

f(x) 1 0

1

x

The function is discontinuous at x = 1.

WorkeD exampLe 5

a By first factorising the numerator, simplify the rational function

x2 + 5 x + 6 , stating the value for which the function does not exist x+3 (that is, is discontinuous). b Find lim f ( x ), where a is the value at which f (x) is discontinuous. f ( x) =

tUtoriaL eles-1437 Worked example 5

x→ a

think

Write

x 2 + 5x + 6 x+3 ( x + 3)( x + 2) = ( x + 3)

a f (x) =

a 1 Factorise the numerator.

2

f ( x ) = x + 2, x ≠ − 3 so f (x) is discontinuous at x = –3

Cancel (x + 3) from the numerator and denominator.

b lim x + 2 = − 3 + 2

b Write an expression for lim f ( x ) and evaluate by

substituting x =

−3.

x→a

x→ − 3

= −1

Limits of hybrid functions Hybrid functions are functions that have different rules for different parts of the domain.

WorkeD exampLe 6

x 2 , x ∈( − ∞ , 2] . x − 1, x ∈(2, ∞ )

a Sketch the graph of the hybrid function f ( x ) = b Find

i lim f ( x ) x→ 2 −

ii lim f ( x ) x→ 2 +

iii lim f ( x ) if it exists. x→ 2

c Find solutions to parts a and b using a CAS calculator. 394

Maths Quest 11 Mathematical Methods CAS

think

Write/DraW

a 1 Sketch f (x) = x2 over the domain (−∞, 2].

y

a

4

Sketch (on the same axes) f (x) = x − 1 over the domain (2, ∞).

2

f(x)

1 0 i Substitute x = 2 into f (x) = x 2.

b

x

2

b i lim f ( x ) = lim x 2 x → 2−

x → 2−

x → 2+

x → 2+

ii Substitute x = 2 into f (x) = x − 1.

=4 ii lim f ( x ) = lim ( x − 1)

iii Are these limits equal?

=1 iii lim f ( x ) does not exist x→2

(as left limit ≠ right limit). c 1 Using the hybrid function feature of the CAS

calculator, define the functions and their domains. 2

2 − x ∞ < x ≤ 2 Define f ( x ) = x − 1 2 ≤ x < ∞

Plot the function using a suitable window on the CAS calculator.

y 4

f(x)

1 0 3

Use the limit feature of the CAS calculator to find the limit from the left.

x → 2−

4

Record the result.

4

5

Use the limit feature of the CAS calculator to find the limit from the right.

x → 2+

6

Record the result.

1

7

Write the answers.

x

2

lim ( x 2 )

lim ( x − 1)

c i lim ( f (x)) = 4 x → 2−

ii lim ( f (x)) = 1 x → 2+

iii lim( f (x)) does not exist as the left x→2

limit does not equal the right limit.

Limits of discontinuous, rational and hybrid functions exercise 9B

Which of the following graphs are discontinuous?

1 a

b

y

y

c

3

y

0 0

3

x

0

4

x

x

Chapter 9 • Differentiation

395

y

d

f

y

4

x

0

−1

y

e

0

2

0

x

x

3

For each discontinuous function above, state the value of x for which it is discontinuous.

2

x2 − 4x . x →4 x − 4

a Find lim

3 We4

x2 − 4x , stating the value of x for which it is discontinuous. x−4 x2 + x a Evaluate f (x) when x = 0 if f ( x ) = . Comment on this result. x For what value of x is f (x) discontinuous? Factorise the numerator of f (x). Now simplify f (x). Sketch the graph of f (x). Evaluate lim f ( x ) if it exists.

b Sketch the graph of f ( x ) = 4 b c d e f

x →0

By first factorising the numerator, simplify the following rational functions, stating the value for which the function does not exist (is discontinuous). 6 x − 18 x 2 − 5x x 2 + 5x + 4 x 2 + 3x b f (x) = c f (x) = d f (x) = a f (x) = x−3 x x+4 x

5 We5a

e f (x) =

x 2 − 7x + 6 x−6

f

f (x) =

x3 + 8 x+2

g f (x) =

x 2 + 3x − 4 x −1

h f (x) =

x 3 − 27 x−3

For each rational function in question 5, find lim f ( x ), where a is the value at which f (x) is x →a discontinuous. 7 We6a Sketch the graphs of the following hybrid functions. 6 We5b

4 − x , x ∈(−∞ , −1] − 3 x + 1, x ∈( 1, ∞ )

x + 3, x ∈(−∞ , 2) 4, x ∈[2, ∞ )

b g( x ) =

2 x , x ∈(−∞ , 0) 2 x , x ∈[0, ∞ )

d p( x ) =

a f (x) =

2 − x + 1, x ∈( ∞ , 1] x + 2, x ∈(1, ∞ )

c h( x ) = 8 We6b

For each of the corresponding functions in question 7, evaluate the following. ii lim f ( x ) iii lim f ( x )

a i lim f ( x ) x → 2−

x → 2+

x →2

b i lim g( x ) x→ 1

ii lim g( x ) x→ 1

iii lim− g( x )

c i lim h( x )

ii lim h( x )

iii lim h( x )

d i lim p( x )

ii lim p( x )

iii lim p( x )

− −

x →0− x →1−

x→ 1

− +

x →0+

x →0

x →1+

x →1

9 We6c Investigate whether the following limits exist. For those that do exist, state the limit.

x + 2, x ∈(− ∞ , 2) x → 2 3, x ∈[2, ∞ )

b lim

5 − x , x ∈(− ∞ , 3) x →3 2 x + 1, x ∈[3, ∞ )

d lim

2 − x , x ∈( ∞ , 2] x →1 x + 2, x ∈ (2, ∞ )

f

a lim

c lim

e lim

396

Maths Quest 11 Mathematical Methods CAS

x − 4, x ∈(− ∞ , 0] x → 2 x − 4, x ∈ (0, ∞ )

2 x − 3, x ∈(− ∞ , 1] x →1 2 x + 1, x ∈ (1, ∞ ) 2 − − 4 − x , x ∈( ∞ , 2) lim x → 2 x + 2, x ∈[ − 2, ∞ ) −

10 Evaluate the following. a lim( x 2 + 3 x + 2)

b lim

x →2

x→

3 x + 1, x ∈(− ∞ , 1) x →1 x 2 + 3, x ∈[1, ∞ )

− 1 2

14 x + 7 2x + 1

x 2 − 9 x − 10 1 x +1

c lim

d lim−

e lim(12 − 3 x − 2 x 2 )

f lim

x→

2 x − 2 x + 1, x ∈(− ∞, 0] x→ 0 x + 3, x ∈(0, ∞ )

x →2

x 2 + 3x + 2 x→ 1 x+2 ( x + 3)( x − 3) i lim x →3 x−3

h lim( x 3 + x 2 − 5 x )

g lim−

x →1

j

x 2 + 7x + 6 1 x +1

9C

( x + 2)(2 x − 3) x+2

x 2 − 6x + 8 x →4 x−4

k lim− x→

lim

x→ − 2

l lim

Differentiation using first principles

The gradient function is the rule for the instantaneous rate of change of a given function at any point. The gradient at any point (x, y) can be found by substitution into the gradient function. Consider the secant PQ drawn to the curve f (x) at right. The coordinates of P are [x, f (x)] and the coordinates of Q are [x + h, f (x + h)]. So the gradient of the secant PQ =

y

rise run

y = f(x) Q

f(x + h)

f ( x + h) − f ( x ) x+h−x f ( x + h) − f ( x ) = h

=

Tangent at P

P

f(x) 0

x+h x

x

As h approaches zero, that is, as Q draws as close as possible to P along the curve, PQ effectively becomes a tangent to the curve at P. We can therefore say that the gradient of the tangent at P is

lim

h→ 0

f ( x + h) − f ( x ) or h

f ′( x ) = lim

h→ 0

y Q

f ( x + h) − f ( x ) , h ≠ 0, h

where f ′(x) denotes the gradient of a tangent at any point, x, on the graph of f (x). That is, f ′(x) is the gradient function of f (x). The process of finding the gradient function lim

h→ 0

Q

Q moves closer to P as h approaches 0.

