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Active Maths 1

Michael Keating, Derek Mulvany and James O’Loughlin

Special Advisors: Oliver Murphy, Colin Townsend and Jim McElroy

2πr

Junior certificate

Project Maths – Strand 5

y 2 -

y 1x 2

- x

1

m =

πr2

Editor: Priscilla O’Connor, Sarah Reece

Designer: Liz White

Layout: Compuscript

Illustrations: Compuscript, Denis M. Baker, Rory O’Neill

ISBN: 978-1-78090-165-71383

© Michael Keating, Derek Mulvany, James O’Loughlin and Colin Townsend, 2012

Folens Publishers, Hibernian Industrial Estate, Greenhills Road, Tallaght, Dublin 24, Ireland

Acknowledgements

The authors would like especially to thank Jim McElroy for his work on the written solutions and his invaluable advice.

The authors and publisher wish to thank Thinkstock for permission to reproduce photographs.

All rights reserved. No part of this publication may be reproduced or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without prior written permission from the publisher. The publisher reserves the right to change, without notice, at any time, the specification of this product, whether by change of materials, colours, bindings, format, text revision or any other characteristic.

Strand 5 Chapters

Chapter 30 Functions ...............................................498 30.1 Functions ...............................................499 Idea of a Function ..................................499 Naming Functions ..................................500 Mapping Diagrams .................................500 Couples ..................................................500 Input–Output Table ................................501 Domain ..................................................501 Range .....................................................501 Codomain ..............................................501 30.2 Function Notation .................................505 Codomain ..............................................506

Chapter 31 Graphing Functions ..............................511 31.1 Linear Functions ....................................512 31.2 Quadratic Functions .............................515 31.3 Applications of Graphs .........................516 Solve f(x) = 0 .........................................516 Solve f(x) = q, Where q Is a Number ......517 Find the Value of f(p) ..............................517 Solve f(x) = g(x) ......................................518 31.4 Transformation of Linear Functions ......523 Linear Functions .....................................523 31.5 Transformations of Quadratic

Functions ...............................................527 Graphs of the Form y = ax2 ....................527 Graphs of the Form y = x2 + b ...............528 Graphs of the Form y = (x + b)2 .............529

Contents

Strand 5 Activities

Chapter 30 Functions ...............................................203

Chapter 31 Graphing Functions ..............................205

Answers

Strand 5 Chapters

30ch

apte

rch

apte

r

 Engage with the concept of a function, domain, codomain and range

 Make use of function notation f (x) = f: x →f: x y =

Learning OutcomesIn this chapter you will learn to:

Functions

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30.1 FunctionsIn mathematics, a relation is a rule that describes how certain things are linked to each other. For example, the harder you kick a ball, the further it will travel.

There are many examples of relations in real life where one thing is related to another.

� Your fitness is related to how much exercise you do.

� Your arm length is related to your height.

� The area of a circle is related to its radius.

idea of a FunctionA function is a special type of relation. When thinking about functions, it can be easier to think about them as a type of machine or calculator.

In a function, the number you put in (the input) is changed into another number (the output) according to a certain relationship (the rule).

3

6

Rule: Double input

(Input)

(Output)

In the above function, each input is doubled.

In a function, each input gives back one output only. For example, an input of 4 into the function machine shown will only ever give an answer of 8 (2 × 4). It can never give back any other answer.

A function does allow for two inputs to give the same output.

–2 (Input)

(Output) 4

Rule: (input)2

2 (Input)

(Output) 4

Rule: (input)2

In the function machine shown, every input is squared. Inputs of –2 and 2 will both have the same output, 4.

KEY W ORDS

YOU SHOULD REMEMBER.. .

function input

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For an everyday example of functions, consider the barcodes used to scan goods in shops.

In a shop, if a barcode is scanned (inputted) at a checkout (function), then only one price (output) will appear. The register will not show two prices for the same barcode.

Input: Barcode Output: Shows that the item costs €1

However, when the checkout shows that the price of an item is €1, there could be many goods in the shop that cost €1.

Output: €1 Input: Numerous items in the shop could cost €1.

naming FunctionsIt is useful to give a function a name. The most common name is f, but other names can also be used, such as g or h (or any other name we want).

Mapping DiagramsOne method of showing a function is to draw a mapping diagram. In a mapping diagram, each input of the function is mapped to its output.

The mapping diagram below shows the function g. In the function g, each input is multiplied by 3 and then 1 is added.

An input of 4 will result in an output of 13. So 4 is mapped onto 13. This is illustrated by the arrows in the diagram.

� When a function is shown using a mapping diagram, no input will map onto two or more outputs.

� However, a function does allow for two inputs to be mapped to the same output.

Input Output

f is not a function

f

f is a function

Input Outputf

couplesA function can also be written as a set of couples or ordered pairs. In each couple, the first element (input) is related to the second element (output).

Each couple or ordered pair is written as (Input, Output).

f = {(a,1), (b,4), (c,3), (d,5)} are the couples or ordered pairs of the function f shown on the right.

When a function is written as a set of couples, no two couples will have the same input.

� The couples: {(1,3), (3,5), (2,4)} represent a function, as no two couples have the same input.

� The couples: {(1,3), (1,2), (3,4)} do not represent a function, as the input ‘1’ appears twice.

ab

c

f

d

1

4

3

5

4

Input Outputg

1

0

13

4

1

Rule: Multiply by 3and add 1

Rule: Multiply by 3 and add 1

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input–output tableAn input–output table can be used to show functions.

Input Rule: Double the input and add 4 Output1 2(1) + 4 63 2(3) + 4 104 2(4) + 4 12

We can then write this information as a set of couples: {(1,6), (3,10), (4,12)}.

Consider the function h as shown.

145

h

36789

DomainThe domain is the set of values that you can put into a function. It is the set of all inputs. It is the set of all the first elements of the ordered pairs.

In the diagram above, {1, 4, 5} is the domain of the function h.

RangeThe range is the set of values that you get out of a function when you input the domain values. It is the set of all actual outputs. It is the set of all the second elements of the ordered pairs.

In the diagram above, {3, 6, 7} is the range of the function h.

codomainIn the diagram above, {3, 6, 7, 8, 9} is the codomain of the function h. It is the set of all elements into which the function maps.

ACTIVITY 30.1

range

image

Worked Example 30.1

A mapping diagram of the relation R is shown.

–1

0

1

2

3

R

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6

(i) Identify the domain.

(ii) Identify the range.

(iii) Identify the codomain.

(iv) List the ordered pairs.

(v) Is the relation R a function? Explain your answer.

Solution (i) Domain = {–1, 0, 1}

(ii) Range = {2, 3, 4, 5}

(iii) Codomain = {2, 3, 4, 5, 6}

(iv) Ordered pairs: {(–1,2), (0,3), (1,4), (1,5)}

(v) The relation R is not a function because the input 1 has two outputs: 4 and 5.

codomain

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Worked Example 30.2

(i) From the following ordered pairs, draw a mapping diagram to illustrate the relation T.

T = {(–5,25), (1,1), (2,4), (6,36), (5,25)}

(ii) Is the relation T a function?Explain your answer.

Solution (i) RangeTDomain

–5

5

1

2

6

25

1

4

36

(ii) T is a function. Each input has only one output.

Worked Example 30.3

A function machine is shown.

Find the range of this function and write the function as a set of ordered pairs.

SolutionIt may be easier to use an input–output table to show this function.

Input Rule: Square the input and subtract 1

Output

1 (1)2 – 1 0

–3 (–3)2 – 1 8

2 (2)2 – 1 3

4 (4)2 – 1 15

0 (0)2 – 1 –1

Range: {0, 8, 3, 15, –1}

Ordered pairs: {(1,0), (–3,8), (2,3), (4,15), (0,–1)}

{1, –3, 2, 4, 0}

Rule: Square inputand subtract 1

Exercise 30.1

1. Each of the following mapping diagrams represents a relation. Identify the mapping diagrams that also represent a function. Give a reason for each of your answers.

(i)

(ii)

(iii)

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(iv)

(v)

(vi)

2. Identify which of the following relations are functions. In each case give a reason for your answer.

A = {(1,4), (1,5), (3,5), (2,4)}

B = {(1,3), (2,6), (5,15), (–1,–3)}

C = {(1,2), (2,4), (3,4), (5,6)}

D = {(1,1), (1,2), (2,5), (3,5)}

3. A function machine has a rule that states, ‘Add 4 to the input.’ The numbers 4, 5, 6 and 8 are inputted into the function machine. What are the outputs?

4. A function has a rule that states, ‘Multiply the input by 2 and add 3.’ The inputs for the function are {0, 1, 3, – 4}. What are the outputs?

5. Write down the domain, codomain and range of each of the following relations, and identify which relations are functions:

(i) 0

123 9

6340

(ii) 5

279 11

6487

(iii) –3

3

4

5

8

20

13

(iv)

194

1

2

6

1571158

6. A relation R = {(1,10), (2,11), (3,12), (5,14)}.

(i) Write out the domain and range of this relation.

(ii) Draw a mapping diagram of this relation.

(iii) Is this relation a function? Give a reason for your answer.

7. In each of the following relations, change one couple so that the relation represents a function:

A = {(0,6), (6,5), (6,4), (9,1)}

B = {(–1,5), (–3,10), (5,15), (–1,–3)}

C = {(1,4), (–1,4 ), (2,12), (2,10)}

D = {(3,1), (3,3), (4,2), (10,8)}

8. A function has a rule that states, ‘Divide the input by 2.’ The inputs for this function are {10, –8, 20, 50}. Write down the range of this function.

9. A rule for a function states, ‘Square the input and subtract 1.’ The inputs are {1, –3, –5, 5}. Write the inputs and outputs of this function as a set of ordered pairs.

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10. State a rule for each of the functions shown:

(i) 4

11

25 31

17

10

(ii) 10

50

25

30 –9

11

–14

–29

(iii) 0

5

10 100

25

0

(iv) 4

8

15

10 16

12

26

4

11. A relation S = {(0,0), (1,1), (3,–27), (–3,–27), (1,–1)}.

(i) Write out the domain and range of this relation.

(ii) Draw a mapping diagram of this relation.

