Review of Essential Skills
and Knowledge Solutions
R–1
1. a)
b)
c)
d)
e)
f)
g)
h)
i)
Copyright © 2009 by Nelson Education Ltd.
j)
k)
l)
m)
Exponent Laws, pp. 540–541
x 2x 7 ⫽ x 9
( y 3 ) 5 ⫽ y 15
(x ⫹ 2) 4 (x ⫹ 2) ⫽ (x ⫹ 2) 5
a9
⫽ a6
a3
y4
1
⫽ 3
y7
y
(x 2y) 2
x 4y 2
⫽
(xy 3 ) 4
x 4y 12
1
⫽ 10
y
80 ⫽ 1
(20 ⫹ 2) 0 ⫽ (1 ⫹ 2) 0
⫽ 30
⫽1
1
4⫺1 ⫽ 1
4
1
⫽
4
⫺1
⫺4⫺5 ⫽ 5
4
1
⫽⫺
1024
1 ⫺1
a b ⫽ 10
10
4 ⫺2
3 2
a b ⫽a b
3
4
9
⫽
16
1
(a 3bc 0 ) ⫺2 ⫽ 3 0 2
(a bc )
1
⫽ 6 2
a b
n) a
x 2y ⫺5
x ⫺2y
b
⫺3
⫺4
⫽ (x 4y ⫺2 ) ⫺4
x 4 ⫺4
b
y2
y2 4
⫽ a 4b
x
y8
⫽ 16
x
3
(y ⫹ 1) (y ⫹ 2) 4
(y ⫹ 2) 3
⫽
(y ⫹ 1) 5 (y ⫹ 2)
(y ⫹ 1) 2
1
1
27⫺3 ⫽ 1
273
1
⫽
3
3
2252 ⫽ 兹153
⫽ 153
⫽ 3375
⫺1000.5 ⫽ ⫺兹100
⫽ ⫺10
1
3
643 ⫻ 162 ⫽ 4 ⫻ 兹163
⫽ 4 ⫻ 43
⫽ 4 ⫻ 64
⫽ 256
⫽a
o)
2. a)
b)
c)
d)
R–2
Expanding and Simplifying
Polynomial Expressions,
pp. 542–543
3x(5x 2 ⫹ 3x ⫺ 4) ⫽ 15x 3 ⫹ 9x 2 ⫺ 12x
b) (2x ⫹ 7) 2 ⫽ (2x) 2 ⫹ 2(2x)(7) ⫹ 72
⫽ 4x 2 ⫹ 28x ⫹ 49
2
c) 3(x ⫺ 4) ⫺ (2 ⫺ x)(2 ⫹ x)
⫽ 3 3 (x) 2 ⫺ 2(x)(4) ⫹ (4) 2 4 ⫺ 3 (2) 2 ⫺ (x) 2 4
⫽ 3(x 2 ⫺ 8x ⫹ 16) ⫺ (4 ⫺ x 2 )
⫽ 3x 2 ⫺ 24x ⫹ 48 ⫺ 4 ⫹ x 2
⫽ 4x 2 ⫺ 24x ⫹ 44
1. a)
Appendix R: Review of Essential Skills and Knowledge
573
⫽ (x 2 )(x 2 ) ⫺ (x) 2 (6x) ⫹ (x 2 )(9) ⫺ (6x)(x 2 )
⫹ (6x)(6x) ⫺ (6x)(9) ⫹ (9)(x 2 ) ⫺ (9)(6x)
⫹ (9)(9)
⫽ x 4 ⫺ 6x 3 ⫹ 9x 2 ⫺ 6x 3 ⫹ 36x 2 ⫺ 54x ⫹ 9x 2
⫺ 54x ⫹ 81
⫽ x 4 ⫺ 12x 3 ⫹ 54x 2 ⫺ 108x ⫹ 81
e)
f)
2. a)
b)
c)
d)
e)
f)
574
R–3
Factoring Polynomial
Expressions, p. 544
6x 2 ⫺ 5x ⫽ x(6x ⫺ 5)
b) 28x ⫺ 14xy ⫽ 14x(2 ⫺ y)
c) x 2 ⫺ x ⫺ 6 ⫽ (x ⫺ 3)(x ⫹ 2)
d) 3y 2 ⫹ 18y ⫹ 24 ⫽ 3(y 2 ⫹ 6y ⫹ 8)
⫽ 3(y ⫹ 2)(y ⫹ 4)
e) x 2 ⫺ 64 ⫽ (x ⫹ 8)(x ⫺ 8)
f ) x 4 ⫺ 81 ⫽ (x 2 ⫹ 9)(x 2 ⫺ 9)
⫽ (x 2 ⫹ 9)(x ⫹ 3)(x ⫺ 3)
2
2. a) 6y ⫺ y ⫺ 2
Multiply 6(⫺2) ⫽ ⫺12
Two numbers whose product is ⫺12 and whose sum is ⫺1
are 3 and ⫺4.
6y 2 ⫺ y ⫺ 2 ⫽ 6y 2 ⫹ 3y ⫺ 4y ⫺ 2
⫽ (6y 2 ⫹ 3y) ⫺ (4y ⫹ 2)
⫽ 3y(2y ⫹ 1) ⫺ 2(2y ⫹ 1)
⫽ (3y ⫺ 2)(2y ⫹ 1)
b) 12x 2 ⫹ x ⫺ 1
Multiply 12(⫺1) ⫽ ⫺12
Two numbers whose product is ⫺12 and whose sum is 1
are ⫺3 and 4.
12x 2 ⫹ x ⫺ 1 ⫽ 12x 2 ⫺ 3x ⫹ 4x ⫺ 1
⫽ (12x 2 ⫺ 3x) ⫹ (4x ⫺ 1)
⫽ 3x(4x ⫺ 1) ⫹ 1(4x ⫺ 1)
⫽ (3x ⫹ 1)(4x ⫺ 1)
c) 5a 2 ⫹ 7a ⫺ 6
Multiply 5(⫺6) ⫽ ⫺30
Two numbers whose product is ⫺30 and whose sum is 7
are ⫺3 and 10.
5a 2 ⫹ 7a ⫺ 6 ⫽ 5a 2 ⫹ 10a ⫺ 3a ⫺ 6
⫽ (5a 2 ⫹ 10a) ⫺ (3a ⫹ 6)
⫽ 5a(a ⫹ 2) ⫺ 3(a ⫹ 2)
⫽ (5a ⫺ 3)(a ⫹ 2)
d) 12x 2 ⫺ 18x ⫺ 12 ⫽ 6(2x 2 ⫺ 3x ⫺ 2)
Multiply 2(⫺2) ⫽ ⫺4
Two numbers whose product is ⫺4 and whose sum is ⫺3
are ⫺4 and 1.
