Chapter 1 Graphs and Polynomials Solutions

8/9/2019 Chapter 1 Graphs and Polynomials Solutions

1/24

CHAPTER 1 Graphs and polynomials • EXERCISE 1A | 1

Maths Quest 12 Mathematical Methods CAS Second Edition Solutions Manual

Exercise 1A — The binomial theorem

1 a ( x + 3)2 = 2

2

0 x

⎛ ⎞⎜ ⎟⎝ ⎠

+ 1 12

31

x⎛ ⎞⎜ ⎟⎝ ⎠

+ 22

32

⎛ ⎞⎜ ⎟⎝ ⎠

= x2 + 6 x + 9

b ( x + 4)5 = 55

0 x⎛ ⎞⎜ ⎟

⎝ ⎠ + 4

54

1 x⎛ ⎞⎜ ⎟

⎝ ⎠ + 3 2

54

2 x⎛ ⎞⎜ ⎟

⎝ ⎠ + 2 3

54

3 x⎛ ⎞⎜ ⎟

⎝ ⎠

+ 45

44 x

⎛ ⎞⎜ ⎟⎝ ⎠

+ 55

45

⎛ ⎞⎜ ⎟⎝ ⎠

= x5 + 5 x44 + 10 x316 + 10 x264 + 5 x256 + 1 × 1024

= x5 + 20 x

4 + 160 x

3 + 640 x

2 + 1280 x + 1024

c ( x − 1)8 = 8

8

0 x

⎛ ⎞⎜ ⎟⎝ ⎠

+ 78

( 1)1 x

⎛ ⎞−⎜ ⎟

⎝ ⎠ + 6 2

8( 1)

2 x

⎛ ⎞−⎜ ⎟

⎝ ⎠

+ 5 38

( 1)3 x

⎛ ⎞−⎜ ⎟

⎝ ⎠ + 4 4

8( 1)

4 x

⎛ ⎞−⎜ ⎟

⎝ ⎠ + 3 5

8( 1)

5 x

⎛ ⎞−⎜ ⎟

⎝ ⎠

+ 2 68

( 1)6 x

⎛ ⎞−⎜ ⎟

⎝ ⎠ + 7

8( 1)

7 x

⎛ ⎞−⎜ ⎟

⎝ ⎠ + 8

8( 1)

8

⎛ ⎞−⎜ ⎟

⎝ ⎠

= x8 − 8 x7 + 28 x6 − 56 x5 + 70 x4 − 56 x3 + 28 x2

− 8 x + 1

d (2 x + 3)4 = 44

(2 )0

x⎛ ⎞⎜ ⎟⎝ ⎠

+ 34

(2 ) 31

x⎛ ⎞⎜ ⎟⎝ ⎠

+ 2 24

(2 ) 32

x⎛ ⎞⎜ ⎟⎝ ⎠

+ 34

(2 )33

x⎛ ⎞⎜ ⎟⎝ ⎠

+ 44

34

⎛ ⎞⎜ ⎟⎝ ⎠

= 16 x4 + 96 x3 + 216 x2 + 216 x + 81

e (7 − x)4 = 44

70

⎛ ⎞⎜ ⎟⎝ ⎠

+ 3 2 24 4

7 ( ) 7 ( )1 2

x x⎛ ⎞ ⎛ ⎞

− + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

+ 3 44 4

7( ) ( )3 4

x x⎛ ⎞ ⎛ ⎞− + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= 2401 − 1372 x + 294 x2 − 28 x3 + x4

f (2 − 3 x)5 = 55

20

⎛ ⎞⎜ ⎟⎝ ⎠

+ 45

2 ( 3 )1

x⎛ ⎞

−⎜ ⎟⎝ ⎠

+ 3 25

2 ( 3 )2

x⎛ ⎞

−⎜ ⎟⎝ ⎠

+ 2 35

2 ( 3 )3

x⎛ ⎞

−⎜ ⎟⎝ ⎠

+ 45

2( 3 )4

x⎛ ⎞

−⎜ ⎟⎝ ⎠

+ 55

( 3 )5

x⎛ ⎞

−⎜ ⎟⎝ ⎠

= 32 − 240 x + 720 x2 − 1080 x

3 + 810 x

4 − 243 x

5

2 a

31

x x

⎛ ⎞+⎜ ⎟

⎝ ⎠ = x3 + 3 x2

1

x

⎛ ⎞⎜ ⎟⎝ ⎠ + 3 x

21

x

⎛ ⎞⎜ ⎟⎝ ⎠

+

31

x

⎛ ⎞⎜ ⎟⎝ ⎠

= x3 + 3 x + 3 x

+ 3

1 x

b

72

3 x x

⎛ ⎞−⎜ ⎟

⎝ ⎠ = (3 x)7 − 7(3 x)6

2

x

⎛ ⎞⎜ ⎟⎝ ⎠ + 21(3 x)

5

22

x

⎛ ⎞⎜ ⎟⎝ ⎠

− 35(3 x)4

32

x

⎛ ⎞⎜ ⎟⎝ ⎠ + 35(3 x)

3

42

x

⎛ ⎞⎜ ⎟⎝ ⎠

− 21(3 x)2

52

x

⎛ ⎞⎜ ⎟⎝ ⎠ + 7(3 x)

6 72 2

x x

⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= 2187 x7 − 10 206 x5 + 20 412 x3 − 22 680 x

+ 3

15 120 6048

x x− +

5 7

1344 128

x x−

c

62 3 x

x

⎛ ⎞+⎜ ⎟

⎝ ⎠= ( x2)6 + 6( x2)5

3

x

⎛ ⎞⎜ ⎟⎝ ⎠ + 15( x

2)4

23

x

⎛ ⎞⎜ ⎟⎝ ⎠

+ 20( x2)3

33

x

⎛ ⎞⎜ ⎟⎝ ⎠ + 15( x

2)2

43

x

⎛ ⎞⎜ ⎟⎝ ⎠

+ 6( x2)

5

3 x

⎛ ⎞⎜ ⎟⎝ ⎠

+

6

3 x

⎛ ⎞⎜ ⎟⎝ ⎠

= x12 + 18 x9 + 135 x6 + 540 x3 + 1215 + 3

1458

x

+ 6

729

x

d

5

2

32 x

x

⎛ ⎞−⎜ ⎟

⎝ ⎠ =

5 4

2 2

3 35 (2 ) x

x x

⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ +

32

2

310 (2 ) x

x

⎛ ⎞⎜ ⎟⎝ ⎠

−

23

2

310 (2 ) x

x

⎛ ⎞⎜ ⎟⎝ ⎠

+4 5

2

35 (2 ) (2 ) x x x

⎛ ⎞−⎜ ⎟

⎝ ⎠

= 10 7

243 810

x x− +

4

1080 720

x x− + 240 x2 − 32 x5

3 (r + 1)th

term isn

r

⎛ ⎞⎜ ⎟⎝ ⎠

(ax)n − r b

r

a ( x − 7)3

i x2is the 2nd term

⇒ = 1

Coefficient = 3

1

⎛ ⎞⎜ ⎟⎝ ⎠

x2 (−7)

1

= −21

ii x3 is first term ⇒ r = 0

term =

3

0

⎛ ⎞

⎜ ⎟⎝ ⎠ x3

70

Coefficient = 1

iii x4

Coefficient = 0

b (2 x + 1)5

i x2 is the 4

th term ⇒ r = 3

term = 5

3

⎛ ⎞⎜ ⎟⎝ ⎠

(2 x)213

Coefficient = 40

ii x3 is the third term

⇒ r = 2

term = 5

2

⎛ ⎞⎜ ⎟⎝ ⎠

(2 x)31

2

Coefficient = 80

iii x4 is the 2nd term ⇒ r = 1

term = 5

1

⎛ ⎞⎜ ⎟⎝ ⎠

(2 x)411

Coefficient = 80

c

52

3 x x

⎛ ⎞+⎜ ⎟

⎝ ⎠

i Coefficient of x2 = 0

Chapter 1 — Graphs and polynomials


2/24

2 | CHAPTER 1 Graphs and polynomials • EXERCISE 1B


ii x3 is the 5th term ⇒ r = 4

term =

15 2

4 x

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠(3 x)4

Coefficient = 810

iii Coefficient of x4 = 0

d

62 3 x

x

⎛ ⎞−⎜ ⎟

⎝ ⎠


ii x3 is the 4th term ⇒ r = 3

term = 6

3

⎛ ⎞⎜ ⎟⎝ ⎠

( x2)3

33

x

⎛ ⎞−⎜ ⎟

⎝ ⎠

Coefficient = −540

iii Coefficient of x4 = 0

e

6

2

37 x

x

⎛ ⎞+⎜ ⎟

⎝ ⎠


ii x3 is the 2nd term ⇒ r = 1

term = 6

1

⎛ ⎞⎜ ⎟⎝ ⎠

(7 x)5

1

2

3

x

⎛ ⎞⎜ ⎟⎝ ⎠

Coefficient = 302 526iii Coefficient of x4 is 0.

4

32 5

3 x x

⎛ ⎞−⎜ ⎟

⎝ ⎠

x3 is the 2nd term ⇒ r = 1

term = 3

1

⎛ ⎞⎜ ⎟⎝ ⎠

(3 x2)2

15

x

−⎛ ⎞⎜ ⎟⎝ ⎠

= 3 × 9 x4 × 1

5

x

−

= −135 x3 The answer is A.

5 When the expression for C is expanded it does not contain an x5 term. The first three terms contain x8, x6 and x4 respectively.

All the other expressions contain an x5

term.The answer is C.

6

53

2

2 x

x

⎛ ⎞+⎜ ⎟

⎝ ⎠ = ( x

3)5 + 5( x

3)4

2

2

x

⎛ ⎞⎜ ⎟⎝ ⎠ + 10( x

3)

3

2

2

2

x

⎛ ⎞⎜ ⎟⎝ ⎠

+ 10( x3)2

3

2

2

x

⎛ ⎞⎜ ⎟⎝ ⎠ + 5( x

3)

4

2

2

x

⎛ ⎞⎜ ⎟⎝ ⎠

+

5

2

2

x

⎛ ⎞⎜ ⎟⎝ ⎠

= 1 x15 + 10 x10 + 40 x5 + 80 + 3

80

x +

2

32

x

∴ = 1 + 10 + 40 + 80 + 80 + 32

= 243

The answer is D.

7 (2 x − 3)4

= (2 x)4 − 4(2 x)33 + 6(2 x)232 − 4(2 x)33 + 34

= 16 x4 − 96 x

3 + 216 x

2 − 216 x

3 + 81

The answer is D.