(iii) Is this relation a function? Give a reason for your answer.

12. Rule: Add 5 to the input. Domain: {0, 2, 4, 6}

13. Rule: Subtract 5 from the input. Domain: {10, 5, 50, –5}

14. Rule: Double the input and then add 3. Domain: {–3, 1, 4, 7}

15. Rule: Triple the input and then subtract 3. Domain: {0, 4, –6, 6}

16. Rule: Square the input and then add 1. Domain: {3, 1, –4, –3}

17. A function has a rule that states, ‘Subtract twice the input from 1.’ The inputs of the function are {3, 5, 6, –3, –1}. Draw a mapping diagram to show the inputs and outputs of this function.

18. A rule for a function is to square the input, multiply the resulting answer by 2 and then subtract 3. The domain of this function is {0, 8, –2, 5, –10}. Draw a mapping diagram to show the domain and range of this function.

19. A relation T = {(2,1), (4,8), (–4,8), (6,18)}.

(i) Write out the domain and range of this relation.

(ii) Draw a mapping diagram of this relation.

(iii) Is this relation a function? Give a reason for your answer.

20. Rule: Square the input and then add 4. Domain: {0, –5, –12, 2}

21. Rule: Square the input and then subtract 6. Domain: {0, 7, –5, –14}

22. Rule: Cube the input and then subtract 10. Domain: {0, 1, 2, –2}

23. Rule: Square the input, multiply the answer by 2, and then add 5 times the input. Domain: {0, 1, 4, 8}

24. A function machine has a rule that states, ‘Add 10 to the input.’ The machine gives outputs of {14, 25, 0, –20}. What is the domain of this function?

25. A function rule states that the input should be multiplied by 2 and then 2 subtracted from the answer. The range of this function is {4, 12, –4, 0}. Find the domain of this function.

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30.2 Function notationA function is usually written using mathematical notation. When writing a function, we usually use the letter:

� f to denote the function (though other letters may be used)

� x to denote the input

� y or f(x) to denote the output

For example, consider the function f, which has the function rule ‘Twice the input plus 1’. This could be written as:

f(x) = 2x + 1 Or y = 2x + 1 Or f: x → 2x + 1 Or f: x 2x + 1

Function name Output

Input

If we want to find the value of the output when we input 5 into this function, we are being asked to find f(5).

26. To convert from degrees Celsius (°C) to degrees Fahrenheit (°F), the following rule can be used: ‘Multiply the temperature in degrees Celsius by 1.8 and then add 32.’

For example, 10°C is equal to 50°F.

(a) Convert the following temperatures to degrees Fahrenheit:

(i) 0°C (ii) 16°C (iii) –2°C

(b) Convert the following temperatures to degrees Celsius:

(i) 32°F (ii) 104°F (iii) 23°F

27. The area of land flooded (in metres squared) by water from a burst pipe is given by the function rule ‘input squared’, where the input is the number of hours the pipe has been burst.

(a) Find the area of land flooded by this burst pipe in:

(i) 1 hour (ii) 3 hours (iii) 5 hours

(b) How long would it take for an area of 225 m2 of land to be flooded by this burst pipe?

Worked Example 30.4

A function f is defined as f(x) = 2x + 1. Find:

(i) f(4) (ii) f(–3) (iii) f(–3) + 2f(4)

Solution (i) f(4)

We input 4 into the function.

f(4) = 2(4) + 1

= 8 + 1

∴ f(4) = 9

(ii) f(–3)

f(–3) = 2(–3) + 1

= – 6 + 1

∴ f(–3) = –5

(iii) f(–3) + 2f(4)

We already know the values for f(–3) and f(4).

2f(4) means that we first find f(4) and then multiply our answer by 2.

f (–3) + 2f (4) = –5 + 2(9)

= –5 + 18

∴ f(–3) + 2f(4) = 13

ACTIVITY 30.2

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Worked Example 30.5

g(x) = x2 + 2x – 5 is a function defined on the domain {1, 2, 3, 4}.

Find the range and write the function as a set of couples.

SolutionFor this example, use an input–output table. Our inputs will be from the domain {1, 2, 3, 4}.

x x2 + 2x – 5 y

1 (1)2 + 2(1) – 5 –22 (2)2 + 2(2) – 5 33 (3)2 + 2(3) – 5 104 (4)2 + 2(4) – 5 19

Range: {–2, 3, 10, 19}

Couples: {(1,–2), (2,3), (3,10), (4,19)}

Worked Example 30.6

h is the function h: x → 4x – 5, as shown on the mapping diagram.

Find the values of p, q and r.

2

h

–3

r 7

q

p

Solution From the diagram, we know that p is the output when we have an input of 2.

∴ h(2) = p

h(2) = 4(2) – 5

= 8 – 5

∴ p = 3

q is the output when we have an input of –3.

∴ h(–3) = q

h(–3) = 4(–3) – 5

= –12 – 5

= –17

∴ q = –17

When r is the input, 7 is the output.

∴ h(r) = 7

⇒ 4r – 5 = 7

4r = 12

∴ r = 3

codomainLet set A = {–2, 0, 2}. Let set B = {0, 1, 5}.

Let f be the function that maps set A to set B.

This can be written as f: A → B, where f(x) = x2 + 1.

We now find f (–2), f (0) and f (2).

� f (–2) = (–2)2 + 1

= 4 + 1

= 5

� f (0) = (0)2 + 1

= 0 + 1

= 1

� f (2) = (2)2 +1

= 4 + 1

= 5

List the couples: {(–2,5), (0,1), (2,5)}

Set A = the domain = {–2, 0, 2}.

Set B = the codomain (the set into which A is mapped) = {0, 1, 5}.

{1, 5} = the range (the set of actual outputs).

This is illustrated using the mapping diagram shown.

BfA

–2

0

2 5

0

1

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Exercise 30.2

1. A function is defined as f(x) = 2x. Find:

(i) f(0) (ii) f(3) (iii) f(4)

2. If f(x) = 4x, find:

(i) f(–1) (ii) f(4) (iii) f(12)

3. f: x → x is a function. Find:

(i) f(1) (iii) f(–2)

(ii) f(5) (iv) 3f(–2)

4. A function f is defined as f(x) = – 4x. Find:

(i) f(2) (iii) f(6)

(ii) f(–3) (iv) 2f(1)

5. f(x) = –x. Find:

(i) f(0) (iii) f(1)

(ii) f(–1) (iv) [f(2)]2

6. Rule: ‘Add 5 to the input.’

7. Rule: ‘Subtract 5 from the input.’

8. Rule: ‘Double the input and then add 3.’

9. Rule: ‘Triple the input and then subtract 3.’

10. Rule: ‘Multiply the input by 4 and then subtract 1.’

11. f(x) = x – 3

12. g: x 2x + 1

13. h(x) = 3 – 4x

14. y = 4x – 11

15. p: x → x2 + 3

16. q(x) = 2x2 – 4

17. s(x) = 5x2 + x

18. t: x → x2 – 3x + 2

19. g(x) = 5x + 1

20. g: x → 3x + 6

21. g(x) = –x + 4

22. g: x 5x – 12

23. g(x) = 4 – 2x

24. Rule: ‘Square the input.’

25. Rule: ‘Square the input and then subtract 6.’

26. Rule: ‘Cube the input and then subtract 10.’

27. Rule: ‘Square the input and then multiply the answer by 2.’

28. Rule: ‘Square the input and then add 2 times the input to the answer.’

29. h(x) = x2 + 1

30. h: x → x2 + x + 3

31. h(x) = 2x2 – 4x + 5

32. h: x 4x2 – 2x – 9

33. h(x) = 3x – x2

34. A function f is defined as f(x) = 4x – 8. Find:

(i) f(0) (iii) f(0) + f(–2)

(ii) f(–3) (iv) 2[f (1)]2

35. A function f is defined as f: x → x2 – 2.

(i) Find f(4).

(ii) Find f(2).

(iii) Is f(0) < f(–1)?

(iv) Find 3f(1) – 2f(–3).

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36. A function f is defined as f: x 3x2 + x – 1.

(a) Find:

(i) f (0)

(ii) f (3)

(iii) 5[f (0)]2 – f (–4)

(b) Investigate if f (2) – f (–2) = 0.

37. A function f is defined as f (x) = √__ x .

Find:

(i) f (0) (iii) f (64)

(ii) f (100) (iv) f (25) + 2f (4)

38. If y = 4x, find the values of y when x ∈ {1, 3, –4, 0}.

39. g(x) = 3 – 2x is a function defined on the domain {0, –1, 5, –6, 7}. Find the range and write the function as a set of couples.

40. p: x → x2 + 2x – 5 is a function defined on the domain {1, 2, 3, 4}. Find the range and write the function as a set of couples.

41. y = x2 – 25 is a function defined on the domain {0, 4, 5, 6}. Find the range and write the function as a set of couples.

42. h(x) = (x + 1)(x – 3) is a function defined on the domain {–3, –4, 0, 4}. Find the range and write the function as a set of couples.

43. f: x 2x is a function, as shown in the diagram.

5

7

–3

d 16

b

c

a

Write down the values of a, b, c and d.

44. f is a function such that f (x) = 3x – 4.

0

2

f

g 11

e

–10

d

Write down the values of d, e, f and g.

45. f: x → x3 – 1 is a function.

r

s

1

4 p

7

q

–2

Write down the values of p, q, r and s.

46. f(x) = x2 – 4 is a function, as shown in the diagram.

a

b0

ed

0

c

21

Write down the values of a, b, c, d and e, where a, d > 0.

47. The range of the function f (x) = 2x + 4 is {4, 6, 14, 36}. Find the domain of the function f.

48. The range of the function f: x → 3 – 5x is {–2, – 47, 13, 3}. Find the domain of the function f.

49. Set A = {–1, 0, 1} and set B = {0, 1, 2}. Let f be the function that maps set A to set B. If f: A → B, where f (x) = x2 + 1, list the elements of:

(i) The range of the function f

(ii) The codomain of the function f

50. A function f is defined as f(x) = 6x – 12.

(i) If f(x) = 18, find the value of x.

(ii) Find the value of x if f(x) = 0.