6(2x 2 ⫺ 3x ⫺ 2) ⫽ 6(2x 2 ⫺ 4x ⫹ x ⫺ 2)
⫽ 63 (2x 2 ⫺ 4x) ⫹ (x ⫺ 2) 4
⫽ 632x(x ⫺ 2) ⫹ 1(x ⫺ 2) 4
⫽ 6(2x ⫹ 1)(x ⫺ 2)
1. a)
Appendix R: Review of Essential Skills and Knowledge Solutions
Copyright © 2009 by Nelson Education Ltd.
1
1
b ax ⫹ b
3
2
1
1
1
1
⫽ 6 c (x)(x) ⫺ (x) ⫹ (x) ⫺ a b a b d
3
2
2
3
1
1
⫽ 6ax 2 ⫹ x ⫺ b
6
6
⫽ 6x 2 ⫹ x ⫺ 1
5x(2x ⫺ 4) 2 ⫽ 5x 3 (2x) 2 ⫺ 2(2x)(4) ⫹ (4) 2 4
⫽ 5x(4x 2 ⫺ 16x ⫹ 16)
⫽ 20x 3 ⫺ 80x 2 ⫹ 80x
2
3 2x(x ⫺ 1) 4 ⫽ (2x 2 ⫺ 2x) 2
⫽ (2x 2 ) 2 ⫺ 2(2x 2 )(2x) ⫹ (2x) 2
⫽ 4x 4 ⫺ 8x 3 ⫹ 4x 2
5(x ⫺ 1)(x ⫹ 1)(x ⫹ 2)
⫽ 5(x 2 ⫺ 1)(x ⫹ 2)
⫽ 53 (x 2 )(x) ⫺ (1)(x) ⫹ (2)(x 2 ) ⫺ (1)(2) 4
⫽ 5(x 3 ⫺ x ⫹ 2x 2 ⫺ 2)
⫽ 5x 3 ⫹ 10x 2 ⫺ 5x ⫺ 10
2(x 2 ⫺ x ⫹ 3)(x ⫺ 7)
⫽ 23 (x 2 )(x) ⫺ (x 2 )(7) ⫺ (x)(x) ⫹ (x)(7)
⫹ (3)(x) ⫺ (3)(7) 4
⫽ 2(x 3 ⫺ 7x 2 ⫺ x 2 ⫹ 7x ⫹ 3x ⫺ 21)
⫽ 2(x 3 ⫺ 8x 2 ⫹ 10x ⫺ 21)
⫽ 2x 3 ⫺ 16x 2 ⫹ 20x ⫺ 42
4(x ⫺ 2) 3
⫽ 4(x ⫺ 2) 2 (x ⫺ 2)
⫽ 43 x 2 ⫺ 2(x)(2) ⫹ 22 4 (x ⫺ 2)
⫽ 4(x 2 ⫺ 4x ⫹ 4)(x ⫺ 2)
⫽ 43 (x 2 )(x) ⫺ (x 2 )(2) ⫺ (4x)(x)
⫹ (4x)(2) ⫹ (4)(x) ⫺ (4)(2) 4
⫽ 4(x 3 ⫺ 2x 2 ⫺ 4x 2 ⫹ 8x ⫹ 4x ⫺ 8)
⫽ 4(x 3 ⫺ 6x 2 ⫹ 12x ⫺ 8)
⫽ 4x 3 ⫺ 24x 2 ⫹ 48x ⫺ 32
(x ⫺ 5)(x ⫺ 2)(x ⫹ 5)(x ⫹ 2)
⫽ (x ⫺ 5)(x ⫹ 5)(x ⫺ 2)(x ⫹ 2)
⫽ (x 2 ⫺ (5) 2 )(x 2 ⫺ (2) 2 )
⫽ (x 2 ⫺ 25)(x 2 ⫺ 4)
⫽ (x 2 )(x 2 ) ⫺ (x 2 )(4) ⫺ (25)(x 2 ) ⫹ (25)(4)
⫽ x 4 ⫺ 29x 2 ⫹ 100
(3x ⫺ 4) 2 (2x ⫹ 3)
⫽ 3 (3x) 2 ⫺ 2(3x)(4) ⫹ (4) 2 4 (2x ⫹ 3)
⫽ (9x 2 ⫺ 24x ⫹ 16)(2x ⫹ 3)
⫽ (9x 2 )(2x) ⫹ (9x 2 )(3) ⫺ (24x)(2x) ⫺ (24x)(3)
⫹ (16)(2x) ⫹ (16)(3)
⫽ 18x 3 ⫹ 27x 2 ⫺ 48x 2 ⫺ 72x ⫹ 32x ⫹ 48
⫽ 18x 3 ⫺ 21x 2 ⫺ 40x ⫹ 48
(x ⫺ 3) 4
⫽ (x ⫺ 3) 2 (x ⫺ 3) 2
⫽ 3x 2 ⫺ 2(x)(3) ⫹ (3) 2 4 ⫻ 3x 2 ⫺ 2(x)(3) ⫹ (3) 2 4
⫽ (x 2 ⫺ 6x ⫹ 9)(x 2 ⫺ 6x ⫹ 9)
d) 6ax ⫺
3. (x ⫺ y)(x 2 ⫹ xy ⫹ y 2 )
⫽ (x)(x ) ⫹ (x)(xy) ⫹ (x)(y ) ⫺ (y)(x ) ⫺ (y)(xy)
⫺ (y)(y 2 )
⫽ x 3 ⫹ x 2y ⫹ xy 2 ⫺ yx 2 ⫺ xy 2 ⫺ y 3
⫽ x3 ⫺ y3
2
R–4
2
2
Working with Rational
Expressions, pp. 545–548
x⫽0
b) no restrictions
c) x ⫽ 2
d) x ⫽ ⫺6, 6
2ab
a
⫽
2. a)
4b
2
b⫽0
x(x ⫹ 5)
x 2 ⫹ 5x
b) 2
⫽
(x ⫺ 1)(x ⫹ 5)
x ⫹ 4x ⫺ 5
x
⫽
x⫺1
x ⫽ ⫺5, 1
3h(h ⫹ 2)
3h 2 ⫹ 6h
⫽
c)
2
h ⫹ 4h ⫹ 4
(h ⫹ 2) 2
3h
⫽
h⫹2
h ⫽ ⫺2
b(b 2 ⫺ a 2 )
b 3 ⫺ a 2b
⫽
d) 2
b ⫺ 2ab ⫹ a 2
(b ⫺ a) 2
b(b ⫺ a)(b ⫹ a)
⫽
(b ⫺ a) 2
b(b ⫹ a)
⫽
b⫺a
b⫽a
(x ⫹ 1)(x ⫹ 2)
x 2 ⫹ 3x ⫹ 2
⫽
e)
2
(x ⫹ 3)(x ⫹ 2)
x ⫹ 5x ⫹ 6
x⫹1
⫽
x⫹3
x ⫽ ⫺3, ⫺2
t 2 (t ⫹ 1)
t3 ⫹ t2
⫽
f)
t ⫺ t3
t(1 ⫺ t 2 )
t(t ⫹ 1)
⫽
(1 ⫺ t)(1 ⫹ t)
t
⫽
1⫺t
t ⫽ ⫺1, 0, 1
2y 2
y
6x
⫻
⫽
3. a)
8y
3x
2
x ⫽ 0, y ⫽ 0
b)
c)
Copyright © 2009 by Nelson Education Ltd.
