BLM Extra Practice KEY - Ms Jeong...

BLM 1–4 Section 1.1 Extra Practice

1. a) arithmetic; t1 = 4, d = 3; 16, 19, 22 b) arithmetic; t1 = 12, d = –5; –8, –13, –18 c) not arithmetic d) not arithmetic e) arithmetic; t1 = x, d = 2; x + 8, x + 10, x + 12 2. a) –5, –7, –9, –11 b) 10, 9.5, 9, 8.5

c) 3, 3 + x, 3 + 2x, 3 + 3x d) 7 8 9 10, , , 3 3 3 3

3. a) 10, 7, 4, 1

BLM 1–10 (continued)

b) 1 14 , 5, 5 , 62 2

4. a) tn = 4n + 2; t50 = 202 b) tn = 7 12 2

− n ;

t50 =1212

−

5. a) 77 b) 26

6. a) 4, 8 , 12 , 16 b) 10 , 8, 6 , 4 , 2

c) 20, 14 , 8 , 2 , 4 , 10− −

7. t1 = 12, tn = 5n + 7, t40 = 207 8. a) t1 = –15, d = 4, tn = 4n – 19 b) t1 = 93, d = –3, tn = 96 – 3n

9. 10 25 23

; , 8,3 3 3

=x

10. a) 15, 18 b) tn = 3n + 3 c) 63 asterisks d) 41st diagram


1. a) –936 b) 232.5 c) 252.5 or 1

2522

2. a) 378 b) 0 c) 400x 3. a) 15 b) 25 c) 21

4. a) t12 = –41, S12 = –228 b) t12 = 47

5, S12 = 60

5. a) 413 b) 95 3

6. 71 071 7. 2850 8. t1 = 8, t9 = 40 9. a) S1 = 7, S2 = 20, S3 = 39, S4 = 64, S5 = 95 b) t1 = 7, t2 = 13, t3 = 19, t4 = 25, t5 = 31 c) S5 = 3(5)2 + 4(5) = 95 10. 6 + 11 + 16 + ⋯ + t20 = $1070. Therefore, the arithmetic series method pays more money.


1. a) geometric, r = 3, tn = 11(3)n – 1

b) not geometric c) geometric, r = 2, tn = 1

3(2)n – 1

d) geometric, r = 0.4, tn = (0.5)(0.4)n – 1

2. a) 7, –21, 63, –189 b) –8, –4, –2, –1 c) 3, 1.8, 1.08, 0.648 d) –4, 16, –64, 256 3. a) 10 b) 14 c) 7 d) 12 4. a) tn = 2(7)n – 1 b) tn = 6(–3) n – 1

c) tn = 7(4)n – 1 d) tn =

11

40964

n−

5. a) 126, 882 b) 4

3, 12, 36 c) ±10, 20, ±40

6. 4

7. a) t1 = 9 × 1010, r = ±0.01,

tn = (9 × 1010 )(±0.01)n – 1

b) t1 = –48, r = –6, tn = (–48)(–6)n – 1

c) t1 = 1.75, r = ±2, tn = (1.75)(±2) n – 1

d) t1 = ±6, r = ±0.5, tn = (6)(±0.5) n – 1

8. a) x = 2 b) y = 610

or 35

9. 384 10. a) $211 200, $185 856, $163 553 b) tn = 240 000(0.88) n – 1, tn = value of digger, in dollars, n – 1 = years since purchase c) $98 082 d) 6 years


1. a) geometric series, the common ratio is 1.2 b) geometric series, the common ratio is –0.2

c) geometric series, the common ratio is 23

d) not geometric, no common ratio

2. a) t1 = 0.43, r = 0.01, S6 = 4399

b) t1 = 5, r = –1, S10 = 0

c) t1 = –100, r = –0.5, S7 = 107516

−

3. a) 232.05 b) –4092 c) 15516

− d) 12 285

4. a) 531 440 b) 4095 c) 3367128

5. a) 1.2 b) 3 6. a) 6 b) 9 7. 1916.25 8. 4 9. a) 10, 30, 90, 270 b) 12, 6, 3, 1.5 10. 94.2 m


1. a) convergent b) convergent c) convergent d) divergent

2. a) –20 b) 6 c) does not exist d) 5

4 e) 24 f)

8

7−

3. a) 2 3

63 63 63 7 + + + =

100 11(100) (100)⋯

b) 5 5 5 41

7.4 + + + + = 7100 1000 10 000 90

⋯

c)

2 3 4

456 456 456 41 1110.123 + + + + =

333 000(1000) (1000) (1000)⋯

4. 21

2 5.

1

2− 6.