8 Fourth term = 6C 3 x

3 × (3 y)

3

= 20 × x3 × 27 y3

= 540 x3 y3

9 Term 3 ⇒ r = 2

=

2

79 32 4

x⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠

= 278 732

16

x

= 219 683

4

x

10 x6, x3, x0

3rd term is independent of x. r = 2

=

2

4

2

6 2(3 )

2 x

x

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠

= 4860

11 Powers of x are ( x2)5, ( x2)4 31

x

⎛ ⎞⎜ ⎟⎝ ⎠ , ( x

2)3

2

3

1

x

⎛ ⎞⎜ ⎟⎝ ⎠

x10, x5, x0, …

The third term is independent of x.

term = 5C 2( x

2)3

2

3

4

x

⎛ ⎞−⎜ ⎟

⎝ ⎠

= 10 × (+16)

= 160

12

4

2

2

3 x

x

⎛ ⎞+⎜ ⎟

⎝ ⎠

Powers of x are ( x2)4, ( x2)3 21

x

⎛ ⎞⎜ ⎟⎝ ⎠ , ( x

2)2

2

2

1

x

⎛ ⎞⎜ ⎟⎝ ⎠

x8, x4, x0, …The third term is independent of x.

term =

4

C 2( x

2

)

2

2

2

3

x

⎛ ⎞⎜ ⎟⎝ ⎠

= 6 × 9

= 54

13 Expand ( p + 3)5

= p5 + 5 p43 + 10 p332 + 10 p233 + 5 p34 + 35

= p5 + 15 p4 + 90 p3 + 270 p2 + 405 p + 243

∴ (2 p − 5)( p5 + 15 p4 + 90 p3 + 270 p2 + 405 p + 243)

⇒ 2 p6 + 30 p5 + 180 p4 + 540 p3 + 810 p2 + 486 p − 5 p5 − 75 p4 − 450 p3 − 1350 p2 − 2025 p − 1215

Coefficient of p4 term = 180 − 75 = 105

14 (2a − 1)n

2nd term is nC 1(2a)n − 1(−1)1

coefficient: −n × 2n − 1 = −192

n × 2n × 12

= 192

n × 2n = 384

= 3 × 27

= 3 × 2 × 26

= 6 × 26

n = 6

Exercise 1B — Polynomials

1 Polynomial expressions consist of terms which have non-negative integer powers of x only.

Not Polynomial:

ii x4 + 3 x2 − 2 x + x

iii x7

+ 3 x6

− 2 xy + 5 x

vi 2 x5 + x

4 − x

3 + x

2 + 3 x − 2

x

Polynomial:

i x3 − 2 x

iv 3 x8 − 2 x5 + x2 − 7

v 4 x6 − x3 + 2 x − 3

2 a P ( x) + Q( x) = 8 − 3 x + 2 x2 + x4 + x5 − 3 x4 − 4 x2 − 1

= x5 − 2 x

4 − 2 x

2 − 3 x + 7

b Q( x) − R( x) = x5 − 3 x4 − 4 x2 − 1 − (8 x3 + 7 x2 − 4 x)

= x5 − 3 x

4 − 4 x

2 − 1 − 8 x

3 − 7 x

2 + 4 x

= x5 − 3 x4 − 8 x3 − 11 x2 + 4 x − 1

c 3 P ( x) − 2 R( x)

3 P ( x) = 3(8 − 3 x + 2 x2 + x4)


3/24

CHAPTER 1 Graphs and polynomials • EXERCISE 1B | 3


= 24 − 9 x + 6 x2 + 3 x4

2 R( x) = 2(8 x3 + 7 x2 − 4 x)

= 16 x3 + 14 x2 − 8 x

∴ 3 P ( x) − 2 R( x) = 24 − 9 x + 6 x2 + 3 x4 − (16 x3 + 14 x2 − 8 x)

= 24 − 9 x + 6 x2 + 3 x4 − 16 x3 − 14 x2 + 8 x

= 3 x4 − 16 x3 − 8 x2 − x + 24

d 2 P ( x) − Q( x) + 3 R( x)

2 P ( x) = 2(8 − 3 x + 2 x2 + x4)

= 16 − 6 x + 4 x2 + 2 x4

3 R( x) = 3(8 x3 + 7 x2 − 4 x)= 24 x3 + 21 x2 − 12 x

2 P ( x) − Q( x) + 3 R( x)

= 16 − 6 x + 4 x2 + 2 x

4− ( x5 − 3 x

4 − 4 x

2 − 1) + 24 x

3 + 21 x

2

− 12 x

= 16 − 6 x + 4 x2 + 2 x4 − x5 + 3 x4 + 4 x2 + 1 + 24 x3 + 21 x2 − 12 x

= 17 − 18 x + 29 x2 + 24 x

3 + 5 x

4 − x

5

3 a P ( x) = x6 + 2 x5 − x3 + x2

i degree = 6

ii P (0) = 06 + 2 × 05 − 03 + 02

= 0

iii P (2) = 26 + 2 × 25 − 23 + 22

= 124

iv P (−1) = −16 + 2 × −15 − (−1)3 + (−1)2

= 1

b P ( x) = 3 x7 − 2 x6 + x5 − 8

i degree = 7

ii P (0) = 3 × 07 − 2 × 06 + 05 − 8

= −8

iii P (2) = 3 × 27 − 2 × 2

6 + 2

5 − 8

= 280

iv P (−1) = 3 × (−1)7 − 2 × (−1)6 + (−1)5 − 8

= −3 − 2 − 1 − 8

= −14

c P ( x) = 5 x6 + 3 x4 − 2 x3 − 6 x2 + 3

i degree = 6

ii P (0) = 5 × 06 + 3 × 04 − 2 × 03 − 6 × 02 + 3

= 3iii P (2) = 5 × 26 + 3 × 24 − 2 × 23 − 6 × 22 + 3

= 331

iv P (−1) = 5 × (−1)6 + 3 × (−1)4 − 2 × (−1)3 − 6 × (−1)2 + 3

= 5 + 3 + 2 − 6 + 3

= 7

d P ( x) = −7 + 2 x − 5 x2 + 2 x3 − 3 x4

i degree = 4

ii P (0) = −7 + 2 × 0 − 5 × 02 + 2 × 03 − 3 × 04

= −7

iii P (2) = −7 + 2 × 2 − 5 × 22 + 2 × 23 − 3 × 24

= −55

iv P (−1) = −7 + 2 × (−1) − 5 × (−1)2 + 2 × (−1)3 − 3(−1)4

= − 7 − 2 − 5 − 2 − 3

= −19

4 P ( x) = x8 − 3 x6 + 2 x4 − x2 + 3

P (−2) = (−2)8 − 3 × (−2)6 + 2 × (−2)4 − (−2)2 + 3

= 95

The answer is B.

5 P ( x)= 2 x7 + ax

5 + 3 x

3 + bx − 5

P (1)= 4

∴ 4 = 2 × 17 + a × 15 + 3 × 13 + b × 1 − 5

4 = 2 + a + 3 + b − 5

4 = a + b [1]

P (2) = 163

163 = 2 × 27 + a × 2

5 + 3 × 2

3 + b × 2 − 5

= 256 + 32a + 24 + 2b − 5

−112 = 32a + 2b [2]

[1] × 2 8 = 2a + 2b [3]

[2] − [3] −120 = 30a

a = −4 b = 8

6 f ( x) = ax4 + bx3 − 3 x2 − 4 x + 7

f (1) = −2

∴ −2= a × (1)4 + b × (1)3 − 3 × 12 −4 × 1 + 7

−2= a + b − 3 − 4 + 7

−2= a + b

∴ −2 − b = a

[1] f (2) = −5

∴ −5= a × 24 + b × 2

3 − 3 × 2

2 − 4 × 2 + 7

−5= 16a + 8b − 12 − 8 + 7

−5= 16a + 8b − 13

8 = 16a + 8b

8 = 8(2a + b)

1 = 2a + b [2]Substitute [1] into [2]

1 = 2(−2 − b) + b

1 = −4 − 2b + b

1 = −4 − b

b = −5

If b = −5, then [1] −2 − −5 = a.

3 = a∴ f ( x) = 3 x4

− 5 x3

− 3 x2

− 4 x + 7

7 Q( x) = x5 + 2 x

4 + ax3 − 6 x + b

Q(2) = 45

∴ 45 = 25 + 2 × 24 + 23a − 6 × 2 + b

45 = 52 + 8a + b

−7 = 8a + b

−7 − 8a = b [1]

Q(0) = −7

∴ −7 = 05 + 2 × 04

+ a × 03

− 6 × 0 + b

−7 = b [2]Substitute [2] into [1].

−7 − 8a = −7

−8a = 0

a = 0∴ Q( x) = x5

+ 2 x4

− 6 x − 7.

8 P ( x) = ax6 + bx4

+ x3

− 6

If 3 P (1) = −24

then 3 P ( x) = 3(ax6 + bx4

+ x3

− 6)

−24 = 3(a × 16 + b × 14

+ 13

− 6)

−8 = a + b + 1 − 6

−8 = a + b − 5

−3 = a + b

−3 − a = b [1]

If 3 P (−2) = 102

then 3 P ( x) = 3(ax6 + bx4

+ x3

− 6)

102 = 3[a(−2)6 + b(−2)4

+ (−2)3

− 6)

34 = 64a + 16b − 8 − 6

34 = 64a

+ 16b

− 1448 = 64a + 16b

(÷ 16) 3 = 4a + b [2]Substitute [1] into [2]

3 = 4a + (−3 − a)

3 = 4a − 3 − a

6 = 3a

2 = a

If 2 = a then b = −3 − a

b = −3 − 2

b = −5

∴ P ( x) = 2 x6 − 5 x4

+ x3

− 6

9 a P ( x) = ax4 − x3

+ 3 x2

− 5

If P (1) = −1


4/24

4 | CHAPTER 1 Graphs and polynomials • EXERCISE 1C


then −1 = a × (1)4 − (1)3

+ 3 × (1)2 − 5

−1 = a − 1 + 3 − 5

−1 = a − 3

2 = a

The answer is C.

b f ( x) = xn

− 2 x3 + x2

− 5 x

If f (2) = 10

then 10 = 2n

− 2 × 23 + 22

− 5 × 2

10 = 2n

− 16 + 4 − 10

10 = 2n − 22

32 = 2n

25 = 2n

∴ n = 5

The answer is D.