(iii) Find the value of a if f(a) = –12.

51. The amount of money saved by Luca is given by the function f(x) = 8x + 30. f(x) represents the amount of money saved in euros and x represents the number of weeks.

(a) Find the amount of money saved by Luca after:

(i) 3 weeks

(ii) 6 weeks

(iii) 11 weeks

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(b) How much money had Luca saved initially?

(c) How long will it take Luca to save €350?

(d) How much money does Luca save per week?

52. The profit (y) for different mobile phones can be written as a function y = 2x – 100, where x is the selling price of the phone.

(a) Find the profit for each of the following phones:

(i) Phone A: Selling price €100

(ii) Phone B: Selling price €50

(iii) Phone C: Selling price €75

(iv) Phone D: Selling price €125

(b) The company selling these mobile phones wants each phone to be profitable. State which phone should have its selling price increased.

53. The temperature of a lake can be found using the following function: f(x) = x2 – 3x + 2, where f(x) is the temperature in degrees Celsius and x is the time of day. The function is valid for the period 10 p.m. to 4 a.m.

The temperature was read between the hours of midnight (x = 0) and 4 a.m. (x = 4).

(a) Find the temperature of the lake at:

(i) Midnight (iv) 3 a.m.

(ii) 1 a.m. (v) 4 a.m.

(iii) 2 a.m.

(b) What was the temperature of the lake at 10 p.m. on the previous day?

Revision Exercises

1. f: x → 5x + 2 is a function. Find f(3).

2. g: x → 7x – 1 is a function. Evaluate g(2).

3. f: x 10x + 3. Find f(7).

4. g: x → 6x – 10. Find g(5).

5. f: x → 2x + 1 is a function.

Copy and complete the diagram, replacing the question marks with the appropriate values.

6. g: x 4x – 3 is a function.

Calculate the values of a, b and c.

7. f: x → 9 – x is a function, as shown in the diagram.

Write down the values of p, q, r and s.

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8. f: x → 5x – 2 is a function. The domain of f is {1, 2, 3}. Find the range of f.

9. f is a function such that f(x) = 10 – x.

(i) Find f(6).

(ii) If the domain of f is {6, 7, 8, 9, 10, 11}, find the range.

10. f: x → x2 + 1 is a function defined on the domain {–2, –1, 0, 1, 2}. Find the range.

11. f: x → 3x + 11 is a function. The domain is {1, 2, 3}. Find the range.

12. f: x → 7x – 10 is a function. The domain is {0, 1, 2, 3}. Find the range.

13. y = x2 + 4 is a function. The domain is {–5, 0, 5}. Find the range.

14. f: x → 2x2 is a function. Evaluate:

(i) f(3) (ii) f(5) (iii) f(10)

15. f: x → 3x2 – 5 is a function. If the domain is {2, 3, 4, 5}, find the range.

16. f: x 10x2 – 3 is a function. Evaluate:

(i) f(1) (ii) f(2) (iii) f(3)

17. f: x → 5x – 8 is a function. Find:

(i) f(7)

(ii) f(–7)

(iii) The value of p if f(p) = 17

(iv) The value of q if f(q) = q

18. g: x → 2x + 3 is a function defined on the domain N (the set of natural numbers).

(i) Find the values of g(1), g(2) and g(3).

(ii) Investigate if g(3) = g(1) + g(2).

(iii) Find the value of x if g(x) = 53.

19. f is a function such that f(x) = 3x3 – 5. Fill in the missing numbers in the diagram.

20. f: x → 5x2 + 1 is a function. Fill in the missing numbers in the diagram.

21. A = {0, 1, 2} and B = {0, 1, 2, 3, 4, 5} are two sets. f: A → B, where f(x) = 2x + 1.List the elements of these sets:

(i) The range of f

(ii) The codomain of f

22. A = {–1, 0, 1}, B = {0, 1, 2, 3}. f: A → B, where f(x) = x2 + 2. List the elements of these sets:

(i) The range of f

(ii) The codomain of f

23. f: x → x2 + 2x + 3 is a function. Evaluate f(5).

24. f: x 100 – 2x2 is a function. Evaluate:

(i) f(5) (ii) f(7) (iii) f(2)

25. f: x → x2 – x – 6. Evaluate f(3) and f(–2).

26. f: x → (2x – 1)(x – 3). Evaluate:

(i) f(5) (ii) f(3) (iii) f ( 1 __ 2 ) 27. f: x → (6 – x)(x + 1). Evaluate:

(i) f(4) (ii) f(6) (iii) f(–1)

28. The number of hours homework which a student should do per day is estimated

to be x +1 _____ 2 , where x = the number of the

years in secondary school. How many hours

of homework should be done by a student:

(i) In Third Year? (ii) In Second Year?

29. f: x → 2x + k. If f(6) = 22, find k.

30. f: x → ax + 7.

(i) If f(5) = 22, find the value of a.

(ii) Show on the numberline the solution of f(x) ≤ 16, x ∈ N.

 Interpret simple graphs

 Plot points and lines

 Draw graphs of the following functions and interpret equations of the form f(x) = g(x) as a comparison of functions

 f(x) = ax, where a ∈ Z; x ∈ R

 f(x) = ax + b, where a, b ∈ Z; x ∈ R

 f(x) = ax2 + bx + c, where a ∈ N; b, c ∈ Z; x ∈ R

 Use graphical methods to find approximate solutions where f(x) = g(x) and to interpret the results

Learning OutcomesIn this chapter you will learn to:

Graphing Functions ch

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31.1 Linear Functions

A constant is a value that does not vary.

Examples: 5 is a constant, 3.5 is a constant.

A graph is a pictorial representation of information showing how one quantity varies with another related quantity. The graph of a linear function is a straight line. The graph of f(x) = 2x + 1 is shown below.

–4 –3 –2 –1

–1

1

2

3

4

5

–2

–3

f(x) = 2x + 1

1 2 3 4 50

y

x

KEY W ORDS

YOU SHOULD REMEMBER.. .

ACTIVITY 31.1

Worked example 31.1

Draw the graph of the function f: x → 2x – 3 in the domain –3 ≤ x ≤ 2, x ∈ R.

Solution‘f: x → 2x – 3’ is the function to be graphed.

The phrase ‘in the domain –3 ≤ x ≤ 2’ tells us that the graph will be drawn between x = –3 and x = 2.

‘x ∈ R’ means that x can have any value within the given domain.

We will now use an input–output table. By substituting into the function all the whole number values between –3 and 2 (including both of these numbers), we can calculate the corresponding y-values and hence plot the line.

x 2x – 3 f (x) or y

–3 2(–3) – 3 –9–2 2(–2) – 3 –7–1 2(–1) – 3 –50 2(0) – 3 –31 2(1) – 3 –12 2(2) – 3 1

The set of couples of this linear function is:

{(–3, –9), (–2, –7 ), (–1, –5), (0, –3), (1, –1), (2, 1)}.

Plot these couples on a graph and join them with a straight line.

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–1–1

–2

–3

–4

–5

–6

–7

–8

–9(–3,–9)

(–2,–7)

(–1,–5)

f(x) = 2x – 3

(0,–3)

(1,–1)

(2,1)1

2

1 2 3 4 50–2–3–4–5

x

Worked example 31.2

Graph the function f: x 2x + 4 in the domain –2 ≤ x ≤ 3, x ∈ R.

SolutionWe only need two points in order to draw a straight line, but it is good practice to choose at least three points, as the third point can act as a check. You should always choose the end values of the given domain.

x 2x + 4 y

– 2 2(– 2) + 4 0

0 2(0) + 4 4 3 2(3) + 4 10

Couples: {(– 2,0), (0,4), (3, 10)}

Now plot these co-ordinates on a graph and join them with a straight line.

10(3,10)y = 2x + 4

(0,4)

(–2,0)

9

8

7

6

5

4

3

2

1

–1 1 2 3 4

x

y

–2–3–4 0

General rules to be applied when drawing graphs of linear functions

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exercise 31.1

1. Graph the function f(x) = x + 2 in the domain –5 ≤ x ≤ 4, x ∈ R by completing the following input–output table.

x x + 2 y

–5

0

4

2. A function is defined by f(x) = 2x – 4. Find:

(i) f(0) (iii) f(5)

(ii) f(3)

Hence, graph the function in the domain 0 ≤ x ≤ 5, x ∈ R.

3. Graph the function y = 3x – 6 in the domain –1 ≤ x ≤ 5, x ∈ R.

4. Draw the graph of the function f: x 2x + 7 in the domain – 6 ≤ x ≤ 0, x ∈ R.

5. Draw the graph of the function f(x) = 2x – 3 in the domain –3 ≤ x ≤ 2, x ∈ R.

6. Draw the graph of the function g(x) = 4x – 5 in the domain – 4 ≤ x ≤ 2, x ∈ R.

7. Draw the graph of the function h: x → –2x + 4 in the domain –1 ≤ x ≤ 6, x ∈ R.

8. Graph the function y = x in the domain –3 ≤ x ≤ 3, x ∈ R.

9. Graph the function f(x) = 3 – x in the domain – 4 ≤ x ≤ 4, x ∈ R.

10. Draw the graph of the function f(x) = 3x in the domain –2 ≤ x ≤ 4, x ∈ R.

11. Draw the graph of the function f: x → –2x in the domain –3 ≤ x ≤ 3, x ∈ R.

12. Graph the function y = –x in the domain –5 ≤ x ≤ 4, x ∈ R.

Worked example 31.3

Graph the function f(x) = – 4x in the domain – 4 ≤ x ≤ 3, x ∈ R.

Solution

x – 4x y

– 4 – 4(– 4) 16

–1 – 4(–1) 4

3 – 4(3) –12

Couples: {(– 4,16), (–1,4), (3, –12)}

When trying to plot this line, we encounter problems with the scale. The highest y-value is 16 and the lowest y-value is –12. We need to increase our scale on the y-axis to ensure the graph fits on the page. Note that we still scale in single units on the x-axis.

