L1-maths-graphing-workbook.pdf

Study Tip:

Sing Songs It’s corny but it really works! Take facts or ideas you need to learn. Transform them into a goofy poem, song or rap. It’s: • Easier to learn • Easier to recall • Not so boring • Best for learning facts

PAGE 38

GRAPHING

GRAPHING INVESTIGATE TABLES, EQUATIONS & GRAPHS 4 CREDITS (91028)

THE SKILLS YOU NEED TO KNOW: LINEAR FEATURES

p 42

There are two features for any linear graph that are important: 1. The y-intercept This is the value when the graph starts and is the initial condition. 2. The gradient This is the slope of the graph and gives the speed/rate etc. of what is being graphed.

GRAPHING LINEAR EQUATIONS

p 47

• Method 1 ( y = mx + c ) 1. Plot the y-intercept 2. Then plot the points according to the gradient • Method 2 - Cover-up method 1. To find the x-intercept make y = 0 and solve. 2. To find the y-intercept make x = 0 and solve. 3. Plot these two points and draw a line through them.

LINEAR SEQUENCES

1. Draw a 5 columned table 2. The x column is always 1, 2, 3, 4, ... or the identifier of your sequence. 3. The y column is the terms of your sequence 4. The m column is the difference between the y values (they should all be the same) 5. The mx column is the difference multiplied by the x value. 6. The final column c is the difference between the mx and y values. It should always be the same number. 7. The final step is to put the m and c values into the equation;

y = mx + c

WRITING LINEAR EQUATIONS

p 52

1. The general equation is: y = mx + c 2. c is the y-intercept 3. m is the gradient =

INTERPRETING LINEAR GRAPHS

Drawing a Graph 1. Plotting the points 2. Discrete or continuous

p 45

rise run

p 55

Interpreting a Graph 1. y-intercept 2. Gradient 3. Other interesting features • State the reasoning behind any answer giving actual values from the graph such as the gradient(s), starting point(s), or value(s) at a specific point.

Note: Problems may rely on knowledge from earlier graphing sections

PAGE 39

QUADRATIC SEQUENCES p 63 1. Draw a 9 columned table 2. The x column is always 1, 2, 3, 4, ... or the identifier of your sequence. 3. The y column is the terms of your sequence 4. The d1 column is the difference between the y values 5. The d2 column is the difference between the d1 values (they should all be the same). a is half d2. 6. The ax2 column is half the second difference multiplied by the x value squared. 7. The L column is the difference between the y and ax2 values. This gives us a linear sequence. 8. The final three columns are solving a linear sequence. 9. The final step is to put the a, m and c values into the equation 2

y = ax + mx + c

TRANSLATING PARABOLAS

p 77

1. To translate an equation by (c,d) simply replace x with (x - c) and y with (y - d) and simplify.

SCALING PARABOLAS p 80

• Parabolas can be made taller or shorter by putting a number in front of the equation. • Steps: 1. Write equation using either vertex or x-intercept method 2. Find a point not used to write the equation and substitute it into the equation with k 3. Rearrange to find the value of k 4. Write out the final equation

PAGE 40

GRAPHING PARABOLAS: X-INTERCEPTS p 65 1. The general equation is: y = ( x + a )( x + b)

2. a and b are the two x-intercepts. They are the negative of the numbers in the equation. The vertex of the parabola is half way between the x-intercepts. 3. Substitute the x value of the vertex into the original equation to find the v value of the vertex. Note: A negative sign at the front of the equation means the parabola is upside down.

GRAPHING PARABOLAS: VERTEX p 71 1. The general equation is y = ( x − c) 2 + d

2. c is how far the vertex has been horizontally shifted from the origin. 3. d is how far the vertex has been vertically shifted from the origin. 4. Use this form if there are no x-intercepts or if you are given a vertex and the x-intercepts are unclear. Note: A negative sign at the front of the equation means the parabola is upside down.

QUADRATIC PROBLEMS p 84 These are almost always Excellence questions. These common themes often exist in the questions: 1. Write an equation 2. Substitute in known values 3. Solve 4. Put in context Note: There are often different ways of coming to the same answer.

GRAPHING

Study Tip:

Difficult Areas If you are struggling: Don’t spend hours trying to understand Do: • Write the problem down as precisely as possible so someone can help • Ask a teacher – they will be pleased to help, it’s their job! • Or ask a parent, sibling or friend Don’t be afraid to ask!

PAGE 41

LINEAR FEATURES SUMMARY

There are two features for any linear graph that are important: • The y-intercept This is the value when x = 0 and is the initial condition. e.g. y = 2. • The gradient This is the slope of the graph and gives the speed/rate etc. of what is being graphed. Gradient =

rise run

= e.g. Gradient

rise 2 1 = = run 4 2

For a complete tutorial on this topic visit www.learncoach.co.nz

PRACTICE QUESTIONS

1.

Sam needs to hire a car and he gets quotes from two different companies. He uses the graph below to work out the cost of a trip for either company.

3.

a. How far is Amanda’s home from the school? b. How fast (in meters/min) did Jamie ride on her

a. How

much does Rob’s Rides charge per kilometre? b. Charges by Cam’s Cars include a flat charge for every trip. How much is it?

2.

James is doing some home renovations and needs a concrete mixer. Two companies hire them out: Hire Co. and Garden Equipment. The cost for up to 30 days is shown.

a. How much does Garden Equipment charge per day?

b. What is the initial fixed fee at Hire Co.?

PAGE 42

Jamie and Amanda both leave school at the same time, Jamie rides a bike and Amanda skates. The following graph shows their distances from home and the time it takes them to get home.

4.

c.

way home? How long did it take Jamie to ride home?

A dairy company is testing two different types of pump for pumping milk from its tankers into the factory tanks. By attaching the pumps to the milk tankers and measuring how long it takes to empty them it can be determined how well they perform. The graph below shows what happened with the first pump.

a. How many litres of milk did the tanker hold initially?

b. How many litres were emptied in one minute?

GRAPHING 5.

Will and Leah were travelling overseas. They wanted to go online to confirm bookings and the next stage of their travel. There were three internet cafes near their hostel: Cyber Time, Gateway, and Cafe Cyber. The graph below shows the connection costs for Cyber Time and Gateway Cafes for up to two hours (120 minutes).

Cafe Cyber costs 7 cents per minute plus an initial connection fee of 50 cents. The cost to use Cafe Cyber can be described by the equation C = 7t + 50 where t = time the internet was used (in minutes), and C = cost of the time on the internet (in cents). a. Which one of the three cafes has the highest connection fee? b. How is this shown by the graphs?

6.

Brian and Lisa are at a mountain bike park where there are various tracks, jumps and drop-offs. At one point two different jumps are side by side: Hang Time and Baby Steps. They can be seen below in the graph where d = distance from the start (in meters) and h = height above the ground (in meters).

a. What is the height of the Baby steps jump at the start?

b. What is the increase in height of the Hang Time over the 5 m?

Study Tip:

Believe in Yourself Self belief is the biggest indicator of achievement: Set yourself a goal that you really want • Then understand and genuinely believe it is possible to pass (or get Excellence) • Stop worrying, just patiently follow the steps outlined for you • When you meet barriers, don’t be discouraged, simply find a way through If you can’t make progress yourself, just ask for help.

PAGE 43

ANSWERS

PRACTICE 1.

a. It goes through the points (0,0) and (40,60) rise 60 − 0 60 = = 1.5 gradient = = run 40 − 0 40

2.

c. 4.

100 = $20 per day (from gradient) 5 (Achieved)

5.

b. $150 (from y-intercept)



a. 1200 m or 1.2 km (y-intercept)



(Achieved)

(Achieved)

(Achieved)

a. 25000 L (y-intercept) (Achieved) 25000 b. = 2500 litres per minute

10

6.

(Achieved)

6 minutes (x-intercept)



Therefore there is a charge of $1.50 per kilometre. (Achieved) b. Flat charge is $30 (y-intercept). (Achieved)

a.

3.

1800 = 300 m per minute. 6

b.

(Achieved)

a. Gateway (Achieved) b. It has the highest C-intercept. i.e. the cost is

highest when time is zero.

(Achieved)

a. 1 m (h-intercept) b. 3 − 0.5 = 2.5 m

(Achieved) (Achieved)

Study Tip:

Study Location Choose an ideal study place: • Can be the library, your room, anywhere quiet • Should have none of the distractions that could slow you down • Should have all the resources you need • Should enable more effective study and give better results

PAGE 44

GRAPHING

LINEAR SEQUENCES SUMMARY y = mx + c

• • • •

When given a linear sequence, e.g. 10, 13, 16, 19, a 5 columned table is drawn The x column is always 1, 2, 3, 4, ... or the identifier of your sequence. The y column is the terms of your sequence x y m The m column is the difference between the y values 1 10 (they should all be the same) 3 2 13 • The mx column is the difference multiplied by the x 3 3 16 value. In this case 3 × 1, 3 × 2, ... 3 19 • The final column c is the difference between the mx and 4 y values. It should always be the same number. • The final step is to put the m and c values into the equation; y = mx + c In this case giving: y = 3x + 7

mx

c

3

7

6

7

9

7

12

7

For a complete tutorial on this topic visit www.learncoach.co.nz

OLD NCEA QUESTIONS

1.

Sarah starts making a pattern of houses using toothpicks as shown in the diagram below. Design (n)

Number of toothpicks used in the design (T)

1

5

2

9

She begins a table for the number of toothpicks she uses for the number of houses in the pattern. Give the rule for calculating the number of toothpicks T that Sarah will need to make the ‘nth’ design.

2.