Exercise 1C — Division of polynomials

1 a2

3 2

3 2

2

2

2 13

2 5 24

4

2 5

2 8

13 2

13 52

50

x x

x x x x

x x

x x

x x

x

x

+ +

⎡ − + −− − ⎢

−⎢⎣

⎡ +− ⎢

−⎢⎣

−⎡− ⎢

−⎣

Q( x) = x2 + 2 x + 13

R( x) = 50

b

4 3 2

5 4 3 2

5 4

4 3

4 3

3 2

3 2

3 6 18 58

0 3 0 4 33

3

3 3

3 9

6 0

6 18

x x x x

x x x x x x

x x

x x

x x

x x

x x

− + − +

⎡ + − + + ++ − ⎢

+⎢⎣

⎡− −− ⎢− −⎢⎣

⎡ +− ⎢

+⎢⎣

2

2

18 4

18 54

58 3

58 174

171

x x

x x

x

x

⎡− +− ⎢

− −⎢⎣

+⎡− ⎢

+⎣

−

Q( x) = x4 − 3 x3 + 6 x2 − 18 x + 58

R( x) = −171

c

3 2

4 3 2

4 3

6 17 53 155

6 2 4 03

6 18

x x x

x x x x x

x x

+ + +⎡ − + − +

− − ⎢−⎢⎣

3 2

3 2

2

17 2

17 51

53 4

53 159

155 0

155 465

465

x x

x x

x x

x x

x

x

⎡ +− ⎢

−⎢⎣

⎡ −− ⎢

−⎣

+⎡− ⎢

−⎣

Q( x) = 6 x3 + 17 x2 + 53 x + 155

R( x) = 465

d

3 2

4 3 2

4 3

3 2

3 2

7 7 101

3 9 27

3 6 0 12 03 1

3

7 0 7

73

x x x

x x x x x

x x

x x

x x

− + +

⎡ − + + ++ − ⎢

+⎢⎣

⎡− +⎢−⎢− −⎢⎣

2

2

712

3

7 7

3 9

1010

9

101 101

9 27

101

27

x x

x x

x

x

⎡+⎢

⎢−⎢

+⎢⎣

⎡+⎢

− ⎢⎢ +⎢⎣

−

Q( x) = x3 − 7

3 x2 +

7

9 x +

101

27

R( x) = 20

327

−101

27

−⎛ ⎞=⎜ ⎟

⎝ ⎠

2 a i P ( x)= x3 − 2 x

2 + 5 x − 2

P (4) = 43 − 2 × 42 + 5 × 4 − 2

= 50

ii P ( x)= x5 − 3 x3 + 4 x + 3

P (−3) = (−3)5 − 3 × (−3)3 + 4 × (−3) + 3

= −171iii P ( x) = 6 x4 − x3 + 2 x2 − 4 x

P (3) = 6 × 34 − 33 + 2 × 32 − 4 × 3

= 465

iv P ( x) = 3 x4 − 6 x3 + 12 x

1

3 P

⎛ ⎞−⎜ ⎟

⎝ ⎠ = 3 ×

41

3

⎛ ⎞−⎜ ⎟

⎝ ⎠− 6 ×

31

3

⎛ ⎞−⎜ ⎟

⎝ ⎠+ 12 × −

1

3

= −320

27

b The values obtained in 2 were the same as the remainder

values obtained in 1.

3 a P (3) = 33 + 9 × 32 + 26 × 3 − 30

= 156

Since P (3) ≠ 0, x − 3 is not a factor

b P (−2) = (−2)4 − (−2)3 − 5 × (−2)2 − 2 × (−2) − 8

= 0

Since P (−2) = 0 then x + 2 is a factor.

c P (+ 4) = 4 − 9 × + 4 + 6 × 42 − 13 × (+ 4)

3

− 12 × (+ 4)4 + 3 × (+ 4)5

= 4 − 36 + 94 − 832 − 3072 + 3072

= −768

Since P (−4) ≠ 0 then 4 − x is not a factor.


5/24

CHAPTER 1 Graphs and polynomials • EXERCISE 1C | 5


d P 1

2

⎛ ⎞−⎜ ⎟

⎝ ⎠ = 4 ×

61

2

⎛ ⎞−⎜ ⎟

⎝ ⎠+ 2 ×

51

2

⎛ ⎞−⎜ ⎟

⎝ ⎠− 8 ×

41

2

⎛ ⎞−⎜ ⎟

⎝ ⎠

− 4 ×

31

2

⎛ ⎞−⎜ ⎟

⎝ ⎠+ 6 ×

21

2

⎛ ⎞−⎜ ⎟

⎝ ⎠− 9 × −

1

2 − 6

= 0.0625 + − 0.0625 − 0.5 + 0.5 + 1.5 + 4.5 − 6

= 0

Since P 1

2

⎛ ⎞−⎜ ⎟

⎝ ⎠ = 0 then 2 x + 1 is a factor.

4 a f ( x) = x4 − 4 x3 − x2 + 16 x − 12

A x + 1 ⇒ f (− 1)= (−1)4 − 4 × (−1)3 − (−1)2 + 16 × (−1) − 12

= 1 + 4 − 1 − 16 − 12

= −24

B x ⇒ f (0)= −12

C x + 2 ⇒ f (− 2)= (−2)4 − 4 × (−2)3 − (−2)2 + 16 × (−2) − 12

= 16 + 32 − 4 − 32 − 12

= 0

Since f (−2) = 0 then ( x + 2) is a factor.

D x + 3 ⇒ f (−3)= (−3)4 − 4 × (−3)3 − (−3)2 + 16 × (−3) − 12

= 120E x − 4 ⇒ f (4)

= 44 − 4 × 43 − 42 + 16 × 4 − 12

= 36The answer is C.

b

3 2

4 3 2

4 3

3 2

3 2

2

2

6 11 6

4 16 122

2

6

6 12

11 16 11 22

6 12

6 12

0

x x x

x x x x x

x x

x x

x x

x x x x

x

x

− + −

⎡ − − + −+ − ⎢

+⎢⎣

⎡− −− ⎢

− −⎢⎣

⎡+− ⎢+⎢⎣

− −⎡− ⎢

− −⎣

Test x = 1 into x3 − 6 x2 + 11 x − 6

= 13 − 6 × 1 + 11 − 6

= 0

∴ x − 1 is a factor.

2

3 2

3 2

2

2

5 6

6 11 61

5 11

5 5

6 6

6 6

0

x x

x x x x

x x

x x

x x

x

x

− +

⎡ − + −− − ⎢

−⎢⎣

⎡− +− ⎢

− +⎢⎣

−⎡− ⎢

−⎣

x2 − 5 x + 6 = ( x − 3)( x − 2) f ( x) factorises to

( x + 2)( x − 1)( x − 3)( x − 2)

The answer is B.

5 a P ( x) = x3 + 4 x

2 − 3 x − 18

Test x = ± 1 P ( x) ≠ 0

x = 2, P ( x) = 0

∴ ( x − 2) is a factor

2

3 2

3 2

2

2

6 9

4 3 182

2

6 3

6 12

9 18

9 18

0

x x

x x x x

x x

x x

x x

x

x

+ +

⎡ + − −− − ⎢

−⎢⎣

⎡ −− ⎢

−⎢⎣

−⎡

− ⎢ −⎣

∴ ( x − 2)( x2 + 6 x + 9)

= ( x − 2)( x + 3)2

b P ( x) = 3 x3 − 13 x2 − 32 x + 12

Test x = ± 1 P ( x) ≠ 0

x = ± 2

when x = −2, P ( x) = 0

∴ ( x + 2) is a factor.

2

3 2

3 2

2

2

3 19 6

3 13 32 122

3 6

19 32

19 38

6 12

6 12

0

x x

x x x x

x x

x x

x x

x

x

+ − +

⎡ − − ++ − ⎢

+⎢⎣⎡− −

− ⎢− −⎢⎣

+⎡− ⎢

+⎣

∴ ( x + 2)(3 x2 − 19 x + 6)

= ( x + 2)(3 x − 1)( x − 6)

c P ( x) = x4 + 2 x

3 − 7 x2 − 8 x + 12

Test x = −2, P ( x) = 0

∴ ( x + 2) is a factor.

3 2

4 3 2

4 3

2

2

0 7 6

2 7 8 122

2

0 7 8

7 14

6 12

6 12

0

x x x

x x x x x

x x

x x

x x

x

x

+ − +

⎡ + − − ++ − ⎢

+⎢⎣

⎡ − −− ⎢

− −⎢⎣

+⎡− ⎢

+⎣

∴ ( x + 2)( x3 + 0 x

2 − 7 x + 6)

Test x = 2, P ( x) = 0

∴ ( x − 2) is a factor.2

3 2

3 2

2

2

2 3

0 7 62

2

2 7

2 4

3 6

3 6

0

x x

x x x x

x x

x x

x x

x

x

+ −

⎡ + − +− − ⎢

−⎢⎣

⎡ −−⎢

−⎢⎣

− +⎡−⎢

− +⎣

∴ ( x + 2)( x − 2)( x2 + 2 x − 3)

( x + 2)( x − 2)( x + 3)( x − 1)


6/24

6 | CHAPTER 1 Graphs and polynomials • EXERCISE 1C


d P ( x) = 4 x4 + 12 x3 − 24 x2 − 32 x

Test x = −1, P ( x) = 0

∴ x + 1 is a factor3 2

4 3 2

4 3

3 2

3 2

2

2

4 8 32

4 12 24 32 01

4 4

8 24

8 8

32 32

32 32

0

x x x

x x x x x

x x

x x

x x

x x

x x

+ −

⎡ + − − ++ − ⎢

+⎢⎣

⎡ −− ⎢

+⎢⎣⎡− −

− ⎢− −⎢⎣

∴ ( x + 1)(4 x3 + 8 x2 − 32 x)

Take out factor of 4 x.

4 x( x + 1)( x2 + 2 x − 8)

∴ 4 x( x + 1)( x − 2)( x + 4)

6 a 3 x3 + 3 x

2 − 18 x = 0

3 x is a common factor

∴ 3 x( x2 + x − 6) = 0

3 x( x + 3)( x − 2) = 0

x = 0, −3, 2b 2 x

4 + 10 x3 − 4 x

2 − 48 x = 0

Take out 2 x as a common factor

∴ 2 x( x3 + 5 x2 − 2 x − 24) = 0

Factorise x3 + 5 x2 − 2 x − 24

Test x = 2, f (2) = 0

∴ ( x – 2) is a factor2

3 2

3 2

2

2

7 12

5 2 242

2

7 2

7 14

12 24

12 24

0

x x

x x x x

x x

x x

x x

x

x

+ +

⎡ + − −− − ⎢

−⎢⎣

⎡ −− ⎢

−⎢⎣

−⎡− ⎢

−⎣

∴ 2 x( x − 2)( x2 + 7 x + 12)

= 2 x( x − 2)( x + 3)( x + 4) = 0

∴ x = 2, −3, 0, and − 4

c 2 x4 + x

3 − 14 x2 − 4 x + 24 = 0

Test x = 2, f ( x) = 0


3 2

4 3 2

4 3

3 2

3 2

2

2

2 5 4 12

2 14 4 242

2 45 14

5 10

4 4

4 8

12 24

12 24

0

x x x

x x x x x

x x x x

x x

x x

x x

x

x

+ − −

⎡ + − − +− − ⎢

−⎢⎣⎡ −

− ⎢−⎢⎣

⎡− −− ⎢

− +⎢⎣

− +⎡− ⎢

− +⎣

∴ ( x − 2)(2 x3 + 5 x2 − 4 x − 12)

Test x = −2, f (−2) = 0

∴ ( x + 2) is a factor

2

3 2

3 2

2 6

2 5 4 122

2 4

x x

x x x x

x x

+ −

⎡ + − −+ − ⎢

+⎢⎣

2

2

4

2

6 12

6 12

0

x x

x x

x

x

⎡ −− ⎢

−⎢⎣

− −⎡− ⎢

− −⎣

∴ ( x − 2)( x + 2)(2 x2 + x − 6)

( x − 2)( x + 2)(2 x − 3)( x + 2)

x = 2, −2, or3

2

d x4 − 2 x

2 + 1 = 0

Test x = + 1, f ( x) = 0

∴ ( x − 1) is a factor3 2

4 3 2

4 3

3 2

3 2

2

2

1

0 2 0 11

2

0

1

1

0

x x x

x x x x x

x x

x x x x

x x

x x

x

x

+ − −

⎡ + − + +− − ⎢

−⎢⎣

⎡−− ⎢−⎢⎣

⎡− +− ⎢

− +⎢⎣

− +⎡− ⎢

− +⎣

∴ ( x − 1)( x3 + x2 − x − 1)