1. a)
d)
e)
f)
2ab
6ac
2ab
10b
⫼
⫽
⫻
5bc
10b
5bc
6ac
2b
⫽ 2
3c
a ⫽ 0, b ⫽ 0, c ⫽ 0
(x ⫹ 1) 2
x 2 ⫺ 2x ⫹ 1
⫻
x 2 ⫹ 2x ⫺ 3
x 2 ⫹ 4x ⫹ 3
2
(x ⫺ 1) 2
(x ⫹ 1)
⫻
⫽
(x ⫺ 1)(x ⫹ 3)
(x ⫹ 1)(x ⫹ 3)
(x ⫹ 1)(x ⫺ 1)
⫽
(x ⫹ 3) 2
x ⫽ ⫺3, ⫺1, 1
x⫺7
2x ⫺ 14
x⫺7
25
⫼
⫽
⫻
10
25
10
2x ⫺ 14
x⫺7
25
⫽
⫻
10
2(x ⫺ 7)
5
⫽
4
x⫽7
x 2 ⫺ 5x ⫹ 6
x 2 ⫺ 4x ⫺ 5
x⫺5
⫻
⫼ 2
2
2
x ⫺1
x ⫺4
x ⫹ 3x ⫹ 2
x 2 ⫺ 5x ⫹ 6
x 2 ⫺ 4x ⫺ 5
x 2 ⫹ 3x ⫹ 2
⫽
⫻
⫻
x⫺5
x2 ⫺ 1
x2 ⫺ 4
(x ⫺ 3)(x ⫺ 2)
(x ⫺ 5)(x ⫹ 1)
⫽
⫻
(x ⫹ 1)(x ⫺ 1)
(x ⫹ 2)(x ⫺ 2)
(x ⫹ 2)(x ⫹ 1)
⫻
x⫺5
(x ⫺ 3)(x ⫹ 1)
⫽
x⫺1
x ⫽ ⫺2, ⫺1, 1, 2, 5
3m 2 ⫺ 7m ⫺ 6
9m ⫺ 6
⫻
2
2
6m ⫹ 3m
2m ⫺ 5m ⫺ 3
9m 2 ⫺ 4
⫼
4m 2 ⫹ 4m ⫹ 1
3m 2 ⫺ 7m ⫺ 6
9m ⫺ 6
⫽
⫻
2
2
6m ⫹ 3m
2m ⫺ 5m ⫺ 3
4m 2 ⫹ 4m ⫹ 1
⫻
9m 2 ⫺ 4
3(3m ⫺ 2)
(3m ⫹ 2)(m ⫺ 3)
⫻
⫽
3m(2m ⫹ 1)
(2m ⫹ 1)(m ⫺ 3)
(2m ⫹ 1) 2
⫻
(3m ⫹ 2)(3m ⫺ 2)
1
⫽
m
2 1
2
m ⫽ ⫺ , ⫺ , 0, , 3
3 2
3
Appendix R: Review of Essential Skills and Knowledge
575
b)
c)
d)
e)
576
4
2
12
10
⫹
⫽
⫹
5x
3x
15x
15x
22
⫽
15x
x⫽0
5(x ⫹ 1)
5
2
⫺
⫽
x⫺1
x⫹1
(x ⫺ 1)(x ⫹ 1)
2(x ⫺ 1)
⫺
(x ⫹ 1)(x ⫺ 1)
5x ⫹ 5 ⫺ 2x ⫹ 2
⫽
(x ⫺ 1)(x ⫹ 1)
3x ⫹ 7
⫽
(x ⫹ 1)(x ⫺ 1)
x ⫽ ⫺1, 1
(2x ⫹ 1)(x ⫺ 2)
2x ⫹ 1
5
15
⫹
⫽
⫹
3
x⫺2
3(x ⫺ 2)
3(x ⫺ 2)
2x 2 ⫺ 3x ⫺ 2 ⫹ 15
⫽
3(x ⫺ 2)
2
2x ⫺ 3x ⫹ 13
⫽
3(x ⫺ 2)
x⫽2
6x
3x
⫺ 2
x 2 ⫺ 5x ⫹ 6
x ⫹ x ⫺ 12
6x
3x
⫽
⫺
(x ⫺ 3)(x ⫺ 2)
(x ⫺ 3)(x ⫹ 4)
6x(x ⫹ 4)
⫽
(x ⫺ 3)(x ⫺ 2)(x ⫹ 4)
3x(x ⫺ 2)
⫺
(x ⫺ 3)(x ⫹ 4)(x ⫺ 2)
6x 2 ⫹ 24x ⫺ 3x 2 ⫹ 6x
⫽
(x ⫺ 3)(x ⫺ 2)(x ⫹ 4)
3x 2 ⫹ 30x
⫽
(x ⫺ 3)(x ⫺ 2)(x ⫹ 4)
x ⫽ ⫺4, 2, 3
x⫹3
x⫺1
⫺
x⫺4
x⫹2
(x ⫹ 3)(x ⫹ 2)
(x ⫺ 1)(x ⫺ 4)
⫽
⫺
(x ⫺ 4)(x ⫹ 2)
(x ⫹ 2)(x ⫺ 4)
2
x ⫹ 5x ⫹ 6
x 2 ⫺ 5x ⫹ 4
⫽
⫺
(x ⫺ 4)(x ⫹ 2)
(x ⫹ 2)(x ⫺ 4)
x 2 ⫹ 5x ⫹ 6 ⫺ x 2 ⫹ 5x ⫺ 4
⫽
(x ⫺ 4)(x ⫹ 2)
10x ⫹ 2
⫽
(x ⫺ 4)(x ⫹ 2)
x ⫽ ⫺2, 4
Appendix R: Review of Essential Skills and Knowledge
x⫹1
x⫹2
⫺ 2
x ⫹ 2x ⫺ 3
x ⫹ 4x ⫺ 5
x⫹1
x⫹2
⫽
⫺
(x ⫹ 3)(x ⫺ 1)
(x ⫹ 5)(x ⫺ 1)
(x ⫹ 1)(x ⫹ 5)
⫽
(x ⫹ 3)(x ⫺ 1)(x ⫹ 5)
(x ⫹ 2)(x ⫹ 3)
⫺
(x ⫹ 5)(x ⫺ 1)(x ⫹ 3)
x 2 ⫹ 6x ⫹ 5
⫽
(x ⫹ 3)(x ⫺ 1)(x ⫹ 5)
x 2 ⫹ 5x ⫹ 6
⫺
(x ⫹ 5)(x ⫺ 1)(x ⫹ 3)
x 2 ⫹ 6x ⫹ 5 ⫺ x 2 ⫺ 5x ⫺ 6
⫽
(x ⫹ 3)(x ⫺ 1)(x ⫹ 5)
x⫺1
⫽
(x ⫹ 3)(x ⫺ 1)(x ⫹ 5)
1
⫽
(x ⫹ 3)(x ⫹ 5)
x ⫽ ⫺5, ⫺3, 1
4x(x ⫺ 3)
2
x⫺3
2
5. 4x ⫹ 1 ⫹
⫽
⫹
⫹
x⫺3
x⫺3
x⫺3
x⫺3
4x 2 ⫺ 12x ⫹ x ⫺ 3 ⫹ 2
⫽
x⫺3
4x 2 ⫺ 11x ⫺ 1
⫽
x⫺3
x⫽3
f)
R–5
2
Slope and Rate of Change of a
Linear Function, pp. 549–550
y2 ⫺ y1
x2 ⫺ x1
⫺9 ⫺ (⫺5)
⫽
⫺4 ⫺ 1
4
⫺4
⫽
⫽
⫺5
5
y2 ⫺ y1
b) m ⫽
x2 ⫺ x1
4⫺4
⫽
7 ⫺ (⫺1)
0
⫽
8
⫽0
1. a)
m⫽