P Q 0

h

x

f ( x + h) − f ( x ) h

is called differentiation from first principles. Two different forms of notation are commonly used to represent a function and its derivative. 1. The European notation of Leibniz is: 2. The alternative notation is: (a) y for the function (a) f (x) for the function dy d (b) for the derivative. [ f (x)] or Dx ( f ) for the derivative. (b) f ′(x) or dx dx Chapter 9 • Differentiation 397

WorkeD exampLe 7

Find the derivative of x2 − 2 x from first principles: a by hand b using a CAS calculator. think

a 1 Define f (x).

3

The derivative is equal to: f ( x + h) − f ( x ) lim . h→ 0 h Find f (x) and f (x + h).

4

Simplify the numerator f (x + h) − f (x).

5

Factorise the numerator f (x + h) − f (x).

2

6

f ( x + h) − f ( x ) by h→ 0 h cancelling the common factor of h. Simplify lim

Write

a f (x) = x2 − 2x

f ′(x) = lim

h→ 0

f ( x + h) − f ( x ) h

f (x) = x2 – 2x f (x + h) = (x + h)2 – 2(x + h) f (x + h) − f (x) = (x + h)2 − 2(x + h) − (x2 − 2x) = x2 + 2xh + h2 − 2x − 2h − x2 + 2x = 2xh + h2 − 2h = h(2x + h − 2) lim

h→ 0

f ( x + h) − f ( x ) h h(2 x + h − 2) = lim h→ 0 h = lim(2 x + h − 2), h ≠ 0 h→ 0

Evaluate the limit by substituting h = 0.

= 2x − 2

b 1 Using the CAS calculator, define the function.

b Define f (x) = x2 − 2x

7

2

Define the formula used to evaluate the derivative.

f ( x + h) − f ( x ) h

3

Record the result.

2x + h − 2

4

Determine the limit as h → 0.

5

Record the result.

6

Write the answer.

lim(2 x + h − 2)

h→ 0

2x − 2 f ( x + h) − f ( x ) h f ′( x ) = 2 x − 2 f ′( x ) = lim h→ 0

WorkeD exampLe 8

If g (x) = 2 x2 + 5x − 2, find: a g ′(x) using first principles b the value(s) of x where the gradient equals 0. think

a 1 Let g (x) = 2x 2 + 5x − 2. 2

3

398

tUtoriaL eles-1438 Worked example 8

Write

a g(x) = 2x2 + 5x − 2

g( x + h) − g( x ) h

The derivative is equal to: g( x + h) − g( x ) . lim h→ 0 h

g ′(x) = lim

Find g(x) and g(x + h).

g(x) = 2x2 + 5x – 2 g(x + h) = 2(x + h)2 + 5(x + h) – 2

Maths Quest 11 Mathematical Methods CAS

h→ 0

4

Simplify the numerator g(x + h) − g(x).

5

Factorise.

6

g( x + h) − g( x ) by cancelling h→ 0 h the common factor of h.

g(x + h) − g(x) = 2(x + h)2 + 5(x + h) − 2 − (2x2 + 5x − 2) = 2(x2 + 2xh + h2) + 5x + 5h − 2 − 2x 2 − 5x + 2 = 2x2 + 4xh + 2h2 + 5x + 5h − 2 − 2x2 − 5x + 2 = 4xh + 2h2 + 5h = h(4x + 2h + 5) g( x + h) − g( x ) h→ 0 h h(4 x + 2h + 5) = lim h→ 0 h = lim(4 x + 2h + 5), h ≠ 0

Simplify lim

lim

h→ 0

7

Evaluate the limit by substituting h = 0.

b Solve g ′(x) = 0.

= 4x + 5 So g ′(x) = 4x + 5. b g ′(x) = 0

4x + 5 = 0 4x = −5 − x = 45 So the gradient equals 0 when x = − 45 .

Note: For any polynomial function, f (x), when the expression f (x + h) − f (x) is simplified, all of its terms have h as a factor.

exercise 9C

Differentiation using first principles

1 We7a Find the derivative of the following from first principles. a 5x − 7 b x2 + 10x c x2 − 8x

dy . dx b y = x2 − 3x + 1 e y = 6x − 2x2

d x3 + 2x

2 We7b Use first principles to find a y=x+3 d y = 9 − x2

c y = 4x2 f y = x3 + 5x − 4

DiGitaL DoCS doc-9787 SkillSHEET 9.2 Differentiating from first principles doc-9788 WorkSHEET 9.1

If g(x) = x2 − 6x, find: a g ′(x) using first principles. b the value(s) of x where the gradient equals 0.

3 We8

a If f (x) = x3 − 8, find f ′(x) using first principles.

4

b Hence, determine the value(s) of x where the gradient function is equal to 12. 5 By first deriving the gradient function f ′(x), evaluate f ′(3) when f (x) is equal to: a 7x + 5 b x2 + 4x c x2 − 3x + 2

d x3 − 5.

6 mC Which of the following do not denote the gradient at any point on a graph? (One or more answers

may be correct.) a f ′(x)

B lim

h→ 0

f ( x + h) − f ( x ) h

C lim

h→∞

f ( x + h) − f ( x ) h

f ( x + h) − f ( x ) h 7 mC The most accurate method for finding the gradient when x = 3 for the function f (x) = x2 + 2x is by: a sketching the graph and drawing a tangent at x = 3 to find the gradient B finding the gradient of the secant to the curve joining the points where x = 3 and x = 3.1 C finding f ′(x) using first principles and evaluating f ′(3) D guessing e finding the gradient of the line from the origin to the point (3, 15) 8 mC Given that f ′(x) = 4x if f (x) = 2x2 and g ′(x) = 3x2 + 1 if g(x) = x3 + x, then the derivative 3 + 2x 2 + x must be equal to: of x a 3x2 + 4x + 1 B 12x3 + 4x C 3x2 + 4x D 2x5 + 2x3 e 5x2 + 1 D

dy dx

e

Chapter 9 • Differentiation

399

9D

Finding derivatives by rule

Fortunately, the tedious process of finding derivatives from first principles need not be applied once rules are established. For polynomial functions, the following rules apply. Rule 1. If f (x) = x n, then f ′(x) = nx n − 1. Rule 2. If f (x) = ax n, then f ′(x) = nax n − 1. Rule 3. If f (x) = c, then f ′(x) = 0 (where c is constant). Rule 4. If f (x) = g(x) + h(x), then f ′(x) = g ′(x) + h′(x). WorkeD exampLe 9

