261 Lesson 5.2 Algebra 2 Worked-Out Solution Key

Chapter 5,
continued
5.2 Guided Practice (pp. 338–341)
Lesson 5.2
Investigating Algebra Activity 5.2 (p. 336)
1.
1. The function is a polynomial function written as
f (x) 5 22x 1 13 in standard form. It has degree 1
(linear) and a leading coefficient of 22.
As x approaches 2`,
f(x) approaches 2`.
As x approaches 1`,
f (x) approaches 1`.
2. The function is not a polynomial function because the
term 5x 22 has an exponent that is not a whole number.
3. The function is a polynomial function written as
h(x) 5 6x 2 2 3x 1 : in standard form. It has a degree
of 2 (quadratic) and a leading coefficient of 6.
2.
As x approaches 2`,
4. f (x) 5 x 4 1 2x 3 1 3x 227
f(x) approaches 1`.
f (22) 5 (22)4 1 2(22)3 1 3(22)2 2 7
As x approaches 1`,
5 16 2 16 1 12 2 7
f (x) approaches 2`.
55
5. g(x) 5 x 3 2 5x 2 1 6x 1 1
3.
g(4) 5 (4)3 2 5(4)2 1 6(4) 1 1
As x approaches 2`,
5 64 2 80 1 24 1 1
f(x) approaches 1`.
59
As x approaches 1`,
f(x) approaches 1`.
4.
6. 2
As x approaches 2`,
5
f(x) approaches 2`.
As x approaches 1`,
f (x) approaches 2`.
3
21
7
10
26
50
13
25
57
5
f (2) 5 57
7. 21
22
21
0
4
25
2
21
1
25
1
21
5
210
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
5. a. f (x) 5 x n, where x is odd: As x approaches 1`,
f (x) approaches 1`. As x approaches 2`,
f (x) approaches 2`.
b. f (x) 5 2x n, where n is odd: As x approaches 2`,
f (x) approaches 1`. As x approaches 1 `,
f (x) approaches 2`.
22
f (21) 5 210
8. The degree is odd and the leading coefficient is negative.
9.
c. f (x) 5 x n, where n is even : As x approaches 2`,
f (x) approaches 1`. As x approaches 1 `,
f (x) approaches 1`.
x
23
22
21
0
1
2
y
132
37
4
23
4
37 132
2
y
d. f (x) 5 2x n, where x is odd : As x approaches 2`,
f (x) approaches 2`. As x approaches 1 `,
f (x) approaches 2`.
1
21
6. As x approaches 2`, f (x) approaches 1`. As x
approaches 1`, f (x) approaches 1`. f(x) is going to
resemble its highest power. Because x 6 is the highest
power, the end behavior will be the same as f (x) 5 x 6.
3
10.
x
x
23
22
21
0
1
y
32
9
0
21
0
3
23 216
y
1
22
x
Algebra 2
Worked-Out Solution Key
261
Chapter 5,
continued
11.
x
23
22
21
0
1
y
58
20
6
4
2
2
10. f (x) 5 5x 4 2 x 3 2 3x 2 1 8x
3
f (2) 5 5(2)4 2 (2)3 2 3(2)2 1 8(2)
212 250
5 80 2 8 2 12 1 16
5 76
y
g(x) 5 22x 5 1 4x 3
11.
g(23) 5 22(23)5 1 4(23)3 5 486 2 108 5 378
12. h(x) 5 6x 3 2 25x 1 20
h(5) 5 6(5)3 2 25(5) 1 20 5 750 2 125 1 20 5 645
1
21
x
s
0
5
10
15
20
25
30
E
0
3.2
51
258.2
816
1992.2
4131
The wind speed needed to generate the wave is about 25
miles per hour.
5 128 1 48 2 4 1 10
5 182
g(x) 5 4x 5 1 6x3 1 x 2210x 1 5
14.
g(22) 5 4(22)5 1 6(22)3 1 (22)2 2 10(22) 1 5
5 2128 2 48 1 4 1 20 1 5
3000
5 2147
2000
15. 3
1000
0
3
1
h(24) 5 }2(24)4 2 }4(24)3 1 (24) 1 10
0
10
20
30
5
s
Wind speed
(miles per hour)
5.2 Exercises (pp. 341–344)
5
leading coefficient is 25 while the constant term is 6.
f (x) 5 2x 2 1 8 in standard form. It has a degree 2
(quadratic) and a leading coefficient of 21.