16

14

12

10

8

6

4

2

–2–1 1 2 3 4 5–2–3–4

–4

–6

–8

–10

–12

0

f(x) = – 4x

(–1,4)

(–4,16)

(3,–12)

x

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s31.2 Quadratic FunctionsA quadratic function (in x) involves an x2 term and is of the form y = ax2 + bx + c, where a (a ≠ 0), b and c are constants. The graph of a quadratic function takes the form of a curve, known as a parabola. The graph of a quadratic function can be drawn by making a table of values for x and finding the corresponding values for y. Then plot the resultant couples.

The graph of f(x) = x2 + 3x – 1 is shown here.

ACTIVITY 31.2 0 1–1–2–3–4–5–6 2 3 4 5

f(x) = x2 + 3x – 1

–1

1

2

3

4

5

6

–2

–3

x

Worked example 31.4

Graph the function f(x) = x2 – 6x + 8 in the domain 0 ≤ x ≤ 6, x ∈ R.

Solution

x x2 – 6x + 8 y0 (0)2 – 6(0) + 8 81 (1)2 – 6(1) + 8 32 (2)2 – 6(2) + 8 03 (3)2 – 6(3) + 8 –14 (4)2 – 6(4) + 8 05 (5)2 – 6(5) + 8 36 (6)2 – 6(6) + 8 8

Because we are dealing with a curve, we use every couple to graph the function.

Couples: {(0,8), (1,3), (2,0), (3, –1), (4,0), (5,3), (6,8)}

(0,8)

(5,3)

(6,8)

f(x) = x2 – 6x + 8

(3,–1)

(2,0) (4,0) x0 1–1

–1

1

2

3

4

5

6

8

2 3 4 5 6

(1,3)

Worked example 31.5

Draw the graph of the function y = 3x2 + 2x – 3 in the domain – 2 ≤ x ≤ 2, x ∈ R.

Solution

x 3x2 + 2x – 3 y

–2 3(–2)2 + 2(–2) – 3 5

–1 3(–1)2 + 2(–1) – 3 –2

0 3(0)2 + 2(0) – 3 –3

1 3(1)2 + 2(1) – 3 2

2 3(2)2 + 2(2) – 3 13

Couples: {(–2,5), (–1,–2), (0, –3), (1,2), (2,13)}(0,–3)

(1,2)

(–2,5)

(2,13)

(–1,–2)

14

12

10

8

6

4

2

–2

–1 1 2 3 4–2–3

–4

0

y

x

y = 3x2 + 2x – 3

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sexercise 31.2

1. By completing the following input–output table, graph the function f(x) = x2 – 4x + 1in the domain –1 ≤ x ≤ 5, x ∈ R.

x x2 – 4x + 1 y

–1 (–1)2 – 4(–1) + 1 6

0

1

2

3

4

5 (5)2 – 4(5) + 1 6

2. Graph the function f(x) = x2 + 2x – 3 in the domain –4 ≤ x ≤ 2, x ∈ R.

3. Draw the graph of the function f: x x2 – 5x + 4 in the domain 0 ≤ x ≤ 5, x ∈ R.

4. Draw the graph of the function f(x) = x2 – x + 2 in the domain –2 ≤ x ≤ 3, x ∈ R.

5. Draw the graph of the function y = x2 + 2x in the domain – 4 ≤ x ≤ 2, x ∈ R.

6. Draw the graph of the function g(x) = x2 – 3 in the domain –3 ≤ x ≤ 3, x ∈ R.

7. Graph the function h(x) = 2x2 + x – 2 in the domain –2 ≤ x ≤ 2, x ∈ R.

8. Graph the function f(x) = 2x2 – 2x – 3 in the domain –2 ≤ x ≤ 3, x ∈ R.

9. Draw the graph of the function y = 3x2 – 9x + 4 in the domain –1 ≤ x ≤ 4, x ∈ R.

10. Draw the graph of the function y = 4x2 + 5x – 1 in the domain –2 ≤ x ≤ 2, x ∈ R.

11. Graph the function f: x → 3x2 – 4 in the domain –2 ≤ x ≤ 2, x ∈ R.

12. Draw the graph of the function g(x) = 2x2 + 1 in the domain –2 ≤ x ≤ 2, x ∈ R.

31.3 appLications oF GraphsGraphs of functions can be used to solve a variety of problems. In problems involving graphs, we are required to read off certain values from the graph. This means that the answers are usually estimates.

There are several types of questions that require reading values from our graph. For each type of question, we must also be able to interpret what the results signify.

When asked to use a graph to answer questions, we must always remember to:

� Use only the graph to find our answer.

� Draw the lines/points on the graph that we use to find our answers.

solve f(x) = 0We need to find the value(s) of x when f(x) or y = 0.

On our graph, this is the point(s) where the graph hits (or crosses) the x-axis ( y = 0). We read off the x-value(s) at this/these point(s).

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sWorked example 31.6

Using the graph of the function f(x) = x2 + 3x – 4, estimate the values of x for which f(x) = 0.

4y

x

f(x) = x2 + 3x – 4

3

2

1

–1–1 1 2 3–2–3–4–5

–2

–3

–4

–5

–6

0

SolutionOur graph crosses the x-axis at x = – 4 and at x = 1.

solve f(x) = q, Where q is a numberWe are being asked to find the value(s) of x when f(x) or y = q (a certain number).

On our graph, we draw the horizontal line y = q and mark off the point(s) where our graph hits this line. We then read off the x-value(s) of this/these point(s).

Worked example 31.7

Using the graph of the function f(x) = 3x – 2, estimatethe value of x for which f(x) = 4.

Find the Value of f(p)We are being asked to find the value of y when x = p (a certain number).

On our graph, we draw the vertical line x = p and read off the y-value of the point where our graph hits this line.

y = 4

12

10

8

6

4

2

–1

–2

–4

–6

–8

1 2 3 4 5 6–2–3–4 0

(4,10)

(–2,–8)

f(x) = 3x – 2

y

x

SolutionWe first draw the horizontal line y = 4 (parallel to the x-axis). We then read off the x-value of the point where the given graph crosses this line.

The graph crosses this line at x = 2.

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sWorked example 31.8

Using the graph of the function f(x) = x2 – 3x – 2, estimate the value of f(3).

–5

–4

–3

–2

–1

1

2

3

4

5f(x) = x2 – 3x – 2

6

–1 1 2 4 5 6 73

x = 3

0–2

y

x

SolutionWe first draw the vertical line through x = 3 (parallel to the y-axis). We then read off the y-value of the point where this line cuts the given quadratic graph.

∴ f(3) = –2

solve f(x) = g(x)We are being asked to read off from our graph the point(s) where the two graphs intersect.

Worked example 31.9

Using the same axes and scales, graph the functions g(x) = 4 – x and f(x) = 2x + 1 in the domain –2 ≤ x ≤ 4, x ∈ R. Hence, find the value of x for which g(x) = f(x).

SolutionStep 1 g(x) = 4 – x

x 4 – x y–2 4 – (–2) 60 4 – (0) 44 4 – 4 0

Couples: {(–2,6), (0,4), (4,0)}

Step 2 f(x) = 2x + 1

x 2x + 1 y–2 2(–2) + 1 –30 2(0) + 1 14 2(4) + 1 9

Couples: {(–2,–3), (0,1), (4,9)}

ACTIVITY 31.3

Step 3 Graph these two lines and mark the point of intersection.

–3

–2

–1

1

2

3

4

5

6

7

8

9

–1 1 2 3 4

(4,0)

(0,4)

(–2,–3)

(4,9)

(0,1)

f(x) = 2x + 1

(–2,6)

g(x) = 4 – x

5 6 70–2–3–4

y

x

(1,3)

From our graph, the point of intersection is (1,3).

∴ x = 1

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sWorked example 31.10

The graph of the function f : x → x2 + 5x – 1 is shown.

–7

–6

–5

–4

–3

–2

–1

–1 1 2–2–3–4–5–6

1

0

2

3

4

5

6f(x) = x2 + 5x – 1

y

x

Using your graph, estimate:

(i) The values of x for which f(x) = 0

(ii) The values of x for which f(x) = 5

(iii) The value of f(–3)

Solution (i) The values of x for which f(x) = 0

–7

–6

–5

–4

–3

–2

–1

1

2

3

4

f(x) = x2 + 5x – 1

5

6

–1 1 2–2–3–4–5–6

0

y

x

The graph intersects the x-axis at approximately x = –5.2 and x = 0.2.

(ii) The values of x for which f(x) = 5

We draw the horizontal line y = 5 and read off the x-values of the points of intersection of this line and the given graph.

–1–1

–2

–3

–4

–5

–6

–7

1

2

3

4

5

6

1 2 30–2

y = 5f(x) = x2 + 5x – 1

–3–4–5–6–7–8

y

x

x = – 6 Or x = 1

(iii) The value of f(–3)

We draw the vertical line x = –3 and read off the y-value of the point where this line crosses the given graph.

x = –3

–8

–6

–5

–4

–3

–2

–1

1

2

3

4

5

–1 0 1 2–2–3–4

f(x) = x2 + 5x – 1

–5–6

–7

y

x

∴ f(–3) = –7

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sexercise 31.3

1. The graph of the function f(x) = 2x + 2 is shown.

–6

–5

–4

–3

–2

–1

1

2f(x) = 2x + 2

3

4

5

6

–1 0 1 2 3 4–2–3–4

y

x

Estimate (from your graph):

(i) The value of f(–3)

(ii) The value of x for which f(x) = 0

(iii) The value of x for which f(x) = 4

2. The graph of the function y = – x + 5 is shown.

y = –x + 5

y

x

6

5

4

3

2

1

–1–1 0 1 2 3 4 5 6 7 8

–2

Use the graph to find:

(i) The value of y when x = 7

(ii) The value of y when x = 0

(iii) The value of y when x = 2

3. Draw the graph of the function f: x 3x – 9 in the domain –1 ≤ x ≤ 5, x ∈ R.

Use your graph to find:

(i) The value of x for which f(x) = 0

(ii) The value of x for which f(x) = –3

(iii) The value of f(1)

(iv) The value of f(1) – f(0)

4. The graph of the function f: x x is shown.

f(x) = x

–2

–1

1

2

3

4y

x

–3

–3 –2 –1 0 1 2 3 4 5–4–5

–4

Use the graph to find:

(i) The value of f(2.5)

(ii) The value of x for which f(x) = 0

(iii) The value of x for which f(x) = –3.5

5. Draw the graph of the function y = 1 – 3x in the domain –2 ≤ x ≤ 3, x ∈ R.

Estimate (from your graph):

(i) The value of x for which 1 – 3x = 0

(ii) The value of x for which 1 – 3x = 1

(iii) The value of 1 – 3x when x = –1.5

6. Draw the graph of the function f: x → –2x in the domain – 4 ≤ x ≤ 3, x ∈ R.

Estimate (from your graph):

(i) The value of x for which –2x = 0

(ii) The value of x for which –2x = – 4

(iii) The value of –2x when x = 0

Check your answers obtained from the graph by using algebra.