3

4π 7. 14 m 8. |x| < 1

9. 2

3 10. a) 125 761 m3 b) 480 000 m3


1. a) Example: b) Example:

c) Example: d) Example:

2. a) 55° b) 25° c) 75° d) 5° 3. a) 140°, 220°, 320° b) 108°, 252°, 288° c) 92°, 268°, 272° d) 177°, 183°, 357° 4. a) 150° b) 225° c) 300°


5. a) No b) No c) Yes d) No

6. a) a = 10, b = 20 3 b) DE = 2 3 m – 2 2 m

7. 12 3 cm


1. a) b)

c) d)

2. a) sin θ =5

26; cos θ =

1

26; tan θ = 5

b) sin θ = –3

5; cos θ =

4

5; tan θ =

–3

4

c) sin θ = 12

13; cos θ =

–5

13; tan θ =

12

–5

d) sin θ = 0; cos θ = 1; tan θ = 0

3. a) sin θ = 1

2; cos θ =

–1

2; tan θ = –1

b) sin θ = – 3

2; cos θ =

–1

2; tan θ = 3

c) sin θ = –1

2; cos θ =

3

2; tan θ =

–1

3

4. a) positive b) negative c) negative d) negative

5. a) cos θ = –4

5; tan θ =

3

4

b) sin θ = – 5

3; tan θ =

– 5

2

c) sin θ = 5

13; cos θ =

–12

13

6. a) 225°, 315° b) 30°, 210° c) 30°, 330° d) 270° 7. a) 51°, 129° b) 144°, 216° c) 138°, 318° d) 260°, 280°

8. a) False. Sin 120° is in quadrant II so it is positive, and cos 210° is in quadrant III so it is negative. b) False. Cos 170° is in quadrant II so it is negative, and cos 350° is in quadrant IV so it is positive.

c) True. The reference angle for both sin 200° and sin 340° is 20°. Both are negative. d) True. The reference angles are not equal, but both ratios, cos 300° and sin 150° are equal to 0.5. Both are positive since the cosine ratio is positive in quadrant IV and the sine ratio is positive in quadrant II.

BLM 2–6 Section 2.3 #27 Concept Map

Example:


1. a) 4.0 cm b) 5.3 m 2. a) 43° b) 125° 3. a) 31° b) 6.4 cm

4. a) ∠F = 105°; DF = 8.3 cm; EF = 11.7 cm

b) ∠N = 77°; ∠M = 43°; NO = 6.3 m 5. a) no solution

b) ∠Q = 7°; ∠R = 70°; PR = 1.6 cm

c) First triangle: ∠F = 59°; ∠E = 81°; DF = 9.2 cm

Second triangle: ∠F = 121°; ∠E = 19°; DF = 3.0 cm

d) First triangle: ∠T = 77°; ∠S = 38°; RT = 2.7 mm

Second triangle: ∠T = 103°; ∠S = 12°; RT = 0.9 mm 6. 435 cm2


1. a) 7.9 mm b) 7.3 m 2. a) 73° b) 20° 3. a) 7 cm b) 40°

4. a) 5.9 m; ∠E = 85°; ∠F = 52°

b) ∠G = 119°; ∠H = 35°; ∠I = 26°

5. a) 21 cm b) 45 –18 2 cm 6. 17°


1. a) The graph can be obtained by applying a change in width about the x-axis by a factor of 3. The graph opens upward, has a minimum y-value of 0, and the range is y ≥ 0. b) The graph can be obtained by applying a change in width about the x-axis by a factor of 5, and then a reflection in the x-axis. The graph opens downward, has a maximum y-value of 0, and the range is y ≤ 0. c) The graph can be obtained by applying a vertical translation up 8 units. The graph opens upward, has a minimum y-value of 8, and the range is y ≥ 8. d) The graph can be obtained by applying a vertical translation down 5 units. The graph opens upward, has a minimum y-value of –5, and the range is y ≥ –5. 2. a) The graph of y = (x – 6)2 can be obtained from y = x2 by applying a horizontal translation 6 units to the right.

vertex: (6, 0); axis of symmetry: x = 6;

domain: x ∈ R; range: y ≥ 0; x-intercept: x = 6; y-intercept: y = 36

b) The graph of y = (x + 1)2 can be obtained by applying a horizontal translation 1 unit to the left.

vertex: (–1, 0); axis of symmetry: x = –1;

domain: x ∈ R; range: y ≥ 0; x-intercept: x = –1; y-intercept: y = 1 c) The graph of y = (x + 4)2 + 3 can be obtained by applying a horizontal translation 4 units to the left and a vertical translation 3 units up.

vertex: (–4, 3); axis of symmetry: x = –4;

domain: x ∈ R; range: y ≥ 3; x-intercept: none; y-intercept: y = 19


d) The graph of y = (x – 2)2 – 1 can be obtained by applying a horizontal translation 2 units to the right and a vertical translation 1 unit down.

vertex: (2, –1); axis of symmetry: x = 2;

domain: x ∈ R; range: y ≥ –1; x-intercepts: x = 1 and 3; y-intercept: y = 3 3. a) y = 0.5x

2 b) y = –0.5x

2 c) y = 0.5(x + 6)2

d) y = 0.5x2 – 3

4. a) The graph can be obtained from the graph of f (x) = x2 by applying a horizontal translation 7 units to the left, and a vertical translation 3 units down. b) The graph can be obtained from the graph of f (x) = x2 by applying a change in width about the x-axis by a factor of 2, a reflection in the x-axis, and a vertical translation 5 units up. c) The graph can be obtained from the graph of f (x) = x2 by applying a change in width about the

x-axis by a factor of 1

,3

a reflection in the x-axis, and

a horizontal translation 3 units to the right. d) The graph can be obtained from the graph of f (x) = x2 by applying a change in width about the x-axis by a factor of 4, a horizontal translation 2 units to the left, and a vertical translation 1 unit down.