Copyright © 2009 by Nelson Education Ltd.
4. a)
y2 ⫺ y1
x2 ⫺ x1
⫺4 ⫺ (⫺2)
⫽
5⫺5
⫺2
⫽
0
The slope is undefined. (The line is vertical.)
y2 ⫺ y1
d) m ⫽
x2 ⫺ x1
9⫺5
⫽
⫺2 ⫺ (⫺3)
4
⫽ ⫽4
1
2. a) The graph is a vertical line 3 units left of the y-axis.
b) The graph is a horizontal line 6 units above the x-axis.
3. The slope of the height function is 1.2. In this situation,
that means the plant grows 1.2 cm per week.
4. a) 3x ⫹ 5y ⫹ 10 ⫽ 0
5y ⫽ ⫺3x ⫺ 10
3
y⫽⫺ x⫺2
5
Therefore, the slope is ⫺35 and the y-intercept is ⫺2.
b) Ax ⫹ By ⫹ C ⫽ 0
By ⫽ ⫺Ax ⫺ C
A
C
y⫽⫺ x⫺
B
B
c) m ⫽
A
C
Therefore, the slope is ⫺B and the y-intercept is ⫺B .
Copyright © 2009 by Nelson Education Ltd.
R–6
The Zeros of Linear and
Quadratic Functions,
pp. 551–552
y ⫽ ⫺3(x ⫹ 4)
To solve for the zeros, let y ⫽ 0:
0 ⫽ ⫺3(x ⫹ 4)
x ⫽ ⫺4
b) y ⫽ ⫺2(x ⫹ 3)(x ⫺ 7)
To solve for the zeros, let y ⫽ 0:
0 ⫽ ⫺2(x ⫹ 3)(x ⫺ 7)
0 ⫽ x ⫹ 3 or 0 ⫽ x ⫺ 7
x ⫽ ⫺3 or x ⫽ 7
c) y ⫽ 4(x ⫹ 9) 2
To solve for the zeros, let y ⫽ 0:
0 ⫽ 4(x ⫹ 9) 2
x ⫽ ⫺9
1. a)
d) y ⫽ 2x 2 ⫹ x ⫺ 15
y ⫽ (2x ⫺ 5)(x ⫹ 3)
To solve for the zeros, let y ⫽ 0:
0 ⫽ (2x ⫺ 5)(x ⫹ 3)
0 ⫽ 2x ⫺ 5 or 0 ⫽ x ⫹ 3
x ⫽ 2.5 or x ⫽ ⫺3
1
2. a) y ⫽ x ⫺ 2
3
To solve for the zeros, let y ⫽ 0:
1
0⫽ x⫺2
3
1
2⫽ x
3
x⫽6
b) The line with y-intercept 3 and slope 12 has equation
y ⫽ 12x ⫹ 3.
To solve for the zeros, let y ⫽ 0:
1
0⫽ x⫹3
2
1
⫺3 ⫽ x
2
x ⫽ ⫺6
3. a) f (x) ⫽ x 2 ⫹ x ⫺ 42
To solve for the zeros, let f (x) ⫽ 0:
0 ⫽ x 2 ⫹ x ⫺ 42
0 ⫽ (x ⫹ 7)(x ⫺ 6)
0 ⫽ x ⫹ 7 or 0 ⫽ x ⫺ 6
x ⫽ ⫺7 or x ⫽ 6
b) y ⫽ 16x 2 ⫺ 8x ⫹ 1
To solve for the zeros, let y ⫽ 0:
0 ⫽ 16x 2 ⫺ 8x ⫹ 1
0 ⫽ (4x ⫺ 1) 2
0 ⫽ 4x ⫺ 1
1
x⫽
4
c) g(x) ⫽ x 2 ⫺ 3x ⫹ 1
To solve for the zeros, let g(x) ⫽ 0:
0 ⫽ x 2 ⫺ 3x ⫹ 1
Use the quadratic formula to solve for x:
⫺(⫺3) ⫾ 兹(⫺3) 2 ⫺ 4(1)(1)
x⫽
2(1)
3 ⫾ 兹5
x⫽
2
Appendix R: Review of Essential Skills and Knowledge
577
R–7
1. a)
Exponential Functions, p. 553
x
⫺2
⫺1
0
1
2
b)
x
⫺2
⫺1
0
1
2
578
y ⴝ 3x
1
9
1
3
1
3
9
y ⴝ 10x
1
100
1
10
1
10
100
8
6
4
2
y
x
–4 –3 –2 –1 0
–2
8
6
4
2
–4 –3 –2 –1 0
–2
1
2 3
y
x
1
2
Appendix R: Review of Essential Skills and Knowledge
c)
1 x
yⴝa b
2
x
⫺2
⫺1
0
1
2
d)
x
⫺2
⫺1
0
1
2
8
6
4
2
4
2
1
1
2
1
4
x
–3 –2 –1 0
–2
y ⴝ 1.5x
4
9
2
3
1
3
2
9
4
y
2.5
2.0
1.5
1.0
0.5
–2
0
–1
–0.5
1
2 3
y
x
1
2
1 x
y ⫽ 3x
3
The graphs are mirror images in the y-axis. The y-values of
the points of each graph are the reciprocals of the other.