Test x = −1, f ( x) = 0

∴ ( x + 1) is a factor2

3 2

3 2

1

11

0

1

1

0

x

x x x x x x

x

x

−

⎡ + − −+ − ⎢+⎢⎣

− −⎡− ⎢

− −⎣

∴ ( x − 1)( x + 1)( x2 − 1) = 0

( x − 1)( x + 1)( x − 1)( x + 1) = 0

x = ±1AlternativelyLet u = x2

u2 – 2u + 1 = 0

(u – 1)(u – 1) = 0u = 1 ∴ x2 = 1

x = ±1

7 If ( x − 2) is a factor then

when x = 2, f ( x) = 0

0 = x3 + ax

2 − 6 x − 4

f (2) = 0 = 23 + a2

2 − 6 × 2 − 4

0 = 8 + 4a − 12 − 4

0 = 4a − 8

8 = 4a

2 = a

8 Let

P ( x) = x3 + x2

− ax + 3

P (1) = 1 + 1 − a + 3 = 0


7/24

CHAPTER 1 Graphs and polynomials • EXERCISE 1D | 7


( x − 1) is a factor

a = 5

9 If ( x + 3) is a factor then

when x = −3, f ( x) = 0

f (−3) = 0 = 2(−3)4 + a(−3)3 − 3 × (−3)

+ 18

0 = 162 − 27a + 9 + 18

0 = 189 − 27a

27a = 189

a = 710 If ( x + 1) is a factor then

when x = −1, f ( x) = 0

f (−1)= 0 = −a − 4 − b − 12

0 = −a − b − 16

a = −b − 16 [1]

If ( x − 2) is a factor then

when x = 2, f ( x) = 0

f (2) = 0 = 8a − 16 + 2b − 12

0 = 8a + 2b − 28

28 = 8a + 2b

14 = 4a + b [2]Sub [1] into [2]

14 = 4(−b − 16) + b

14 = −4b − 64 + b14 = −3b − 64

78 = −3b

−26 = b

∴ a = + 26 − 16

a = 10

11 (2 x − 3) and ( x + 2) are factors of

2 x3 + ax2

+ bx + 30

P ( x) = 2 x3 + ax2

+ bx + 30

P (−2) = 2(−2)3 + a(−2)2

+ b(−2)

+ 30 = 0

−16 + 4a − 2b + 30 = 0

4a − 2b = −14 [1]

P

3

2

⎛ ⎞

⎜ ⎟⎝ ⎠ = 2

33

2

⎛ ⎞

⎜ ⎟⎝ ⎠ + a

23

2

⎛ ⎞

⎜ ⎟⎝ ⎠ + b3

2

⎛ ⎞

⎜ ⎟⎝ ⎠

+ 30 = 0

2 × 27

8 +

9

4

a +

3

2

b + 30 = 0

27 + 9a + 6b + 120 = 0

9a + 6b = −147

3a + 2b = −49 [2]

[1] + [2] 7a = −63

a = −9

Substitute into [1] 4 × −9 − 2b = −14

−2b = 22

b = −11

a = −9, b = −11.

Exercise 1D — Linear graphs

1 a 2 x + 3 y = 12

x-intercept when y = 0

2 x = 12

x = 6

y-intercept when x = 0

3 y = 12

y = 4

b 2 y − 5 x − 10 = 0


−5 x = 10

x = −2


2 y = 10

y = 5

c 2 x − y = 1


2 x = 1

x = 1

2


− y = 1

y = −1

2 a y = mx + c

y = 3 x + c

find c using the point (2, 1)

1 = 3 × 2 + c

−5 = c

∴ y = 3 x − 5

−3 x + y + 5 = 0

b y = mx + c

y = −2 x + c

find c, sub in (−4, 3)3 = −2 × −4 + c

3 = 8 + c

−5 = c

∴ y = −2 x − 5

2 x + y + 5 = 0

3 a (−3, −4), (−1, −10)

m = 10 4

1 3

− +

− +

= 6

2

−

= −3

y = −3 x + c

sub in (−3, −4)

−4 = −3 × −3 + c

−13 = c

y = −3 x − 13

3 x + y + 13 = 0

b (7, 5), (2, 0)

m = 5 0

7 2

−

−

= 5

5

= 1

y = x + csub in (2, 0)

0 = 2 + c

−2 = c

y = x − 2

− x + y + 2 = 0

4 2 y − 3 x − 6 = 0

A 2 × 6 − 3 × 2 − 6 = 0

12 − 6 − 6 = 0

B 2 × 0 − 3 × − 2 − 6 = 0

0 + 6 − 6 = 0

C 2 × 3 − 3 × 0 − 6 = 0

6 − 0 − 6 = 0

D 2 × 2 − 3 × 1 − 6 = 0

4 − 3 − 6 ≠ 0

E 2 × 9 − 3 × 4 − 6 = 0

18 − 12 − 6 = 0

The answer is D.

5 a i −2 = 5

1 2

b −

+

−2 = 5

3

b −

−6 = b − 5

−1 = b

ii y

− x

= 7

∴ y = x + 7

m = 1

1 = 5

1 2

b −

+

3 = b − 5

8 = b

b parallel to y = 3 x − 4

∴ m = 3

y = 3 x + c

sub in (4, 5)

5 = 3 × 4 + c

−7 = c

∴ y

= 3 x

− 7

0 = 3 x − y − 7

c 2 y − x + 1 = 0

2 y = x − 1

y = 1

2 x −

1

2

m = −2 gradient of perpendicular line

y − y1 = m( x − x1)

Sub in (−2, 4) y − 4 = −2( x + 2)

y − 4 = −2 x − 4

2 x + y = 0

6 i x + 2 y + 4 = 0

• x-intercept when y = 0

x = − 4• y-intercept when x = 0

2 y = − 4

y = − 2

Graph e

ii y = 3 − Graph f

iii y − 2 x − 2 = 0


−2 = 2 x

−1 = x

• y-intercept when x = 0

y = 2Graph a

iv 3 y + 2 x = 6


8/24

8 | CHAPTER 1 Graphs and polynomials • EXERCISE 1E



2 x = 6

x = 3

• y-intercept when x = 0

3 y = 6

y = 2

Graph c

v y − 2 x = 0

• x- and y-intercepts occur at the

origin.Graph b

vi x = −2 − Graph d.

7 a y ≥ −2 or [−2, ∞)

b y > −5 or (−5, ∞)

c −2 ≤ y < 3 or [−2, 3)

d −2 ≤ y ≤ 3 or [−2, 3]

e R

f −∞


9/24

CHAPTER 1 Graphs and polynomials • EXERCISE 1E | 9


c f ( x) = 10 + 3 x − x2

y-intercept x = 0

y = 10

x-intercept y = 00 = (5 − x)(2 + x)

x = 5 or −2

d f ( x) = 6 x2 − x − 12

y-intercept x = 0

y = −12

x-intercept(s) y = 0

0 = (3 x + 4) (2 x − 3)

x = −4

3 or

3

2

3 a f ( x) = x2 − 6 x + 8

= x2 − 6 x + 32 − 32 + 8

= ( x − 3)2 − 9 + 8

= ( x − 3)2 − 1TP is (3, −1)

b f ( x) = x2 − 5 x + 4

= x2 − 5 x +

2 25 5

2 2

⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ + 4

=

25

2 x

⎛ ⎞−⎜ ⎟

⎝ ⎠−

25

4 + 4

=

25

2 x

⎛ ⎞−⎜ ⎟

⎝ ⎠−

25

4 +

16

4

=

25

2 x

⎛ ⎞−⎜ ⎟

⎝ ⎠−

9

4

TP is5 9

,2 4

⎛ ⎞−⎜ ⎟

⎝ ⎠

c f ( x)= 10 + 3 x − x2

= −( x2 − 3 x − 10)

= −

2 22 3 33 10

2 2 x x

⎡ ⎤⎛ ⎞ ⎛ ⎞− + − −⎢ ⎥⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

= −

23 9

102 4

x⎡ ⎤⎛ ⎞

− − −⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦

= −

23 9 40

2 4 4 x

⎡ ⎤⎛ ⎞− − −⎢ ⎥⎜ ⎟

⎝ ⎠⎢ ⎥⎣ ⎦

= −

23

2 x

⎛ ⎞−⎜ ⎟

⎝ ⎠ +

49

4

TP is3 49

,2 4

⎛ ⎞+⎜ ⎟

⎝ ⎠ =

1 11 , 12

2 4

⎛ ⎞⎜ ⎟⎝ ⎠

d f ( x)= 6 x2 − x − 12

= 2

6 26

x x

⎛ ⎞− −⎜ ⎟

⎝ ⎠

=

2 22 1 16 2

6 12 12 x x⎡ ⎤⎛ ⎞ ⎛ ⎞− + − −⎢ ⎥⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

=

21 1 288

612 144 144

x⎡ ⎤⎛ ⎞

− − −⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦

=

21 289

612 144

x⎡ ⎤⎛ ⎞

− −⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦

=

21

612

x⎛ ⎞

−⎜ ⎟⎝ ⎠

−289

24

TP is1 1

, 1212 24

⎛ ⎞⎜ ⎟⎝ ⎠

4 a i y = 2 − x2

a = −1, h = 0, k = 2

TP = (0, 2)

ii Domain = R

iii Range = (−∞, 2]

b i y = ( x − 6)2

a = 1, h = 6, k = 0

TP = (6, 0)

ii Domain = R

iii Range = [0, ∞)

c i y = −( x + 2)2

a = −1, h = −2, k = 0

TP = (−2, 0)

ii Domain = R

iii Range = (−∞, 0]

d i y = 2( x + 3)2 − 6

a = 2, h = −3, k = −6

TP = (−3, −6)

ii Domain = R

iii Range = [−6, ∞)

5 Using y = A ( x + B)2 + C

a i TP = (1, − 2)

∴ B = −1 and C = −2

Assume A = 1

⇒ y = 1( x − 1)2 − 2 y = x2 − 2 x + 1 − 2

y = x2 − 2 x − 1

ii Domain = R

iii Range [−2, ∞)

b i TP = (2, −3)

∴ B = −2

C = −3

Assume A = 1

⇒ y = 1 ( x − 2)2 − 3

= x2 − 4 x + 4 − 3

= x2 − 4 x + 1

ii Domain = [−1, ∞)

iii Range = [−3, ∞)

c i TP = (1, 9)

∴ B = −1 and C = 9

Assume A = −1

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

y = − x2 + 2 x − 1 + 9

y = − x2 + 2 x + 8

ii Domain = [−4, 4)

iii Range = [−16, 9]

6 a y = 2 x2 + 3

TP = (0, 3)


y = 3 x-intercepts when y = 0

0 = 2 x2 + 3

There are no x-intercepts.

b y = (2 x − 5) (2 x − 3)

= 4 x2 − 16 x + 15

= 4 215

44

x x⎛ ⎞

− +⎜ ⎟⎝ ⎠

= 4 215

4 4 44

x x⎛ ⎞

− + − +

⎜ ⎟⎝ ⎠

= 4 21

( 2)4

x⎛ ⎞

− −⎜ ⎟⎝ ⎠

= 4( x − 2)2 − 1

TP = (2, −1)

y-int = (0, 15)

x-int = (1, 5, 0), (2, 5, 0)

c y = (2 x − 3)2 − 8

TP = 3

, 82

⎛ ⎞−⎜ ⎟

⎝ ⎠


y = (−3)2 − 8

= 1

x-intercepts when y = 0

0 = (2 x − 3)2 − 8

= 4 x2 − 12 x + 9 − 8

= 4 x2 − 12 x + 1From the graphics calculator,

x = 2.91 and x = 0.09

7 a y = x2 − 2 x − 3

x-intercepts y = 0

0 = ( x − 3)( x + 1)

x = 3 or −1

(3, 0)(−1, 0)The answer is B.