5. a) b) c) d)

Vertex (5, 1) (–2, 0) (–4, –5) (0, 3)

Axis of

symmetry x = 5 x = –2 x = –4 x = 0

Direction upward downward upward downward

Max/min min y = 1 max y = 0 min y = –5 max y = 3

Domain x ∈ R x ∈ R x ∈ R x ∈ R

Range y ≥ 1 y ≤ 0 y ≥ –5 y ≤ 3

Number of

x-intercepts 0 1 2 2

6. a) y = 3(x – 2)2 b) y = –2(x + 2)2 + 3

c) y = 1

2(x – 3)2 – 2 d) y = –1(x – 4)2 + 1

7. a) f (x) = –2(x – 5)2 b) f (x) = 2

3(x – 2)2 – 6


1. a) Yes. The function fits the standard form of a quadratic function with a = 1, b = –15, and c = 0. b) y = x2 – 16 Yes. The function fits the standard form of a quadratic function with a = 1, b = 0, and c = –16. c) Yes. The function fits the standard form of a quadratic function with a = –4.9, b = 0, and c = 400. d) No. The function does not fit the standard form of a quadratic function.

2. a) b)

Vertex (–1, –4) (–1, 9)

Axis of symmetry x = –1 x = –1

x-intercepts –3 and 1 –4 and 2

y-intercept –3 8

Direction upward downward

Max/min min y = –4 max y = 9

Domain x ∈ R x ∈ R

Range y ≥ –4 y ≤ 9

3. a) y = x2 + 14x + 39 b) f (x) = –6x2 – 3x + 30

c) h(t) = –9t2 – 18t + 41 d) y = 8x

2 + 26x + 15

4. a) b)

c) d)

a) b) c) d)

Vertex (4, –1) (–4, 9) (2, –4) (1, 5)

Axis of

symmetry x = 4 x = –4 x = 2 t = 1

x-intercepts 3 and 5 –1 and –7 0 and 4 0 and 2

y-intercept 15 –7 0 0

Direction upward downward upward downward

Max/min min: –1 max: 9 min: –4 max: 5

Domain x ∈ R x ∈ R x ∈ R t ∈ R

Range y ≥ –1 f (x) ≤ 9 y ≥ –4 h(t) ≤ 5


5. a) w = width; 2 width + length = 200 m of fencing, so length = 200 – 2w

b) A(w) = w(200 – 2w) or A(w) = –2w2 + 200w

c)

d) 5000 m2 e) 50 m by 100 m

6. a)

b) 100 m; this represents the initial height of the projectile c) 21.9 s; this represents the time that the projectile is in the air d) 651.25 m; occurs at 10.5 s


1. a) 25; (x – 5)2 b) 16; (x + 4)2 c) 36; (x – 6)2 d) 1; (x + 1)2 2. a) y = (x + 1)2 – 5; (–1, –5) b) y = (x – 3)2 + 4; (3, 4) c) y = (x + 4)2 – 10; (–4, –10) d) y = (x + 12)2 – 90; (–12, –90) 3. a) y = 3(x – 2)2 + 1 b) y = –2(x + 5)2 – 6 c) y = 6(x – 4)2 – 96 d) y = –4(x + 7)2 4. a) y = (x + 3)2 – 5; min of –5 when x = –3

b) y = 2(x – 4)2 – 1; min of –1 when x = 4

c) y = –3(x + 2)2 + 5; d) y = –1(x – 9)2 + 81;

max of 5 when x = –2 max of 81 when x = 9

5. a) b) c) d)

Vertex (–5, –9) (–1, 6) (–7.5, 4.5)

1 2,

3 3

Axis of

symmetry x = –5 x = –1 x = –7.5 x =1

3

Max/min min y = –9 max y = 6 min y = 4.5 min y =

2

3

Domain x ∈ R x ∈ R x ∈ R x ∈ R

Range y ≥ –9 y ≤ 6 y ≥ 4.5 y ≥

2

3

6. a) R(x) = (1200 + 100x)(6.00 – 0.30x) b) 4 weeks; $7680 c) Example: Assume that yield increases will remain constant at 100 bushels per week; assume price will decrease at 30¢ each week.

2 675

122x

− =


1. a) 2 b) none c) 2 d) 1 2. a) –3, 2 b) no real roots c) – 8.2, 1.2 d) 3 3. a) no solution

b) –2

c) 0, –5

d) 2, –2

4. a) 1.1, –3.5 b) –3.9, 3.9 c) no solution d) –2.8, 1.8 5. Example: a) –10, 15 b) – 20, 20 c) – 0.7, 0.1

d) no solution 6. a) m = 16 b) m < 16 c) m > 16 7. 4.5 s 8. 5 cm, 12 cm, 13 cm


1. a) (x + 4)(x – 5) b) 3(x – 3)(x – 7)

c) –4(x + 1)(x + 2) d) 12

( 3)( 4)x x+ −

2. a) (2x – 1)(7x + 5) b) (x + 5)(3x – 4) c) (4x + 3y)(x + y) d) (2x – 3)(3x – 4) 3. a) 4(3x + 2y)(x – y) b) 3y(2x + 5)(x + 2) c) 10(7x – 5y)(2x – 5y) d) 7x(3x + y)(2x + 3y) 4. a) (x – 7y)(x + 7y) b) (5x – 3)(5x + 3)

c) 5

2x y

+

5

2x y

−

or 14

(2 5 )(2 5 )x y x y+ −

d) 16(x – 3) 5. a) (x + 4)(x – 8) b) (6x + 7)(4x – 3) c) 2(7x + 4)(7x – 3) d) (2x