3. D ⫽ R
R ⫽ {y苸R冷 y ⬎ 0 }
y-intercept ⫽ 1
no x-intercept
asymptote: y ⫽ 0
4. T ⫽ 20 ⫹ 76(0.92) t
a) T ⫽ 20 ⫹ 76(0.92) 0
T ⫽ 96 ⬚C
b) T ⫽ 20 ⫹ 76(0.92) 10
T ⫽ 53 ⬚C
c) T ⫽ 20 ⫹ 76(0.92) 60
T ⫽ 20.5 ⬚C
d) The equation of the horizontal asymptote is T ⫽ 20.
It means the coffee is approaching 20 ⬚C (room
temperature).
e) 76 ⬚C is the difference between the initial temperature
of the coffee and the temperature of the room.
2. y ⫽ a b
Copyright © 2009 by Nelson Education Ltd.
d) y ⫽ 3x 2 ⫹ 5x ⫹ 4
To solve for the zeros, let y ⫽ 0:
0 ⫽ 3x 2 ⫹ 5x ⫹ 4
Use the quadratic formula to solve for x:
⫺5 ⫾ 兹(5) 2 ⫺ 4(3)(4)
x⫽
2(3)
⫺5 ⫾ 兹⫺23
x⫽
6
Since the square root of ⫺23 is not a real number, the
function has no zeros.
4. Since the function is quadratic and has zeros of ⫺2 and 2,
we know it is of the form y ⫽ C(x ⫹ 2)(x ⫺ 2) where C
is a constant. Expanding the right side of the equation
yields y ⫽ Cx 2 ⫺ 4C. To find C, substitute the
coordinates of the y-intercept, (0, 8), in for x and y:
8 ⫽ ⫺4C
C ⫽ ⫺2
Therefore the equation of the function is
y ⫽ ⫺2(x ⫹ 2)(x ⫺ 2).
5. a)
y ⫽ 16x 2 ⫺ 40x ⫹ 25
2
b ⫺ 4ac ⫽ (⫺40) 2 ⫺ 4(16)(25)
⫽0
The discriminant is zero, so there is only 1 zero.
b)
y ⫽ 16x 2 ⫺ 40x ⫹ 23
2
b ⫺ 4ac ⫽ (⫺40) 2 ⫺ 4(16)(23)
⫽ 128
The discriminant is positive, so there are two zeros.
R–8
Transformations of Functions,
pp. 554–556
3. A vertical stretch of factor 3 and a reflection in the x-axis
1. a)
Comparing the transformed function with the general
form y ⫽ af (k(x ⫺ d )) ⫹ c, we have a ⫽ 3, k ⫽ 1,
d ⫽ 0, So the transformations involved are a vertical
stretch of factor 3 and a vertical translation 2 units
down.
b) Comparing the transformed function with the general
form y ⫽ af (k(x ⫺ d )) ⫹ c, we have a ⫽ 1, k ⫽ 12,
d ⫽ ⫺3, c ⫽ 0. So the transformations involved are a
1
⫽ 2 and a horizontal
horizontal stretch of factor
Copyright © 2009 by Nelson Education Ltd.
A12B
translation 3 units to the left.
c) Comparing the transformed function with the general
form y ⫽ af (k(x ⫺ d )) ⫹ c, we have a ⫽ 1, k ⫽ 2,
d ⫽ 0, c ⫽ 7. So the transformations involved are a
horizontal compression of factor 12 and a vertical
translation 7 units up.
d) Comparing the transformed function with the general
form y ⫽ af (k(x ⫺ d )) ⫹ c, we have a ⫽ ⫺3,
k ⫽ 2, d ⫽ 1, c ⫽ ⫺2. So the transformations
involved are a vertical stretch of factor 3, a horizontal
compression of factor 12, a reflection in the x-axis, a
horizontal translation 1 unit to the right, and a
vertical translation 2 units down.
e) Comparing the transformed function with the general
form y ⫽ af (k(x ⫺ d )) ⫹ c, we have a ⫽ ⫺1,
k ⫽ ⫺1, d ⫽ 0, c ⫽ 4. So the transformations
involved are a reflection in the x-axis, a reflection in
the y-axis, and a vertical translation 4 units up.
f ) Comparing the transformed function with the general
form y ⫽ af (k(x ⫺ d )) ⫹ c, we have a ⫽ ⫺15,
k ⫽ ⫺1, d ⫽ 0, c ⫽ ⫺3. So the transformations
involved are a vertical compression of factor 15, a
reflection in the x-axis, a reflection in the y-axis, and a
vertical translation 3 units down.
2. a) This transformation is a horizontal compression of
factor 13, so the new point is A23, 5B .
b) The transformations involved are a vertical stretch of
factor 2 and a reflection in the x-axis, so the new point
is (2, ⫺10).
c) The transformation involved is a horizontal
translation 4 units to the right. So the new point
is (6, 5).
d) The transformation involved is a vertical translation
7 units up. So the new point is (2, 12).
means that a ⫽ ⫺3, a horizontal translation 3 units to the
right means that d ⫽ 3, and a vertical translation 2 units
up means that c ⫽ 2. k ⫽ 1 since it was unspecified, so
the new equation is f (x) ⫽ ⫺3(x ⫺ 3) 2 ⫹ 2.