7. The graph of the function f: x → x2 + 2x – 3 is shown.

8

7

6

5

4

3

2

1

–1–1

–2

–3

–4

1 2 3–2–3–4 0

f(x) = x2 + 2x – 3

y

x

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s Using the graph, estimate:

(i) The value of f(–1)

(ii) The values of x for which f(x) = 0

(iii) The values of x for which f(x) = –3

(iv) The value of f(2)

(v) The value of x2 + 2x – 3 when x = 1

(vi) The values of x for which x2 + 2x – 3 = 4

(vii) The value of x for which x2 + 2x – 3 = – 4

(viii) The value of x for which f(x) = f(2), x ≠ 2

8. Draw the graph of the function y = x2 – 6x + 4 in the domain –1 ≤ x ≤ 6, x ∈ R.

Use your graph to find:

(i) The value of y when x = 3

(ii) The value of y when x = 1.5

(iii) The values of x when y = 4

(iv) The values of x for which y = 0

(v) The value of x when y = –5

9. The graph of the function f: x → 3x2 + 2x – 3 is shown.

f(x) = 3x2 + 2x – 3

1

2

3

4

5

6y

x

–1

–2

–2 –1 0 1 2

–3

–4

Use the graph to estimate:

(i) The value of f(0.5)

(ii) The values of x for which f(x) = 3

(iii) The values of x for which f(x) = –2

(iv) The value of x for which f(x) = f(–1), x ≠ –1

10. Draw the graph of the function f(x) = 2x2 – 5x + 4 in the domain –1 ≤ x ≤ 3, x ∈ R.

(i) Estimate from your graph the value of f(2.5).

(ii) Estimate from your graph the values of x for which f(x) = 4.

(iii) Can you find values of x where f(x) = 0? Give a reason for your answer.

(iv) Show that f(0) + f(2) = 6.

11. The graph of the function f: x → x2 – 2 is shown.

f(x) = x2 – 26

y

x

5

4

3

2

1

–1–1 0 1 2 3–2–3

–2

–3

Estimate from the graph:

(i) The value of x2 – 2 when x = 0

(ii) The values of x for which x2 – 2 = 0

(iii) The value of x for which x2 – 2 = –2

12. Draw the graph of the function y = 2x2 + 5x in the domain – 4 ≤ x ≤ 1, x ∈ R.

Use your graph to find:

(i) The value of 2x2 + 5x when x = –1

(ii) The values of x for which 2x2 + 5x = 0

(iii) The values of x for which 2x2 + 5x = –3

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g(x) = 3x + 1 are shown.

Estimate the value of x for which f(x) = g(x).

6

y

x

5

4

3

2

1

–1

–2

–3

–4

–4 –3 –2 –1 0 1 2 3

–5

–6f(x) = 2x + 2 g(x) = 3x + 1

14. (i) Using the same scales and axes, draw the graphs of the two functions y = –2x and y = 3x – 5 in the domain –1 ≤ x ≤ 3, x ∈ R.

(ii) Use your graphs to solve the simultaneous equations y = –2x and y = 3x – 5.

15. Using the same axes and scales, draw the graphs of the functions f(x) = 4 – 2x and g(x) = x – 5 in the domain –1 ≤ x ≤ 6, x ∈ R.

Estimate from your graph:

(i) The value of f(1)

(ii) The value of g(1)

(iii) The value of x for which g(x) = 0

(iv) The value of x for which f(x) = 2

(v) The value of x for which f(x) = g(x)

16. Draw the graph of the function f: x → x2 – 4 in the domain –3 ≤ x ≤ 3, x ∈ R.

Estimate (from your graph):

(i) The value of x2 – 4 when x = 1.5

(ii) The values of x for which x2 – 4 = 0

(iii) The values of x for which x2 – 4 = 3

(iv) The value of x for which x2 – 4 = – 4

17. The graphs of the functions f(x) = 2x – 1 and g(x) = x2 are shown.

Estimate the value of x for which f(x) = g(x).

Verify your answer by using algebra.

18. (a) Using the same axes and scales, draw the graphs of the functions f(x) = 2x and g(x) = x2 + 3x in the domain – 4 ≤ x ≤ 1, x ∈ R.

(b) Estimate (from your graph):

(i) The value of f(–2)

(ii) The value of g(–2)

(iii) The value of x for which f(x) = 0

(iv) The values of x for which g(x) = 3

(v) The values of x for which f(x) = g(x)

19. The graphs of the functions f(x) = x2 – 2x and g(x) = x2 + 3x – 5 are shown.

Estimate the value of x for which f(x) = g(x).

5

y

x

4

3

2

1

–1

–2

–5 –4 –3 –2 –1 1 2 3 40

–3

–4

–5

–6

–7

f(x) = x2 – 2xg(x) = x2 + 3x – 5

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s 20. The graphs of the functions f(x) = x2 – 1 and g(x) = x2 – 3x are shown.

Estimate from the graph:

(i) The values of x for which g(x) = 0. Explain what this answer signifies.

(ii) The values of x for which f(x) = 0. Explain what this answer signifies.

(iii) The value of x for which f(x) = g(x). Explain what this answer signifies.

31.4 transFormations oF Linear Functions

This section will cover what happens to a graph of a function when one or more parts of the function change. This is called a transformation of the graph of the function. When we graph a function under a transformation, the graph changes shape and/or location.

Linear FunctionsWhen we transform a linear function, the graph can shift up or down and/or change slope.

When we are transforming graphs of linear functions, it is best if they are in the form y = mx + c, where m is the slope and c is the y-intercept (the y-value where the line crosses the y-axis).

A change in the value of the slope m will result in the slope of the line increasing or decreasing.

–2

–1

1

2

3

4

5

6

7y

x–1 1 2 3 4 5 6 7 8 90–2–3

f(x) = 3x + 2g(x) = x + 2

The slope of f(x) is 3. The slope of g(x) is 1.

A change in the value of the y-intercept c will result in the graph of the function moving vertically up or down the y-axis. This transformation will result in a line that is parallel to the original line.

0–2 –1 1 2 3 4 5 6 7 8–3–4–5

g(x) = x + 4

–1

1

2

3

4

5

6

7

y

x

f(x) = x + 2

For g(x), the y-intercept is 4.

For f(x), the y-intercept is 2.

� �

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sWorked example 31.11

The graph of the function y = 3x – 4 is shown.

y = 3x – 4

–1–1

1

2

3

4

5

6

7y

1 2 3 4 5 6 7 8 9–2–3 0

–2

–3

–4

–5

–6

x

Sketch the graphs of the following functions:

(i) y = 3x + 1

(ii) y = 4x – 4

Solution (i) y = 3x + 1

The slope has not changed but the y-intercept is now (0,1).

The line will be parallel to the original line but will now move vertically upwards so that it goes through the point (0,1).

y = 3x – 4y = 3x + 1

–1

–1

1

2

3

4

5

6

7y

x

1 2 3 4 5 6 7 8 90

–2

–3

–4

–5

(ii) y = 4x – 4

The y-intercept has not changed, but the slope has changed.

A slope of 4 or 4 _ 1 means that we move up 4 units for every 1 unit we go to the right.

We draw a line which goes though the y-intercept (0,–4) with a slope of 4.

y = 3x – 4

–5

–4

–3

–2

–1

1

2

3

4

5

6

7

y

x–1 1 2 3 4 5 6 7 80–2

y = 4x – 4

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sexercise 31.4

1. The graph of the linear function f(x) = x + 2 is shown.

–11 –10 –9

–8

–7

–6

–5

–4

–3

–2

–1

1

2

3

4

5

6

7

8y

x–8 –7

f(x) = x + 2

–6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9

Use your graph to match the following functions with the functions shown on the graph:

(i) g(x) = x + 5

(ii) h(x) = x – 1

2. The graph of the linear function f: x → 2x – 3 is shown.

f(x) = 2x – 3

–4

–5

–4

–3

–2

–1

1

2

3

4

5

6

7

8

9

10

–3 –2 –1 0 1 2 3 4 5 6 7 8 9 10 11 12

x

y Use your graph to match the following functions with the functions shown on the graph:

(i) h: x → x – 3

(ii) g: x → 3x – 3

3. The graph of the linear function f: x → 3x + 1 is shown.

f(x) = 3x + 1

–1–1 0 1 2 3 4 5 6 7 8 9 10

x

y

–2–3–4

1

2

3

4

5

6

7

8

9

10 Use your graph to match

the following functions with the functions shown on the graph:

(i) g: x → 3x + 4

(ii) h: x → x + 1

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s 4. The graph of the linear function y = –2x + 5 is shown.

y = –2x + 5

–3 –2 –1 0

1

–1

–2

–3

–4

2

3

4

5

6

7

8

1 2 3 4 5 6 7 8 9

x

y

Use your graph to match the following functions with the functions shown on the graph:

(i) y = –4x + 5 (ii) y = –x + 5

5. The graph of the function y = 2x + 1 is shown. Copy this graph into your copybook and sketch the graphs of the following functions:

y = 2x + 1

8

7

6

5

4

3

2

1

–1

–2

–3

–4

–4 –3 –2 –1 0 1 2 3 4 5 6

–5

–6

x

y

(i) y = 2x + 4

(ii) y = 2x – 3

(iii) y = 3x + 1

6. Graph the function y = 3x – 6 in the domain –2 ≤ x ≤ 5, x ∈ R. Hence, sketch the graph of the following functions:

(i) f(x) = 3x – 3 (ii) g(x) = x – 6 (iii) h(x) = –3x – 6

7. Graph the function y = –2x in the domain – 4 ≤ x ≤ 3, x ∈ R. Hence, sketch the graph of the following functions:

(i) y = – 4x (ii) y = –x (iii) y = –2x + 3

8. Graph the function f(x) = 6 – x in the domain –3 ≤ x ≤ 9, x ∈ R. Hence, sketch the graph of the following functions:

(i) g(x) = 6 – 2x (ii) h(x) = 6 – 5x (iii) j(x) = 6 + x

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s31.5 transFormations oF Quadratic

FunctionsConsider the function f(x) = x2.