2 + 3)(x2 – 3)

6. a) –3, 5 b) 4, –8 c) 3, 6 d) 5±

7. a) 1

2− ,

4

3 b) 5,

1

7− c)

1

5− , 2 d)

3

2, –6

8. a) 13

8,

13

8− b)

7

3,

7

3− c)

1

4 ,

1

4− d) 8, –10

9. a) –1, 2

3 b)

1

2, 4 c)

1

3− ,

1

2 d) 6,

7

2−

10. a) 1

3− b)

3

2 c)

5

2− d)

4

7


1. a) 36 b) 100 c) 49

4 d)

4

25

2. a) (x + 3)2 = 5 b) (x – 4)2 = 11 c) d) (x + 5)2 = 33 3. a) –1, 9 b) 0, –1 c) 0.9, –0.7 d) –7.5, –6.5

4. a) 31− ± b) 5 13

2

± c) 0.2, –0.8 d)

3

7

5. a) 3

4, –1 b)

2

31− ± c) 62 2− ± d) 1, 5

6. a) –0.21, 4.71 b) –0.26, 1.26 c) –9.47, –0.53 d) –0.88, 0.38 7. 6, 16


1. a) two real roots b) no real roots c) one real root d) no real roots 2. a) none b) 1 c) 2 d) 2

3. a) 5 ± 2 b) 7 3

2

± c)

2

3 d) 0,

3

2

4. a) 0.50, 0.33 b) no solution c) –0.59, 2.26 d) –4.46, 1.12


5. a) 52− ± b) 1 2 2

2

± c)

5 3

4

− ± d) 72 ±

6. a) No solution;

b) 0, –7; Factor method: can be factored quickly because x is a common factor

c) 5

;2

− Factor method: a perfect square trinomial

d) 34 ;− ± Complete the square method: already in

the form (x + a)2 = b

e) 1 7

6;

− ±Quadratic formula: exact values are

required for the answer

7. a) –3 b) 1

2−

BLM 4–8 Chapter 4 Review #22

( )

2 2

2 2

2

2

2

2

2

2

2

2

4 4

4

2

4

2 4

4

2

b c

a a

b b b c

a aa a

b b ac

a a

b b ac

a a

b b ac

a

ax bx c

x x

x x

x

x

x

2

−

4

−

− ± −

+ = −

+ = −

+ + = −

+ =

+ = ±

=

Subtract c from both sides. Divide both sides by a. Complete the square. Factor the perfect square trinomial. Take the square root of both sides. Solve for x.


1. a) 3 6 b) 5 14 c) 7 2x d) 211 3x y xy

2. a) 80 b) 6877 c) 581x d) 2 3175x y

3. a) 14, 2 3, 3 5, 50− − b) 2, 18, 32, 6 2

c)5

25 3, 60, 3 , 16− − − − d)

12, 20, 2 6, 4 3

4. a)12 11 b) 5 3 5 2x − c)16 8 6+

d)10 5 4 7 y−

5. a) 21 5d b) 6 10e c)13 3 d) 7 4 3− −

6. a) 2 ,x x ≥ 0 b) 24 ,x x x ≥ 0

c) 2 ,x x x− − x ≥ 0 d) 27 ,x y x ≥ 0

7. a) 32 7 b) 232x x y c) 34 5− d) −2x

8. 9 2 3 6+


1. a)12 15 b) 5 2 c) 21

22

d) 9x6

2. a) 20 5 2+ b)15 3 3 5−

c) 2 2 4+ d) 6 3 2 , 0x x x− ≥

3. a) 6 4 3+ b)146 23 7+ c) −49 d) 93 24 15+

4. a) 2 3, 0x x x− − ≥ b) x2 − 5

c) 2 3 2,x x+ − x ≥ 0 d) 4 4 1, 0x x x+ + ≥


5. a) 2 b) 6 c)2

3

x d)

2 3

4

x

6. a)30

3 b)

4 30

5 c)

5 3

3 d)

30

14

7. a)

3 3

3

− b)

2 2 3

3

+

c) 2 d)

2 2x x

x

+

8. a) 2 3 2− b)

1 5

4

− −

c) 30 3 5− − d)

8 3 2 10 12 30

19

+ − −


1. a) x ≥ −3, x = 46 b) x ≥ 0,16

5x =

c)5 5

3 3,x x≤ = d) x ≤ 0, x = −288

2. a) x = 28, x ≥ 0 b) y = 0, y ∈ R

c) v = 8, v ≥ −1 d) x = 17, x ≥152

−

3. a) m = 1, m ≤ 43

b) x = −1, 5; x = −1, x ≥ 1

c) n =9 17

2

+, n ≥ 0 d) x = 2 5,− + x ∈ R

4. a) x = 8, x ≥ 32

b) y = −1, y = 5; y = −1, y ≥ 1

c) no solution d) p = 3, p ≥32

5. a) w =9

4, w ≥ 0 b) x = 0, x = 16; x ≥ 0

c) y = 4, y ≥ 0 d) x = 6, x ≥ 5

6. a) x = 169, x ≠ 0 b) x = 8, x ≠ 0, 2

7. The value x = −5 is extraneous. x ≥ −6

8. 90.6 m


1. a) multiplication; (xy2) b) multiplication; (x + 5)

c) division; (a) d) division; (x − 2)