Differentiate each of the following. a y = x8 b y = 3 x2 c y = 7x + 3

d y = 2 x5 +

think

3 5

x2 − 6 x

Write

a Write the expression for y. Apply rule 1

to find the derivative.

b Apply rule 2.

c Apply rules 2, 3 and 4. Remember that x0 = 1.

d Differentiate the 3 terms separately (that is,

apply rules 2 and 4).

a y = x8

dy = 8 x 8−1 dx = 8x 7 b y = 3x2 dy = 2(3 x 2−1 ) dx = 6x c y = 7x + 3 dy = 7 x1−1 + 0 dx = 7x 0 = 7 d y = 2x5 + 35 x2 – 6x

dy = 5(2 x 5−1 ) + 2( 35 x 2−1 ) − 1(6 x1−1 ) dx = 10 x 4 + 65 x − 6 x 0 = 10 x 4 + 65 x − 6

WorkeD exampLe 10

Find f ′(x) if f (x) = 3x(x − 2). think

Write

1

Write down f (x).

f (x) = 3x(x − 2)

2

Expand the brackets.

f (x) = 3x2 − 6x

3

Differentiate by rule.

f ′(x) = 6x − 6

WorkeD exampLe 11

If g( x ) =

4 x3 + 3 x2 , find g ′(x): a by hand x

think

a 1 Factorise the numerator, because at this

stage we can only differentiate a constant denominator.

400

Maths Quest 11 Mathematical Methods CAS

b using a CAS calculator. Write 3 2 a g( x ) = 4 x + 3 x

=

x

x x + 3) x

2 (4

2

Simplify g(x).

= x(4x + 3), x ≠ 0

3

Expand the brackets.

= 4x2 + 3x

4

Differentiate g(x) by rule.

b 1 Using the CAS calculator, define the function.

g ′(x) = 8x + 3 4 x 3 + 3x 2 x

b Define g( x ) =

2

Use the differentiation feature of the CAS calculator to find g′(x).

d ( g( x )) dx

3

Record the result.

8x + 3

4

Write the answer.

Given g(x) =

4 x 3 + 3x 2 , g ′(x) = 8x + 3 x

WorkeD exampLe 12

Differentiate each of the following. Express all answers with a positive index. 1 1 4 a f ( x) = x 3 b f ( x) = 7 c f ( x) = x 3 d f ( x) = . x x −

think

a 1 Write down f (x).

Write –3

a f ′(x) = x

−3 − 1

f ′(x) = −3x

−4

2

Differentiate by rule 1.

= −3x

3

Express the answer with a positive index.

=

b 1 Write down f (x). 2

Bring the x-term to the numerator using the index laws, as we can only differentiate a constant denominator.

3

Differentiate by rule 1.

4

Express the answer with a positive index.

c 1 Write down f (x). 2

Differentiate by rule 1.

b f (x) =

Express the answer with a positive index.

d 1 Write down f (x).

x4 1 x7 −7

= 1x

−7 − 1

f ′(x) = −7(1x − = −7x 8 =

−

)

7 x8 1

c f (x) = x 3 1 −1

f ′(x) = 13 (x 3 ) =

3

−3

=

x

−2 3

3 1 2

3x 3 4 d f (x) = x 4

2

Convert x to index form.

f (x) =

3

Bring the x-term to the numerator using the index laws.

f (x) = 4 x

1

x2 −1 2

Chapter 9 • Differentiation

401

4

f ′(x) =

Differentiate by rule 2.

−1 2

(4 x − 12 − 1)

= − 2x 5

6

=

Express with a positive index.

=

Express the power of x back in surd form.

−3 2

−2 3

x2 −

2

x3

To evaluate the gradient of a curve at a given point, substitute the given value of x into the gradient function or derivative. For the function f (x), the gradient at the point (a, f (a)) is found by evaluating f ′(a). This gives the gradient of the tangent at x = a, which equals the gradient of the curve only at that point.

WorkeD exampLe 13

For the function f (x) = 13 x3 − x2 − 5x + 3, evaluate f ′(−1), f ′(0) and f ′(4). think

Write 1

f (x) = 3 x3 – x2 – 5x + 3 f ′(x) = x2 − 2x − 5

1

Write the original function and then find the gradient function, f (x).

2

Substitute x = −1 into the gradient function.

f ′(−1) = (−1)2 − 2(−1) − 5 =1+2−5 = −2 − ∴ f ′( 1) = −2

3

Substitute x = 0 into the gradient function.

f ′(0) = (0)2 − 2(0) − 5 = −5 ∴ f ′(0) = −5

4

Using the CAS calculator, define the function.

5

Differentiate f (x) and evaluate the derivative at x = 4.

6

Record the result.

3

7

Write the answer.

f ′(4) = 3

1

Define f ( x ) = 3 x 3 − x 2 − 5 x + 3 d ( f ( x )) | x = 4 dx

tangents and normals As mentioned earlier in this chapter, the derivative f ′(x) is actually the gradient function. This means that the value of the gradient at any particular point on a curve is equal to the numerical value of the derivative at that point. Recall that if the gradient of a tangent to a curve at point P is m T, then the normal, mN, is a straight line perpendicular − 1 and passing (at right angles) to the tangent such that mN = mT through the point P as shown at right.

y Normal

Tangent

P 0

x

Also recall that the equation of a straight line is given by y − y1 = m(x − x1) where (x1, y1) is the point P, above, and m is the gradient. 402

Maths Quest 11 Mathematical Methods CAS

WorkeD exampLe 14

a Find the equation of the tangent to the curve f (x) = x2 + 6x − 8 at the point where the gradient

has a value of 8.

b Hence, find the equation of the normal at this point. c Find the equation of the tangent at this point using a CAS calculator. think

Write

a 1 Find the gradient function of the curve,

f ′(x).

2

Find x1, the value of x where f ′(x) = 8; that is, solve 2x + 6 = 8.

Alternatively, use the solve and differentiation features of a CAS calculator to find x1 when f ′(x1) = 8.

a f (x) = x2 + 6x – 8

f ′(x) = 2x + 6

For gradient = 8 2x + 6 = 8 2x = 2 x=1

So x1 = 1.

dy solve ( x 2 + 6 x − 8) = 8, dx

x

Record the result.

x=1

3

Find f (x1) to determine the value of y1.

y1 = f (x1) = f (1) = (1)2 + 6(1) − 8 = −1

4

Simplify the equation y − y1 = mT (x − x1) to find the equation of the tangent.

The equation of the tangent at the point (1, −1) is y − −1 = 8(x − 1) y + 1 = 8x − 8 y = 8x − 9

b 1 Find the gradient of the normal using

mN = 2

−1

mT

b mN =

−1 8

.

Simplify the equation y − y1 = mN (x − x1) to find the equation of the normal.