4. a)
x
f (x)
⫺2
⫺1
0
1
2
⫺8
⫺1
0
1
8
b) Comparing the transformed function with the general
form y ⫽ af (k(x ⫺ d )) ⫹ c, we have a ⫽ 12, k ⫽ 1,
d ⫽ 4, c ⫽ 5. So the transformations involved are a
vertical compression of factor 21, a horizontal
translation 4 units to the right, and a vertical
translation 5 units up.
c)
Point on f Point on g
(⫺2, ⫺8)
(⫺1, ⫺1)
(0, 0)
(1, 1)
(2, 8)
(2, 1)
a3,
9
b
2
(4, 5)
a5,
11
b
2
(6, 9)
5. Y2 can be rewritten as Y2 ⫽ 兹⫺(x ⫺ 4), which shows
that Y2 is the image of Y1 under a reflection in the y-axis
and a horizontal translation 4 units to the right.
R–9
Families of Functions,
pp. 557–558
1. a)
All non-vertical lines can be written in the form
y ⫽ mx ⫹ b; those that have slope 3 are therefore of
the form y ⫽ 3x ⫹ b.
b) Substituting (4, 7) into the equation from part a)
gives
7 ⫽ 3(4) ⫹ b
b ⫽ ⫺5
So the equation of the line is y ⫽ 3x ⫺ 5.
2. a) The quadratic function y ⫽ (x ⫺ 2)(x ⫺ 4) has zeros
at the required points. Multiplying this equation by any
non-zero factor a will not change the zeros of the
function, so the equation of the family is
y ⫽ a(x ⫺ 2)(x ⫺ 4).
Appendix R: Review of Essential Skills and Knowledge
579
4.
5.
6.
R–10 Trigonometric Ratios and
Special Angles, pp. 559–561
1. a) Using exact values, cos u ⫽
1
2
and sin u ⫽ 兹3
2 , so
1 2
兹3 2
cos u ⫹ sin u ⫽ a b ⫹ a
b
2
2
1
3
⫽ ⫹ ⫽1
4
4
2
580
2
Appendix R: Review of Essential Skills and Knowledge
兹2
b) Using exact values, cos u ⫽ 兹2
2 and sin u ⫽ 2
兹2 2
兹2 2
b ⫹a
b
2
2
2
2
⫽ ⫹ ⫽1
4
4
cos2 u ⫹ sin2 u ⫽ a
2. a)
2.0
1.5
1.0
0.5
0
–1
–0.5
y
x
u
1
2
In the triangle drawn on the graph, x ⫽ 1 and y ⫽ 1,
so tan u ⫽ 1.
This means that u ⫽ 45⬚.
b)
y
4
3
2
1
x
u
0
–2
–1
1
–1
In the triangle drawn on the graph, the leg adjacent
to the angle u has length 1 and the leg opposite u
opposite
has length 3. Since tan u ⫽ adjacent , tan u ⫽ 3.
This means that u ⫽ tan ⫺1 (3) ⬟ 71.6⬚.
opposite
opposite
Since cos u ⫽ hypotenuse and sin u ⫽ hypotenuse, they
will have equal values when the adjacent leg and
opposite leg to the angle are the same length. This
happens only in an isosceles right triangle, which gives
one solution of u ⫽ 45⬚. cos u and sin u have the same
sign only in quadrants I and III, so the only other
place where they are the same is the corresponding
value of u in quadrant III, which is u ⫽ 225⬚.
y
b) By definition, sin u ⫽ r , so sin u ⫽ 1 means that
y ⫽ r. Since x 2 ⫹ y 2 ⫽ r 2, x ⫽ 0. The following
diagram illustrates this situation.
3. a)
(0, 1)
y
y=r=1
90°
x=0
x
Copyright © 2009 by Nelson Education Ltd.
3.
b) Since the y-intercept is ⫺4, the function passes
through the point (0, ⫺4). Substituting this into the
equation from part a) gives
⫺4 ⫽ a(⫺2)(⫺4)
1
a⫽⫺
2
So the desired equation is y ⫽ ⫺12 (x ⫺ 2)(x ⫺ 4).
a) When x ⫽ 0, the value of kx is 0 and does not change
no matter what k is. The value of y at this point
is 20 ⫽ 1, so all the members of this family meet
at (0, 1).
b) The parameter k determines the horizontal
stretch/compression factor and whether the graph is
reflected in the y-axis.
3
c) Substituting x ⫽ 4 into the equation y ⫽ 24x gives
3
y ⫽ 24 ⫻4 ⫽ 8, so (4, 8) is a point on
this curve.
The family of quadratic functions that have vertex (⫺2, 5)
is described by the general equation y ⫽ a(x ⫹ 2) 2 ⫹ 5.
The value of a can be determined by substituting in the
point (1, 8):
8 ⫽ a(1 ⫹ 2) 2 ⫹ 5
1
a⫽
3
So the equation of the desired quadratic is
y ⫽ 13 (x ⫹ 2) 2 ⫹ 5.
The family of quadratic equations that have x-intercepts at
5 and ⫺1 are described by the general equation
y ⫽ a(x ⫺ 5)(x ⫹ 1). The value of a can be determined
by substituting in the point (7, ⫺40):
⫺40 ⫽ a(7 ⫺ 5)(7 ⫹ 1)
5
a⫽⫺
2
So the equation of the desired quadratic is
y ⫽ ⫺52 (x ⫺ 5)(x ⫹ 1).
The given information states that
3 ⫽ f (2) ⫽ a(2) 2 ⫺ 6(2) ⫺ 7
11
a⫽
2
So the quadratic function has equation
f (x) ⫽ 112x 2 ⫺ 6x ⫺ 7.
The CAST rule states that sine values are negative in
the third and fourth quadrants, so the only value of u
for which sin u ⫽ ⫺1 is u ⫽ 270⬚.
4. a) The following diagram shows that a 135⬚ angle is
related to a 45⬚ acute angle.
d) The following diagram shows that a 300⬚ angle is
related to a 60⬚ acute angle.
y
y
x
300°
60°
135°
45°
x
So 冷 csc 300⬚ 冷 ⫽ csc 60⬚ ⫽
R–11 Graphing y ⴝ sin x and
y ⴝ cos x, p. 562
y
1. Examine the graph:
1.5
1.0
0.5
x
Copyright © 2009 by Nelson Education Ltd.
30°
So 冷 cos 210⬚ 冷 ⫽ cos 30⬚ ⫽ 兹3
2 .
Since 135⬚ is in quadrant III, where cosine is negative,
cos 210⬚ ⫽ ⫺兹3
2 .
c) The following diagram shows that a 225⬚ angle is
related to a 45⬚ acute angle.
y
225°
⫽ 2兹3
3 .
Since 300⬚ is in quadrant III, where cosecant is
negative, csc 300⬚ ⫽ ⫺2兹3
3 .