3 4 5 6

25

Kiri decides to make a different pattern involving separate houses. She begins with the same design as Sarah, as shown in the diagram to the right. Each new shape adds one more toothpick to each side of the previous design, as shown in the diagram to the far right. a. Give the rule for the number of toothpicks required to make the ‘nth’ house in Kiri’s pattern. b. Use this rule to find the number of toothpicks needed for the 6th house in the pattern.

PRACTICE QUESTIONS

For each of the following patterns find the linear equation that represent them.

3.

6, 15, 24, 33, 42

4.

14, 16, 18, 20, 22

5.

2, -2, -6, -10, -14

6.

3, 28, 53, 78, 103

7.

Jo grouped marbles together into groups. Each group had a different number, the first had 4, the second had 11, the third had 18, the fourth had 25, and the fifth had 32. Write an equation which models this pattern.

8.

Sam had a bag of lollies. If in the first minute he had 37 and each minute thereafter it decreased in the following pattern: 37, 29, 21, 13, 5. Write an equation which models the number of lollies, N, and the time passed, t.

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ANSWERS

5.

NCEA 1.

n

T

1

5

2

9

3

13

4

17

5

21

Equation is:

2.

a.

m

3

-6

8

1

4

-10

12

1

5

-14

16

1

20

1

T = 4n + 1

2

10

3

15

4

20

5

25

(Achieved)

m

mx

c

5

0

10

0

15

0

20

0

25

0

5 5 5 5

6.

T = 5n (Achieved) b. T = 5 × 6 = 30 (Merit)

PRACTICE x

y

1

6

2

15

3

24

4

33

5

42

Equation is:

4.

y

1

14

2

16

3

18

4

20

5

22

PAGE 46

mx

c

9

-3

18

-3

27

-3

36

-3

45

-3

9 9 9 9

y = 9 x − 3

x

Equation is:

m

y

m

(Achieved)

mx

c

2

12

4

12

6

12

8

12

2 2 2 2

= 2 x + 12

10

12 (Achieved)

x

y

1

3

2

28

3

53

4

78

5

103

x

y

1

4

2

11

3

18

4

25

5

32

t

N

1

37

2

29

3

21

4

13

5

5

Equation is:

c

-4

6

-8

6

-12

6

-16

6

-20

6

-4 -4 -4 -4

m

(Achieved)

mx

c

25

-22

50

-22

25 25 25 25

75

-22

100

-22

125

-22

m

(Achieved)

mx

c

7

-3

14

-3

21

-3

28

-3

35

-3

7 7 7 7

y = 7 x − 3

Equation is:

8.

mx

y = 25 x − 22

Equation is:

7.

m

y = −4 x + 6

Equation is:

Equation is:

3.

-2

1

4

5

2

4

4

1

2

c

4

T

y

1

mx

4

n

x

m

(Achieved)

mx

c

-8

45

-16

45

-24

45

-32

45

-40

45

-8 -8 -8 -8

N = −8t + 45

(Achieved)

GRAPHING

GRAPHING LINEAR EQUATIONS SUMMARY

• Method 1 ( y e.g.

y=

= mx + c )

3 x+3 4

1. Plot the y-intercept (in this case it is +3) 2. Then plot the points according to the gradient (in this case move up 3 and across 4) • Method 2 - Coverup method e.g. 4 y − 3 x = 12 1. To find the x-intercept make y = 0 and solve.

4 × 0 − 3 x = 12 −3 x = 12 x = −4

2. To find the y-intercept make x = 0 and solve.

4 y − 3 × 0 = 12 4 y = 12 y=3

3. Plot these two points and draw a line through them.

For a complete tutorial on this topic visit www.learncoach.co.nz

OLD NCEA QUESTIONS

1. 2.

Sketch graphs of the equations below on the grids given.

a. y = 6 − 2 x

b. 3 y − 2 x + 6 = 0

Sketch graphs of the equations below on the grids given.

1 2

a. y = x + 2

b. 3x + 2 y = 18

There are more blank sets of axes on the next page.

PAGE 47

PRACTICE QUESTIONS

Draw the following graphs:

3. x = 5 6.

y = −7

9. 2 y + 3 x = 9

PAGE 48

4.

y = 2x − 5

7.

y = 4x − 2

10. 4 y − 8 x = −8

1 x−2 2 5 8. y = − x − 6 2 1 11. y = x + 5 3 5.

y=

GRAPHING

PAGE 49

ANSWERS

b. Not in the form y = mx + c therefore cover-up

NCEA 1.

method: y = 0:

a. Rearranged: y = −2 x + 6 y-intercept = c = 6 gradient = m = -2

x = 0:

(Achieved) b. Not in the form y = mx + c therefore cover-up method: y = 0: 3× 0 − 2x + 6 = 0

x = 0:

−2 x + 6 = 0 −2 x = −6 x=3 3y − 2× 0 + 6 = 0 3y + 6 = 0 3 y = −6 y = −2



(Achieved)

3.

4.

2.

3 × 0 + 2 y = 18 2 y = 18 y=9

PRACTICE





3 x + 2 × 0 = 18 3 x = 18 x=6

(Achieved)

y-intercept = c = -5 gradient = m = 2

(Achieved)

a. y-intercept = c = 2

gradient = m = 1/2





PAGE 50

(Achieved)

(Achieved)

GRAPHING 5.

9.

y-intercept = c = -2 gradient = m = 1/2

Not in the form y = mx + c therefore cover-up method: 2 × 0 + 3x = 9 y = 0:

x = 0:



3x = 9 x=3 2 y + 3× 0 = 9 2y = 9 y = 4.5

(Achieved)

6.



10.

7.



8.

x = 0:

−8 x = −8 x =1 4 y − 8 × 0 = −8 4 y = −8 y = −2

(Achieved)

y-intercept = c = -6 gradient = m = -5/2



11.



Not in the form y = mx + c therefore cover-up method: y = 0: 4 × 0 − 8 x = −8

(Achieved)

y-intercept = c = -2 gradient = m = 4

(Achieved)

(Achieved)

y-intercept = c = 5 gradient = m = 1/3

(Achieved)



(Achieved)

PAGE 51

WRITING LINEAR EQUATIONS SUMMARY

• The general equation is: y = mx + c • c is the y-intercept e.g. from graph y-intercept = 2 • m is the gradient

=

rise run

e.g. from any two points on the graph

m=

2 1 = 4 2

• Combined together the equation for this line is y =

1 x+2 2

For a complete tutorial on this topic visit www.learncoach.co.nz

OLD NCEA QUESTIONS

1.

Jake is hiring a builder to help with some odd jobs around his house. Jake agrees to pay the builder for his travel costs and an hourly rate for work on the job. A graph is drawn that shows how much Jake pays for the work. Find the equation for the line on the graph that shows the total cost for doing the work

2.

Write the equation of the graph shown below.

3.

Write the equations of the lines drawn on the grid to the right.

PAGE 52

GRAPHING

PRACTICE QUESTIONS

Write the equations of the lines drawn below.

4.

5.

6.

7.

Study Tip:

Follow the Steps Think logically to increase your achievement level: • Learn the steps to achieving every type of question • Follow the steps EVERY time!

PAGE 53

ANSWERS

5.

NCEA 1.

2. 3.

y-intercept is at y = 40. Another point in the line is (4, 140)

a. y-intercept is at y = -6. Gradient is -2/3

2 y = − x − 6 3

rise 140 − 40 100 = = 25 gradient = = run 4−0 4 therefore equation is C = 25h + 40 (Merit) h

b. No y-intercept as line is vertical.

It is a vertical line that crosses at x = -3 so the equation is x = -3. (Achieved)

c.

a. y-intercept is at y = 0.

another point in the line is (4, -3)

rise −3 − 0 −3 = = run 4−0 4 −3 x (Merit) h therefore equation is y = 4 another point in the line is (-5, 0)

rise 0 − 4 −4 4 gradient = = = = run −5 − 0 −5 5 4 therefore equation is y = x + 4 (Merit) h 5

PRACTICE 4.

Gradient is -1

y = 4x + 8

(Merit) h

b. y-intercept is at y = 1.

c.

1 x + 1 2

y-intercept is at y = -5. Gradient is 0 y = −5

PAGE 54

b. y-intercept is at y = 0. c.

Gradient is 3 y = 3 x (Merit) No y-intercept as line is vertical. Gradient is infinite (Achieved) x = 6

7. a. y-intercept is at y = 2. Gradient is -4/3

(Merit) h

(Achieved)

(Merit)

b. y-intercept is at y = -8. Gradient is 3

c.

Gradient is 1/2

y=

y = − x + 2 (Merit)

4 y = − x + 2 3

a. y-intercept is at y = 8. Gradient is 4

Gradient is infinite. (Achieved) x = −6 y-intercept is at y = 6. Gradient is -1 y = − x + 6 (Merit)

6. a. y-intercept is at y = 2.

gradient =

b. y-intercept is at y = 4.

(Merit)

y = 3 x − 8 (Merit)

y-intercept is at y = 0. Gradient is 1/5

1 y = x (Merit) 5

GRAPHING

INTERPRET LINEAR GRAPHS SUMMARY

Drawing a Graph 1. Plotting the points 2. Discrete or continuous

Interpreting a Graph 1. y-intercept 2. Gradient 3. Other interesting features • State the reasoning behind any answer giving actual values from the graph such as the gradient(s), starting point(s), or value(s) at a specific point.

For a complete tutorial on this topic visit www.learncoach.co.nz

OLD NCEA QUESTIONS

1.

b. Kiri

Sarah starts making a pattern of houses using toothpicks as shown in the diagram below.

decides to make a different pattern involving separate houses. She begins with the same design as Sarah, as shown in the diagram to the near right. Each new shape adds one more toothpick to each side of the previous design, as shown in the diagram to the far right. The equation for the new pattern is T = 5n . Describe how the graph for the number of toothpicks Kiri used for n houses relates to Sarah’s graph.