15
39
93
13
31
109
12
6
25
9
216
8
228
66
24
14
233
75
8
2. The end behavior of a polynomial is the behavior the
3. The function is a polynomial function written as
16
8
1. f (x) is a degree 4 (quartic) polynomial function. The
function demonstrates as it approaches 6`.
28
f (3) 5 109
16. 22
Skill Practice
22
f (22) 5 75
17. 26
4. The function is a polynomial function written as
f (x) 5 8x 1 6x 2 3 in standard form. It has a degree
4 (quartic) and a leading coefficient of 8.
4
5. The function is a polynomial that is already written in
standard form. It has degree 4 (quartic) and a leading
coefficient of :.
18. 4
f (x) 5 5x 3 2 2x 2 1 10x 2 15
f (21) 5 5(21)3 2 2(21)2 1 10(21) 2 15
5 25 2 2 2 10 2 15
5 232
262
Algebra 2
Worked-Out Solution Key
212
114
2
219
149
0
14
235
232
2128
2456
232
2114
2491
f (4) 5 2491
5
9.
26
28
28
7. The function is a polynomial function already written
8. The function is not a polynomial function because the
2
term }x has an exponent that is not a whole number.
35
f (26) 5 149
5x 22 has an exponent that is not a whole number.
coefficient of 2}2 .
27
1
6. The function is not a polynomial function because the term
in standard form. It has degree 3 (cubic) and a leading
8
1
19. 2
22
22
f(2) 5 211
3
0
28
13
24
22
24
224
21
22
212
211
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
Energy per square foot
(foot-pounds)
E
3
1
h(x) 5 }2x 4 2 }4x3 1 x 1 10
13.
12.
Chapter 5,
20. 23
0
10
0
0
227
218
54
2192
576
21728
218
64
2192
576
21755
6
6
continued
37. Sample answer: f (x) 5 2x 5 2 4x 2 1 3x 2 1
y
x
1
21
f (23) 5 21755
21. 3
27
27
11
4
0
221
230
278
210
226
278
38.
f (3) 5 278
22. 4
1
1
0
0
3
220
4
16
64
268
4
16
67
248
x
23
22
21
0
1
2
3
y
227 28
21
0
1
8
27
y
2
22
x
f (4) 5 248
23. The error is that a zero was not placed as the coefficient of
the missing x 3 term.
22
24
24
39.
0
9
221
7
8
216
14
14
8
27
27
21
x
23
22
21
0
y
281 216
21
0
1
2
3
21 216 281
y
2
22
x
f (22) 5 21 Þ 117
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
24. A; The ends approach opposite directions so the function
must be odd. It imitates an odd function whose coefficient
is positive.
25. The degree is even and the leading coefficient is positive.
26. The degree is odd and the leading coefficient is negative.
40.
x
27. The degree is even and the leading coefficient is negative.
y
28. f (x) l 1 ` as x l2`
23
22
2240 229
f (x) l 1 ` as x l 1 `
21 0 1
2
2
3
3 4 35 246
y
29. f (x) l 2` as x l2`
f (x) l 2` as x l 1 `
30. f (x) l 1 ` as x l2`
1
f (x) l 2` as x l 1 `
22
x
31. f (x) l 2` as x l2`
f (x) l 1 ` as x l 1 `
32. f (x) l 1 ` as x l2`
f (x) l 1 ` as x l 1 `
41.
x
23 22 21
y
79
14
33. f (x) l 1 ` as x l2`
0
1
2
3
21 22 21 14 79
y
f (x) l 2` as x l 1 `
34. f (x) l 2` as x l 2`
f (x) l 1 ` as x l 1 `
35. f (x) l 1 ` as x l2`
f (x) l 1 ` as x l 1 `
36. f (x) l 1 ` as x l2`
4
22
x
f (x) l 2` as x l 1 `
Algebra 2
Worked-Out Solution Key
263
Chapter 5,
42.
continued
x
23
22 21 0 1
y
32
13
2
47.
3
5 4 23 222
6
x
y
23
22
21
0
1
2
3
2238 232 24 24 22 32 248
y
y
1
21
x
1
21
43.
x
23
y
212
x
22 21 0
2
1
2
3
48.