10/24

10 | CHAPTER 1 Graphs and polynomials • EXERCISE 1E


b y = x2 − 2 x − 3

y = x2 − 2 x + 12 − 12 − 3

y = ( x − 1)2 − 4

TP = (1, −4)The answer is C.

8 f ( x) = −( x + 3)2 + 4

TP = (−3, 4)

∴ range ( ,4]−∞

The answer is D.

9 y = ( x − 4)2

x ∈ [0, 6]TP = (4, 0)

When x = 0 y = (−4)2 = 16

∴ range [0, 16]

When x = 6 y = (6 − 4)2 = 22 = 4

But x = 4 y = 0The answer is A.

10 a f ( x) = ( x − 2)2

− 4TP = (2, −4)

y-int x = 0

y = (0 − 2)2 − 4

y = (−2)2 − 4

y = 0

x-int y = 0

0 = ( x − 2 − 2)( x − 2 + 2)

0 = ( x − 4)( x)

∴ x = 4 or 0

b f ( x) = −( x + 4)2 + 9

TP = (−4, 9)

y-int x = 0

y = −(0 + 4)2 + 9

y = −16 + 9

y = −7

x-int y = 0

0 = 9 − ( x + 4)2

0 = (3 − ( x + 4))(3 + ( x + 4))

0 = (3 − x − 4) (3 + x + 4)

0 = (− x − 1)(7 + x)

x = −1 or −7

c y = x2 + 4 x + 3

y = x2 + 4 x + 4 − 4 + 3

y = ( x + 2)2 − 1

TP = (−2, −1)

y-int x = 0

y = 3

x-int y = 0

0 = ( x + 2 − 1)( x + 2 + 1)

0 = ( x + 1)( x + 3)

∴ x = −1 or −3

d y = 2 x2 − 4 x − 6

y = 2[ x2 − 2 x − 3]

= 2[ x2 − 2 x + 1 − 1 − 3]

= 2[( x − 1)2 − 4]

y = 2( x − 1)2 − 8

TP = (1, −8)

y-int x = 0

y = −6

x-int y = 0

0 = 2[( x − 1 − 2)( x − 1 + 2)]

0 = 2( x − 3)( x + 1)

∴ x = 3 or −1

11 a y = x2 − 2 x + 2 x ∈ [−2, 2]

y = x2 − 2 x + 1 − 1 + 2

y = ( x − 1)2 + 1

∴ TP = (1, 1)

i Domain = [−2, 2]ii Range:

When x = −2

y = 10When x = 2 y = 2

but TP = (1, 1)

∴ [1, 10]

b y = − x2 + x − 1 x ∈ R+

y = −( x2 − x + 1)

y = −

2 2

2 1 1 12 2

x x⎛ ⎞⎛ ⎞ ⎛ ⎞⎜ − + − + ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

y = −

21 3

2 4 x

⎡ ⎤⎛ ⎞− +⎢ ⎥⎜ ⎟

⎝ ⎠⎢ ⎥⎣ ⎦

y = −

21

2

x⎛ ⎞

−⎜ ⎟

⎝ ⎠

−3

4

∴ TP = 1 3

,2 4

⎛ ⎞−⎜ ⎟

⎝ ⎠

i Domain = R+

ii Range = (−∞, −3

4]

c f ( x)= x2 − 3 x − 2 x ∈ [−10, 6]

f ( x)= x − 3 x +

23

2

⎛ ⎞⎜ ⎟⎝ ⎠

−

23

2

⎛ ⎞⎜ ⎟⎝ ⎠

− 2

=

23

2 x

⎛ ⎞−⎜ ⎟

⎝ ⎠ −

9

4 −

8

4

=

23

2 x

⎛ ⎞−⎜ ⎟

⎝ ⎠−

17

4

∴ TP = 3 17

,

2 4

⎛ ⎞−⎜ ⎟

⎝ ⎠

i Domain = [−10, 6]ii Range:

When x = −10 y = 128

∴ 17 ,1284

⎡ ⎤−⎢ ⎥⎣ ⎦

d f ( x) = −3 x2 + 6 x + 5 x ∈ [−5, 3)

=

2 5

3 2 3 x x

⎡ ⎤

− − −⎢ ⎥⎣ ⎦

= 25

3 2 1 13

x x⎡ ⎤

− − + − −⎢ ⎥⎣ ⎦

= 28

3 ( 1)3

x⎡ ⎤

− − −⎢ ⎥⎣ ⎦

= −3( x − 1)2 + 8

TP = (1, 8)

i Domain = [– 5, 3)ii Range:

When x = −5, y = −100

∴ [– 100, 8]

12 V (t ) = 2t 2 − 16t + 40 t ∈ [0, 10]

V (t ) = 2(t 2 − 8t + 20)

= 2[t 2 − 8t + 16 − 16 + 20]

= 2[(t − 4)2 + 4]

= 2(t − 4)2 + 8

∴ TP = (4, 8)

When t = 0 V (t ) = 40

When t = 10

V (t ) = 2 × 62 + 8

= 80

a minimum V = 8 m3

b maximum V = 80 m3

13 h(t ) = −3t 2 + 12t + 36

h(t ) = −3[t 2 − 4t − 12]


11/24

CHAPTER 1 Graphs and polynomials • EXERCISE 1F | 11


= −3[t 2 − 4t + 4 − 4 − 12]

= −3[(t − 2)2 − 16]

= −3(t − 2)2 + 48

∴ TP = (2, 48)

a maximum height = 48 m

b When h(t ) = 0

0 = −3[(t − 2 − 4)(t − 2 + 4)]

0 = −3(t − 6)(t + 2)

∴ t = 6 or −2

Since time 0≥ ⇒ 6 secondsc Domain [0, 6]Range [0, 48]

14 a h(t ) = t 2 − 12t + 48, t ∈ [0, 11]The lowest point is the

y-coordinate of the turning pointh(t ) = t 2 − 12t + 36 − 36 + 48

= (t − 6)2 + 12

TP = (6, 12)Lowest point is 12 m above theground.

b Time taken is the x-coordinate ofthe turning point.

t = 6 secondsc Check the end points of the domain.

h(0) = 48

h(11) = 112 − 12 × 11 + 48

= 37The highest point above the ground

is 48 m.d Domain = [0, 11]

Range = [12, 48]e

Exercise 1F — Cubic graphs1 a Positive cubic so a = 1. Goes

through origin so x is a factor. y = x( x − a)( x − b)

= x( x + 6)( x − 5)b Positive cubic in form

y = a( x − m)( x − n)2

∴ a = 1, m = 1, n = −2

∴ y = 1( x − 1)( x + 2)2

2 a Positive cubic in form

y = ( x − l )( x − m)( x − n)

l = −3, m = 1, n = 4

∴ y = ( x + 3)( x − 1)( x − 4)

∴ (v)

b Negative cubic in form

y = a( x − m)( x − n)2

∴ a = −1, m = 5, n = − 2

∴ y = −1( x − 5)( x + 2)2

y = (5 − x)( x + 2)2 (iv)

c Negative cubic in form

y = a( x − l )( x − m)( x − n)

a = −1, l = −3, m = 1, n = 4

∴ y = −1( x + 3)( x − 1)( x − 4)

y = ( x + 3)(1 − x)( x − 4)

(ii)d Positive cubic in form

y = a( x − t )3

a = 1, t = 3

∴ y = ( x − 3)3 (i)

e Positive cubic in form

y = a( x − l )( x − m)( x − n)

a = 1, l = −4, m = −2, n = 1

∴ y = ( x + 4)( x + 2)( x − 1)(vi)

f Positive cubic in form

y = a( x − m)( x − n)2

a = 1, m = 5, n = −2

y = ( x − 5)( x + 2)2 (viii)

g Negative cubic in form

y = a( x − t )3

a = −1, t = 3

∴ y = −1( x − 3)3

y = (3 − x)3 (vii)

h Negative cubic in form

y = a( x − l )( x − m)( x − n)

a = −1, l = − 4, m = −2, n = 1

∴ y = −1( x + 4)( x + 2)( x − 1)

y = ( x + 4)( x + 2)(1 − x)(iii)

3 a y = x3 + x2 − 4 x − 4

y-intercept

x = 0 y = −4Factorise to find x-intercepts

Test x = −1, y = 0

∴ x + 1 is a factor2

3 2

3 2

4

4 41

4 4

4 4

0

x

x x x x

x x

x

x

−

⎡ + − −+ − ⎢

+⎢⎣

− −⎡− ⎢

− −⎣

∴ y = ( x + 1)( x2 − 4)

y = ( x + 1)( x − 2)( x + 2)

If y = 0, x = −1, 2, or −2

b y = 2 x3 − 8 x2 + 2 x + 12

y-int x = 0

y = 12Factorise to find x-intercepts

Test x = −1 so y = 0

∴ ( x + 1) is a factor

2

3 2

3 2

2

2

2 10 12

2 8 2 121

2 2

10 2

10 10

12 12

12 12

0

x x

x x x x

x x

x x

x x

x

x

− +

⎡ − + ++ − ⎢

+⎢⎣

⎡− +− ⎢

− −⎢⎣

+⎡− ⎢

+⎣

∴ y = ( x + 1)(2 x2 − 10 x + 12)

y = 2( x + 1)( x − 2)( x − 3)

If y = 0, then x = −1, 2 or 3

c y = −2 x3 + 26 x + 24

y-int x = 0

y = 24

Factorise to find x-interceptsTest x = −1 so y = 0

∴ ( x + 1) is a factor.2

3 2

3 2

2

2

2 2 24

2 0 26 241

2 2

2 26

2 2

24 24

24 24

0

x x

x x x x

x x

x x

x x

x

x

− + +

⎡− + + ++ − ⎢

− −⎢⎣

⎡ +− ⎢

+⎢⎣

+⎡− ⎢

+⎣

∴ y = ( x + 1)(−2 x2 + 2 x + 24)

y = 2( x + 1)(− x − 3)( x − 4)

If y = 0, then x = −1, −3 or 4.

d y = − x3 + 8 x2 − 21 x + 18

y-int x = 0

y = 18

x-intercept

Factorise:Test x = 3 so y = 0

∴ ( x − 3) is a factor.2

3 2

3 2

2

2

5 6

8 21 183

3

5 21

5 15

6 18

6 18

0

x x

x x x x

x x

x x

x x

x

x

− + −

⎡− + − +− − ⎢

− +⎢⎣

⎡ −− ⎢

−⎢⎣

− +⎡− ⎢

− +⎣


12/24

12 | CHAPTER 1 Graphs and polynomials • EXERCISE 1F


∴ y = ( x − 3)(− x2 + 5 x − 6)

= −( x − 3)( x2 − 5 x + 6)

y = −1( x − 3)( x − 3)( x − 2)

∴ x = 3 or 2

4 a x3 + 6 x2 + 12 x + 8 = y

Test x = −2 so y = 0

∴ ( x + 2) is a factor.2

3 2

3 2

2

4 4

6 12 82

2

4 12

4 8

4 8

4 8

0

x x

x x x x

x x

x x

x x

x

x

+ +

⎡ + + ++ − ⎢

+⎢⎣

⎡ +− ⎢

+⎣

+⎡− ⎢

+⎣

∴ y = ( x + 2)( x2 + 4 x + 4)

= ( x + 2)( x + 2)2

y = ( x + 2)3

The answer is B.

b In form y = a( x − t )3

a = 1, t is intercept

The answer is E.