She writes a table for the number of toothpicks she uses for the number of houses in the pattern and works out an equation as: T = 4n + 1 Design (n) 1 2 3 4 5 6

Number of toothpicks used in the design (T) 5 9 13 17 21 25

a. On the grid below, sketch a graph showing the number of toothpicks required for up to the 10th design.

2.

The table below gives the adult single train fares for travel from the centre of a city. Number of stations, n 1–3 4–6 7–9 10 – 12

Stage number 1 2 3 4

Adult single fare $2.00 $3.25 $4.50 $5.75

a. On the grid below, sketch the graph of the adult

train fares against the number of stations from the centre of the city.

PAGE 55

b. A child’s fare is $1.50 for the first stage.

Each additional stage, for a child, increases the fare by 75 cents. If a graph was drawn for the child’s fares, describe the similarities and differences between the graphs of the child’s fare and the adult’s fare.

3.

Blake receives a copy of his bank statement and finds he is overdrawn (he has a negative amount in the bank). He starts a saving plan. The graph below shows the amount of money Blake hopes to have in his bank account, S, if he follows his savings plan for n weeks.

a. How much does Blake plan to bank each week? b. Give the equation for the graph of Blake’s c.

4.

saving plan in terms of S, the amount in Blake’s account, and n, the number of weeks after the start of his saving plan. Blake’s grandmother thinks he should be saving more. At the end of 4 weeks she tells him that if the amount in his bank account at the end of 9 weeks is $300, she will give him $50. He increases the fixed amount he saves each week from the end of week 4. He reaches his grandmother’s target of $300 in his account and banks the $50 from his grandmother. He continues saving at the increased rate after banking the $50 from his grandmother. Describe how the graph changes from week 4 onwards. Hints: You can do this by giving equations for some parts of the graph. You may find it helpful to sketch the graph using the grid above.

Emma is employing Ian to build a deck at her house. She provides all the building material. She pays Ian $P for the number of hours, h, that he works. She also pays for Ian’s travel to her home each day. Ian works for 8 hours each day. He knows the deck will take more than 4 hours to build.

PAGE 56

To help Emma know how much she can expect to pay, Ian provides the following table: Number of hours worked (h) 4 5 6 7 8 9 10

Payment (P) $160 $185 $210 $235 $260 $345 $370

a. On the grid below, plot a graph showing the payment required for the number of hours worked.

b. How much does Ian charge for his travel each

day? Explain why the graph rises more steeply after 8 hours. d. What would Emma expect to pay if the work took 30 hours? Explain your calculation. e. Zarko lives next door to Emma and says he could build the deck for her. He does not need to be paid for his travel, but he charges $35 an hour. How long would the work take if the payments to Zarko and Ian were the same? Explain how you calculated your answer. Hint: there may be more than one solution. f. Another builder gives Emma a graph, showing the amount she would charge for 8 hours work.

c.

Give the rule for the payment that this builder would receive for the first 8 hours that she worked.

GRAPHING 5.

Brad and Zara were arranging hexagonal shaped tables for their wedding. They try different arrangements of tables to see how many people they can fit when the tables are put next to each other, as shown in the diagram. They mark where each person could sit with an X.

6.

The graph below shows Mark and Katie’s journeys from their homes to school. They leave home at the same time. Katie rides her bike and Mark walks.

Brad says the equation for the total number of people, p, if there are n tables, is given by the equation p = 4n + 2 . Zara plots a graph of the number of people, p, against the number of tables.

a. How much further does Katie live from school than Mark?

b. Write a full detailed comparison of Mark and Katie’s journeys from their homes to school.

Explain how the equation and graph relate to the number of people at the tables.

PRACTICE QUESTIONS

7.

Sam needs to hire a car and he gets quotes from two different companies. He uses the graph below to work out the cost of a trip for either company.

8.

James is doing some home renovations and needs a concrete mixer. Two companies hire them out: Hire Co. and Garden Equipment. The cost for up to 30 days is shown.

a. For what distance would the cost of hiring be the same? out the equation for the Cam’s Cars graph.

b. Write

James figures out that he will need the concrete mixer for 27 days. Which hire place would be cheaper? Explain your answer.

PAGE 57

9.

Jamie and Amanda both leave school at the same time, Jamie rides a bike and Amanda skates. The following graph shows their distances from home and the time it takes them to get home.

11.

a. Find

the equation of the growth trend line, where y = Sarah’s height, in cm and x = the number of years since 2003. b. What does the gradient of the trend line show about how Sarah is growing? c. From this trend line, estimate Sarah’s height in the year 2000.

a. Write an equation for Amanda’s distance from home, d, with respect to the time taken, t.

b. They both left school at 3.17 pm. At what time

would the both be the same distance away from home?

10.

A dairy company is testing two different types of pump for pumping milk from its tankers into the factory tanks. By attaching the pumps to the milk tankers and measuring how long it takes to empty them it can be determined how well they perform. The graph below shows what happened with the first pump.

a. What is the equation of the graph shown? b. The pump can be operated at a slower speed c.

if necessary. Add a line to the graph above to show this. A similar test, with the other pump on a different tanker was also performed. What happened can be modelled by the equation:

y = 30000 − 3000 x

Sketch the graph of this equation

PAGE 58

Sarah has been measuring her height since 2003. Below is a graph showing her height, a trend line has also been added to the graph.

12.

Sam was a painter and noticed that the more tins of pain he put in the back of his Ute the more it sagged. He measured this for a few different quantities of paint tins and measured the gap between the tyre and the body of the Ute.

a. What is the equation of the line? Use y = gap size, in cm and x = number of paint tins.

b. With how many paint tins would it take for the c.

Ute to sag and hit the tyre? Sam changed to using paint tins that are twice the weight of the old ones. Draw the graph of what you would expect to happen on the above graph.

GRAPHING 13.

Will and Leah were travelling overseas. They wanted to go online to confirm bookings and the next stage of their travel. There were three internet cafes near their hostel: Cyber Time, Gateway, and Cafe Cyber. The graph below shows the connection costs for Cyber Time and Gateway Cafes for up to two hours (120 minutes).

Cafe Cyber costs 7 cents per minute plus an initial connection fee of 50 cents. The cost to use Cafe Cyber can be described by the equation C = 7t + 50 where t = time the internet was used (in minutes), and C = cost of the time on the internet (in cents). a. Draw the graph for Cafe Cyber from t = 0 to t = 120 minutes.

14.

Brian and Lisa are at a mountain bike park where there are various tracks, jumps and drop-offs. At one point two different jumps are side by side: Hang Time and Baby Steps. They can be seen below in the graph where d = distance from the start (in meters) and h = height above the ground (in meters).

a. What is the equation of the graph of the Hang Time jump?

b. What is the equation of the graph of the Baby Steps jump?

b. i. How long would Will and Leah need to use

the internet for, for the cost to be the same in Cyber Time and Gateway. ii. How is this shown by the graphs?

c. i. Use the graph to write the equation for the Cyber Time cafe.

ii. Use the graph to write the equation for the Gateway cafe.

Study Tip:

Motivation Bored? If you do feel bored, spend some time focusing on what you want to achieve. If a grade isn’t enough, think of a reward that you will treat yourself to if you reach your target grade.

PAGE 59

ANSWERS

NCEA

both increase as the stage number increases (every three stations). However, the adult single fare increases by $1.25 per stage while the child single fare increases by 75 c per stage. This results in increasing vertical distances between the points plotted as the stage number increases (every three stations). (Excellence)

1. a.

3. a. When n increases by one, S increases by $40, i.e. Blake plans to bank $40 per week.

b. S-intercept= -200, gradient = 40 so equation is c. (Achieved: Correct graph as a line) (Merit: Correct graph as a line, starting at 1 and finishing at 10) (Excellence: Correct Graph as discrete variables, starting at 1 and finishing at 10) b. The graph for the number of toothpicks Kiri uses for the nth house will differ from Sarah’s in the gradient. The gradient will be steeper, having increased from 4 in Sarah’s to 5 in Kiri’s graph. Both graphs will still begin at the same place (1,5). (Excellence)

rise 300 − (−40) = run 9−4 340 = = 68 5

New Gradient =

So Blake increases his savings to $68 each week and the graph will have a steeper gradient from n = 5. Let S = 68n + c . When n = 9, we want s = 68 × 9 + c = 300 . i.e. c = 300 − 68 × 9 = −312 . Therefore S = 68n − 312 for n = 5, ..., 9 When n = 10, Blake banks the $50 from his grandmother and continues to save at the increased rate. This will result in a steeper rise in the graph at n = 10, but the same gradient of 68 thereafter. Therefore: S = 68n − 312 + 50 = 68n − 262 for n ≥ 10 (Achieved - a correct) (Merit - b correct) (Excellence - c correct)

2. a.

b.

(Achieved: Sloping line connecting steps) (Merit: Show horizontal sets of points) (Excellence: Correct Graph as discrete variables)

The graphs of the adult and child single fares are similar in that they remain constant for the stations within each of the four stages and they

PAGE 60

S = 40n − 200

At the end of four weeks, Blake’s bank balance is 40 × 4 − 200 = −$40 . At the end of 9 weeks, Blake wants his bank balance to be $300. We calculate a new gradient using the points (4,-40) and (9,300):

4. a.