The equation of the normal at the point (1, −1) y − −1 =

is

y+1=

−1 (x − 1) 8 − ( x − 1)

8 8y + 8 = −x + 1 x + 8y + 7 = 0 c 1 The gradient is 8 at the point (1, −1).

c solve(y − −1 = 8(x − 1), y)

Substitute these values into the equation y − y1 = m(x − x1) and solve for y. 2

Record the result.

y = 8x − 9

3

Write the answer.

The equation of the tangent to the curve f (x) = x2 + 6x − 8, at the point (1, −1) with a gradient of 8 is: yT = 8x − 9.

Chapter 9 • Differentiation

403

WorkeD exampLe 15

For the function f (x) = x3 − x2 + 1, find the coordinates of the points where the gradient is perpendicular to the line y = −x + 3. Hence, find the equations of the tangents and normals at these points. think 1

Find the derivative.

2

Find the gradient of the tangent. As the tangent is perpendicular to the line y = −x + 3, the product of their gradients is −1, that is, m1m2 = −1. Let m1 equal the gradient of the tangent and let m2 equal the gradient of the line y = −x + 3.

3

Put f ′(x) = 1 and solve for x to find the point where the gradient of f(x) = 1.

Write

f (x) = x3 − x2 + 1 ∴ f ′(x) = 3x2 − 2x. − m1 = 1 where m2 = −1 m2

=

−1

(− 1) =1 ∴ mtangent = 1 3x2 − 2x = 1 − 2x − 1 = 0 (3x + 1)(x − 1) = 0 3x2

∴x= 4

Substitute x = 1 and x = − 13 into f (x) to find their y-coordinates, and state the points.

Find the equations of the tangent lines.

−1

or x = 1

3

f (1) = (1)3 − (1)2 + 1 = 1 f

−1 3

−1 3

=

∴ (1, 1) and 5

3

−1 3

–

−1 23 , 3 27

2

are the coordinates of the points.

23 27

(

= x

− − 1 3 1

y=x+3+ y=x+ Find the equations of the normal lines. −1 Use mnormal = to find the gradient mtangent of the normal.

23 27

+1=

The equation of the tangent at the point (1, 1) is y − 1 = 1(x − 1) y=x−1+1 ∴ y=x − The equation of the tangent at the point 1 , 23 is y–

6

)

23 27

32 27

−

23 27

−

(

= x

− − 1 3

y = −x – 13 + y = −x +

Maths Quest 11 Mathematical Methods CAS

3 27

1 1 = −1 The equation of the normal at the point (1, 1) is y − 1 = −1(x − 1) y = −x + 1 + 1 ∴ y = −x + 2 − The equation of the normal at the point 1 , 23 is mnormal =

y–

404

tUtoriaL eles-1440 Worked example 15

14 27

)

23 27

3 27

WorkeD exampLe 16

a Graph the function f (x) = x3 − 2x and f ′(x) on the same set of axes. b How can f (x) be used to predict the location and nature of the stationary points of f (x)? think

Write/DraW

a f1(x) = x3 − 2x

a 1 Using the CAS calculator, define f (x)

and f ′(x).

2

f2 ( x ) =

Plot f (x) and f ′(x) on the same axes.

d ( f1 ( x )) dx

f '(x)

y

f(x)

x

b The x-intercepts of f ′(x) predict the location

and nature of the stationary points of f (x).

exercise 9D

b At x = −0.816, f ′(x) goes from positive to negative

(left to right), indicating a maximum turning point for f (x) = x3 − 2x. At x = 0.816, f ′(x) goes from negative to positive (left to right), indicating a minimum turning point for f (x).

Finding derivatives by rule

Differentiate each of the following. d y = 4x4 + 23 x2 − 5x b y = 7x2 c y = 5x + 2 dy 2 Find if y is: dx a 3x4 b 8x7 c 5x 5 d −4x6 − 3 − e 2x f 7x. 3 Match the correct derivative from the set a to G below to each of the following. 1 We9

a y = x6

a x8 e

x4

+

2x3

−3

a 5x4 + 18x 2 − 4 e 6x2 − 4

b x4 + x 2 f x5 + 6x3 − 4x

c 2x3 − 4x + 7 g xp

d x 2 + 6x − 5

B 4x 3 + 2x F 2x + 6

C px p −1 G 4x3 + 6x 2

D 8x7

Differentiate the following.

4

a y = x6 + 3x 2 − 4 d y = 10x5 − 3x4 + 2x3 − 8x

b y = 5x4 − 7x3 + 6x e y=6

c y = x11 − 3x6 + 4x5 + 3x 2 f y = 3x4 + 5x4

Find f ′(x) if f (x) is:

5 a d

2 x4 3 3 x3 4

b

+ 12 x 2 − 3x

6 We10

e

a g( x ) =

+ 16 x3

c

6 x7 7

f

4x3 − 47 x 2 − 65 x + 8.

Find f ′(x) if f (x) is:

a f (x) = x(x + 3) d f (x) = 9(8 − 3x)2 7 We11

−5 2 x 8 5 2 x + 3 x4 5 4

b f (x) = 3x(2x − 5) e f (x) = (x + 2)3

c f (x) = (x + 4)2 f f (x) = (2x − 5)3.

Find g ′(x) by first simplifying g (x). x 3 + 5x x

b g( x ) =

8x3 − 6x 2x

c g( x ) =

3x 3 + 2 x 2 − 5x x

d g( x ) =

5x 4 + x 3 + 7 x 2 x2 Chapter 9 • Differentiation

405

Differentiate each of the following. Express all answers with a positive index.

8 We12 −4

−7

a x e

b x

−4x−6

i

5 x3 1

−4

−8

c 3x

d 5x

1 x4

f

−3x−5

g

j

10 x6

k 2x 2

1

2

m 4x 4

n 3x 5

o

q 4 x

r

3

s

x

x

h

1 x9

l

x3

p

2 x

2

1 x

3

Evaluate i f ′(1) ii f ′(−2) and iii f ′(0) for each of the following.

9 We13

a f (x) = 5x2 + 3x − 1 b f (x) = 13 x3 + 2x 2 − 4 c f (x) = 3x 2 − 2x + 6 d f (x) = x3 + 7x − 8 10 We13 a Find the x-intercepts of the parabola y = x 2 − 5x + 6. b Find the gradient of the parabola at the points where it crosses the x-axis. c Determine the value of x for which the gradient of the parabola is: i 0 ii 7 iii −3.

DiGitaL DoCS doc-9779 Gradient at a point doc-9789 tangent and normal

11 a Find the x-intercepts of the curve y = 2x 2 + 5x − 3. b Find the gradient of the curve at these points. c Find the coordinates of the point where the gradient is 0. 12

Find the coordinate(s) of the points on the curve x 3 − 3x2 where the tangent: a is parallel to the x-axis b is parallel to the line y = −3x + 2.