5 Function graph is a negative cubic so

a = −1

Point of inflection (2, 2)

The answer is C.

y = (2 − x)3 + 2

6 f ( x) = 5( x + 1)3

− 3Point of inflection (−1, −3)

Graph is a positive cubic.

Τhe answer is A.

7 Positive cubic

Turning point at (1, 0) because ofrepeated factor

x-intercept at (−3, 0)

y-intercept at (0, 6)

The answer is B.

8 Negative cubic B or D

Point of inflection is (a, b)

a < 0 so,

y = −( x + a)3 + b

The answer is D.9 y = −h( x − a)2 ( x − c)

x = 0, y = −ha2(−c)

b = ha2c

h = 2

b

a c

y = 2

b

a c− ( x − a)2( x − c)

The answer is E.

10 a f ( x) = x3 + x

2 − 10 x + 8 x ∈ [2, ∞)

a > 1 positive

b > 1 so 2 turning points.

y-intercept x = 0

y = 8

When x = 2

y = 23 + 22 − 20 + 8

= 0

Closed end point = (2, 0)

i Domain [2, ∞)

ii range [0, ∞)

b f ( x) = 3 x3 − 5 x2 − 4 x + 4 for

x ∈ [−2, −1]

a > 1 so positive

b ≠ 0 ∴ 2 turning points.

y-intercept x = 0

y = 4

When x = −2

y = 3 × (−2)3 − 5 × (−2)2 − 4 × −2

+ 4

= −32

When x = −1

y = 3 × (−1)3 − 5 × (−1)2 − 4 × −1

+ 4

= 0

Closed end point (−2, −32)

Closed end point (−1, 0)

i Domain [−2, −1]

ii Range [–32, 0]

c f ( x) = −3 x3 + 4 x2 + 27 x − 36

x ∈ (0, 1]

a


13/24

CHAPTER 1 Graphs and polynomials • EXERCISE 1G | 13


0 = 4a + 2a − 36

0 = 6a − 36

36 = 6a

6 = a

∴ b = 12 − 36

= −24

12 0 = −1 − 2 − a + 10

0 = 7 − a [1]

∴ a = 7

∴ y = 6 + (7 + b) x − 4 x2 − x3 0 = 6 + (7 + b) × (−1) − 4 × 1

− (−1)

0 = 6 − 7 − b − 4 + 1

0 = −4 − b

b = −4

13 a f ( x) = a( x + b)3 + c

point of inflection (3, 3)

⇒ b = −3 and c = 3

f ( x) = a( x − 3)3 + 3

When x = 2, f ( x) = 0

0 = a(2 − 3)3 + 3

0 = a × (−1)3 + 3

0 = −a + 3∴ a = 3

∴ f ( x) = 3( x − 3)3 + 3

b Point of inflection due to

reflection = (−3, 3)

∴ g ( x) = −3( x + 3)3 + 3

domain [−4, −2]

range [0, 6]

c When f ( x) = 3.375

3.375 = 3( x − 3)3 + 3

x = 3.5

∴ width = 3.5 × 2

= 7 cm

14 d (t ) = at 2(b − t )

a (2, 3) and (5, 0)

∴ 3 = 4a(b − 2) [1]

0 = 25a(b − 5) [2]

3 = (4ab − 8a) × 25

0 = (25ab − 125a) × 4

75 = 100ab − 200a

0 = 100ab − 500a

∴ 75 = 300a

1

4 = a

Sub1

4 = a into [1]

3 = 4 × 1

4 (b − 2)

3 = 1(b − 2)

5 = b

b ∴ Rule: d (t ) = 2

4

t (5 − t ) for

domain = [0, 5]

c

d d (t ) =2 3

5

4 4

t t −

d ′(t ) = 2

10 3

4 4

t t −

Let d ′(t ) = 0 = 2

10 3

4

t t −

0 = 10t − 3t 2

0 = t (10 − 3t )

∴ t = 0 or 10 − 3t = 010 = 3t

∴ time is1

33

hours

When time is1

33

,

d (t ) = 2(3.3)

4

× 1.6

Maximum distance = 4.6 km

Exercise 1G — Quartic graphs

1 a y = ( x − 2)( x + 3)( x − 4)( x + 1)

y-intercept when x = 0 y = −2 × + 3 × −4 × + 1

y = 24


x = −3, −1, 2 and 4

b y = 2 x4 + 6 x3 − 16 x2 − 24 x + 32


y = 32

x-intercept when y = 0Test x = 1, y = 0


Test x = + 2 so y = 0


( x − 1)( x − 2) = x2 − 3 x + 2

2

4 3 22

4 3 2

3 2

3 2

2

2

2 12 16

2 6 16 24 323 2

2 6 4

12 20 24

12 36 24

16 48 32

16 48 32

0

x x

x x x x x x

x x x

x x x

x x x

x x

x x

+ +

⎡ + − − +− + −⎢

− +⎢⎣

⎡ − −− ⎢

− +⎢⎣

⎡ − +− ⎢ − +⎢⎣

∴ ( x − 1)( x − 2)(2 x2 + 12 x + 16) = y

( x − 1)( x − 2)2( x2 + 6 x + 8) = y

2( x − 1)( x − 2)( x + 4)( x + 2) = y

When using N.F.L, x = 1, 2, −4, −2

then y = 0.

c y = x4 − 4 x2 + 4


y = 4


Let a = x2

y = a2 − 4a + 4

y = (a − 2)2

Sub a = x2 back in

y = ( x2 − 2)2

∴ x = ± 2

d y = −2 x4 + 15 x3 − 37 x2 + 30 x

y-intercept, when x = 0

y = 0


Test x = 2 ∴ y = 0

Test x = 3 ∴ y = 0

( x − 2)( x − 3) are factors

( x − 2)( x − 3) = x2

− 5 x + 62

4 3 22

4 3 2

3 2

3 2

2 5

2 15 37 305 6

2 10 12

5 25 30

5 25 30

0

x x

x x x x x x

x x x

x x x

x x x

− +

⎡− + − +− + − ⎢

− + −⎢⎣

⎡ − +− ⎢

− +⎢⎣

∴ ( x − 2)( x − 3)(−2 x2 + 5 x)= y

x( x − 2)( x − 3)(−2 x + 5) = y

Using N.F.L, x = 0, 2, 3 or5

2

e y = 6 x4 + 11 x3 − 37 x2 − 36 x + 36


y = 36


Test x = −3 ∴ y = 0

Test x = 2

( x + 3)( x − 2) are factors

( x + 3)( x − 2) = x2

+ x − 62

4 3 22

4 3 2

3 2

3 2

2

2

6 5 6

6 11 37 36 366

6 6 36

5 36

5 5 30

6 6 36

6 6 36

0

x x

x x x x x x

x x x

x x x

x x x

x x

x x

+ −

⎡ + − − ++ − −⎢

+ −⎢⎣

⎡ − −− ⎢

+ −⎢⎣

⎡− − +− ⎢

− − +⎢⎣

∴ ( x + 3)( x − 2)(3 x − 2)(2 x + 3) = y

Using N.F.L


14/24

14 | CHAPTER 1 Graphs and polynomials • EXERCISE 1G


x = −3, 2,2

3, −

3

2

2 a y = x2( x − 2)( x − 3)

y = 0, x2( x − 2)( x − 3) = 0

Turning point (0, 0)

Intercepts at x = 2 and x = 3


y = 0(−2)(−3)

= 0Positive quarticMaximum turning point at (1.16, 2.08)

Minimum turning points at (0, 0)

and (2.59, −1.62)

b y = −( x + 1)2( x − 1)

2

x = 0, y = 1 y-intercept is 1

y = 0, ( x + 1)2( x − 1)2 = 0

x = −1, x = 1

Minimum turning points at (−1, 0)and (1, 0).

Maximum turning point (0, 1)

c y = ( x − 1)2( x + 1)( x + 3)


y = 0, ( x − 1)2( x + 1)( x + 3) = 0

x = 1, −1, −3

Positive quarticMinimum turning points

(−2.28, −9.91) and (1, 0)Maximum turning point

(−0.22, 3.23)

d y = ( x + 2)3(1 − x)


y = 0, ( x + 2)3(1 − x) = 0

x = −2, 1

Point of inflection (2, 0)

x-intercept is 1

Negative quartic (1 − x) = −( x − 1)

Maximum turning point (0.25, 8.54)

3 a f ( x) = x4 − 8 x2 + 16

Let x2 = a

f ( x) = a2 − 8a + 16

= (a − 4)2

Substitute x2 = a back in:

f ( x)=

( x

2

−

4)

2

∴ ( x − 2)2( x + 2)

2

The answer is E.

b f ( x) = x4 − 8 x2 + 16

y-int when x = 0

y = 16 (0, 16)

x-int when y = 0

2, −2

∴ The answer is B.

c range = [0, 16]

The answer is A.

d 25 = x4 − 8 x

2 + 16

0 = x4 − 8 x2 − 9

Let x2 = a

0 = a2

− 8a − 90 = (a − 9)(a + 1)

Substitute x2 = a back in

0 = ( x2 − 9)( x2 + 1)

∴ x = ± 3

∴ (−3, +3) is the restricted domain.The answer is D.

e f (−1) = 1 − 8 + 16 = 9

f (0) = 16Range: (9, 16)

Answer is D.

f f (0) = 16

f (2) = 0

Range is y ≥ 0 or [0, ∞ )

Answer is C.4 a x = −2, −1, 1, 3

y = a( x + 2)( x + 1)( x − 1)( x − 3) y-int (0, 6)

6 = a(0 + 2)(0 + 1)(0 − 1)(0 − 3)

6 = a × 6

a = 1

y = ( x + 2)( x + 1)( x − 1)( x − 3)

b x = 4, 2, −1

Repeated factor at x = −2.

y = a( x − 4)( x − 2)2( x + 1)

y-int (0, 8)