(Achieved: 4 - 8 hrs plotted somehow) (Merit: 4 - 8 and 8 - 12 plotted somehow) (Excellence: Continuous line for 4 - 8 and 8 - 12, make sure 8 is $260 not $320)

GRAPHING b. Between h = 4 and h = 8, the payments increase

by $25. Therefore Ian charges $25 per hour. Let P = 25h + c be the equation of the line. When h = 4 we want P = 25 × 4 + c = 160 Therefore c = 160 − 25 × 4 = 60 so P = 25h + 60 When h = 0, P = 60 i.e. Ian’s travel cost must be $60. Alternative Solution: When h = 9, the payment increases by 345-260 = $85. Since this includes one hours work ($25) and a second days travel, the cost of the travel must be: 85-25 = $60. c. The steeper rise in the graph after 8 hours reflects the cost of travel included for a second day’s work. d. 30 hrs = 3 × 8 hrs + 6 hrs , so Ian will make 4 trips to Emma’s house. Cost of 30 hours work at $25 per hour

5.

6.

Cost of 4 trips at $60 per trip = 4 × 60 = $240 Therefore total cost = 750 + 240 = $990 To model Zarko’s payments at $35 per hour and no charge for travel, let P = 35h . To equate these payments solve:

35h = 25h + 60 35h − 25h = 60 10h = 60 h=6

So Zarko’s and Ian’s payments would be the same if they worked 6 hours. NB: There is more than one solution. As Ian works for a lower hourly wage but charges a travel allowance, the graph of his costs zigzags above and below Zarko’s giving multiple times where their costs are the same. See the graph of Ian’s and Zarko’s payments in the next column. This graph suggests that other solutions are at 12 and 18 hours. This is because the graphs coincide at these h values (and also at h = 6).

a. Katie’s y-intercept is y = 6

Mark’s y-intercept is y = 1.5. Difference = 6 - 1.5 = 4.5 km. (Achieved) b. Mark and Katie both leave home at the same time for school. Mark has only 1.5 km to travel (y-intercept) while Katie has 6 km to travel (y-intercept). As Katie is riding her bike she travels faster than Mark. This can be seen in the gradients of their respective journeys: Katie has a gradient of -1/3 or a speed of 3 minutes per kilometre while Mark has a gradient of -1/12 or a speed of 12 minutes per kilometre. Mark’s speed is a quarter of Katie’s. They both arrive at school at the same time 18 minutes after leaving home (x-intercept). (Excellence)

= 30 × 25 = $750

e.

We can see in the diagram that for each arrangement of tables, four people are seated at the top and bottom sides of each table, and two people are seated at the ends. This gives the equation p = 4n + 2 , where 4 represents the increase of 4 people each time a table is added, while the 2 represents the two people on the ends of the row of tables. This is also mirrored in the graph where n = 1, p = 4 ×1 + 2 = 6 , and each time n increases by 1 the p value is increased by 4. (Excellence)

PRACTICE

7. a. The charges would be the same for a journey of 60 km.

(Achieved)

b. It goes through the points (0,30) and (60,90). rise 90 − 30 60 = =1 = run 60 − 0 60 therefore equation is C = d + 30 gradient =

(Merit)

8.

Hire Co. would be cheaper because their cost for 27 days is lower than Garden Equipment’s. The y value for 27 days is lower for Hire Co. than for Garden Equipment. (Merit)

9. a. y-intercept = 1200.

rise 0 − 1200 = run 8−0 −1200 = = −150 8 therefore the equation is d = −150t + 1200 gradient =

f.

The graph passes through (0,60) and (6,300) y-intercept = 60.

rise 300 − 60 240 = = 40 = run 6−0 6 therefore P = 40h + 60 gradient =



b. The graphs cross at 4 minutes so 4 minutes after they leave they are the same distance from home. 3.17 pm + 4 minutes = 3.21 pm. (Achieved - a or b correct) (Merit - a and b correct)

(Achieved - b or c correct) (Merit - d correct) (Excellence - e and f correct)

PAGE 61

10. a. y-intercept = 25000. b.

c.



c.

(Achieved - a or c correct) (Merit - two of a, b or c correct) (Excellence - all correct)

13. a.



(Achieved - b or c correct) (Merit - two of a, b or c correct) (Excellence - all correct)

b. i. 50 minutes (Achieved) ii. It is shown by the intersection of the Cyber

11. a. y-intercept = 100 and another point on the line

Time and Gateway graphs.

is (4,140)

c. i. C-intercept is (0,0)

rise 140 − 100 40 = = 10 = run 4−0 4 Equation: y = 10 x + 100 gradient =

rise 1200 − 0 = run 120 − 0 1200 = = 10 120 Therefore C = 10t (Merit)

cm per year. The year 2000 would have a x value of -3 so substituting that into the equation should give Sarah’s height in 2000.

ii. C-intercept is (0,100)

y = 10 × −3 + 100 = 70 cm

occurs at (30,5)

rise 5 − 20 −15 1 gradient = = = =− run 30 − 0 30 2 1 Equation of the line is y = − x + 20 2

b. It would hit when y = 0 1 − x + 20 = 0 2 1 − x = −20 2 x = 40

40 paint tins would be required to make it hit the wheel.

PAGE 62

rise 900 − 100 = run 100 − 0 800 = =8 100 Therefore C = 8t + 100 (Merit) gradient =

(Achieved - a or b correct) (Merit - a and b correct) (Excellence - all correct)

12. a. y-intercept occurs at (0,20) and another point

(Achieved)

gradient =

b. Sarah has average growth of approximately 10 c.



Gradient = -2500 Therefore M = −2500t + 25000

14.

a. h-intercept is at (0,0.5)

rise 2.5 1 = = run 5 2 1 Therefore h = d + 0.5 (Merit) 2 gradient =

b. h-intercept = 1.

rise 1 = run 5 1 Therefore h = d + 1 5 gradient =

(Merit)

GRAPHING

QUADRATIC SEQUENCES SUMMARY y = ax 2 + mx + c

• • • • • • • • • •

When given a quadratic sequence, e.g. 3, 12, 25, 42, 63. We then draw a 9 columned table: The x column is always 1, 2, 3, 4, ... or the identifier of your sequence. The y column is the terms of your sequence x y d1 d2 ax2 L m mx c The d1 column is the difference between the 1 3 2 1 3 -2 9 3 y values 2 12 4 8 4 6 -2 13 3 The d2 column is the difference between the 3 25 4 18 7 9 -2 d1 values (they should all be the same). 17 3 4 42 4 32 10 12 -2 The a value is half d2 in this case a = 2. 21 3 5 63 50 13 15 -2 Therefore the ax2 column is half the second 2 2 difference multiplied by the x value squared. In this case 2 × 1 , 2 × 2 , ... The L column is the difference between the y and ax2 values. This gives us a linear sequence. The final three columns are solving a linear sequence. 2 The final step is to put the a, m and c values into the equation y = ax + mx + c 2 In this case giving: y = 2 x + 3 x − 2 For a complete tutorial on this topic visit www.learncoach.co.nz

OLD NCEA QUESTIONS

1.

Sarah starts making a pattern of houses using toothpicks as shown in the diagram below. She made a table for the number of toothpicks she uses for the number of houses in the pattern. Give the rule for the total number of toothpicks that Sarah would need if she was to continue following the pattern and complete ‘n’ designs.

PRACTICE QUESTIONS

The following patterns are quadratic in nature. Find the quadratic equation that represent them.

2. 3. 4. 5. 6.

1, 3, 9, 19, 33 2, 18, 40, 68, 102 2, 6, 12, 20, 30 -13, -25, -45, -73, -109 Graeme was doing a running drill at hockey practice where he had to run a 20 m distance a set number of times in each set. Set 1: 22, Set 2: 20, Set 3: 16, Set 4: 10, Set 5: 2. The number of times he has to run can be modelled by a quadratic equation. Find the equation from the information given, let s be the set number and T be the times run.

Use this rule to find the total number of toothpicks needed to complete the first 12 designs using S a r a h ’s of toothpicks pattern. Design (n) Number used in the design (T)

7.

1

5

2

9

3

13

4

17

5

21

6

25

Lisa was plaing a game on the computer where she had to level up. Each time she went up a level it got harder, and she needed more experience than the previous level to do it. Using the information given work out the quadratic formula for the experience needed for each level. Let l be her level and E be the experience needed.

Level (l) Experience (E) 5

82

6

117

7

158

8

205

9

258

PAGE 63

ANSWERS

5.

NCEA 1.

x

y d1 d2 ax2

L

m mx c

1

5

2

3

3

0

2

14

4

8

6

6

0

3

27

4

18

9

9

0

4

44

4

32

12

12

0

5

65

50

15

15

0

9 13 17 21

3 3 3 3

Therefore the equation for the total number of toothpicks can be written: T = 2n

2

+ 3n

-13

2

-25

3

-45

4

-73

5

-109

6.

25

-2

-16

26

2

3

4

8

-5

-8

3

5

2

3

9

4

18

-9

-12

3

4

19

4

32

-13

-16

3

5

33

50

-17

-20

3

7.

7

6

117

6

108

9

6

147

11

6

192

13

243

15

1

2

3

-1

7

-8

2

18

6

12

6

14

-8

9

258

3

40

6

27

13

21

-8

4

68

6

48

20

28

-8

5

102

35

-8

y d1 d2 ax2

L

1

2

1

1

2

6

2

4

2

3

12

2

9

3

4

20

2

16

4

5

30

4 6 8 10

25

5 2

m mx c 1 1 1 1

1

0

2

0

3

0

4

0

5

0

Equation is therefore: y = x + x (Merit)

PAGE 64

0 0 0 0

0

-9

0

-9

0

-9

0

-9

0

-9

35 41 47 53

2

m mx c 1 1 1 1

1

22

2

22

3

22

4

22

5

22

(Excellence)

75

205

x

m mx c

+ s + 22

82 158

Equation is therefore: y = 3 x + 7 x − 8 (Merit)

27

5

8

27

-25

L

7

2

-8

E d1 d2 ax2

m mx c

75

-6

l

L

7

-4

Equation is therefore: T = − s

y d1 d2 ax2

34

-2

2

x

7

-9

-9

10

28

-100

-2

4

7

-9

24

3

22

-64

-4

-4

7

-8

-2

-1

16

-9

20

2

-4

-36

2

1

14

-8

23

1

-4

-9

-1

16

10

-16

22

3

-4

-36

-8

1

m mx c

6

-28

-9

L

L

-4

-20

-4

T d1 d2 ax2

y d1 d2 ax2 2

-12

L

s

x

Equation is therefore: y = 2 x − 4 x + 3 (Merit)

4.