–2

2

4

6

8

10

12

–2 –1 1

y = x2

2 3 4 5

y

x0–3–4

� The graph of the function goes through the point (0,0).

� The lowest point on the U-shaped curve is also (0,0).

� The curve is symmetrical about the y-axis.

When the graph of a quadratic function is transformed, the graph moves vertically and/or horizontally.

� When the graph is shifted (translated) vertically, it moves parallel to the y-axis.

� When the graph is shifted (translated) horizontally, it moves parallel to the x-axis.

Graphs of the Form y = ax2

We have seen the graph of y = x2 above.

If we change the coefficient of x2 (the number in front of the x2), the graph will become either narrower or wider.

x y = x 2 y = 3x 2

–3 9 27

–2 4 12

–1 1 3

0 0 0

1 1 3

2 4 12

3 9 27

y = 3x2

y = x2

0

1

–1–2

–1 1 2 3 4

y

x–2–3

23456789

Compare the y-values of y = 3x2 and y = x2. We note that the y-values of y = 3x2 have been multiplied by a factor of 3.

If we graph y = 3x2, we notice the graph is narrower than y = x2. Therefore, as we increase the coefficient of x2, the graph narrows.

x y = x 2 y = 2x 2

–3 9 18

–2 4 8

–1 1 2

0 0 0

1 1 2

2 4 8

3 9 18

y = 3x2 y = 2x2

–2

–1–1 1 2 3 4–2–3–4

1

2

34

5

6

7

8

9

10

0

y = x2

y

x

Compare the y-values of y = 2x2 and y = x2. We note that the y-values of 2x2 have been multiplied by a factor of 2.

The graph of y = 2x2 (red graph) is narrower than the graph of y = x2 (black graph) but not as narrow as the graph of y = 3x2 (green graph) as 2 < 3.

ACTIVITY 31.4

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s

Graphs of the Form y = x2 + bThe graph of y = x2 + 3 is the graph of a function y = x2 under a certain transformation. We have translated the graph y = x2 three units vertically upwards.

x y = x2 y = x2 + 3

–3 9 12

–2 4 7

–1 1 4

0 0 3

1 1 4

2 4 7

3 9 12

–1

1

2

3

4

5

6

7

8

9

10

–1 1 2 3 40–2–3–4

y = x2 + 3 y = x2y

x

All points of the graph have been shifted vertically upwards by three units.

The graph of y = x2 – 4 is the graph of y = x2 but shifted four units vertically downwards.

x y = x2 y = x2 – 4

–3 9 5

–2 4 0

–1 1 –3

0 0 –4

1 1 –3

2 4 0

3 9 5

y = x2 – 4

–1–1 1 2 3 4–2–3

1

2

3

4

5

–2

–3

–4

–5

0

y = x2 y

x

All points of the graph have been shifted vertically downwards by four units.

The constant b in a function of the form y = x2 + b will determine where the function intercepts the y-axis.

ACTIVITY 31.5

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sGraphs of the Form y = (x + b)2

The graph of y = (x + 3)2 is the graph of a function y = x2 under a certain transformation. We have shifted the graph y = x2 three units to the left, horizontally along the x-axis.

x y = x2

–3 9

–2 4

–1 1

0 0

1 1

2 4

3 9

–1 1 2 3 4 5–1

123456789

1011

0–2–3–4–5–6

y = (x + 3)2 y = x2

y

x

x y = (x + 3)2

–6 9

–5 4

–4 1

–3 0

–2 1

–1 4

0 9

All points of the graph have been shifted horizontally to the left by three units.

Alternatively, we can solve the equation (x + 3)2 to determine where the function hits the x-axis:

(x + 3)2 = 0 ⇒ x + 3 = 0 ∴ x = –3

We can also determine where the graph hits the y-axis. The value of the constant b helps to determine where the function intersects the y-axis. As the constant b is squared, the graph will hit the y-axis at (b)2.

∴ The y-intercept of the function y = (x + 3)2 is (0,9).

A graph of y = (x – 4)2 is similar to the graph of y = x2, but this time shifted horizontally along the x-axis to the right by four units.

The y-intercept of the function is (– 4)2. Therefore, the y-intercept is (0,16).

x y = x2

–3 9

–2 4

–1 1

0 0

1 1

2 4

3 9

y = (x – 4)2

–5 –4 –3 –2 –1 1 2 3 4 5 6 7 80

2

4

6

8

10

12

14

16

18

y = x2

y

x

x y = (x – 4)2

1 9

2 4

3 1

4 0

5 1

6 4

7 9

All points of the graph have been shifted horizontally to the right by four units.

ACTIVITY 31.6

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sWorked example 31.12

The graph of the function y = 3x2 is shown. Sketch the graphs of the following functions:

–2

2

4

6

8

10

12

–1 1 2–2 0

y = 3x2

y

x

(i) y = 2x2

(ii) y = 3x2 – 2

Worked example 31.13

Graph the function f(x) = (x + 2)2 in the domain –5 ≤ x ≤ 1, x ∈ R. Use your graph to sketch the graph of:

(i) f(x) = (x – 1)2 (ii) f(x) = (x + 2)2 – 5

Solution

We first draw the function f(x) = (x + 2)2 in the domain –5 ≤ x ≤ 1, x ∈ R.

x (x + 2)2 y–5 (–5 + 2)2 9–4 (–4 + 2)2 4–3 (–3 + 2)2 1–2 (–2 + 2)2 0–1 (–1 + 2)2 1

0 (0 + 2)2 41 (1 + 2)2 9

Couples: {(–5,9), (–4,4), (–3,1), (–2,0), (–1,1), (0,4), (1,9)}

–5 –4 –3 –2 –1–1

1

2

3

4

5

6

7

8

9

0 1 2 3

y

x

y = (x + 2)2

Solution

(i) As 2 is less than 3, the graph of y = 2x2 will be wider than y = 3x2. Two points on the graph can be found to aid in sketching, such as (–1,2) and (1,2).

5

6

4

3

2

1

0 1 2

–1

–1–2

y = 3x2

y

x

y = 2x2

(–1,2) (1,2)

(ii) y = 3x2 – 2

The graph of y = 3x2 is shifted two units vertically downwards.

y = 3x2

y = 3x2 – 2

21–1–2–2

2

4

6

8

10

12

0

y

x

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s (i) f(x) = (x – 1)2

Touch x-axis: Put x – 1 = 0

∴ x = 1 ⇒ Graph touches x-axis at (1,0).

To find the y-intercept: (–1)2 = 1

∴ The y-intercept is (0,1).

Sketch the function through the point (1,0) and the point (0,1).

y = (x + 2)2 y = (x – 1)2

–6 –5 –4 –3 –2 –1 0

1

2

3

4

5

6

7

8

9

–11 2 3 4 5

y

x

(ii) f(x) = (x + 2)2 – 5

This is the function y = (x + 2)2, which has been shifted vertically downwards five units.

Find the y-intercept:

(0,4) → (0,–1) (Subtract 5 from y-value)

Also, find the new minimum point: (–2,0) → (–2,–5)

y = (x + 2)2

y = (x + 2)2 – 5

543210–1–2–3–4–5

–5

–4

–3

(0,–1)

(–2,–5)

–2

–1

1

2

3

4

5y

x

exercise 31.5

1. The graph of the function y = x2 is shown. Use your graph to match the following functions with the functions shown on the graph:

–4 –3 –2 –1–1

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

0 1 2 3 4 5

x

y

y = x2

(i) y = 3x2

(ii) y = 5x2

2. The graph of the function y = x2 is shown. Use your graph to match the following functions with the functions shown on the graph:

y = x2

14

13

12

11

10

9

8

7

6

5

4

3

2

1

–1

–2

–4 –3 –2 –1 0 1 2 3 4 5

x

y

–3

–4

–5

(i) y = x2 – 3

(ii) y = x2 + 4

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s 3. The graph of the function y = x2 is shown.

Use your graph to match the following functions with the functions shown on the graph:

(i) y = (x + 1)2 (ii) y = (x – 5)2

4. The graph of the function y = 6x2 is shown. Use your graph to match the following functions with the functions shown on the graph:

y = 6x2

–3 –2 –1 0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

1 2 3

x

y

(i) y = 2x2 (ii) y = 4x2

5. The graph of the function y = x2 + 2 is shown. Use your graph to match the following functions with the functions shown on the graph:

9

8

7

6

5

4

3

2

1

–1

–2

–3

–3 –2 –1 0 1 2

x

y

3 4

–4

–5

y = x2 + 2

(i) y = x2 – 2 (ii) y = x2 – 4

6. The graph of the function y = (x – 5)2 is shown. Use your graph to match the following functions with the functions shown on the graph:

10987654321

–1–2–3

–3 –2 –1 0 1 2 3 4 5 6 7 8

x

y

y = (x – 5)2

(i) y = (x – 2)2 (ii) y = x2

7. The graph of the function y = x2 is shown. Copy this graph into your copybook and sketch the graphs of the following functions:

–2

–1–1 0 1 2 3–2–3

1

2

3

4

5

6

7

8

9

10y

x

y = x2

(i) y = 2x2 (iii) y = 2x2 + 4

(ii) y = 4x2

8. Graph the function f(x) = 4x2 in the domain –2 ≤ x ≤ 2, x ∈ R.

Hence, sketch the graphs of the following functions:

(i) g(x) = 5x2 (ii) h(x) = 2x2

9. Graph the function f: x x2 + 4 in the domain –3 ≤ x ≤ 3, x ∈ R.

Hence, sketch the graphs of the following functions:

(i) g(x) = x2 + 1 (ii) h: x x2 – 2

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s 10. The graph of the function f(x) = (x – 1)2 is

shown. Copy this graph into your copybook and sketch the graphs of the following functions:

f(x) = (x – 1)2

10

9

8

7

6

5

4

3

2

1

–1

–2

–3

–3–4–5 –2 –1 0 1 2 3 4 5 6 7 8

x

y

(i) g: x (x – 5)2 (ii) h: x (x + 2)2

11. Graph the function f(x) = (x + 4)2 in the domain –7 ≤ x ≤ –1, x ∈ R.

Hence, sketch the graph of the following functions:

(i) g(x) = (x + 3)2 (ii) h: x (x + 4)2 + 3

12. Graph the function f(x)= x2 – 6x + 9 in the domain 0 ≤ x ≤ 6, x ∈ R.

Hence, by factorising x2 – 6x + 9, sketch the graph of the following functions:

(i) h(x) = (x – 4)2 (ii) g: x (x – 3)2 + 2

revision exercises

1. Draw the graph of the function: f: x → 2x + 1 in the domain –1 ≤ x ≤ 4.

2. Draw the graph of the function: y = 3x + 1 in the domain –1 ≤ x ≤ 5.

3. The graph of the function f: x → 3x – 3 is shown.

f(x) = 3x – 3

–1–1

1

2

3

4

5

6

7

8

–2

–3

–4

0 1 2 3 4 5

x

y

From the graph, estimate the value of:

(i) x when f(x) = 0 (iii) f(3)

(ii) x when f(x) = 3

4. The graph of the function f: x → 2x + 3 is shown.

f(x) = 2x + 3

y

x1

–1

–2

–3

–3 –2 –1 0 1 2 3 4 5 6

2

3

4

5

6

7

8

9

10

Estimate from the graph:

(i) The value of f(1.5)

(ii) The value of x for which f(x) = 8

(iii) The value of x for which f(x) = –1

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s 5. The graph of the function y = 5 – x is shown.

–2

–4 –3 –2 –1 0 1 2 3 4 5 6–1

1

2

3

4

5

6

7

8

9y = 5 – x

x

y

Estimate from the graph:

(i) The value of y if x = 3

(ii) The value of x if y = 7

6. (a) Draw the graph of the function y = 6 – 2x in the domain –2 ≤ x ≤ 4,

x ∈ R.

(b) Estimate from your graph:

(i) The value of y if x = 1.3

(ii) The value of x if y = 1.4

7. (a) Draw the graph of the function f: x → 4x – 5 in the domain –2 ≤ x ≤ 3,

x ∈ R.

(b) Estimate from your graph:

(i) The value of f(0.2)

(ii) The value of x if f(x) = 6

8. The graph of the function y = x2 – 4x + 1 is shown.

y = x2 – 4x + 1

x

y

–2

–2 –1 0 1 2 3 4 5 6–1

1

2

3

4

5

6

7

8

9

–3

Estimate from the graph:

(i) The value of y when x = 3

(ii) The values of x if y = 6

9. The graph of the function f: x x2 – 6x + 5 is shown.

y = x2 – 6x + 5y

x

7

6

5

4

3

2

1

–1–1 0 1 2 3 4 5 6 7

–3

–4

–5

–2

Estimate from the graph:

(i) The value of y when x = 6

(ii) The values of x if y = 0

10. Graph the function f: x → x2 – 3x – 4 in the domain –2 ≤ x ≤ 5, x ∈ R.

Find from your graph:

(i) The value of f(4.5)

(ii) The solution to x2 – 3x – 4 = 0

(iii) The approximate solutions to x2 – 3x – 4 = –2

11. Draw the graph of the function y = x2 + x – 12 in the domain – 5 ≤ x ≤ 4, x ∈ R.

Use your graph to find:

(i) The value of y when x = 1.5

(ii) The values of x if y = 0

(iii) The approximate values of x for which y = – 4

12. Draw the graph of the function y = x2 – x – 5 in the domain –3 ≤ x ≤ 4, x ∈ R.

(i) The value of y when x = 1.8

(ii) The values of x if y = 0

(iii) The values of x for which y = 4

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s 13. The graph of the function y = x2 – 2x + 1 is

shown.

y = x2 – 2x + 1

x

y7

6

5

4

3

2

1

–1–2 –1 0 1 2 3 4 5

Estimate from the graph:

(i) The value of x2 – 2x + 1 when x = –1

(ii) The value of x if x2 – 2x + 1 = 0

14. The graph of the function f: x 2x2 – 3x + 1 is shown.

1

–1–1.5 –0.5 0 0.5 1 1.5 2 2.5 3 3.5–1

2

3

4

5

6

7f(x) = 2x2 – 3x + 1

y

x

Estimate from the graph:

(i) The value of f(1)

(ii) The values of x for which f(x) = 6

(iii) The values of x for which f(x) = 2

15. The graph of the function y = 2x2 + 2x – 4 is shown.

y = 2x2 + 2x – 4

y

x

7

6

5

4

3

2

1

–1–3 –2.5 –2 –1.5 –1 0 0.5 1 1.5 2 2.5 3.53–0.5

–2

–3

–4

–5

Estimate from the graph:

(i) The value of 2x2 + 2x – 4 when x = –1

(ii) The roots of the function y = 2x2 + 2x – 4

(iii) The solutions of the equation 2x2 + 2x – 4 = –3

16. Draw the graph of f: x → 2x2 – 3x – 7 in the domain –2 ≤ x ≤ 3, x ∈ R.

Find, from your graph:

(i) The value of f(1.7)

(ii) The values of x for which f(x) = 0

17. Draw the graph of the function y = 3x2 – 3x – 4 in the domain –2 ≤ x ≤ 3, x ∈ R.

Use your graph to estimate:

(i) The values of x for which 3x2 – 3x – 4 = 0

(ii) The values of x for which 3x2 – 3x – 4 = 11

18. Draw the graph of the function f: x → 4x2 + 6x – 7 in the domain – 4 ≤ x ≤ 2, x ∈ R.

Find, from your graph:

(i) The value of f(–2.8)

(ii) The values of x for which f(x) = –1

19. The graph of the function y = x2 + 4x is shown.

y = x2 + 4x

y

x–0.5

–1

1

2

3

4

5

6

7

–2

–3

–4

0.5 1 1.5 20–1–1.5–2–2.5–3–3.5–4–4.5–5.5 –5

Estimate from the graph:

(i) The value of y when x = 0.3

(ii) The values of x if y = 0

(iii) The values of x for which y = –2

(iv) The least value of y and the value of x at which it occurs

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s 20. Draw the graph of the function y = x2 – 6 in

the domain – 4 ≤ x ≤ 4, x ∈ R.

Use your graph to find:

(i) The value of y when x = 1.4

(ii) The values of x if y = 0

(iii) The values of x for which y = 5

21. The graphs of the functions y = 3 – x and y = x – 1 are shown. Use the graphs to find the point of intersection of the two functions.

y = x – 1

y = –x + 3

–3

–2

–1

1

2

3

4

5

6

7

8

9

10

–1 1 2 3 4 5 6

y

x

7–2–3 0

22. The graphs of the functions f(x) = –2x and g(x) = 3x – 5 are shown. Estimate the value of x for which f(x) = g(x).

g(x) = 3x – 5y

x

f(x) = –2x

654321–1–1

1

2

3

4

5

6

7

8

–2

–3

–4

–5

–6

–2–3 0

23. (i) Using the same scales and axes, draw (in the domain –3 ≤ x ≤ 1) the graphs of the two functions y = –x and y = 2x + 3.

(ii) Use your graphs to solve the simultaneous equations y = –x and y = 2x + 3.

24. (i) Solve the simultaneous equations y = 2 – x and y = 3x – 10 using an algebraic method.

(ii) Solve these equations by graphing the functions y = 2 – x and y = 3x – 10 in the domain 0 ≤ x ≤ 5, x ∈ R.

25. The graphs of the two functions f: x → x2 – x + 1 and g: x → x + 4 are shown.

g(x) = x + 4

f(x) = x2 – x – 1

–1

123

45

6789

10

y

x

11

12

–1 1 2 3 4 5 60–2–3

Use the graphs to find the point of intersection of the two functions.

26. On the same page of graph paper and using the same axes and scales, draw the graphs of the two functions:

f: x → 2x2 + x – 5 and g: x → 2x + 1 in the domain –3 ≤ x ≤ 2, x ∈ R.

Use the graphs to find the values of x for which:

(i) f(x) = 0 (ii) f(x) = g(x)

27. The graphs of the two functions f: x → x2 – 3x – 1 and g: x → 2 – x are shown.

–3

–2

–1–0.5 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.50–1.5 –1–2

1

2

3

4

5

7y

x

6

g(x) = 2 – x

f(x) = x2 – 3x – 1

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which:

(i) f(x) = –3 (iii) f(x) = g(x)

(ii) g(x) = 0

28. Using the same axes and scales, draw the graphs of the two functions:

f: x → x2 – 4x and g: x → x2 – 8x + 16 in the domain 0 ≤ x ≤ 6, x ∈ R.