8 = a(0 − 4)(0 − 2)2(0 + 1)

8 = −16a

a = −1

2

y = 1

2

−( x − 4)( x − 2)2( x + 1)

5 a y = (2 − x)( x2 − 4)( x + 3) x ∈ [2, 3]

y-int when x = 0

y = 2 × −4 × 3

= −24

x-int when y = 0

x = 2, −2, −3.

when x = 2, y = 0 closed end point

x = 3, y = −30 closed end point

i Domain [2, 3]

ii Range [−30, 0]

b y = 9 x4 − 30 x

3 + 13 x

2 + 20 x + 4 x ∈

(−2, −1]

y-int when x = 0

y = 4 x-int when y = 0

Test x = 2 y = 0

∴ ( x − 2) factor

3 2

4 3 2

4 3

3 2

3 2

2

2

9 12

9 30 13 20 42

9 18

12 13

12 24

11 20

11 22

2 4

2 4

0

x x

x x x x x

x x

x x

x x

x x

x x

x

x

− −

⎡ − + + +− − ⎢

−⎢⎣

⎡− +− ⎢

− +⎢⎣

⎡− +− ⎢

− +⎢⎣

− +⎡− ⎢

− +⎣

11 2 x −

y = ( x − 2)(9 x3 − 12 x2 − 11 x − 2)

Try for 2nd factor of x − 2

Test x = 2, 9 x3 − 12 x2 − 11 x − 2 = 0

x − 2 is a factor. So turning pointat (2, 0)

2

3 2

3 2

2

2

9 6 1

2 9 12 11 2

9 18

6 11

6 12

2

x x

x x x x

x x

x x

x x

x

+ +

− − − −

−

−

−

−

y = ( x − 2)2(9 x2 + 6 x + 1)

y = ( x − 2)2(9 x2 + 6 x + 1)

= ( x − 2)2(3 x + 1)2

Turning points at (2, 0) and

1, 0

3

⎛ ⎞−⎜ ⎟

⎝ ⎠

when x = −2, y = 400 open end point

when x = −1, y = 36 closed end point


15/24

CHAPTER 1 Graphs and polynomials • EXERCISE 1H | 15


i Domain (−2, −1]ii Range [36, 400)

c y = −( x − 2)2( x + 1)2 x ∈ (−∞, −2]

y-int when x = 0 y = −1 × 4 × 1

y = −4

x-int when y = 0

0 = −1( x − 2)2( x + 1)

2

∴ x = −1 or 2

When x = −2, y = −16.

y = −(−2 − 2)2(−2 + 1)2

y = −(−4)2(−1)

2

y = −16 × 1

y = −16

i Domain (−∞,−2]

ii Range (−∞,−16]

d y = − x4 + 4 x2

x ∈ [−3, −2]

y-int when x = 0

y = 0

x-int when y = 0

0 = x2(− x2 + 4)

0 = x2(4 − x2)

0 = x2(2 − x)(2 + x)

∴ x = 0 ± 2When x = −3, y = −45

x = −2, y = 0

Both of these are closed end points.

i Domain [−3, −2]

ii Range [−45, 0]

6 f ( x) = x4 + ax3 − 4 x2 + bx + 6

(2, 0):

0 = 24 + 23a − 4 × 22 + 2b + 6

16 + 8a − 16 + 2b + 6 = 0

6 + 8a + 2b = 0

8a + 2b = −6

Divide both sides by 2:

4a + b = −3

−4a − 3 = b [1]

4a + 3 = − b

(−3, 0):

0 = (−3)4 + (−3)3a − 4 × (−3)2 − 3b + 6

0 = 81 − 27a − 36 − 3b + 6

0 = 51 − 27a − 3b

0 = 17 − 9a − b [2]

Sub [1] into [2]

0 = 17 − 9a + 4a + 3,

0 = 20 − 5a

5a = 20

a = 4

If a = 4 then

b = −4 × 4 − 3

b = −16 − 3

b = −19

7 f ( x) = x4 + ax3 + bx2 − x + 6( x − 1) is a factor

P (1) = 1 + a + b − 1 + 6

a + b = −6 [1]

( x + 3) is a factor

P (−3) = 81 − 27a + 9b + 3 + 6 = 0

27a − 9b = 90

3a − b = 10 [2]

[1] + [2] 4a = 4

a = 1

b = −7

8 y = (a − 2b) x4 − 3 x − 2

Sub in (1, 0):

0 = (a − 2b)1

4

− 3 − 20 = a − 2b − 5

5 = a − 2b

5 + 2b = a [1]

Sub (1, 0) into equation.

y = x4 − x3 + (a + 5b) x2 − 5 x + 7

0 = 1 − 1 + (a + 5b)1 − 5 + 7

0 = a + 5b + 2

−2 = a + 5b [1]

Sub [1] into [2]

−2 = 5 + 2b + 5b

−7 = 7b

−1 = b

If −1 = b then 5 − 2 = a

∴ 3 = a

Exercise 1H — Solving systemsof equations

1 9 0ax y+ = [1]

3 ( 6) 0 x a y+ − = [2]

From [1] 9 y ax= −

9

ax y = − [3]

From [2] ( 6) 3a y x− = −

3

6

x y

a= −

− [4]

Lines which are parallel have no

solutions while lines which are co-incident have infinitely many solutions.In both cases the gradients are thesame. Equate the gradients for [3] and

[4].

2

3

9 6

( 6) 27

6 27 0

( 9)( 3) 0

9 or 3

a

a

a a

a a

a a

a a

− =−

− =

− − =

− + =

= = −

Check: When 3a = − then there are

co-incident lines andsubsequently infinitely manysolutions because equations

become

3 9 0

3 9 0

x y

x y

− + =

− =

3 ( 3 6) 0

3 9 0

x y

x y

+ − − =

− =

When 9a = then there are again

co-incident lines because equations become

9 9 0

0

x y

x y

+ =

+ =

3 (9 6) 0

3 3 0

0

x y

x y

x y

+ − =

+ =

+ =

For a unique solution then

\{ 3,9}.a R∈ −

2 a 5 10mx y− = [1]

3 ( 2) 6 x m y− − = [2]

From [1] 5 10 y mx− = −

25

mx y = − [3]

From [2] 3 6 ( 2) x m y− = −

3 6

2 2

x y

m m= −

− − [4]

Equate gradients

2

3

5 2

( 2) 15

2 15 0( 5)( 3) 0

5 or 3

m

m

m m

m mm m

m m

=−

− =

− − =− + =

= = −

Check: When 3m = − there are

parallel lines and

subsequently no solutions asequations become

3 5 10

3 5 10

x y

x y

− − =

+ = −

3 ( 3 2) 6

3 5 6

x y

x y

− − − =

+ =

When 5m = there are co-incident lines

and subsequently infinitely manysolutions as equations become

5 5 10

2

x y

x y

− =

− =

3 (5 2) 6

3 3 6

0

x y

x y

x y

− − =

− =

− =

b

5 10 [1] 3

3 ( 2) 6 [2]

3 15 30 [3]

3 ( 2) 6 [4]

mx y

x m y m

mx y

mx m m y m

− = ×

− − = ×

− =

− − =


16/24

16 | CHAPTER 1 Graphs and polynomials • EXERCISE 1H


2

3 15 30 [3]

3 ( 2) 6 [4]

15 ( 2) 30 6

( 15 ( 2)) 30 6

( 2 15) 30 6

6( 5)

( 5)( 3)

6 , 53

mx y

mx m m y m

y m m y m

y m m m

y m m m

m y

m m

y mm

− = −

− − =

− + − = −

− + − = −

− − = −

− −=

− +

= − ≠+

Substitute6

3

ym

= −+

into [1]

65 10

3

3010

3

mxm

mxm

⎛ ⎞− − =⎜ ⎟

+⎝ ⎠

+ =+

3010

3

10( 3) 30

3

103

10

( 3)

10, 0, 3

3

mxm

mmx

m

mmxm

m x

m m

x m mm

= −+

+ −=

+

=+

=+

= ≠ ≠ −+

For a unique solution10

,3

xm

=+

6

3 y

m= −

+ providing

\ [ 3, 0, 5}m R∈ −

3 a 9 x y z + + = (1)

2 3 15 x y z − + − = − (2)5 3 29 x y z + + = (3)

(1) + (2) 3 2 6 y z − = − (4)

(2) + (3) 7 14 y =

2 y =

Substitute 2 y = into (4)

3(2) 2 6 z − = −

6 2 6 z − = −

2 12 z − = −

6 z =

Substitute 2, 6 y z = = into (1)

2 6 9 x + + =

8 9 x + =

1 x =

b

1 1 1 9

2 2 3 15

1 5 3 29

x

y

z

⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − = −⎢ ⎥ ⎢ ⎥ ⎢ ⎥

⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

1

2

6

x

y

z

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥

=⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

4 a 5 x y z − − = − (1)

6 2 5 2 x y z + − = − (2)

3 4 13 x y z − + + = (3)

(1) 2× 2 2 2 10 x y z − − = − (4)

(2) (4)+ 8 7 12 x z − = − (5)

(3) 2× 6 2 8 26 x y z − + + = (7)

(2) (7)− 12 13 28 x z − = − (8)

(5) 3× 24 21 36 x z − = − (9)

(8) 2× 24 26 56 x z − = − (10)

(10) (9)− 5 20 z − = − 4 z =

Substitute 4 z = into (5)

8 7(4) 12 x − = −

8 28 12 x − = −

8 12 28 x = − +

8 16 x =

2 x =

Substitute 2, 4 x z = = into (1)

2 4 5 y− − = −

2 5 y− − = −

3 y− = −

3 y =

b

1 1 1 5

6 2 5 2

3 1 4 13

x

y

z

− − −⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥

− = −⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦

2

3

4

x

y

z

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥

=⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

5 3 2 1 x y z + − = [1]

2 x y z + + = [2]

1kx y z + − = [3]

3 2 1

1 1 1

1 1

ma

k

−⎡ ⎤⎢ ⎥ →⎢ ⎥⎢ ⎥−⎣ ⎦

det (ma) = 3k − 5

(using a calculator)

⇒ when5

3k ≠

there is a unique solution.

Therefore, when5

3k = , there is no

solution.

6 a Let a be the number of adult

tickets bought.

Let c be the number of children’s

tickets bought.

Let s be the number of seniors’

tickets bought.

200a c s+ + = (1)

9.5 4.5 3.5 1375a c s+ + = (2)

3c s= (3)

Substitute (3) into (1) and (2)

3 200a s s+ + =

4 200a s+ = (4)

9.5 4.5(3 ) 3.5 1375a s s+ + =

9.5 13.5 3.5 1375a s s+ + =

9.5 17 1375a s+ =

95 170 13 750a s+ =

19 34 2750a s+ = (5)

(4) 19× 19 76 3800a s+ = (6)

(6) (5)− 42 1050 s =

105025

42 s = =

Substitute 25 s = into (3)

3(25) 75c = =

Substitute 25, 75 s c= = into (1)

75 25 200a + + =

100 200a + =

100a =

b

1 1 1 200

9.5 4.5 3.5 1375

0 1 3 0

a

c

s

⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥

=⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦

100

75

25

a

c

s

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥

=⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

There were 100 adult tickets,75 children’s tickets and 25 seniors’tickets sold on the opening night.