1

d1 d2 ax2

Equation is therefore: y = −4 x − 9 (Merit)

(Excellence)

2

3.

y

2

PRACTICE 2.

x

m mx c 2 2 2 2

10

-3

12

-3

14

-3

16

-3

18

-3

Equation is therefore: E = 3l + 2l − 3 (Excellence)

GRAPHING

GRAPHING PARABOLAS: X-INTERCEPTS SUMMARY

• A parabola is a graph of a quadratic equation. • The general equation is: y = ( x + a )( x + b) e.g. y = ( x + 3)( x + 1) • a and b are the two x-intercepts. They are the negative of the numbers in the equation. e.g. x-intercepts are -3 and -1 • The vertex of the parabola is the turning point and is half way between the x-intercepts. e.g. The vertex will lay on the line x = -2 • Substitute the x value of the vertex into the original equation to find the y value of the vertex. e.g. y = (−2 + 3)(−2 + 1) = (1)(−1) = −1 The vertex is at (-2,-1) • An equation written y = x( x + 2) is the same. The x-intercepts are just 0 and -2 this time. Note: A negative sign at the front of the equation means the parabola is upside down.

For a complete tutorial on this topic visit www.learncoach.co.nz

OLD NCEA QUESTIONS

1.

3.

For the graph below give:

a. the intercepts b. the function Sketch the graph of the equation:

y = (4 − x)( x + 2)

2.

4.

For the graph below, give:

a. the x and y intercepts b. the equation of the graph. On the grid below, sketch the graph of

y = −( x − 2)( x + 4)

PAGE 65

PRACTICE QUESTIONS

For the following equations factorise (if necessary), write out the x-intercepts and vertex location, and plot the graphs. There are blank graphs on the next page to draw on.

5.

y = ( x − 1)( x + 3)

6.

y = x( x − 3)

7.

y = x 2 − 10 x + 21

8.

y = 9 − x2

9.

y = ( x + 5)( x + 7)

10. y = (3 − x)( x − 4)

11. y = x 2 + 6 x

12. y = ( x + 3)(2 − x)

For the following graphs give the x-intercepts and write out the equation.

13.

14.

15.

16.

17.

18.

PAGE 66

GRAPHING

PAGE 67

ANSWERS

NCEA 1.

PRACTICE

a. x-intercepts are at x = 1 and x = -2 y-intercept is at y = 2.

5.

(Achieved)

b. We know the intercepts so sub into equation

= −4

giving: y = ( x − 1)( x + 2) . As the graph is upside down it is necessary to add a negative to the front of the equation. The final equation is therefore:

y = −( x − 1)( x + 2) = − x 2 − x + 2

x-intercepts at x = 1 and x = -3 Vertex is midway between x-intercepts at x = -1. The y value is: y = ( −1 − 1)( −1 + 3)

(Merit)

2. a. x-intercepts are at x = -3 and x = 3

y-intercept is at y = -9 (Achieved)

b. From the intercepts:

y = ( x − 3)( x + 3) or y = x 2 − 9



3.

(Achieved)

It is a negative parabola with x-intercepts at x = -2 and x = 4 and a y-intercept at y = 8. Vertex is midway between x-intercepts at x = 1. The y value is: y = ( 4 − 1)(1 + 2)



6.

x-intercepts at x = 0 and x = 3 Vertex is midway between x-intercepts at x = 1.5. The y value is: y = 1.5(1.5 − 3)

= −2.25

= (3)(3) =9



4.

(Achieved)

It is a negative parabola with x-intercepts x = 2 and x = -4 and y-intercept y = 8. Vertex is midway between x-intercepts at x = -1. The y value is: y = −( −1 − 2)( −1 + 4)

= −(−3)(3) =9



PAGE 68

(Achieved)

7.

(Achieved)

y = x 2 − 10 x + 21 = ( x − 7)( x − 3) x-intercepts at x = 7 and x = 3 Vertex is midway between x-intercepts at x = 5. The y value is: y = (5 − 7)(5 − 3)

= −4

(Achieved)



(Achieved)

GRAPHING 8.

y = 9 − x2 = − ( x 2 − 9) = −( x − 3)( x + 3)

11. y = x 2 + 6 x

= x ( x + 6)



x-intercepts at x = 0 and x = -6 Vertex is midway between x-intercepts at x = -3. The y value is: y = −3( −3 + 6)

x-intercepts at x = 3 and x = -3 Vertex is midway between x-intercepts at x = 0. The y value is: y = −(0 − 3)(0 + 3)

= −9

=9



9.

(Achieved)

12.

x-intercepts at x = -7 and x = -5 Vertex is midway between x-intercepts at x = -6. The y value is: y = ( −6 + 5)( −6 + 7)

(Achieved)

x-intercepts at x = 2 and x = -3 Vertex is midway between x-intercepts at x = -0.5. The y value is: y = ( −0.5 + 3)( 2 − ( −0.5))

= 6.25

= −1



10.

= 0.25



13.

x-intercepts at x = -8 and x = -5 Equation is therefore: y = ( x + 5)( x + 8) (Merit)

14.

x-intercepts at x = 2 and x = -4 Equation is therefore: y = ( x − 2)( x + 4) (Merit)

15.

x-intercepts at x = 3 and x = 9 Upside down so negative in front. Equation is therefore: y = −( x − 3)( x − 9) (Merit)

16.

x-intercepts at x = 0 and x = 3 Equation is therefore: y = x ( x − 3) (Merit)

17.

x-intercepts at x = 0 and x = -4 Upside down so negative in front. Equation is therefore: y = − x ( x + 4) (Merit)

18.

x-intercepts at x = -1 and x = 4 Equation is therefore: y = ( x + 1)( x − 4) (Merit)

(Achieved)

x-intercepts at x = 3 and x = 4 Vertex is midway between x-intercepts at x = 3.5. The y value is: y = (3 − 3.5)(3.5 − 4)

(Achieved)

(Achieved)

PAGE 69

Study Tip:

Make Sure You Understand Understanding is important. Wait until you have a clear understanding of the topic before moving on to the next section. When you get stuck, ask for clarification (see your teacher the next day).

PAGE 70

GRAPHING

GRAPHING PARABOLAS: VERTEX SUMMARY • The general equation is y = ( x − c) 2 + d e.g. y = ( x − 1) 2 + 2 ▶▶ c is how far the vertex has been horizontally shifted from the origin. e.g. it has been shifted 1 to the right ▶▶ d is how far the vertex has been vertically shifted from the origin. e.g. it has been shifted 2 up • If given the equation, the vertex is located at the point (c, d). e.g. (1, 2) • Use this form if there are no x-intercepts or if you are given a vertex and the x-intercepts are unclear. • Note: A negative sign at the front of the equation means the parabola is upside down.

For a complete tutorial on this topic visit www.learncoach.co.nz

OLD NCEA QUESTIONS

1.

Sketch the graph of the equation:

y = x2 − 9

2.

Sketch the graph of the equation:

y = x2 −1

PAGE 71

PRACTICE QUESTIONS

For the following equations write out the vertex locations, and plot the graphs. There are blank graphs to draw on, on the next page.

3.

y = ( x + 7) 2 + 3

4.

y = x2 − 5

5.

y = −( x − 3) 2 + 5

6.

y = ( x + 5) 2 − 6

7.

y = ( x − 7) 2

8.

y = −( x − 2) 2 − 2

For the following graphs give the vertex locations and write out the equation.

9.

10.

11.

12.

13.

14.

PAGE 72

GRAPHING

PAGE 73

ANSWERS

5.

NCEA 1.

Vertex is at (3, 5)

Looking at the equation it has no horizontal movement and vertical movement of -9.



2.

(Achieved)

6.

(Achieved)

Vertex is at (-5, -6)

Looking at the equation it has no horizontal movement and vertical movement of -1.



(Achieved)

7.

(Achieved)

Vertex is at (7, 0)

PRACTICE 3.

Vertex is at (-7, 3)



8.

4.

(Achieved)

Vertex is at (2, -2)

(Achieved)

Vertex is at (0, -5)





PAGE 74

(Achieved)

(Achieved)

GRAPHING 9. 10. 11.

Vertex is at (-5, 5) Upside down so negative in front. y = −( x + 5) 2 + 5 (Merit) Vertex is at (0, 7) Upside down so negative in front. y = −( x 2 ) + 7 (Merit) Vertex is at (6, 3)

y = ( x − 6) 2 + 3 (Merit)

12. 13. 14.

Vertex is at (-3, -8) y = ( x + 3) 2 − 8 (Merit) Vertex is at (-4, 0) y = ( x + 4) 2 (Merit) Vertex is at (5, -1) Upside down so negative in front. y = −( x − 5) 2 − 1 (Merit)

Study Tip:

Make it Colourful Colours help you remember! Use highlighters to mark the important points. Use different colours for different categories.

PAGE 75

Exam Tip:

Read the Test Scanning the test before you start can help by: • Giving an overview • Is essential for time planning • Helps prevent basic errors (e.g. not staying on topic)

PAGE 76

GRAPHING

TRANSLATING PARABOLAS SUMMARY

• Translating a quadratic is moving it along and/or up a certain amount. • To translate an equation by (c,d) simply replace x with (x - c) and y with (y - d) and simplify. e.g. Translate y = x 2 − 4 x + 3 by (4,3) y − 3 = ( x − 4) 2 − 4( x − 4) + 3 y − 3 = x 2 − 8 x + 16 − 4 x + 16 + 3 y − 3 = x 2 − 12 x + 35 For a complete tutorial on this topic visit www.learncoach.co.nz y = x 2 − 12 x + 38

OLD NCEA QUESTIONS

1.