0 24 22 12
4
x
23 22 21 0 1 2
3
y
42
42
2
2
6 2 2
y
y
1
21
x
1
21
x
y
23
22
21
0 1
2
3
2105 232 29
0 7
0
257
49.
x
y
23
22
21 0 1 2
3
2158 237 22 1 2 7 22
y
y
2
1
21
45.
x
y
22
x
23
22
21 0
1
2
2246 234 22 0
2
34 246
y
1
21
x
3
x
1
50. B; f (x) 5 2}x 3 1 1
3
From the graph, f (x) l 1` as x l 2`; and
f(x) l 2` as x l 1 `. So, the degree is odd
and the leading coefficient is negative. The
constant term 5 f(0) 5 1.
51. When g(x) 5 2f (x):
g(x) l 2` as x l 2` and g(x) l 1` as x l 1`.
46.
x
23 22 21 0 1 2
y
65
29
11
3
5 5 5 21
52. f(x) 5 a 3x 3 1 a 2x 2 1 a1x 1 a0
When a 3 5 2 and a 0 5 25:
f(x) 5 2x 3 1 a2x 2 1 a1x 2 5
When f(1) 5 0:
2(1)3 1 a2(1)2 1 a1(1) 2 5 5 0
y
a2 1 a1 5 3
When f(2) 5 3:
2(2)3 1 a2(2)2 1 a1(2) 2 5 5 3
1
21
264
Algebra 2
Worked-Out Solution Key
x
4a2 1 2a1 5 28
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
44.
x
Chapter 5,
continued
Solve the system of equations:
a2 1 a1 5 3
56. a. The function is degree 3 (cubic).
b. t
24a2 2 4a1 5 212
3 24
4a2 1 2a1 5 28
4a2 1 2a1 5 28
0
M
2
4
6
8
21,600 21,396 22,800 25,284 28,320
22a1 5 220
a1 5 10
t
Substitute a1 into one of the original equations.
M
Solve for a2.
Indoor movie screens
(thousands)
The cubic function is f (x) 5 2x 3 2 7x 2 1 10x 2 5.
So, f (25) 5 2(25)3 2 7(25)2 1 10(25) 2 5 5 2480.
53. a.
f(x)
g(x)
x
f (x)
g(x)
}
10
1000
840
1.19
20
8000
7280
1.10
50
125,000
120,200
1.04
980,400
1.02
f(x) ø g(x). So, f and g have the same end behavior.
1.46
1.68
2
4
6
10
12
1.59 1.44 1.97 4.41
1
0 1 2 3 4 5 6 7 8 9 10 t
Years since 1992
The number of snowboarders was more than 2 million
in 2002.
8
As t l 2`, P(x) l 1`.
b.
5.5 5.1 4.7 6.7 10.5
S
14
12
10
8
6
4
2
0
8
S
2
Quarterly periodicals
Skateboarding
participants (millions)
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
S
0
6
58. a. As t l 1`, P(x) l 1`.
5 10.9125 carats
t
t
8
12
16
Years since 1987
1.2
54. w 5 0.00071d 2 0.090d 1 0.48d
55.
4
s
2
5 23.9625 2 20.25 1 7.2
0
4
0
w 5 0.0071(15)3 2 0.090(15)2 1 0.48(15)
10
2
Problem Solving
3
20
0
Snowboarding
participants (millions)
f (x)
c. Sample answer: As x l 1`, } l 1 and therefore
g(x)
16
30
t
200 8,000,000 7,920,800 1.01
f (x)
b. As x l 1`, } l1.
g(x)
14
M
40
0
57.
12
31,380 33,936 35,460 35,424
c.
a2 1 10 5 3 l a2 5 27
100 1,000,000
10
P
4000
3000
2000
1000
0
0 1 2 3 4 5 6 7 8 9 10 t
Years since 1992
There were 8 million skateboarders 6 years after
1992 (1998).
c.
0
6
12
18
24 t
Years since 1980
p(t) 5 0.138t 4 2 6.24t 3 1 86.8t 2 2 239t 1 1450
p(30) 5 0.138(30)4 2 6.24(30)3 1 86.8(30)2
2 239(30) 1 1450
5 111,780 2 168,480 1 78,120 2 7170 1 1450
5 15,700 periodicals
It is not appropriate to use this model to predict the
number of periodicals in 2010 beacuse the model was
made for 1980 to 2002.