2. a) x ≠ −3 b) x ≠ 0, y ≠ 0 c) x ≠ −4, 5

d)5

3

−≠x e) a ≠ 0, 3 f ) m ≠ −2, −3

3. a)3

,5−x

x ≠ ±5 b)2

,5

+x

x x ≠ 0, 7

c)3

,3( 3)

+

−

x

x x ≠ ±3 d)

3 1 1

3 1 4 3, ,

−

+≠

x

xx

e)5( 1) 1

2(2 1) 2, , 5

+ −

+≠

x

xx f )

4,

4+

xy

x x ≠ −4, y ≠ 9

4. a)3

,5

r

st r ≠ 0, s ≠ 0, t ≠ 0 b)

3,

2+x x ≠ ± 2

c) ,1+

c

c c ≠ −1, d ≠ 0 d)

1,

4

− c ≠ 7m

e) ,5−

x

x x ≠ −1, 5 f )

1,

+y

y y ≠ 0, 3

5. a)3 1

,4

−

+

x

x x ≠ −4 b)

1,

4(1 )

+

−

a

a a ≠ ±1

c)2 1

,5( 3)

+

−

x

x x ≠ −2, 3 d)

4 1

2 3, 2,

−

−≠

x

xx

e)4 ( 2) 3

3 2 2, , 2

t t

tt

− + −

+≠ f )

25 1 5

2 5 3 2, ,

−

−≠

x

xx

6.1

4 1

x

x

+

+ does not reduce. The non-permissible values

are 1 4

.4 3

, −

≠x


1. a)3

,2

y

x x ≠ 0, y ≠ 0 b)

2

,2( )+

x y

x y x ≠ −y

c)2

,4

+x x ≠ 3 d) x, x ≠ −1, ± 6

2. a)4

,2

−

+

x

x x ≠ ±2 b)

5,

3+y y ≠ ±5, 1, −3

c) x,5

23, 2,

−≠ ±x d)

5(3 1) 5 1

4(3 1) 2 3, , 4,

+

−≠ ± −

x

xx

3. a)3

,a

bc a ≠ 0, b ≠ 0, c ≠ 0 b)

1 5

3 3, , 3

+ −

+≠ −

x

xx

c)3

,4a

a ≠ 0, 4 d)2

2( 3),

3

+x

x x ≠ 0, 3

4. x ≠ −5, −4, −1, 1, 3; The non-permissible values

for the original expression are −5, −4, and 3. The non-permissible values for the reciprocal of the

second term are −1 and 1.

5. a) 4(a + 2), a ≠ 0, ±2, b ≠ 0

b)2 4

2( 1) 5, , 0, 1

− −

−≠

x

xx c)

7,

1

+

+

x

x x ≠ −6, −1, 7

d) 1 − 3y, y ≠ ± 3,1

3

−

6. a)4

,5( 1)( 1)− +

x

x x x ≠ ±1, 0

b)( 2)( 12)

,12 ( 3)

+ −

−

x x

x x x ≠ ±2, −12, 0, 3

c)( 3)

,1

− −

−

x x

x x ≠ ±3, 0, 1, −9 d)

1,

4 x ≠ ±4, −1, 0

7. a) The width is x − 1. b) 1

,2

1≠ − ≠x x

c) The non-permissible values are x ≤ 1 because a value of x of 1 would result in a length of zero, and any smaller value would result in a negative length.


8. a) 2

2

6 9

3

x x

x x

+ +

− b)

3x

x

−

c) Example: Both the product and quotient share two

non-permissible values: x ≠ 0 and 3. This is because both have the same original expressions. To determine the quotient you must multiply the reciprocal of the divisor, so neither the numerator nor denominator of the divisor can equal zero. Therefore,

the quotient has a third non-permissible value: −3.


1. a) 20xy b) (x + 4)(3x + 1) c) (x + 6)(x − 6)

2. a)5 4

,3

−x

x x ≠ 0 b)

23 1,

5

+ +

+

x x

x x ≠ −5

c) 4

,2−x

x ≠ ± 2

3. a)2

2

3 15 28,

21

+ +a a

a a ≠ 0

b)2

2

4 15 5,

5

− − −xy y x x

xy x ≠ 0, y ≠ 0

c)

2

2

56 35 20 7,

35

+ + −x y x

xy x ≠ 0, y ≠ 0

4. a)5 11

,( 5)( 7)

+

− +

x

x x x ≠ −7, 5

b)2 9

,7 ( 3)

+

+

xy x

y y y ≠ 0, −3

c)24 2 1

,( 1)( 1)

− −

+ −

x x

x x x ≠ ± 1

d)22 13 6

,( 2)( 5)( 6)

x x

x x x

+ −

+ − + x ≠ −2, −6, 5

5. a)3

,3

−

−a a ≠ ±3 b)