The parabola shown is moved 3 units to the right and 5 units up. It has an equation of y = − x 2 − x + 2 Give the equation of the parabola in simplified form its new position AND give the y-intercept.

2.

The parabola shown to the right is moved 1 unit to the right and 2 units up. It has an 2 equation of y = x − 9 Give the equation of the parabola in simplified form in its new position AND give the y-intercept.

3.

The equation for the parabola drawn on the graph to the right can be written as y = ( x + 3)( x − 5) If this parabola is moved 3 units to the left and 20 units up, give the equation for the parabola in its new position AND give the coordinates of the y-intercept.

PRACTICE QUESTIONS Translate the following equations by the given amounts, simplify, and also give the y-intercept.

4.

y = x 2 + 3 x − 4 by 6 to the right and 4 down

5.

y = 2 x 2 − 5 x + 5 by 2 to the right and 10 up

6.

y = x 2 + 5 by 1 to the left and 3 down

7.

y = ( x − 4)( x + 3) by 22 to the right and 1 down

8.

y = ( x + 6)( x − 2) by 4 to the left

9.

y = 4 x 2 − 12 x + 20 by 6 to the right and 8 up

10. y = x 2 + 6 x

by 15 to the left and 1 up

11. y = 6 x 2 + x + 8

by 12 down

PAGE 77

ANSWERS

6.

NCEA 1.

The parabola is translated by a vector of (3, 5). Replacing the values and simplifying the equation:

y = − x2 − x + 2 ( y − 5) = −( x − 3) 2 − ( x − 3) + 2 y − 5 = − ( x 2 − 6 x + 9) − x + 3 + 2 y − 5 = − x2 + 6x − 9 − x + 3 + 2 y − 5 = − x2 + 5x − 4 y = − x2 + 5x + 1

y-intercept when x = 0: y = −0 =1

2.

2

7.

+ 5× 0 +1

(Excellence)

The parabola is translated by a vector of (1, 2). Replacing the values and simplifying the equation:

y = x2 − 9 ( y − 2) = ( x − 1) 2 − 9 y − 2 = x2 − 2 x + 1 − 9 y = x2 − 2 x − 6

8. 2

The parabola is translated by a vector of (-3, 20). Replacing the values and simplifying the equation:

y = ( x + 3)( x − 5) ( y − 20) = (( x + 3) + 3)(( x + 3) − 5) y − 20 = ( x + 6)( x − 2) y − 20 = x 2 + 4 x − 12 y = x2 + 4x + 8

y-intercept when x = 0: y = 0 =8

2

9.

+ 4× 0 + 8

(Excellence)

Translated by (6, -4) 2

y = x + 3x − 4 ( y + 4) = ( x − 6) 2 + 3( x − 6) − 4 y + 4 = x 2 − 12 x + 36 + 3 x − 18 − 4 y + 4 = x 2 − 9 x + 14 y = x 2 − 9 x + 10

10.

y-intercept when x = 0: y = 0 − 9 × 0 + 10 = 10 (Excellence) Translated by (2, 10)

y = 2 x2 − 5x + 5 ( y − 10) = 2( x − 2) 2 − 5( x − 2) + 5 y − 10 = 2( x 2 − 4 x + 4) − 5 x + 10 + 5 y − 10 = 2 x 2 − 8 x + 8 − 5 x + 10 + 5 y − 10 = 2 x 2 − 13 x + 23 y = 2 x 2 − 13 x + 33

y-intercept when x = 0: y = 2 × 0 = 33

PAGE 78

2

+ 2× 0 + 3

(Excellence)

Translated by (22, -1)

y = ( x − 4)( x + 3) ( y + 1) = (( x − 22) − 4)(( x − 22) + 3) y + 1 = ( x − 26)( x − 19) y + 1 = x 2 − 45 x + 494 y = x 2 − 45 x + 493

2

− 13 × 0 + 33 (Excellence)

y-intercept when x = 0: y = 0 − 45 × 0 + 493 = 493 (Excellence) Translated by (-4, 0)

y = ( x + 6)( x − 2) y = (( x + 4) + 6)(( x + 4) − 2) y = ( x + 10)( x + 2) y = x 2 + 12 x + 20 2

2

5.

y-intercept when x = 0: y = 0 =3

y-intercept when x = 0: y = 0 + 12 × 0 + 20 = 20 (Excellence)

PRACTICE 4.

y = x2 + 5 ( y + 3) = ( x + 1) 2 + 5 y + 3 = x2 + 2 x + 1 + 5 y + 3 = x2 + 2 x + 6 y = x2 + 2x + 3

2

y-intercept when x = 0: y = 0 − 2 × 0 − 6 = −6 (Excellence)

3.

Translated by (-1, -3)

11.

Translated by (6, 8)

y = 4 x 2 − 12 x + 20 ( y − 8) = 4( x − 6) 2 − 12( x − 6) + 20 y − 8 = 4( x 2 − 12 x + 36) − 12 x + 72 + 20 y − 8 = 4 x 2 − 48 x + 144 − 12 x + 72 + 20 y − 8 = 4 x 2 − 60 x + 236 y = 4 x 2 − 60 x + 244

y-intercept when x = 0: y = 4 × 0 = 244

2

− 60 × 0 + 244 (Excellence)

Translated by (-15, 1)

y = x2 + 6x ( y − 1) = ( x + 15) 2 + 6( x + 15) y − 1 = x 2 + 30 x + 225 + 6 x + 90 y − 1 = x 2 + 36 x + 315 y = x 2 + 36 x + 316

y-intercept when x = 0: y = 02 + 36 × 0 + 316 = 316 (Excellence) Translated by (0, -12)

y = 6x2 + x + 8 ( y + 12) = 6 x 2 + x + 8 y = 6x2 + x − 4

y-intercept when x = 0: y = 6 × 02 = −4

+0−4

(Excellence)

GRAPHING

Exam Tip:

Use the Clues Bullet points and key words are clues in a question. Use them! If you are unsure about the answer, don’t give up. Find the clues and write about closely related word or ideas.

PAGE 79

SCALING PARABOLAS SUMMARY

• Parabolas can be made wider or narrower by putting a number in front of the equation. y = 3( x − 1) 2 + 2 has a 3 in front, making the parabola 3 times narrower e.g. 1 1 1 Or y = ( x + 3)( x − 2) has a in front making it as narrow (4 times wider). 4 4 4 • Steps: y = k ( x − 1) 2 + 2 1. Write equation using either vertex or x-intercept method 2. Find a point not used to write the equation and substitute it into the equation with k 5 = k (0 − 1) 2 + 2 5 = k (−1) 2 + 2 3. Rearrange to find the value of k 5=k +2 k =3 y = 3( x − 1) 2 + 2 4. Write out the final equation • Note: A negative sign at the front of the equation means the parabola is upside down. For a complete tutorial on this topic visit www.learncoach.co.nz

OLD NCEA QUESTIONS

1.

the ball when it is x metres from the point from where it is kicked. If the maximum height of the ball is 2 m, what is the value of k?

In a children’s play park, a ball is kicked so that its flight path can be modelled by the equation h = − kx( x − 6) where h metres is the height of

PRACTICE QUESTIONS For the following equations find the scaling factor using the information given.

2.

Equation: y = k ( x + 3)( x − 4) Passes through the point (3, -12)

4. 6.

3.

Equation: y = k ( x − 2) + 4 Passes through the point (0, 24)

Equation: y = k ( x + 4) − 1 Passes through the point (-10, 11)

5.

Equation: y = k ( x − 2)( x − 5) Passes through the point (6, 16)

Equation: y = k ( x + 1)( x + 4) Passes through the point (-6, 2)

7.

Equation: y = k ( x + 2) + 1 Passes through the point (-4,3)

2

PAGE 80

2

2

GRAPHING For the following graphs find the equations including the scaling factor.

8.

9.

10.

11.

12.

13.

14.

15.

PAGE 81

ANSWERS

8.

NCEA 1.

h is a negative parabola with x-intercepts at x = 0 and x = 6. The maximum height of the ball is reached when

y = k ( x + 7)( x − 1)

0+6 x= =3 2 When x = 3, h = − k (3)(3 − 6) = 9k

Solving for k (substitute in vertex):

−8 = k (−3 + 7)(−3 − 1) −8 = k (4)(−4) −8 = −16k 1 =k 2 1 Equation is therefore: y = ( x + 7)( x − 1) 2 2 OR of the form y = k ( x − c) + d : y = k ( x + 3) 2 − 8

We want this height to be 2m i.e.

9k = 2 ⇒ k =

PRACTICE 2.

3.

4.

5.

6.

7.

2 9

(Merit)

Substituting (3, -12) into the equation:

k ( x + 3)( x − 4) = y k (3 + 3)(3 − 4) = −12 k (6)(−1) = −12 −6k = −12 k =2 (Merit)

solving for k (substitute in x intercept):

0 = k (1 + 3) 2 − 8 8 = 16k 1 =k 2

Substituting (0, 24) into the equation:

k ( x − 2) 2 + 4 = y k (0 − 2) 2 + 4 = 24 k (−2) 2 + 4 = 24 4k = 20 k =5 (Merit) Substituting (-10, 11) into the equation: 2

k ( x + 4) − 1 = y k (−10 + 4) 2 − 1 = 11 k (−6) 2 − 1 = 11 36k = 12 1 k= 3 (Merit)

1 ( x + 3) 2 − 8 2 1 2 7 Both simplify to: y = x + 3 x − (Merit) 2 2 Equation is therefore: y

9.