Algebra 2
Worked-Out Solution Key
265
Chapter 5,
continued
S 5 20.122t3 1 3.49t2 2 14.6t 1 136
S(5) 5 20.122(5)3 1 3.49(5)2 2 14.6(5) 1 136
5 215.25 1 87.25 2 73 1 136
5 135 grams
H 5 20.115t 3 1 3.71t 2 2 20.6t 1 124
H(5) 5 20.115(5)3 1 3.71(5)2 2 20.6(5) 1 124
The equation y 5 }4x passes through the point (24, 23)
Graphing Calculator Activity 5.2 (p. 345)
1. 210axa10, 210aya50
Weight (grams)
2. 220axa50, 22000aya8000
3. 210axa10, 210axa10
S
180
4. 25axa5, 210axa30
H
120
5. 25axa10, 210axa50
6. 25axa10, 240axa40
60
0
7. 25axa10, 250axa50
0
t
2
4
6
8
Days after hatching
c. The chick is more likely to be a Sarus chick than a
hooded chick. At 3 days the Sarus chick’s weight is
about 120 grams, higher than the hooded chicks. Even
though neither chick’s average weight at 3 days equals
130 grams, the Sarus chick is closer.
y 5 0.000304x
3
2.20y 5 0.000304(0.394x)
3
0.000304(0.394x)
y 5 }}
ø (8.452 3 1026)x3
2.2
y
3
y 5 0.000304x 3
2
8. 25axa5, 220axa20
9. Sample answer: The window for g(x) should be the same
x-interval, but the y-interval will be shifted by c units.
10. Sample answer: 210axa20, 0aya3000
Lesson 5.3
5.3 Guided Practice (pp. 346–348)
3
Width (kg and lb)
3
and is perpendicular to the line shown.
W
240
y 5 (8.452 3 1026)x 3
1. (t 2 2 6t 1 2) 1 (5t 2 2 t 2 8)
5 t 2 1 5t 2 2 6t 2 t 1 2 2 8
5 6t 2 2 7t 2 6
2. (8d 2 3 1 9d 3) 2 (d 3 2 13d 2 2 4)
5 8d 2 3 1 9d3 2 d 3 1 13d 2 1 4
5 9d 3 2 d 3 1 13d 2 1 8d 2 3 1 4
5 8d 3 1 13d 2 1 8d 1 1
1
0
3
5 99.375 grams
5 35.625 grams
b.
3
23 5 }4(24) 1 b l b 5 0
New line: y 5 }4x
5 135 2 99.375
60. a.
Intercept of new line: y 5 mx 1 b
5 214.375 1 92.75 2 103 1 124
Difference 5 S(5) 2 H(5)
b.
22 2 2
4
62. G; Slope of given line 5 } 5 2}
320
3
3
Slope of new line 5 }4
0
10
20
30
40
50 x
Length (cm and in.)
The function from part (a) is a vertical shrink of the
original function.
Mixed Review for TAKS
61. C;
Let s 5 number of shirts.
Cost 5 500 1 4.25s
Revenue 5 10.50s
Profit 5 Revenue 2 Cost
0 5 10.50s 2 500 2 4.25s
500 5 6.25s
80 5 s
Amanda must sell 80 shirts before she can make a profit.
3. (x 1 2)(3x 2 2 x 2 5)
5 (x 1 2)3x 2 2 (x 1 2)(x) 2 (x 1 2)(5)
5 3x 3 1 6x 2 2 x 2 2 2x 2 5x 2 10
5 3x 3 1 5x 2 2 7x 2 10
4. (a 2 5)(a 1 2)(a 1 6) 5 (a 2 2 3a 2 10)(a 1 6)
5 (a 2 2 3a 2 10)a 1 (a 2 2 3a 2 10)6
5 a 3 2 3a 2 2 10a 1 6a 2 2 18a 2 60
5 a3 1 3a 2 2 28a 2 60
5. (xy 2 4)3 5 (xy)3 2 3(xy)2(4) 1 3(xy)(4)2 2 (4)3
5 x 3y 3 2 12x 2y 2 1 48xy 2 64
6. T 5 C + D
5
0.542t 2
3
2 7.16t 1
79.4
109t 1
4010
2173.42t 2 2 28711.6t 1 318,394
5.9078t 3 2 780.44t 2 1 8654.6t
59.078t 3 1 1392.98t 2 2 20,057t 1 318,394
T 5 59.078t 3 1 1392.98t 2 2 20,057t 1 318,394
266
Algebra 2
Worked-Out Solution Key
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
59. a.