3 (3 1),

( 2)( 2)( 3)

−

− + +

y y

y y y y ≠ ± 2, −3

c)2

2( 8),

( 2)( 2)

−

− +

x

x x x ≠ ±2 d)

2,

( 4)( 1)− −x x x ≠ 1, 4, 7

6. a)5

,3( 6)−x

x ≠ −1, 6 b)2( 3)

,( 1)( 1)

+

− +

x

x x x ≠ ±1, 7

c) 2 5 22

,( 3)( 2)( 5)

3, 2,5x x

xx x x

+ −≠

+ + −− −

d) 2 19 6

,( 3)( 5)

5,3x x

xx x

+ −≠

− +−

7. a)1

,1

+

−

x

x x ≠ 0, 1 b) ,

3( 4)−

x

x x ≠ 3, 4

c)1

,(4 )(4)

−

+ h h ≠ 0, −4


1. a) 2 b)3

, 02

≠a c)9

, 02

≠x d)1

,6

−2

e)3

,4

−2, x ≠ 0 f ) −6, x ≠ 0

2. a)3

,4

−=x x ≠ −3 b)

3,

5= −x x ≠ 0, 1

c)9

,2

=x 1, x ≠ 0, 3 d) x ≠ ± 2, x = 0

3. a) x = −2,3

5

−≠x b) x = − 5, x ≠ 3, 5

c)4

,7

−=x x ≠ 2, −4, d)

1, 2,

2x = − x ≠ 0, −1

4. a) x = −1, x ≠ ±3 b) x = 7, x ≠ −1, 2

c) x = −1, x ≠ −2, 3 d)2

,3

4, =x3

2, 2≠ −x

5. a) x = ±1.27,3

2

−≠x b) x = −0.71, 4.21, x ≠ 0

c) no solution,3 2

2 30, ,

−≠x

d) x = −20.44, 0.44, x ≠ ±5 6. a) No. 1 is a non-permissible value in the second

term, so x ≠ 1.

b)3

7

−=x

7. 3 and 9 or −6 and 18 8. 60 km/h and 90 km/h


1. a) 42 b) 82

3 c) 3.75 d)

5

61

2. a) | − 3 | , | − 3.9 | , | − 4 | , | − 4.1| , | − 4.5 |

b) 6 6 6 6 6

, , , ,10 25 20 15 5

− − − −

3. a) | − 2.1 | ,3

4,− | − 0.6 | ,

5

3,− − | 1.2 |

b) 46 1 2 1

2 23 46 23, , , 2 , 23

− − − −

4. a) 14 b) 32 c) 13 d) − 2.4

5. a) 16 b) − 6.75 c) 10

3 d) 49

6. a) 8 b) 16 c) 9 d) 8

7. a) 1.5 b) 4 c) 2 d) 3


1. a) x y

0 1

2 0

4 1

6 2

8 3

b) x y

−4 8

−2 0

0 0

2 8

4 24

2. a)

b)


3. a)

x-intercept: 1

2, 0 ;

−

y-intercept: (0, 1);

domain: {x | x ∈ R}; range: {y | y ≥ 0, y ∈ R}

b)

x-intercept: (− 4, 0); y-intercept: (0, 4);


4. a)

x-intercepts: (−5, 0) and (−1, 0); y-intercept: (0, 5);


b)

x-intercepts: 1

2,0

−

and (3, 0); y-intercept: (0, 3);


5. a)

1

5

1

5

5 1, if

5 1, if

+ ≥ −

= − − < −

x x

y

x x

b)

1

2

1

2

4, if 8

4, if 8

−+ ≤

= − >

x x

y

x x

c) 2

2

2( 2) 8, if 4 or 0

2( 2) 8, if 4 < 0

+ − ≤ − ≥=

− + + − <

x x xy

x x

d) 2( 3)( 1), if 3 1

2( 3)( 1), if 3 or 1

− + − − ≤ ≤=

+ − < − >

x x xy

x x x x

6. a) h(x) and k(x) b) all

c) g(x), h(x), and k(x) d) all

7. a) all points where x > 3 b) (0, 0)

c) all points where −4 < x < 0 d) all points


1. a) x = −3 or x = 1 b) no solution

c) x = 5

2± d) x = 0

2. a) yes b) no c) yes d) yes

3. a) x = 1

4 b) no solution c) x > 5 d) n = 8

4. a) x = 1 ± 2 and x = 1

b) x = 4 and x = −1

c) x = 2 and x = −8

d) x = 1 ± 7

2, x =

1

2, and x =

3

2

5. a) x = −5 or x = 5

b) x = 3

2,

− x = −1, x =

1

2,

− and x = 3

c) x = 2 and x = 6.25 d) x = 1 2 2± and x = 1

6. a) yes b) no c) yes d) no

7. a) not possible b) k = 0, k > 4

c) k = 4 d) 0 < k < 4

8. Mark’s solution is incorrect. 0 = (x + 4)(x − 3);

x = −4 or x = 3

9. a) Rearrange the equation 2

| 2 | 0− + − =x

x to

2| 2 | .

xx− + = The graph

2( ) =

xf x is the right side

and g(x) = | −x2 + 2 | is the left side. f (x) = g(x) at the

points of intersection. The intersection points are the solutions to the equation.

b) The solutions are 1.19 and 1.69.