13 We14 Find the equations of the tangent and normal for each of the following curves. a f (x) = 5x2 + 3x − 1 at the point (1, 7) b f (x) = 13 x3 + 2x2 − 4 at x = −2 c f (x) = 3x2 − 2x + 6 at x = 2 d f (x) = x3 − 7x − 8 at the point (0, −8) e f (x) = x2 + 2 at x = a f f (x) = 3x2 − 4x at the point (2a, 12a2 − 8a) 14 We14 a Find the equation of the tangent at the point on the curve x 2 + 4x − 1 where the gradient is 6. b Hence, find the equation of the normal at this point. 15 Find the equation of the normal to the curve y = 2x 2 − 2x + 5 at the point where the curve crosses the

y-axis. 16 Find the equation of the normal to the curve y = −x2 + 4x at: a x=2 b x = 1.

Find the equation of the normal to the curve y = x3 + 2x2 − 3x + 1 at x = −2. 18 If the equation of the normal to the curve y = 2x2 + 4.5x + 1 is y = −2x + c, find: a the coordinates at the point of tangency b the value of c.

17

19 We15 Find the equations of the tangents to the curve f (x) = 2 x3 + 5 x2 − x + 2 that are parallel to the

line y − 2x = 4.

3

2

20

Find the coordinates of the points on the curve f (x) = 2x3 − x2 + 6 where the tangents are parallel to the line −5x + 2y = 12. Hence, find the equation of the tangents at these points.

21

Find the equation of the tangent to the curve f (x) = 2x2 − 4x + 4 that is perpendicular to the line y=

−1 4

x + 4.

Find the coordinates of the points where the tangents to the curve f (x) = x3 − 3x − 1 are perpendicular to the line x + 3y = 3. Hence, find the equations of the normals at these points.

22 We15

406

Maths Quest 11 Mathematical Methods CAS

The tangent to the curve y = ax2 + bx + c at the point (−2, −8) is parallel to the line −7x + y = 10. If the curve also passes through the point (−1, −3.5), find the values of a, b and c. Hence, find the equation of the tangent at the point (−2, −8). 24 a Find the equation of the tangent to the curve y = 2x2 − 2 at x = 2a. b Hence find the x-intercept of the tangent line in terms of a. c A straight line with equation y = 2x − 2 passes through the x-intercept of the tangent line. Find the value of a. d What is the equation of the tangent line? 25 We16 Sketch the graphs of: a f (x) = 4 x − 2 23

b

f (x) = x 2 − 4 x + 2

f ( x ) = x 3 + 3x 2 + 3 and the graph of f ′(x) for each on the same set of axes. How can f ′(x) be used to predict the location and nature of the stationary points of f (x) in each case?

c

9e

rates of change

The rate of change of a function refers to its gradient. For linear functions the gradient is constant; however, the gradient for other functions such as quadratic or cubic polynomials is continually changing.

The rate of change of position with respect to time is velocity.

The rate of change of velocity with respect to time is acceleration.

Differentiation provides us with a tool to describe the gradient of a function and hence determine its rate of change at any particular point. In essence, while average rates of change can be determined from the original function, differentiation of this function provides a new function that describes the instantaneous rate of change. y y = f(x2) Note: The term instantaneous rate of change is often referred to as rate of change. If P(x1, f (x1)) and Q(x2, f (x2)) are two points on the graph Q (x2, f(x2)) of the function with rule y = f (x), then the average rate of change of y with respect to x over the interval x ∈ [x1, x2] is equal to the gradient of the straight line PQ. change in f ( x ) change in x f ( x 2 ) − f ( x1 ) = x 2 − x1

Average rate of change =

P (x1, f(x1)) 0

x1

x2

x

The instantaneous rate of change finds the rate of change at a specific point. dy The instantaneous rate of change of y ( f (x)) with respect to x is given by the derivative ( f ′ (x)). dx Chapter 9 • Differentiation

407

WorkeD exampLe 17

If f (x) = x2 − 2x + 4, determine: a the average rate of change between x = 2 and x = 4 b a new function that describes the rate of change at any point x c the instantaneous rate of change when x = 4 d parts a, b and c using a calculator. think

Write

a f (x) = x2 − 2x + 4

a 1 Write the function. 2

Average rate of change =

change in f ( x ) change in x

f (4) − f (2) 4−2 12 − 4 = 2 =4

Average rate of change =

b Differentiate f (x).

b f ′(x) = 2x − 2

c Substitute x = 4 into f ′(x).

c f ′(4) = 2(4) − 2 = 6

d 1

Define the function on your CAS calculator.

So the rate of change when x = 4 is 6.

d Define f (x) = x2 − 2x + 4

2

Determine f (4) and f (2), and hence the expression f (4) − f (2) . 4−2

f (4) − f (2) 4−2

3

Record the result.

4

4 5

Use the differentiation feature of the CAS calculator to find the derivative of f (x). Record the result.

d ( f ( x )) dx 2x − 2

6

Evaluate the derivative at x = 4.

d ( f ( x )) x = 4 dx

7

Record the result.

6

8

Write the answers.

a The average rate of change between x = 2 and

x = 4 is 4. b f ′(x) = 2x − 2 c The instantaneous rate of change when x = 4 is 6.

WorkeD exampLe 18

A javelin is thrown so that its height, h metres, above the ground is given by the rule h(t) = 20t − 5t2 + 2, where t represents time in seconds.

a Find the rate of change of the height at any time, t. b Find the rate of change of the height when i t = 1 ii t = 2 iii t = 3. c Briefly explain why the rate of change is initially positive, then zero and then negative over the

first 3 seconds.

d Find the rate of change of the height when the javelin first reaches a height of 17 metres. 408

Maths Quest 11 Mathematical Methods CAS

think

a 1 Write the rule. 2

Differentiate h(t).

Write

a h(t) = 20t − 5t2 + 2

h′(t) = 20 − 10t

b 1 Evaluate h′(1).

b i h′(1) = 20 − 10(1)

Evaluate h′(2).

ii h′(2) = 20 − 10(2)

Evaluate h′(3).

iii h′(3) = 20 − 10(3)

2

3

c For rates of change:

Positive means increasing. Zero means neither increasing nor decreasing. Negative means decreasing.

d 1 Find the time at which the javelin is 17 m above

the ground by substituting h = 17 into h(t).

= 10 m/s = 0 m/s

= −10 m/s

c The javelin travels upwards during the first

2 seconds. When t = 2 seconds, the javelin has reached its maximum height. When t > 2 seconds, the javelin is travelling downwards. d 20t − 5t2 + 2 = 17 −5t2

2

Make RHS = 0.

+ 20t − 15 = 0

3

Divide both sides by −5.

4

Factorise, then solve for t. Note: The quadratic formula could also be used to solve for t.

5

The first time it reaches 17 m is the smaller value of t.

The javelin first reaches 17 m when t = 1 s.

6

Evaluate h′(1).

h′(1) = 20 − 10(1) = 10 m/s The rate of change of height is 10 m/s.

t2 − 4t + 3 = 0 (t − 1)(t − 3) = 0 t = 1 or 3

It is worth noting that there are two common ways of writing the derivative as a function. For example, the derivative of the function P(x) = x2 + 5x − 7 may be written as P′(x) = 2x + 5 dP = 2 x + 5. or as dx WorkeD exampLe 19

The shockwave from a nuclear blast spreads out at ground level in a circular manner. a Write down a relationship between the area of ground, A km2, over which the shockwave passes and its radius, r km. b Find the rate of change of A with respect to r. c Find the rate of change of A when the radius is 2 km. d What is the rate of change of A when the area covered is 314 km2?