8 = k (−6 + 8)(−6 + 4) 8 = k (2)(−2) 8 = −4k −2 = k Equation is therefore: y = −2( x + 8)( x + 4) 2 OR of the form y = k ( x − c) + d : y = k ( x + 6) 2 + 8 solving for k (substitute in x intercept):

Substituting (-6, 2) into the equation:

2

k ( x + 2) + 1 = y k (−4 + 2) 2 + 1 = 3 k (−2) 2 + 1 = 3 4k = 2 1 k= (Merit) 2

PAGE 82

Information from the graph: x-intercepts at x = -8 and x = -4 Vertex at (-6, 8) Equation can be of the form y = k ( x − a )( x − b) :

y = k ( x + 8)( x + 4)

k ( x − 2)( x − 5) = y k (6 − 2)(6 − 5) = 16 k (4)(1) = 16 4k = 16 k =4 (Merit)

Substituting (-4, 3) into the equation:

=

solving for k (substitute in vertex):

Substituting (6, 16) into the equation:

k ( x + 1)( x + 4) = y k (−6 + 1)(−6 + 4) = 2 k (−5)(−2) = 2 10k = 2 1 k= 5 (Merit)

Information from the graph: x-intercepts at x = -7 and x = 1 Vertex at (-3, -8) Equation can be of the form y = k ( x − a )( x − b) :

0 = k (−4 + 6) 2 + 8 −8 = k (2) 2 −8 = 4k −2 = k

2

Equation is therefore: y = −2( x + 6) + 8 2 Both simplify to: y = −2 x − 24 x − 64 (Merit)

10.

Information from the graph: y-intercept at y = 6 Vertex at (2, -10) 2 Equation is of the form y = k ( x − c) + d :

y = k ( x − 2) 2 − 10

Continued on next page....

GRAPHING solving for k (substitute in y-intercept):

6 = k (0 − 2) 2 − 10 16 = k (−2) 2 16 = 4k 4=k

13.

y = k ( x + 5) 2 − 8

2 Equation is therefore: y = 4( x − 2) − 10 (Merit)

11.

solving for k (substitute in x-intercept):

0 = k (3 + 5) 2 − 8 8 = k (8) 2 8 = 64k 1 =k 8

Information from the graph: x-intercept at x = 5 and x = -11 Vertex at (-3, 8) 2 Equation can be of the form y = k ( x − c) + d :

1 = ( x + 5) 2 − 8 8 OR of the form y = k ( x − a )( x − b) : y = k ( x − 3)( x + 13)

y = k ( x + 3) 2 + 8

Equation is therefore: y

solving for k (substitute in x-intercept):

0 = k (5 + 3) 2 + 8 −8 = k (8) 2 −8 = 64k 1 − =k 8

solving for k (substitute in vertex):

1 = − ( x + 3) 2 + 8 8 OR of the form y = k ( x − a )( x − b) : y = k ( x − 5)( x + 11) Equation is therefore: y

solving for k (substitute in vertex):

8 = k ((-3) − 5)((-3) + 11) 8 = k (−8)(8) 8 = −64k 1 − =k 8 1 Equation is therefore: y = − ( x − 5)( x + 11) 8 1 3 55 2 Both simplify to: y = − x − x + (Merit) 8 4 8

12.

14.

8 = k (0 + 1)(0 + 5) 8 = k (1)(5) 8 = 5k 8 =k 5

3 = − ( x − 3) 2 + 6 2 15 3 2 Both simplify to: y = − x + 9 x − (Merit) 2 2 Equation is therefore: y

8

Equation is therefore: y = ( x + 1)( x + 5) 5 (Merit)

solving for k (substitute in vertex):

0 = k (1 − 3) 2 + 6 −6 = k (−2) 2 −6 = 4k 3 − =k 2

Information from the graph: x-intercepts at x = -5 and x = -1 y-intercept at y = 8 Equation is of the form y = k ( x − a )( x − b) :

y = k ( x + 1)( x + 5)

y = k ( x − 1)( x − 5)

solving for k (substitute in x-intercept):

−8 = k ((-5) − 3)((-5) + 13) −8 = k (−8)(8) −8 = −64k 1 =k 8 1 Equation is therefore: y = ( x − 3)( x + 13) 8 1 2 5 39 Both simplify to: y = x + x − (Merit) 8 4 8

solving for k (substitute in y-intercept):

Information from the graph: x-intercepts at x = 1 and x = 5 Vertex at (3, 6) Equation can be of the form y = k ( x − a )( x − b) :

6 = k (3 − 1)(3 − 5) 6 = k (2)(−2) 6 = −4k 3 − =k 2 3 Equation is therefore: y = − ( x − 1)( x − 5) 2 2 OR of the form y = k ( x − c) + d : y = k ( x − 3) 2 + 6

Information from the graph: x-intercept at x = 3 and x = -13 Vertex at (-5, -8) 2 Equation can be the form y = k ( x − c) + d :

15.

Information from the graph: vertex at (4, 7) y-intercept at y = 2 2 Equation is of the form y = k ( x − c) + d :

y = k ( x − 4) 2 + 7

solving for k (substitute in y-intercept):

2 = k ( 0 − 4) 2 + 7 −5 = k (−4) 2 −5 = 16k 5 − =k 16 5 Equation is therefore: y = − ( x − 4) 2 + 7 16

(Merit)

PAGE 83

QUADRATIC PROBLEMS SUMMARY

These are almost always Excellence questions. These common themes often exist in the questions: 1. Write an equation (if there isn’t one already) 2. Substitute in known values 3. Solve 4. Put in context Note: There are often different ways of coming to the same answer.

For a complete tutorial on this topic visit www.learncoach.co.nz

OLD NCEA QUESTIONS

1.

A support for a children’s bungy jump is modelled by the function y = − x ( x − 5 ) where y is the height of the support in metres above the ground and x is the distance from the left hand side of the support. a. Sketch the graph of the function for the support on the grid below.

2.

A ball is kicked from the ground and lands at a point 10 m away on the opposite side of a goalpost. The crossbar of the goalpost is 2 m above the ground. When the ball passes over the crossbar, it is at its maximum height of 2.5 m. Give the equation for the height, h metres, of the ball above the ground at a distance, x metres, from where it was kicked, if the path of the ball is modelled by a parabola.

3.

Sam kicks a ball horizontally along a deck that is 5 m above the ground. The ball goes off the edge of the deck. A model for the path of the ball once it leaves the deck is a parabola. 2 The equation of the parabola is y = ax + bx + c The ball hits the ground at a horizontal distance of 3.1 m from the edge of the deck. Find the values of a, b and c.

4.

The cross-section of a ditch on a farm is modelled

b. What is the maximum height of the support? c. A horizontal support beam is put across the support at a height of 4 m above the ground.

How long is the support beam?

2

= y 0.25(5 x − 12 x + 3) by the equation where y is the depth of the ditch in metres and x is the horizontal distance from a given point, P, in metres. At ground level, y = 0, and a pipe crosses the ditch at y = – 0.25 m.

What is the width of the ditch at the level of the pipe?

PAGE 84

GRAPHING 5.

A level path has a flat cover over the top of a parabola-shaped drain. The graph and function show the distance, d metres, from the edge of the path and the depth of the drain, h metres.

The cover on the drain is level with the path and extends 0.7 m each side of the drain. What is the width, w, of the cover?

6.

An animal’s jump can be modelled by a parabola, as shown in the diagram below, where h is the height of the jump in metres, and d is the horizontal distance along the ground in metres.

7.

Huia grows lettuces in a tunnel house. The cross section of the tunnel house can be 2 modelled by the parabola y = − x + 1.69 as shown on the diagram.

a. Find the width of the tunnel house, AB. b. Huia grows tomatoes in a different tunnel

house with a cross section modelled by the parabola, as shown below. The tunnel house is 2 metres high and 4 metres wide. The tomato plants are tied up to a horizontal rail, DC. There are four rows of tomato plants across the tunnel house. The rows are 1 metre apart. How high above the ground is the rail DC?

Find the equation for the parabola, which models a jump that goes a horizontal distance of 9 metres and reaches a maximum height of 2 metres.

PRACTICE QUESTIONS

8.

Jo threw a toy soldier off the top of the high rise building. The height of the figurine can be modelled 2

by the equation: h = 180 − 3t where h is the height of the figurine above Jo’s brother who is waiting on a balcony below, in meters and t is the time since the figurine was let go. a. Draw the graph of h = 180 − 3t 2 for t = 0 to 13. Join the points by a smooth curve. b. Jo had rigged it so that a parachute should have opened 2 seconds after the figurine was let go. This would have caused the figurine to fall at a constant rate until it reached Jo’s brother 12 seconds after it was initially dropped. Add this graph to the one above.

9.

Mary was competing at a national athletics tournament in the Discus. Her coach recorded the flight of the discus during one of her throws. It can be modelled by a parabola and the information can

be seen on the graph below. The maximum height it reaches is 28.125 m.

a. Write

the equation to model Mary’s Discus throw. b. Mary’s mother was watching up in a stand 22 m in the air. Calculate the two occasions where the discus was level with her.

PAGE 85

10.

Jeremy and Jane join an alpine club and get to spend the night in a snow cave they have made. The temperature outside the snow cave from when the sun goes down (6 pm) can be modelled by the

h (h − 16) + 5 , and inside by the 3 equation t = h ( h − 16) + 5 , where t = the 5 equation

t=

11.

Sam was playing cricket. The following graph shows the height of the cricket ball above the pitch after he hit it. y is the height of the cricket ball above the ground in meters, and x is the horizontal distance that the ball has travelled.

temperature (°C), and h = hours since darkness. The temperature inside the snow cave is shown in the graph below.