1.

Function i) Reciprocal ii) Domain iii) Range

a) y = x + 4 {x | x ∈ R} {y | y ∈ R}

1

4xy

+= {x | x ≠ −4, x ∈ R} {y | y ≠ 0, y ∈ R}

b) y = 3x − 9 {x | x∈ R} {y | y ∈ R}

1

3 9xy

−= {x | x ≠ 3, x ∈ R} {y | y ≠ 0, y ∈ R}

c) y = (x + 2)(x − 2) {x | x ∈ R} {y | y ≥ −4, y ∈ R}

1

( 2)( 2)x xy

+ −= {x | x ≠ ±2, x ∈ R} {y | y ≠ 0, y ∈ R}

d) y = x2 + 6x + 9 {x | x ∈ R} {y | y ≥ 0, y ∈ R}

2

1

6 9x xy

+ += {x | x ≠ 3, x ∈ R} {y | y ≥ 0, y ∈ R}

2.

i) Zeros ii) Reciprocal

iii) Non-permissible

Values

iv) Vertical

Asymptote

a) x = −3 1

3 xy

+=

x ≠ −3 x = −3

b) 1

2x =

1

2 1xy

−=

1

2x ≠

1

2x =

c) x = −2

x = 3

1

( 2)( 3)x xy

+ −=

x ≠ −2

x ≠ 3

x = −2

x = 3

d) x = −1

x = −5 2

1

2 12 10x xy

− − −=

x ≠ −1

x ≠ −5

x = −1

x = −5

3. a) x = 5 b) 2

7x =

c) x = −1, x = −0.5 d) x = −4, x = 3

4. There are no x-intercepts, only y-intercepts.

a) 1

5y = b)

1

3y = c)

1

3y

−= d)

1

12y =

5.

Reciprocal

Horizontal

Asymptote

Vertical

Asymptotes Invariant Points Intercepts

a) 1

2xy

+=

y = 0 x = −2 (−1, 1) and (−3, −1) 1

2y =

b) 1

3xy =

y = 0 x = 0 1 1

3 3, 1 and , 1

− −

none

c) 2

1

9xy

−=

y = 0 x = −3

x = 3

(3.16, 1), (−3.16, 1),

(2.83, −1), and (−2.83, −1)

1

9y

−=

d)

2

1

1( )xy

+=

y = 0 x = −1 (−2, 1) and (0, 1) y = 1

BLM 7–9


1. Point (1, −3):

LS = x2 − 4x − y RS = 0

= (1)2 − 4(1) − (−3)

= 0

LS = RS

LS = x − y − 4 RS = 0

= 1 − (−3) − 4

= 0

LS = RS

Therefore, point (1, −3) is a solution.


Point (4, 0):

LS = x2 − 4x − y RS = 0

= (4)2 − 4(4) − (0)

= 0

LS = RS

LS = x − y − 4 RS = 0

= 4 − (0) − 4

= 0

LS = RS Therefore, point (4, 0) is a solution.

2. a) (−2, −4) and (0, 0);

y = x2 + 4x

y = −x2

b) (−1, 2) and (−4, 8);

y = 2x2 + 8x + 8

y = −2x

3. a) (2, −3)

b) (3, −4) and (7, 0)

c) no solution

d) (−1, 8) and (0, 1)

4. a) (−2.50, −1.75)

b) (1.00, 2.00) and (9.00, 154.00)

c) (−0.50, 11.25) and (1.67, 2.22)

d) no solution


5. 78 items, or $195

6. a) x = Max’s age: x + y = 35

y = father’s age: x2 + 5 = y

b)

The two solutions to the system are (−6, 41) and

(5, 30). (−6, 41) is not meaningful because Max

cannot be −6 years old. c) Max is 5 and his father is 30.


1. Point (−1, 11):

LS = 2x + y RS = 9

= 2(−1) + 11

=9

LS = RS

LS = 2x2 − 4x − y RS = −5

= 2(−1)2 − 4(−1) − 11

= −5

LS = RS

Therefore, (−1, 11) is a solution. Point (2, 5):

LS = 2x + y RS = 9

= 2(2) + 5

= 9

LS = RS

LS = 2x2 − 4x − y RS = −5

= 2(2)2 − 4(2) − 5

= −5

LS = RS Therefore, (2, 5) is a solution.

2. Point (−1, −4):

LS = y RS = x2 + 2x − 3

= −4 = (−1)2 + 2(−1) − 3

= −4

LS = RS

LS = y RS = −x2 − 2x − 5

= −4 = −(−1)2 − 2(−1) − 5

= −4

LS = RS

Therefore, (−1, −4) is a solution.

3. a) (3, 7) and (4, 9)

Verify: Point (3, 7)

LS = y RS = 2(3) + 1

= 7 = 7

LS = RS

LS = y RS = (3)2 − 5(3) + 13

= 7 = 7

LS = RS Therefore, (3, 7) is a solution. Point (4, 9)

LS = y RS = 2(4) + 1

= 9 = 9

LS = RS

LS = y RS = (4)2 − 5(4) + 13

= 9 = 9


b) 3 17

2 2,

−

and (2, −2)

Verify:

Point 3 17

2 2,

−

( ) 17322

9 17

2 2

LS (3) 4 RS 0

4

0

LS RS

−

−

= + − =

= + −

=

=

( ) ( ) ( )2

3 3 172 2 2

18 24 34 84 4 44

LS (2) (4) 2 RS 0

0

LS RS

− −

− −

= − − − =

= +

=

=

Therefore, 3 17

2 2,

−

is a solution.