7 a 255 F C P + + = (1)

6 4.5 1067.5 F C P + + = (2)

50C F = + (3)

Substitute (3) into (1)

50 255 F F P + + + =

2 205 F P + = (4)


6 4.5( 50) 1067.5 F F P + + + =

6 4.5 225 1067.5 F F P + + + =

10.5 1067.5 225 F P + = −

10.5 842.5 F P + = 105 10 8425 F P + = (5)

(4) 10× 20 10 2050 F P + = (6)

(5) (6)− 85 6375 F =

637575

85 F = =

Substitute 75 F = into (3)

75 50 125C = + = Substitute 75, 125 F C = = into (1)

75 125 255 P + + =

200 255 P + =

55 P =

b 255 F C P + + = (1)6 4.5 1067.5 F C P + + = (2)

50 F C − + = (3)

1 1 1 255

6 4.5 1 1067.5

1 1 0 50

F

C

P

⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥

=⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦

75

125

55

F

C

P

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥

=⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

75 portions of flake, 125 portions of

chips and 55 potato cakes were soldduring the ‘lunch special’ period.


17/24

CHAPTER 1 Graphs and polynomials • CHAPTER REVIEW | 17


8 a 3 2 y ax bx cx d = + + +

When 0, 4; x y= =

3 24 (0) (0) (0)a b c d = + + +

4 d =

3 2 4 y ax bx cx= + + +

When 2, 62; x y= − = −

3 262 ( 2) ( 2) ( 2) 4a b c− = − + − + − +

66 8 4 2a b c− = − + −

33 4 2a b c− = − + − (1)

When 2, 26; x y= = −

3 226 (2) (2) (2) 4a b c− = + + +

30 8 4 2a b c− = + +

15 4 2a b c− = + + (2)

When 5, 64; x y= =

3 264 (5) (5) (5) 4a b c= + + +

60 125 25 5a b c= + +

12 25 5a b c= + + (3)

(1) (2)+ 48 4b− =

12 b− =

(1) (3)+ 21 21 7a b− = +

3 3a b− = + (4)

Substitute 12b = − into (4)

3 3 12a− = −

9 3a=

3 a=

Substitute 12, 3b a= − = into (3)

12 25(3) 5( 12) c= + − +

12 75 60 c= − +

12 15 c= +

3 c− =

b 4 2 33a b c− + − = − (1)

4 2 15a b c+ + = − (2)

25 5 12a b c+ + = (3)

4 2 1 33

4 2 1 15

25 5 1 12

a

b

c

− − −⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥

= −⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

3

12

3

a

b

c

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥

= −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦

The equation of the cubic function is

3 23 12 3 4. y x x x= − − +

9 a When 0, 2; x y= =

2 20 2 (0) 2 0a b c+ + + + =

4 2 0b c+ + =

2 4b c+ = − [1]

When 6, 2; x y= =

2 26 2 6 2 0a b c+ + + + =

36 4 6 2 0a b c+ + + + =

6 2 40a b c+ + = − [2]

When 3, 1; x y= = −

2 2

3 ( 1) 3 0a b c+ − + − + =

3 10a b c− + = − [3]

From (1) 2 4c b= − − [4]


6 2 2 4 40a b b+ − − = −

6 36a = −

6a = −


3 2 4 10a b b− − − = −

3 3 6a b− = −

2a b− = −

Substitute 6a = − into (5) 6 2b− − = −

4b− =

4b = −

Substitute 4b = − into (1)

2( 4) 4c− + = −

8 4c− + = −

4c =

b 2 4b c+ = − (1)

6 2 40a b c+ + = − (2)

3 10a b c− + = − (3)

0 2 1 4

6 2 1 403 1 1 10

a

bc

−⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥

= −⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦ ⎣ ⎦

6

4

4

a

b

c

−⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥

= −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

10 2 24a b c d e+ − + − =

2 3 2 3 34a b c d e+ − − − =

2 3 2 31a b c d e− + + − + = −

3 5 2 2 3 18a b c d e+ − − + =

4 2 3 5a b c d e− − − + =

1 2 1 1 1 24

2 3 2 1 3 342 1 3 2 1 31

3 5 2 2 3 18

4 2 1 3 1 5

a

bc

d

e

− −⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥

− − −⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥=− − −⎢ ⎥ ⎢ ⎥ ⎢ ⎥

− −⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − −⎣ ⎦ ⎣ ⎦ ⎣ ⎦

7

3

2

5

4

a

b

c

d

e

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥= −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦

Chapter review

Short answer

1 a (2 y − 3 x)5

= (2 y)5 + 5(2 y)4(−3 x) + 10(2 y)3(−3 x)2

+ 10(2 y)2(−3 x)

3 + 5(2 y)(−3 x)

4 + (−3 x)

5

= 32 y5 + 5 × 16 × (−3) y4 x + 10 × 8

× 9 y3 x2 + 10 × 4 × (−27) y2 x3 + 5

× 2 × 81 yx4 − 243 x5

= 32 y5 − 240 y4 x + 720 y3 x2

− 1080 y2 x3 + 810 yx4 − 243 x5

b

82

2

x

x

⎛ ⎞−⎜ ⎟

⎝ ⎠ =

8

2

x⎛ ⎞⎜ ⎟⎝ ⎠

+ 8

72

2

x

x

⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

+ 28

6 22

2

x

x

⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠+ 56

5 32

2

x

x

⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

+ 70

4 42

2

x

x

⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠+ 56

3 52

2

x

x

⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

+ 28

2 62

2

x

x

⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠+ 8

72

2

x

x

⎛ ⎞⎛ ⎞−⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠

+

82

x

⎛ ⎞−⎜ ⎟

⎝ ⎠

= 8 6

256 8

x x− +

47

4

x − 14 x2 + 70

− 2

224

x +

4

448

x −

6

512

x +

8

256

x

2 ( x2 − 1) = ( x − 1)( x + 1)

∴ solutions are x = −1 or 1

∴ If x = −1 then 0 = −7 − a + 5 − 15 + b

0 = −17 − a + b

17 + a = b [1]

If x = 1, then 0 = −7 + a + 5 + 15 + b

0 = 13 + a + b [2]

Sub [1] into [2]

0 = 13 + a + 17 + a

0 = 30 + 2a

−30 = 2a

−15 = a

∴ b = 17 − 15

b = 2

3 a x3 − 12 x2 + 17 x + 90 = y

Test x = −2. y = 0

∴ ( x + 2) is a factor2

3 2

3 2

2

2

14 45

12 17 902

2

14 17

14 28

45 90

45 90

0

x x

x x x x

x x

x x

x x

x

x

− +

⎡ − + ++ − ⎢

+⎢⎣

⎡− +− ⎢

− −⎢⎣+⎡

− ⎢+⎣

∴ ( x + 2)( x2 − 14 x + 45)

( x + 2)( x − 9)( x − 5)

b 2 x4 + 7 x3 − 31 x2 + 0 x + 36 = y

Test x = −1 ∴ y = 0

Test x = 2 ∴ y = 0

∴ ( x + 1)( x − 2) are factors

∴ ( x2 − x − 2) is a factor

2

4 3 22

4 3 2

3 2

3 2

2 9 18

2 7 31 0 362

2 2 4

9 27 0

9 9 18

x x

x x x x x x

x x x

x x x

x x x

+ −

⎡ + − + +− − − ⎢

− −⎢⎣

⎡ − +− ⎢

− −⎢⎣

2

2

18 18 36

18 18 36

0

x x

x x

⎡− + +−⎢

− + +⎢⎣

∴ ( x + 1)( x − 2) (2 x2 + 9 x − 18)

( x + 1)( x − 2) (2 x − 3)( x + 6)


18/24

18 | CHAPTER 1 Graphs and polynomials • CHAPTER REVIEW


4 a (−5, 6), (1, −1)

m = 6 1

5 1

+

− −

= 7

6−

∴ y = 7

6

− x + c

Sub in (1, −1) to find c:

−1 = 7

6− × 1 + c

−1 +7

6 = c

1

6 = c

∴ y = 7

6

− x +

1

6

6 y = −7 x + 1

7 x + 6 y − 1 = 0

b 2 x − y + 10 = 0

2 x + 10 = y

Perpendicular ⇒ m = 1

2

−

∴ y = 1

2− x + c

Sub in point (3, 3)

3 = 3

2

− + c

3 +3

2 = c so c =

9

2.

∴ y = 1

2

− x +

9

2

2 y = − x + 9

x + 2 y − 9 = 0

5 y = − x2 − 2 x + 8

y-int when x = 0∴ y = 8

x-int when y = 0

0 = −1( x2 + 2 x − 8)

0 = −1( x + 4)( x − 2)

∴ x = −4 or 2

TP ⇒ y = −1( x2 + 2 x − 8)= −1( x

2 + 2 x + 1 − 1 − 8)

= −1[( x + 1)2 − 9]

= −1( x + 1)2 + 9

TP = (−1, 9)

Domain = R

Range = (−∞, 9]

6 y = 3 x2 + 8 x − 3

x ∈ [−3, 0)

y-int when x = 0

y = −3


0 = 3 x2 + 8 x − 3

0 = (3 x − 1)( x + 3)

∴ x = 1

3 or −3

TP ⇒ y = 28

3 13

x x⎛ ⎞

+ −⎜ ⎟⎝ ⎠

Now x2 + 8

3 x − 1

=

22 8 8 64 36

3 6 36 36 x x

⎛ ⎞⎛ ⎞+ + − −⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

⎝ ⎠

∴ y =

28 100

36 36

x⎡ ⎤⎛ ⎞

+ −⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦

=

28 100

36 12

x⎛ ⎞

+ −⎜ ⎟⎝ ⎠

TP = 8 100

,6 12

− −⎛ ⎞⎜ ⎟⎝ ⎠

= 4 25

,3 3

− −⎛ ⎞⎜ ⎟⎝ ⎠

Domain [−3, 0)

Range:

When x = 3, y = 0 closed end point

When x = 0, y = −3 open end point

25, 0

3

−⎡ ⎤⎢ ⎥⎣ ⎦

7 a f ( x) = − x3 + bx2 + ax − 18

0 = −(−3)3 + b(−3)2 + a × (−3) − 18

0 = 27 + 9b − 3a − 180 = 9 + 9b − 3a

0 = 3 + 3b − a

a = 3 + 3b [1]

g ( x) = ax2 + bx − 75

0 = a(−3)2 + b × (−3) − 75

0 = 9a − 3b − 75

25 = 3a − b [2]

Sub [1] into [2]

25 = 3(3 + 3b) − b

25 = 9 + 9b − b

16 = 8b

2 = b

∴ a = 3 + 6= 9

b f ( x) = − x3 + 2 x2 + 9 x − 18

y-int when x = 0

∴ y = −18

x-int when y = 0

( x + 3) is a factor2

3 2

3 2

2

2

5 6

2 9 183

3

5 9

5 15

Chapter 1 Graphs and Polynomials Solutions

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Transcript of Chapter 1 Graphs and Polynomials Solutions