So 冷 sin 135⬚ 冷 ⫽ sin 45⬚ ⫽ 兹2
2 .
Since 135⬚ is in quadrant II, where sine is positive,
sin 135⬚ ⫽ 兹2
2 .
b) The following diagram shows that a 210⬚ angle is
related to a 30⬚ acute angle.
210°
1
兹3
2
x
45°
So 冷 tan 225⬚ 冷 ⫽ tan 45⬚ ⫽ 1
Since 225⬚ is in quadrant III, where tangent is
positive, tan 225⬚ ⫽ 1.
–360° –180° 0
–0.5
–1.0
–1.5
y
x
180°
360°
The maximum values occur at x ⫽ ⫺270⬚, 90⬚, 450⬚, etc.
The general equation for such points is
x ⫽ 90⬚ ⫹ k(360⬚), where k is any integer.
2. Examine the graph:
y
1.5
1.0
0.5
x
0
–360° –180°
180° 360°
–0.5
–1.0
–1.5
The graphs intersect at x ⫽ ⫺315⬚, ⫺135⬚, 45⬚, 225⬚, etc.
The general equation for such points is
x ⫽ 45⬚ ⫹ k(180⬚), where k is any integer.
Appendix R: Review of Essential Skills and Knowledge
581
i)
ii)
iii)
iv)
b) i)
ii)
iii)
iv)
Quadrant I values of x are from 0⬚ to 90⬚.
Quadrant II values of x are from 90⬚ to 180⬚.
Quadrant III values of x are from 180⬚ to 270⬚.
Quadrant IV values of x are from 270⬚ to 360⬚.
The graph is increasing from 0 to 1.
The graph is decreasing from 1 to 0.
The graph is decreasing from 0 to ⫺1.
The graph is increasing from ⫺1 to 0.
c) The parent function is y ⫽ sin x. Comparing the
transformed function with the general form
y ⫽ a cos (k(x ⫺ d )) ⫹ c, we have a ⫽ 12, k ⫽ ⫺23,
d ⫽ 60⬚, c ⫽ 0. So the transformations involved are a
vertical compression of factor 12, a horizontal stretch of
1
factor 2 ⫽ 32, a reflection in the y-axis, and a horizontal
3
translation 60⬚ to the right. The amplitude of the new
360⬚
function is 12, its period is 2 ⫽ 540⬚, its axis is y ⫽ 0,
R–12 Transformations of
Trigonometric Functions,
pp. 564–566
3
and there is a horizontal translation of 60⬚ to the right.
y
0.5
1. a) The parent function is f (x) ⫽ sin x. Comparing the
transformed function with the general form
y ⫽ a sin (k(x ⫺ d )) ⫹ c, we have a ⫽ 4, k ⫽ 2,
d ⫽ 0, c ⫽ 4. So the transformations involved are a
vertical stretch of factor 4, a horizontal compression of
factor 12, and a vertical translation 4 units up. The
amplitude of the new function is 4, its period is
360⬚
2 ⫽ 180⬚, its axis is y ⫽ 4, and there is no horizontal
translation.
y
8
6
4
2
x
0
–270° –90°
90° 270°
–2
b) The parent function is f (x) ⫽ cos x. Rewriting the
function so that it is in the correct form:
f (x) ⫽ ⫺cos (3(x ⫺ 30⬚)) ⫺ 2. Comparing this
transformed function with the general form
y ⫽ a cos (k(x ⫺ d )) ⫹ c, we have a ⫽ ⫺1, k ⫽ 3,
d ⫽ 30⬚, c ⫽ ⫺2. So the transformations involved are
a reflection in the x-axis, a horizontal compression of
factor 13, a horizontal translation 30⬚ to the right, and a
vertical translation 2 units down. The amplitude of the
new function is 1, its period is 360⬚
3 ⫽ 120⬚, its axis is
y ⫽ ⫺2, and there is a horizontal translation of 30⬚ to
the right.
y
1
x
0
–270° –90° 90° 270°
–1
–2
–3
–4
582
Appendix R: Review of Essential Skills and Knowledge
–540°–360°–180° 0
x
180° 360° 540°
–0.5
R–13 Solving Trigonometric
Equations in Degrees,
pp. 567–569
1. a)
The value of 兹3
2 is recognizable as a special value from
the 30⬚⫺60⬚⫺90⬚ special triangle.
y
( 3, 1)
2
1 x
30°
3 –1
2
330°
( 3, –1)
x
adjacent
Since cos u ⫽ r ⫽ hypotenuse, we can see that u ⫽ 30⬚
is a solution.
The CAST rule states that cosine values are positive
for first quadrant and fourth quadrant angles. The
angle in quadrant IV is 360⬚ ⫺ 30⬚ ⫽ 330⬚. There are
no other values of u in the domain 30⬚, 360⬚4, so the
solutions are u ⫽ 30⬚ and u ⫽ 330⬚.
Copyright © 2009 by Nelson Education Ltd.
3. a)
y
y
b) By definition, tan u ⫽ x, so we have x ⫽ 25 ⫽ ⫺2
⫺5 . The
points (x, y) ⫽ (5, 2) and (⫺5, ⫺2) can therefore be
used as the points on the terminal arm.
y
(5, 2)
202°
x
22°
(–5, –2)
The CAST rule states that sine values are negative in
the third and fourth quadrants, so the only value of u
for which sin u ⫽ ⫺1 is u ⫽ 270⬚.
e) sin⫺1 (0.554) ⬟ 33.6⬚, so the related acute angle is
about 34⬚. The CAST rule states that sine values are
negative for angles in the third and fourth quadrants:
y
214°
x
326°
A B ⬟ 21.8⬚, so the related angle is about 22⬚.
tan⫺1 25
The CAST rule states that tangent values are positive
for first quadrant and third quadrant angles.
The angle in quadrant III is u ⫽ 180⬚ ⫹ 22⬚ ⫽ 202⬚.
There are no other values of u in the domain 30⬚, 360⬚4 ,
so the solutions are u ⫽ 22⬚ and u ⫽ 202⬚.
c) The value of 1 is recognizable as a special value from
the 45⬚⫺45⬚⫺90⬚ special triangle.
y
(1, 1)
–1
–1
2
225°
x
45°
1
2
(–1, –1)
y
Copyright © 2009 by Nelson Education Ltd.
1
opposite
Since tan u ⫽ x ⫽ adjacent , we can see that u ⫽ 45⬚ is
a solution.