Point (2, −2):

LS = 3(2) + (−2) − 4 RS = 0

= 0

LS = RS

LS = 2(2)2 − 4(2) − (−2) − 2 RS = 0

= 0

LS = RS

Therefore, (2, −2) is a solution.

c) (−4, 10) and (2, 4) Verify:

Point (−4, 10)

LS = y RS = −(−4)2 − 3(−4) + 14

= 10 = 10

LS = RS

LS = y RS = 3(−4)2 + 5(−4) − 18

= 10 = 10

LS = RS

Therefore, (−4, 10) is a solution. Point (2, 4)

LS = y RS = −(2)2 − 3(2) + 14

= 4 = 4

LS = RS

LS = y RS = 3(2)2 + 5(2) − 18

= 4 = 4


d) (5, 0) and (−2, 7) Verify: Point (5, 0)

LS = 4(5) + 0 + 5 RS = 52

= 25 = 25

LS = RS

LS = 52 RS = 5(5) + 2(0)

= 25 = 25


Point (−2, 7)

LS = 4(−2) + 7 + 5 RS = (−2)2

= 4 = 4

LS = RS

LS = (−2)2 RS = 5(−2) + 2(7)

= 4 = 4

LS = RS

Therefore, (−2, 7) is a solution.

4. a) ( )103

1,− and ( )1 263 9

, b) no solution

c) (0, 2) and (3, 1.5) d) ( )1 634 16

,−

and (5, 0)

5. a) (3, 18)

b) (−1.62, −0.21) and (0.62, 0.54)

6. a) k = 7 b) (0, −7)

7. a) k > − 4 b) k = − 4 c) k < − 4

8. a) y = −1(x + 4)2 + 4 and y = (x − 1)2 − 9

b) (−2, 0) and (−1, −5)

9. a) perimeter: 2(3x) + 2(x + 5) = y;

area: (3x)(x + 5) = 3y

b) (5, 50) and (−2, −6) c) The only possible solution is (5, 50). You cannot have a negative perimeter or area.

d) x = 5; perimeter = 50; area = 150 units2

BLM 9–4 Section 9.1 Extra Practice 1. a) B and C b) D c) A, B, C, and D

2. a) i) y ≤2

7x − 2, m =

2

7, y-intercept: −2

ii) solid line

iii)

b) i) y <1

3x −

5

3, m =

1

3, y-intercept:

5

3−

ii) broken line

iii)

c) i) y > −4, m = 0, y-intercept: −4

ii) broken line

iii)

d) i) y ≤5

2− x + 2, m =

5

2− , y-intercept: 2

ii) solid line

iii)

3. a) i) x-intercept:5

2− , y-intercept: 5

ii) broken line

iii)


b) i) x-intercept: 25, y-intercept: −5 ii) solid line iii)

c) i) x-intercept: −2, y-intercept: −6 ii) broken line

iii)

d) i) x-intercept: −5, y-intercept: none ii) broken line iii)

4. a)

b)

c)

d)

5. a) 0.80a + 1.25m ≤ 10.00, where a is the number of apples and m is the number of muffins

b) The number of apples and the number of muffins must be an integer greater than or equal to zero, or

{a | a ≥ 0, a ∈ I} and {m | m ≥ 0, m ∈ I}.

c)

d) You cannot buy 0.8 of a muffin.

6. a) y ≤ 1

2x + 2 b) y ≥ 1 c) y < −2x − 3

BLM 9–5 Section 9.2 Extra Practice 1. a) x = 1, x = −7 b) x < −7 or x > 1 c) −7 < x < 1

2. a) −2, −3 b) −3 < x < −2 c) x < −3 or x > −2 3. a) yes b) yes c) no d) yes

4. a) x < −5 or x > 1 b) −1 ≤ x ≤ 3

c) −1 ≤ x ≤3

2 d) 1 − 3 ≤ x ≤ 1 + 3

5. a) x < 2 or x >9

4 b)

5

4− ≤ x ≤

3

2

c) no solution d) x =3

2

6. a) 2 − 7 < x < 2 + 7 b) 3 29

2

−≤ x ≤

3 29

2

+

c) x ≤ −2 − 10 or x ≥ −2 + 10

d) 11 93

2

− − ≤ x ≤

11 93

2

− +

7. a) x = 0 or x = −6 b) x < −6 or x > 0

c) −5 ≤ x ≤ −1


1. a) B and D b) A and B c) B and C d) B and C

2. a) y ≤ (x − 3)(x + 2) b) y ≥ x2 + 8x + 12

c) y ≤ −2(x + 1)2 + 5 d) y ≥ 2x2 − 3x + 4

3. a)

b)


c)

d)

4. a)

b)

c)

d)

5. 10x2 + 9y − 90 < 0 or y <

10

9− x

2 + 10

(continued)

a) b)

c)

d)

6.

7.

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