Chapter 9 • Differentiation

409

Write

think

a State the formula for the area of a circle.

a A(r) = πr 2

b Differentiate A(r).

b A′(r) = 2πr

c Substitute r = 2 into A′(r).

c A′(2) = 2π(2)

d 1 Substitute A = 314 into the area function A(r)

d A(r) = πr2

Note: The units for the rate of change of A (km2) with respect to r (km) are km2 per km or km2/km.

= 12.57 The rate of change of A when the radius is 2 km is 12.57 km2/km. 314 = πr2 314 r2 = π = 99.95 r = 10 since r > 0

and solve for r.

2

A′(10) = 2π (10) = 62.8 The rate of change of A when area is 314 km2 is 62.83 km2/km.

Find the rate of change when r = 10.

exercise 9e

rates of change

If f (x) = x2 + 5x + 15, find: a the average rate of change between x = 3 and x = 5 b a new function that describes the rate of change c the instantaneous rate of change when x = 5.

1 We17

2 DiGitaL DoCS doc-9793 SkillSHEET 9.3 average rate of change doc-9777 Gradient between two points on a graph

A balloon is inflated so that its volume, V cm3, at any time, t seconds, is: V=

−8 3 t 5

+ 24t 2 , t ∈ [0, 10]

a What is the volume of the balloon when: i t = 0? ii t = 10? b Hence, find the average rate of change between t = 0 and t = 10. c Find the rate of change of volume when: i t=0 ii t = 5 iii t = 10.

The average rate of change between x = 1 and x = 3 for the function y = x2 + 3x + 5 is: B 9 C 5 D 3 e 7 3 2 4 mC The instantaneous rate of change of the function f (x) = x − 3x + 4x when x = −2 is: B −2 a 2 C 28 D 3 e 12 3 mC

a 1

5 mC If the rate of change of a function is described by

dy = 2 x 2 − 7 x , then the function dx

could be: a y = 6x3 − 14x

B y = 23 x 3 − 7 x

C y = 23 x 3 − 72 x 2 + 5

D y = x 3 − 72 x 2 + 2

e 2x2 − 7x + 5

In a baseball game the ball is hit so that its height above the ground, h metres, is h(t) = 1 + 18t − 3t2, t seconds after being struck. a Find the rate of change, h′(t). b Calculate the rate of change of height after: i 2 seconds ii 3 seconds iii 4 seconds. c What happens when t = 3 seconds? d Find the rate of change of height when the ball first reaches a height of 16 metres.

6 We18

DiGitaL DoC doc-9794 SkillSHEET 9.4 instantaneous rate of change

410

Maths Quest 11 Mathematical Methods CAS

7

The position, x metres, of a lift (above ground level) at any time, t seconds, is given by x(t) = −2t2 + 40t. a Find the rate of change of displacement (velocity) at any time, t. b Find the rate of change when: i t=5 ii t = 9 iii t = 11. c What happened between t = 9 and t = 11? d When and where is the rate of change zero?

8 The number of seats, N, occupied in a soccer stadium t hours after the gates are opened is given by:

N = 500t2 + 3500t, t ∈ [0, 5]

a Find N when: i t=1 ii t = 3. b What is the average rate of change between t = 1 and t = 3? c Find the instantaneous rate when: i t=0 ii t = 1 iii t = 3 d Why is the rate increasing in the first 4 hours?

iv t = 4.

9 The weight, W kg, of a foal at any time, t weeks, after birth is given by:

W = 80 + 12t − 103 t 2

where

0 ≤ t ≤ 20.

a What is the weight of the foal at birth? b Find an expression for the rate of change of weight at any time, t. c Find the rate of change after: i 5 weeks ii 10 weeks iii 15 weeks. d Is the rate of change of the foal’s weight increasing or decreasing? e When does the foal weigh 200 kg? 3

10 The weekly profit, P (hundreds of dollars), of a factory is given by P = 4.5n − n 2, where n is the number

of employees.

dP . dn b Hence, find the rate of change of profit, in dollars per employee, if the number of employees is: i 4 ii 16 iii 25. c Find n when the rate of change is zero. a Determine

11 Gas is escaping from a cylinder so that its volume, V cm3, t seconds after the leak starts, is described 1 2 by V = 2000 − 20t − 100 t . a Evaluate the rate of change after: i 10 seconds ii 50 seconds b Is the rate of change ever positive? Why?

iii 100 seconds.

12 We19 Assume an oil spill from an oil tanker is circular and remains that way. a Write down a relationship between the area of the spill, A m2, and the radius, r metres. b Find the rate of change of A with respect to the radius, r. c Find the rate of change of A when the radius is: i 10 m ii 50 m iii 100 m. d Is the area increasing more rapidly as the radius

increases? Why? 13 A spherical balloon is being inflated. a Express the volume of the balloon, V m3, as a function

of the radius, r metres. b Find the rate of change of V with respect to r. c Find the rate of change when the radius is: i 0.1 m ii 0.2 m iii 0.3 m. 14 A rectangular fish tank has a square base, with its height being

equal to half its base length. a Express the length and width of the base in terms of its height, h. b Hence, express the volume, V m3, in terms of the height, h, only. c Find the rate of change of V when: i h=1m ii h = 2 m iii h = 3 m. Chapter 9 • Differentiation

411

15 For the triangular package shown, find: a x in terms of h b the volume, V, as a function of h only c the rate of change of V when: i h = 0.5 m ii h = 1 m.

6 x 30° h

16 A new estate is to be established on the side of a hill. y

80

x

200

Regulations will not allow houses to be built on slopes where the gradient is greater than 0.45. If the equation of the cross-section of the hill is y = −0.000 02x3 + 0.006x2, find: dy a the gradient of the slope dx b the gradient of the slope when x equals: i 160 ii 100 iii 40 iv 20 c the values of x where the gradient is 0.45 d the range of heights for which houses cannot be built on the hill. 17 A bushfire burns out A hectares of land, t hours after it started, according to the rule A = 90t2 − 3t3. a At what rate, in hectares per hour, is the fire spreading at any time, t? b What is the rate when t equals: i 0? ii 4? iii 8? iv 10? v 12? vi 16? vii 20? c Briefly explain how the rate of burning changes during the first 20 hours. d Why isn’t there a negative rate of change in the first 20 hours? e What happens after 20 hours? f After how long is the rate of change equal to 756 hectares per hour?

Sketching graphs containing stationary points 9F

The derivative of a function gives its gradient function — that is, it gives the gradient of a tangent to the curve for any specified value of the independent variable. When the derivative equals zero, the tangent is horizontal. The point or points on the curve where this occurs are called stationary points. In other words, a function f (x) has stationary points when f ′(x) = 0. Stationary points can take the form of: 1. a local minimum turning point 2. a local maximum turning point 3. a stationary point of inflection. y

Local minimum turning point Just to the left of a, the gradient is negative; that is, if x < a, but close to a, then f ′(x) < 0. At the point where x = a, the gradient is zero; that is, at x = a, f ′(x) = 0. Just to the right of a the gradient is positive; that is, if x > a, but close to a, f ′(x) > 0. 412 Maths Quest 11 Mathematical Methods CAS

f '(x) < 0

f '(x) > 0 f '(a) = 0

0

a

x

In other words, for a stationary point at x = a, if the gradient changes from negative to positive as we move from left to right in the vicinity of a, it is a local minimum.