Use the graphs to find the values of x for which:

(i) f(x) = 0 (iii) f(x) = g(x)

(ii) g(x) = 0

29. The graph of the function f(x) = 5x – 1 is shown.

8

6

4

2

–2

–4

–1 1 2 3 4–2–3 0

y = 5x – 1

Use the graph to match the following functions with the functions shown on the graph:

(i) f(x) = 5x + 5 (ii) f(x) = 2x – 1

30. The graph of the linear function f(x) = 2x – 4 is shown.

6

5

4

3

2

1

–1–1 1 2 3 4 5 6 7 8 9–2–3–4–5–6–7

–2

–3

–4

–5

–6

0

f(x) = 2x – 4

Use the graph to match the following functions with the functions shown on the graph:

(i) g(x) = 2x + 3

(ii) h(x) = –2x – 4

31. The graph of the linear function f(x) = –x + 1 is shown.

8

7

6

5

4

3

2

1

–1–1 1 2 3 4 5 6 7 80–2–3–4–5–6–7

–2

–3

y = –x + 1y

x

Use your graph to match the following functions with the functions shown on the graph:

(i) j(x) = 1 – 2x

(ii) g(x) = –4x + 1

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–1

2

4

6

8

10

1 2 3 4–2–3 0

y = 2x2

y

x

Use your graph to match the following functions with the functions shown on the graph:

(i) y = x2 (ii) y = 5x2

33. The graph of the function y = x2 – 4 is shown.

–1

–2

2

4

6

–4

1 2 3 4 5–2–3 0

y = x2 – 4

y

x

Use your graph to match the following functions with the functions shown on the graph:

(i) y = x2 + 1 (ii) y = x2 – 3

34. The graph of the function y = (x – 2)2 is shown.

–3 –2 –1–2

2

4

6

1 2 3 4 50

y = (x – 2)2

y

x

Use your graph to match the following functions with the functions shown on the graph:

(i) y = x2 (ii) y = (x – 1)2

35. Draw the graph of y = x2 in the domain–2 ≤ x ≤ 2. Sketch the graphs of the following functions:

(i) y = 2x2 (ii) y = x2 + 3

36. Draw the graph of the function f: x (x – 4)2 in the domain 0 ≤ x ≤ 8, x ∈ R. Hence, sketch the graph of the following functions:

(i) g(x) = (x – 6)2 (ii) h(x) = (x – 3)2

37. The graph of the function y = x2 is shown.

–3

–2

–1

1

2

3

4

5

6

7

8

–1 1 2

y = x2

y

x3 4 5 60–2–3–4–5–6–7–8

Use your graph to match the following functions with the functions shown on the graph:

(i) y = 3x2 (ii) y = (x + 4)2

38. By factorising and then graphing the function f(x) = x2 – 10x + 25 over the domain 0 ≤ x ≤ 6, x ∈ R, sketch the graph of the following functions:

(i) g(x) = (x – 1)2 (ii) h(x) = (x – 5)2 + 1

Strand 5 Activities

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ACTIVE MATHS 1 – ACTIVITIES

Activity 30.1

(a) (i) Complete the input–output table.

Input Rule: Add 5 to the input

Output

2 2 + 5

3

4

(ii) List the couples for the completed table.

Couples: { ( 2, ) , ( 3, ) , ( 4, ) }

(iii) Draw a mapping diagram for the given relation.

(b) (i) Complete the input–output table.

Input Rule: Twice the input and subtract 2

Output

1 2 ( ) – 2

2

– 4

(ii) List the couples for the completed table.

Couples: { ( 1, ) , ( 2, ) , ( – 4, ) }

(iii) Draw a mapping diagram for the given relation.

(c) (i) Complete the input–output table.

Input Rule: Square the input and add 2

Output

1

–3

4

(ii) List the couples for the completed table.

Couples: { ( , ) , ( , ) , ( , ) }

(iii) Draw a mapping diagram for the given relation.

ACTIVE MATHS 1 – ACTIVITIES

30

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Activity 30.2

For each question:

(i) Write the rule using function notation.

(ii) Complete the input–output table.

(iii) Write down the answers to the remaining questions.

(a) Rule: ‘Add 4 to the input.’ f(x) =

x f(x) = y

1

2

3

∴ f(1) = , f(2) = , f(3) =

(b) Rule: ‘Double the input and then subtract 3.’ f(x) =

x f(x) = y

0

2

–2

∴ f(0) = , f(2) = , f(–2) =

Range = { , , } (c) Rule: ‘Square the input then add 3.’ g(x) =

x g(x) = y

–3

2

5

∴ g(–3) = , g(2) = , g(5) =

Domain = { }

Range = { }

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ACTIVE MATHS 1 – ACTIVITIES

Graphing Functions

Activity 31.1

1. (i) Complete the following input–output table:

x f(x) = 2x + 1 y Couple

–3 2(–3) + 1 –5 (–3,–5)

–2

–1

0

1

2

3

2. (i) Complete the following input–output table:

x f(x) = –x + 4 y Couple

–3 –(–3) + 4 7 (–3,7)

–2

–1

0

1

2

3

(ii) Now graph this line on the diagram above (in Question 1, part (ii)).

(iii) Use your ruler to connect all the points (or couples).

(iv) At what point does the graph of f(x) = 2x + 1 cross the y-axis?

(ii) Now plot these couples on the co-ordinate plane. (The first one has been done for you.)

–5

–5 –4 –3 –2 –1 0

x

y

1 2 3 4 5

–4(–3,–5)

–3

–2

–1

1

2

3

4

5

6

7

ACTIVE MATHS 1 – ACTIVITIES

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Activity 31.2

(i) Complete the following input–output table:

x f(x) = x2 + 2x – 3 y Couple

– 4 (– 4)2 + 2(– 4) – 3 5 (– 4,5)

–3

–2

–1

0

1

2

(ii) Now plot these points on the diagram shown.

(–4,5)

–4

–5

x

y

–4 –3 –2 –1 0 1 2 3 4

–3

–2

–1

1

2

3

4

5

(iii) Join the points with a smooth curve, using the order in which they appear in your table.

(iv) At what point does the graph of f(x) = x2 + 2x – 3 cross the y-axis?

(iii) At what point does the graph of f(x) = –x + 4 cross the y-axis?

(iv) The y-intercept is where the graph of the function crosses the y-axis. What is common about the y-intercept for both functions shown?

(v) At what point do the graphs of the two functions meet?

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ACTIVE MATHS 1 – ACTIVITIES 207

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sActivity 31.3

The graph of the function f(x) = x2 – 5x + 4 is shown.

–1

–1

1

2

3

4

5

6

7

8

9

10

–2

0 1 2 3 4 5 6 7

x

y

(i) Mark on the graph the two points where the function intersects the x-axis. Label these points A and B.

Point A = ( , ) Point B = ( , )

∴ The values of x for which f(x) = 0 are:

x = and x =

(ii) Draw the line y = 4. Mark on the graph the two points where this line intersects the graph of the function f. Label these points C and D.

Point C = ( , ) Point D = ( , )

∴ The values of x for which f(x) = 4 are:

x = and x =

(iii) Draw the line x = 6. Mark on the graph the point where this line intersects the graph of the function f. Label this point E.

Point E = ( , )

∴ The value of f(6) is: y =

What other value of x will give the same answer as f(6)?

ACTIVE MATHS 1 – ACTIVITIES

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sActivity 31.4

(i) Complete the following input–output table:

x f(x) = x2 y Couple–3 (–3)2 9 (–3,9)–2–10123

(ii) Now plot these points on the diagram shown and graph f(x) = x2.

24

28

20

16

12

8

4

0 1 2 3

y

x–1–2–3

(iii) Now complete the following tables:

x f(x) = 2x2 y Couple x f(x) = 3x2 y Couple–3 2(–3)2 18 (–3,18) –3 3(–3)2 27 (–3,27)–2 –2–1 –1

0 01 12 23 3

(iv) Now sketch the graphs of y = 2x2 and y = 3x2 on the same diagram as y = x2.

(v) In general, as the coefficient of the x2 term gets bigger, the graph gets .

What point is common to all three functions?

(vi) Explain why this point is common to all three functions.

31

ACTIVE MATHS 1 – ACTIVITIES 209

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sActivity 31.5

The graph of the function f(x)= x2 is shown.

–3 –2 –1–1

1

2

3

4

5

6

7

8

9

10

–2

(–2,4)

1 2 3 4

x

y(–3,9)

(–1,1)

(0,0)

(1,1)

(2,4)

(3,9)

(i) Complete the following tables:

x f(x) = x2 + 1 y Couple x f(x) = x2 – 2 y Couple

–3 (–3)2 + 1 10 (–3,10) –3 (–3)2 – 2 7 (–3,7)

–2 –2

–1 –1

0 0

1 1

2 2

3 3

(ii) Now draw the graphs of f(x) = x2 + 1 and f(x) = x2 – 2 on the same diagram as f(x) = x2.

(iii) � Every point on the graph of the function f(x) = x2 + 1, when compared to the graph of f(x) = x2, has been

moved by unit.

� Every point on the graph of the function f(x) = x2 – 2, when compared to the graph of f(x) = x2, has been

moved by units.

(iv) � At what point does the graph of the function f(x) = x2 + 1 cross the y-axis?

� At what point does the graph of the function f(x) = x2 – 2 cross the y-axis?

� What is the link between the function and where it crosses the y-axis?

� In general, a change in the value of the constant b of a function of the form f(x) = x2 + b will shift the graph

of f(x) = x2 or .

ACTIVE MATHS 1 – ACTIVITIES

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sActivity 31.6

The graph of the function f(x)= x2 is shown.

(–3,9)

(–2,4)

(–1,1)

(0,0)

(1,1)

(2,4)

(3,9)

–6 –5 –4 –3 –2 –1–1

1

2

3

4

5

6

7

8

9

10

1 2 3 4 5 6

x

y

(i) Complete the following tables:

x f(x) = (x – 3)2 y Couple x f(x) = (x + 2)2 y Couple

0 (0 – 3)2 9 (0,9) –5 (–5 + 2)2 9 (–5,9)1 –42 –33 –24 –15 06 1

(ii) Draw the graphs of f(x) = (x – 3)2 and f(x) = (x + 2)2 on the same diagram as f(x) = x2.

(iii) � Every point on the graph of the function f(x) = (x – 3)2, when compared to the graph of f(x) = x2, has been

moved by units.

� Every point on the graph of the function f(x) = (x + 2)2, when compared to the graph of f(x) = x2, has been

moved by units.

(iv) � At what point does the graph of the function f(x) = (x – 3)2 touch the x-axis?

� At what point does the graph of the function f(x) = (x + 2)2 touch the x-axis?

� What do you notice?

(v) � At what point does the graph of the function f(x) = (x – 3)2 cross the y-axis?

� At what point does the graph of the function f(x) = (x + 2)2 cross the y-axis?

� What do you notice?

� In general, a change in the value of the constant b of a function of the form y = (x + b)2 will shift the graph

or .