The CAST rule states that tangent values are positive
for first quadrant and third quadrant angles. The
angle in quadrant III is u ⫽ 180⬚ ⫹ 45⬚ ⫽ 225⬚.
There are no other values of u in the domain
30⬚, 360⬚4, so the solutions are u ⫽ 45⬚ and
u ⫽ 225⬚.
The angle in quadrant III is
u ⫽ 180⬚ ⫹ 34⬚ ⫽ 214⬚.
The angle in quadrant IV is
u ⫽ 360⬚ ⫺ 34⬚ ⫽ 326⬚.
Therefore, the two solutions in the required domain
are 214⬚ and 326⬚.
x
adjacent
f ) By definition, cos u ⫽ r ⫽ hypotenuse, so
adjacent
hypotenuse ⫽ 1.5. This is impossible, however, because
a leg of a right triangle cannot be longer than the
hypotenuse. Therefore, this equation has no real
solutions.
1
2. a) Solving this equation for cos u gives cos u ⫽ 2. The
1
value of 2 is recognizable as a special value from the
30⬚⫺60⬚⫺90⬚ special triangle.
y
(1, 3 )
2
60°
1
300°
2
3
x
– 3
(1, – 3 )
y
d) By definition, sin u ⫽ r , so sin u ⫽ 1 means that
y ⫽ r. Since x 2 ⫹ y 2 ⫽ r 2, x ⫽ 0. The following
diagram illustrates this situation:
y
(0, 1)
y=r=1
90°
x=0
x
x
adjacent
Since cos u ⫽ r ⫽ hypotenuse, we can see that u ⫽ 60⬚
is a solution.
The CAST rule states that cosine values are positive
for first quadrant and fourth quadrant angles. The
angle in quadrant IV is 360⬚ ⫺ 60⬚ ⫽ 300⬚. There
are no other values of u in the domain 30⬚, 360⬚4 , so
the solutions are u ⫽ 60⬚ and u ⫽ 300⬚.
Appendix R: Review of Essential Skills and Knowledge
583
( – 3 , 1)
2
1
– 3
150°
330°
x
The CAST rule states that cosine values are positive in
the first and fourth quadrants, so the values of u for
which cos u ⫽ 1 are u ⫽ 0⬚ and
u ⫽ 360⬚.
R–14 Proving Trigonometric
Identities, pp. 570–571
1. a)
2
LS ⫽
( 3, –1)
1
, u ⫽ 30⬚ is the related acute angle
Since tan 30⬚ ⫽ 兹3
to the desired angles. The CAST rule states that
tangent values are negative for second quadrant and
fourth quadrant angles.
The angle in quadrant II is u ⫽ 180⬚ ⫺ 30⬚ ⫽ 150⬚.
The angle in quadrant IV is u ⫽ 360⬚ ⫺ 30⬚ ⫽ 330⬚.
There are no other values of u in the domain 30⬚, 360⬚4 ,
so the solutions are u ⫽ 150⬚ and u ⫽ 330⬚.
c) Solving this equation for sin u gives sin u ⫽ ⫺14. Since
sin⫺1 A14B ⬟ 14.4⬚, the related acute angle is about 14⬚.
The CAST rule states that sine values are negative for
angles in the third and fourth quadrants:
y
194°
b)
x
x=r=1
584
LS ⫽ sin4 u ⫺ cos4 u
⫽ (sin2 u ⫺ cos2 u)
⫻ (sin2 u ⫹ cos2 u)
⫽ (sin2 u ⫺ cos2 u)(1)
⫽ sin2 u ⫺ cos2 u
Appendix R: Review of Essential Skills and Knowledge
RS ⫽ sin2 u ⫺ cos2 u
Since the left side and right side are equal, the
equation is an identity.
c)
346°
y = 0 0°
RS ⫽
Since the left side and right side are equal, the
equation is an identity. However, the restriction
cos u ⫽ 0 is necessary to avoid either side being
undefined.
LS ⫽ 1 ⫺ cos2 u
⫽ sin2 u
x
The angle in quadrant III is u ⫽ 180⬚ ⫹ 14⬚ ⫽ 194⬚.
The angle in quadrant IV is u ⫽ 360⬚ ⫺ 14⬚ ⫽ 346⬚.
Therefore, the two solutiobn in the required domain
are 194⬚ and 346⬚.
d) Solving this equation for cos u gives cos u ⫽ 1. By
x
definition, cos u ⫽ r , so cos u ⫽ 1 means that x ⫽ r.
Since x 2 ⫹ y 2 ⫽ r 2, y ⫽ 0. The following diagram
illustrates this situation.
y
1 ⫹ sin u
cos u
1
sin u
⫽
⫹
cos u
cos u
1
⫹ tan u
cos u
sin u
1
⫹
⫽
cos u
cos u
–1
RS ⫽ sin u cos u tan u
sin u
⫽ sin u cos ua
b
cos u
⫽ sin2 u
Since the left side and right side are equal, the
equation is an identity. However, the restriction
cos u ⫽ 0 is necessary to avoid the right side being
undefined.
d)
tan2 u
1 ⫹ tan2 u
tan2 u
⫽
sec2 u
sin2 u
cos2 u
⫽
⫻
1
cos2 u
⫽ sin2 u
LS ⫽
RS ⫽ sin2 u
Since the left side and right side are equal, the
equation is an identity. However, the restriction
cos u ⫽ 0 is necessary to avoid the left side being
undefined.
Copyright © 2009 by Nelson Education Ltd.
b) Solving this equation for tan u gives tan u ⫽ ⫺兹3
3 ⫽
1
1
⫺兹3
. The value of 兹3
is recognizable as a special value
from the 30⬚⫺60⬚⫺90⬚ special triangle.
y
2.
LS ⫽ sin2 x a1 ⫹
1
b
tan2 x
cos2 x
⫽ sin2 x a1 ⫹
b
sin2 x
⫽ sin2 x ⫹ cos2 x
3.
RS ⫽ 1
⫽ cos2 x ⫹ sin2 x
RS ⫽ 2 sin2 x ⫺ sin4 x
Since the left side and right side are equal, the equation is
an identity.
Copyright © 2009 by Nelson Education Ltd.
Since the left side and right side are equal, the equation is
an identity. However, the restriction cos x ⫽ 0 is necessary
to avoid the left side being undefined.
LS ⫽ (1 ⫺ cos2 x)
⫻ (1 ⫹ cos2 x)
⫽ (sin2 x)
⫻ (1 ⫹ (1 ⫺ sin2 x))
⫽ 2 sin2 x ⫺ sin4 x
Appendix R: Review of Essential Skills and Knowledge
585