Local maximum turning point At x < a, but close to a, f ′(x) > 0. At x = a, f ′(x) = 0. At x > a, but close to a, f ′(x) < 0. In other words, for a stationary point at x = a, if the gradient changes from positive to negative as we move from left to right in the vicinity of a, it is a local maximum. The term local maximum or local minimum implies that the function has a maximum or minimum in the vicinity of x = a. This is important because some functions can have more than one stationary point.

y f '(a) = 0 f '(x) < 0

f '(x) > 0 0

x

a

Stationary point of inflection y

y f '(x) > 0 f'(x) < 0 f '(a) = 0

0

f '(x) > 0 a

f '(a) = 0 x

f '(x) < 0 a

0

x

At x < a, but close to a, f ′(x) > 0. At x < a, but close to a, f ′(x) < 0. At x = a, f ′(x) = 0. At x = a, f ′(x) = 0. At x > a, but close to a, f ′(x) > 0. At x > a, but close to a, f ′(x) < 0. In other words, for a stationary point at x = a, if the gradient remains positive or negative in the vicinity of a, it is a stationary point of inflection. Worked Example 20

If f (x) = x3 − 6x2 − 15x, find: a the value(s) of x where the gradient is zero b the stationary point(s). Think

a 1 Write the function.

Write

a f (x) = x3 − 6x2 − 15x

2

Differentiate f (x) to find the gradient function f ′(x).

f ′(x) = 3x2 − 12x − 15

3

Solve f ′(x) = 0 to find the x-values of each stationary point.

For stationary points: f ′(x) = 0 3x2 − 12x − 15 = 0 3(x2 − 4x − 5) = 0 x2 − 4x − 5 = 0 (x − 5)(x + 1) = 0 x = 5 or x = −1

b 1 Substitute each value of x into f (x) to find the

corresponding y-values.

2

Write the coordinates of each stationary point.

b f (5) = (5)3 − 6(5)2 − 15(5)

f (−1)

= −100 = (−1)3 − 6(−1)2 − 15(−1) =8

Stationary points occur at (5, −100) and (−1, 8).

Chapter 9 • Differentiation 413

Worked Example 21

Sketch the graph of the function f (x) = 5 + 4x − x2, labelling all intercepts and stationary points. Think

Write/Draw

1

Write the function.

f (x) = 5 + 4x − x2

2

Find the y-intercept by letting x equal 0.

y-intercept: x = 0, f (0) = 5 + 4(0) − (0)2 =5 so y-intercept is (0, 5).

3

Find the x-intercepts by letting f (x) = 0.

x-intercepts: if f (x) = 0, 5 + 4x − x2 = 0 x2 − 4x − 5 = 0 (x + 1)(x − 5) = 0 x = −1 or x = 5 so x-intercepts are (−1, 0) and (5, 0).

4

Differentiate f (x) to find the gradient function f ′(x).

f (x) = 5 + 4x – x2 f ′(x) = 4 − 2x

5

Solve f ′ (x) = 0 to find the x-value(s) of each stationary point.

For stationary points: f ′(x) = 0 4 − 2x = 0 −2x = −4 x=2

6

Substitute this value of x into f (x) = 5 + 4x − x2 to find the corresponding y-value.

f (2) = 5 + 4(2) − (2)2 =9 so there is a stationary point at (2, 9).

7

Determine the nature of the stationary point at x = 2 by evaluating f ′ (x) to the left and right, say at x = 1 and x = 3.

x < 2: f ′(1) = 4 − 2(1) =2 ∴ f ′(x) > 0 x > 2: f ′(3) = 4 − 2(3) = −2 ∴ f ′(x) < 0

8

Complete a gradient table. Since the gradient changes from positive to negative as we move from left to right in the vicinity of x = 2, the stationary point (2, 9) is a local maximum.

x

1

2

3

Sign of f ′(x)

+

0

−

Slope (2, 9) is a local maximum. 9

Sketch the graph.

y (2, 9) 9 f(x)

5

−1 0

414 Maths Quest 11 Mathematical Methods CAS

2

5 x

WorkeD exampLe 22

a Find the stationary points and determine their nature for the function

f(x) = x3 − x2 − 8x + 8. b Find the coordinates of all intercepts. c Sketch the graph of f (x) showing all stationary points and intercepts. think

Write/DraW

a 1 Write the rule for f (x).

a f (x) = x3 − x2 − 8x + 8

2

Differentiate f (x) to find f ′(x).

f ′(x) = 3x2 − 2x − 8

3

Solve f ′(x) = 0 to find the x-values of each stationary point.

For stationary points, solve f ′(x) = 0 for x. 3x2 − 2x − 8 = 0 (3x + 4)(x − 2) = 0 x = − 43 or x = 2

4

Substitute each value of x into f (x) to find the y-coordinates of the stationary points.

f ( − 43 ) = ( − 43 )3 − ( − 43 )2 − 8( − 43 ) + 8 =

392 27

= 14 14 27

∴ ( − 43 ,14 14 27 ) is one stationary point. f (2) = (2)3 − (2)2 − 8(2) + 8 = −4 ∴(2, −4) is another stationary point. Define f (x) = x3 − x2 − 8x + 8

5

Alternatively, all of this working can be done on a CAS calculator. First, define f (x).

6

Set f ′(x) = 0 and use the solve and differentiation features of the CAS calculator.

d solve ( f ( x )) = 0, x dx

7

Record the result.

x=

8

Evaluate f ( 43) and f (2).

f

9

Record the results.

392 − , 4 27

10

Determine the nature of the stationary point at x = − 43 by evaluating f ′(x) to the left and right. Choose x = −2 and x = −1.

x < 43 : f ′( − 2) = 3( − 2)2 − 2( − 2) − 8 = 8 ∴ f ′(x) > 0

−

11

Complete a gradient table and state the type of stationary point. Since the gradient changes from positive to negative as we move from left to right in the vicinity of x = − 43 , the stationary point ( − 43 , 14 14 27 ) is a local maximum.

12

Determine the nature of the stationary point at x = 2 by evaluating f ′(x) to the left and right. Choose x = 1 and x = 3.

13

Complete a gradient table and state the type of stationary point. Since the gradient changes from negative to positive as we move from left to right in the vicinity of x = 2, the stationary point (2, −4) is a local minimum.

−4 3

−4

, 3

or x = 2 f (2).