The graph has the equation:

y = 0.02(45 − x)( x + 2)

a. What is the value of the y intercept and explain what it means in this situation.

b. What is the greatest height of the cricket ball c.

above the ground? Another time when practicing in his back yard Sam again hits a cricket ball in a parabolic shape.

a. Using the following table add the graph of the temperature outside, the previous graph.

h

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

t=

h (h − 16) + 5 to 3

t

5 0 -4.33 -8 -11 -13.33 -15 -16 -16.33 -16 -15 -13.33 -11 -8 -4.33

b. At what time was the temperature the coldest c.

if the sun went down at 6.00 pm? What temperature difference was there at the coldest time between the outside and where the people were inside the snow cave? Be exact.

PAGE 86

When it leaves the bat the ball has a height of 1.5 m. The ball reaches its maximum height of 5.5 m when it is 8 m away from him. There is a wall 14 m away from Sam that is 2 m high. Form an equation to model the path of the cricket ball and use it to determine whether or not the ball will go over the wall. You must justify your answer with clear mathematical reasoning, including showing exactly how you used your equation to obtain your answer.

GRAPHING 12.

Geoff is at a driving range practicing for a golf tournament coming up. His coach is standing off to the side part way down the fairway. He watches Geoff hit a ball which follows this equation

h=

(70 − d )(d + 90) , where d is the distance in 200

13.

Kahu was playing rugby on the wing. She attempted a little chip kick over a defender 5 meters from the centre line and it travelled as shown in the graph below.

meters from the coach the ball is, and h is the height in meters of the ball above the ground.

a. How far does Geoff hit the ball? b. What is the highest height the ball gets to? Find c.

this as accurately as possible. What is the height of the ball when it passes in front of the coach?

Jane is the defender and is standing 8 meters on the other side of the centre line. She can reach 2.5 m into the air when she jumps. Will she be able to catch hold of the ball or will it go over her head?

Exam Tip:

Answering Questions A basic rule that’s often broken: DON’T WASTE TIME ON DIFFICULT QUESTIONS. By answering one question that you struggle with, you might be missing out on five questions that you can do. If it’s hard, move on and come back to it later if you have time.

PAGE 87

ANSWERS

There are often different way to get to the correct answer. These answers show one or two possible routes.

NCEA

OR The graph is a parabola with x-intercepts at x = 0 and x = 10. Therefore the equation can be written in the form h = s ( x − 0)( x − 10) = sx( x − 10) Since (5,2.5) lies on the graph substituting this into the previous equation will give k.

1. a.

k (5 − 0)(5 − 10) = 2.5 −25k = 2.5 k = −0.1 Therefore, h = −0.1x( x − 10)

These two forms are the same as they both simplify to

3.

b. Maximum

occurs at the vertex midway between the x-intercepts at x = 2.5. Sub into the equation: y = −2.5( 2.5 − 5)

= 6.25

c.

Therefore the support length is 4 - 1 = 3 m. (Achieved - a, b or c correct) (Merit - any two correct) (Excellence - all correct)

(Excellence)

Making the ground and the edge of the deck the zero points of the graph. From the information given the vertex is at (0,5) and one of the x-intercepts is (3.1,0). Substituting the vertex into the standard equation 2

gives: y = a ( x − 0) + 5 = ax using (3.1,0) to find a:

Max height is 6.25 m. To get the distance, we need the two x values that correspond to 4 m high. Solve for y = 4:

4 = − x( x − 5) 4 = − x2 − 5x x2 − 5x + 4 = 0 ( x − 1)( x − 4) = 0 x = 1 and x = 4

h = −0.1x 2 + x

2

0 = a × 3.12 + 5 9.61a = −5 a = −0.52

+5

2

Therefore the equation is y = −0.52 x + 0 x + 5 giving values of a = -0.52, b = 0, and c = 5. (Excellence)

4.

Solve equation for y = -0.25 to get x values to determine the width:

0.25(5 x 2 − 12 x + 3) = −0.25 5 x 2 − 12 x + 3 = 1 5 x 2 − 12 x + 4 = 0 (5 x − 2)( x − 2) = 0

2.

therefore x = 0.4 and x = 2. Width of the ditch is therefore 2 - 0.4 = 1.6 m. (Merit)

5. The graph is a parabola with its vertex at (5,2.5) Therefore, the equation can be written in the form

h = k ( x − 5) 2 + 2.5

Since (0,0) lies on this graph:

k (0 + 5) 2 + 2.5 = 0 25k + 2.5 = 0 25k = −2.5 k = −0.1 2 therefore h = −0.1( x − 5) + 2.5

PAGE 88

6.

The parabola has d-intercepts at d = 5.6 and d = 6.8. So the drain cover width: = 6.8 − 5.6 + 2 × 0.7 (Merit) = 2.6 m The parabola has d-intercepts at d = 0 and d = 9, so h = kd (d − 9) with some value of k. We know that the point (4.5,2) lies on the parabola so by substituting that into the equation we can solve for k.

99  k  − 9 = 2 22  81 − k =2 4 8 k =− 81 8 therefore h = − d ( d − 9) 81

(Excellence)

GRAPHING 7. a. Solve for y = 0 to get width

b. Solve for h = 22

1 − t (t − 15) = 22 2 t 2 − 15t = −44 t 2 − 15t + 44 = 0 (t − 4)(t − 11) = 0 t = 4 and t = 11

− x 2 + 1.69 = 0 need 1.69 = 1.3 x 2 − 1.69 = 0 ( x − 1.3)( x + 1.3) = 0

Therefore x = 1.3 and x = -1.3 The width is then 1.3 - (-1.3) = 2.6 m b. The parabola has its apex at (0,2) and substituting that into the standard equation gives:

The discus was level with her when t = 4 and when t = 11. (Merit - a correct) (Excellence - a and b correct)

y = k ( x − 0) 2 + 2

The point (2,0) also lies along the parabola and substituting that into the equation will find k: 2

k2 + 2 = 0 4k + 2 = 0 4k = −2 k = −0.5

10. a.

The equation for the parabola is therefore

y = −0.5 x 2 + 2

Point C occurs when x = 1.5 so:

y = −0.5 ×1.52 + 2 = 0.875

The rail DC is therefore 0.875 m above the ground. (Merit - a correct) (Excellence - a and b correct)

b. Turning point for both graphs is at h = 8 so 6.00pm + 8 hours = 2.00am the next morning. c. Temp inside at h = 8 is t = 8 (8 − 16) + 5 = −7.8 5 Temp outside at h = 8 is -16.33 (from table) Therefore the difference is: -16.33 − (-7.8) = 8.53°C (Achieved - a correct) (Merit - a and b correct) (Excellence - all correct)

PRACTICE 8. a.

(Achieved) b. Straight line from (2,168) to (12,0) (Achieved)

9. a. The t-intercepts are at (0,0) and (15,0) so the

equation will have the form h = kt (t − 15) The turning point will occur midway between the intercepts at t = 7.5, with a height of 28.125 m. Using this information to find k:

7.5k (7.5 − 15) = 28.125 −56.25k = 28.125 1 k =− 2

The equation is therefore

11.

a. y-intercept when x = 0 so :

y = 0.02(45 − 0)(0 + 2) = 1.8

This means in this situation the ball started at this height (when ) so the ball must have been at 1.8 m above the ground as Sam hit it. b. The turning point is where the cricket ball is at its highest. This is midway between the x-intercepts which from the equation are at x = -2 and x = 45. The midpoint is at x = 21.5. Solving the equation will give the height:

y = 0.02(45 − 21.5)(21.5 + 2) = 11.045 m

1 h = - t (t - 15) 2

PAGE 89

c.

As we are given the vertex location, let the equation of the parabola be:

y = k ( x − c) 2 + d

Let the highest point of the ball’s travel be the location of the y-axis and the ground be the x-axis, the origin is the point where they meet. The vertex is then at the point (0,5.5) and so c = 0 and d = 5.5 i.e. the equation is

y = kx 2 + 5.5

When the ball leaves Sam’s bat it is at the point (-8,1.5) on the parabola we can use this point to find k:

k (−8) 2 + 5.5 = 1.5 64k = −4 4 1 k =− =− 64 16

12. a. From the equation d-intercepts are at d = -90 and d = 70 so the distance hit is :

70 − (−90) = 160 m

c.

between

(70 − (−10))(−10 + 90) = 32 m 200

Solve for d = 0 :

h=

(70 − 0)(0 + 90) = 31.5 m 200

(Achieved - one correct) (Merit - all correct)

PAGE 90

k =−

h=−

The ball will be 3.25 m above the ground as it passes over the 2 m high wall, so it will not be stopped. (Achieved - a correct) (Merit - b correct) (Excellence - c correct)

h=

k (0 − 6)(0 + 10) = 6 −60k = 6

1

1 2 6 + 5.5 = 3.25 16

point is midway x-intercepts which is d = -10

h = k (d − 6)(d + 10)

Since the h-intercept is (0,6), we can solve for k:

1 10

the equation: h = − (d 10 So when d = -8:

To get the height of the ball as it reaches the wall we need to solve the equation for x = 6:

b. Highest

From the graph the vertex is at d = -2 therefore x-intercepts are at d = 6 and d = -10 (equal distances either side of the vertex). As we have both x-intercepts the equation will be of the form: h = k ( d − a )( d − b) a = 6 and b = -10 so equation is:

The flight of the ball can therefore be modelled by

1 2 The equation is now y = − x + 5.5 16 y=−

13.

the

− 6)(d + 10)

1 (−8 − 6)(−8 + 10) = 2.8 m 10

As Jane can only reach 2.5 m into the air the ball will go over her head. (Excellence)