−

−

x > 43 : f ′( − 1) = 3( − 1)2 − 2( − 1) − 8 = − 3 ∴ f ′(x) < 0 x

−2

−4 3

−1

Sign of f ′(x)

+

0

−

Slope ∴ ( − 43 , 14 14 27 ) is a local maximum turning point. x < 2: f ′(1) = 3(1)2 − 2(1) − 8 = −7 ∴ f ′(x) < 0 x > 2: f ′(3) = 3(3)2 − 2(3) − 8 = 13 ∴ f ′(x) > 0 x

1

2

3

Sign of f ′(x)

−

0

+

Slope ∴ (2, −4) is a local minimum turning point.

Chapter 9 • Differentiation

415

b 1 To find the x-intercepts, factorise f (x) by long

division, or use another appropriate method.

2

b f (x) = x3 − x2 − 8x + 8

)= x2(x − 1) − 8(x − 1) )= (x − 1)(x2 − 8) = ( x − 1)( x − 2 2)( x + 2 2)

Solve f (x) = 0 for x-intercepts.

( x − 1) ( x − 2 2) ( x + 2 2) = 0 x = 1 or x = 2 2 or x = − 2 2 The coordinates of the x-intercepts are (1, 0), (2 2, 0) and ( − 2 2, 0).

3

f (0) = (0)3 − (0)2 − 8(0) + 8 = 8 ∴ (0, 8) is the intercept.

Evaluate f (0) to find the y-intercept.

c Sketch the graph of f (x) showing all stationary

c

—) (− —43 , 14 14 27

points and axes intercepts.

y

(0, 8)

(1, 0)

(−2 2, 0) 0

(2 2, 0) (2, −4)

x

WorkeD exampLe 23

The curve with equation y = ax2 + bx + 7 has a stationary point at (−2, 10). Find the values of a and b. think 1

Write the rule.

2

Differentiate y to find the gradient function.

3

Put

4

y = ax2 + bx + 7 dy = 2ax + b dx

dy = 0 and substitute x = −2 into dx

dy , as x = − 2 is a stationary point. dx Substitute x = −2 into y and put y = 10 to get another equation with a and b.

2a(−2) + b = 0 ∴ −4a + b = 0

a(−2)2 + b(−2) + 7 = 10 ∴ 4a − 2b = 3

5

Solve the simultaneous equations [1] and [2]. Add equations [1] and [2] to eliminate a and solve for b.

−2b

6

Substitute b = −3 into equation [1] to find a.

−4a

[1]

[2]

+b=3 −b = 3 ∴ b = −3 −3=0 =3 ∴ a = − 43 −4a

7

Write the values of a and b.

∴ a = − 43 and b = − 3

8

Alternatively, all of this working can be done on a CAS calculator. First, define f (x).

Define f (x) = ax2 + bx + 7

9

Use the solve and differentiation features of the CAS calculator to find a and b.

d solve f (− 2) = 10 and ( f ( x )) = 0 x = − 2, a dx

Record the result.

a=

10

416

Write

Maths Quest 11 Mathematical Methods CAS

−3 4

and b = −3

Sketching graphs containing stationary points exercise 9F

For each of the following functions, determine the value(s) of x where the gradient is zero. f (x) = + 2x b f (x) = x2 − 8x + 5 3 2 f (x) = x − 3x d f (x) = 2x3 + 6x2 − 18x + 1 y = (x + 6)(x − 2) f y = x2(x − 1) 2 1 3 2 y = 10 + 4x − x h y = 3 x − 3x + 5x − 2

1 We20a a c e g

x2

For each function in question 1, determine all of the stationary points.

2 We20b 3

If f (x) = x2 − 8x + 1: a show that there is a stationary point when x = 4 b evaluate f ′(3) and f ′(5) c state which type of stationary point it is.

4

For the function f ′(x) = 5 − x2: a find x when f ′(x) = 0 b state which type of stationary point it has.

5

If f (x) = x3 − 4 then: a show that there is a stationary point when x = 0 only b find f ′ (−1) and f ′(1) c state which type of stationary point it is.

6

If f ( x ) = 13 x 3 − x 2 − 3 x + 5: a show there are stationary points when x = −1 and x = 3 b evaluate f ′(−2), f ′(0) and f ′(4) c state which type of stationary points they are. When x = 1, the curve y = 2x2 − 3x + 1: a is decreasing C has a local minimum e does not exist

DiGitaL DoCS doc-9721 Quadratic graphs doc-9730 Cubic graphs

7 mC

B has a local maximum D is increasing

When x = −1 the function y = x3 − 2x2 − 7x: a is decreasing B has a local maximum C has a local minimum D is increasing e does not exist

8 mC

9 mC

if x >

−2

The graph below that best represents a function with f ′(−2) = 0, f ′(x) < 0 if x < −2, and f ′(x) > 0 is: y

a

y

B

y

C

f(x) f(x)

D

−2

0

x −2

f(x)

y

f(x)

e

y

0 −2

0

x

0

x

−2

0

x

f(x)

x

−2

Chapter 9 • Differentiation

417

f ′(1) = f ′(4) = 0 and f ′(x) < 0 if 1 < x < 4 and f ′(x) > 0 if x < 1 and x > 4. The graph that satisfies these conditions is:

10 mC a

y

B

f(x)

y f(x)

0 0

1

x

4

y

C

D

1

4

x

y

f(x) 0

1

x

4

0

1

4

x

f(x) e

y f(x)

0

4

1

x

11 We21 For each of the following, find the stationary points and determine their nature. a y = x2 + 6x + 2 b y = 8x − 2x2 c y = x3 − x2 1 2 1 3 1 2 f y = (x − 1)3 3 d y = x + x −3 e y = x − x − 2x 2

g y=

x3

+3

h y=

3

2

x3

− 27x + 5

12 We22 Use a CAS calculator to help sketch the graphs of the following functions, labelling all DiGitaL DoCS doc-9795 SkillSHEET 9.5 review of discriminant doc-9796 SkillSHEET 9.6 Solving cubic equations

intercepts and stationary points. f (x) = x2 − 2x − 3 b f (x) = x3 − 3x − 2 c f (x) = x3 − 2x2 + x 2 3 2 f (x) = x (3 − x) e f (x) = x + 4x + 4x f f (x) = x3 − 4x2 − 11x + 30 3 2 3 f (x) = (x + 2) h f (x) = 24 + 10x − 3x − x i f (x) = x3 − 2x2 − x + 2 3 f (x) = 8 − x 13 For the functions y = −2x2 − 5x + 8 and y = 2x3− x2 − 4x + 5: a use a CAS calculator to sketch the graph of each function, showing all coordinates of stationary points b hence, find the x-values for which: dy dy dy i ii iii =0 >0 < 0. dx dx dx a d g j

14 The curve with equation y = 2x2 + bx + 8 has a turning point at x = 3. Find the value of b. 15 We23 The curve with equation y = ax2 + bx has a stationary point at (3, 9). Find the values of a and b. 16 The curve with equation y = ax2 + bx + c passes through the point (2, −4) and has a stationary point at

(1, − 32 ). Find the values of a, b and c. 17 The curve with equation y = ax3 − x2 + bx + 2 has turning points at x = 1 and x = −2. Find the values of a and b.

Solving maximum and minimum problems 9G

There are many practical situations where it is necessary to determine the maximum or minimum value of a function. For quadratic functions, differentiation makes this a relatively simple task because, as we saw in the p