371 Algebra 2 Worked-Out Solution Key

Chapter 7
Prerequisite Skills (p. 476)
16a 1 4b 1 c 5 17
1. The domain of the function is xq2.
3 (21)
216a 2 4b 2 c 5 217
49a 1 7b 1 c 5 56
49a 1 7b 1 c 5 56
2. The range of the function is yq3.
33a 1 3b
3. The inverse of the function is y 5 (x 2 3)2 1 2.
}
5. y 5 Ï x 1 3
}
4. y 5 22Ï x 2 1
Domain: xq0
Domain: xq23
Range: ya21
Range: yq0
1
7a 1 b 5 9
12a
5 12
a51
7(1) 1 b 5 9 l b 5 2
9(1) 1 3(2) 1 c 5 8 l c 5 27
x
21
33a 1 3b 5 39
33a 1 3b 5 39
y
y
5 39
221a 2 3b 5 227
3 (23)
An equation for the parabola is y 5 x 2 1 2x 2 7.
1
x
21
12. y 5 ax 2 1 bx 1 c
(23, 9): 9 5 a(23)2 1 b(23) 1 c l 9a 2 3b 1 c 5 9
(1, 27): 27 5 a(1)2 1 b(1) 1 c l a 1 b 1 c 5 27
3}
6. y 5 Ï x 2 2 1 5
7.
(5, 255): 255 5 a(5)2 1 b(5) 1 c l 25a 1 5b 1 c
y 5 3x 1 5
Domain: all real numbers
5 255
x 5 3y 1 5
Range: all Real numbers
a 1 b 1 c 5 27 l c 5 2a 2 b 2 7
x 2 5 5 3y
9a 2 3b 1 (2a 2 b 2 7) 5
x25
3
y
9 l 8a 2 4b 5 16
25a 1 5b 1 (2a 2 b 2 7) 5 255 l 24a 1 4b 5 248
}5y
32a
a
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
(21) 1 (26) 1 c 5 27 l c 5 0
x
21
y 5 22x3 1 1
9.
An equation for the parabola is y 5 2x 2 2 6x.
1
y 5 }2 x 2, xq0
Lesson 7.1
1
x 5 22y3 1 1
x 5 }2 y 2
x 2 1 5 22y3
7.1 Guided Practice (pp. 479–481)
2x 5 y 2
x21
1. Domain: all real numbers 2. Domain: all real numbers
}
2}
5 y3
2
Ï2x 5 y
Range: y > 0
Î2x 22 1 5 y
}
3
5 21
8(21) 2 4b 5 16 l b 5 26
1
8.
5 232
Range: y > 0
y
}
10. y 5 ax2 1 bx 1 c
y
2
(0, 21): 21 5 a(0)2 1 b(0) 1 c l c 5 21
1
(1, 2): 2 5 a(1) 1 b(1) 1 c l a 1 b 1 c 5 2
2
x
21
(3, 14): 14 5 a(3)2 1 b(3) 1 c l 9a 1 3b 1 c 5 14
a 1 b 1 (21) 5 2 l a 1 b 5 3 l a 5 3 2 b
3. Domain: all real numbers
f (x)
9(3 2 b) 1 3b 5 15 l b 5 2
f(x) 5 3x 1 1 1 2
(0, 5)
a532251
(21, 3)
An equation for the parabola is y 5 x2 1 2x 2 1.
2
2
11. y 5 ax 1 bx 1 c
f(x) 5 3x
(3, 8): 8 5 a(3)2 1 b(3) 1 c l 9a 1 3b 1 c 5 8
(4, 17): 17 5 a(4) 1 b(4) 1 c l 16a 1 4b 1 c 5 17
2
(7, 56): 56 5 a(7)2 1 b(7) 1 c l 49a 1 7b 1 c 5 56
16a 1 4b 1 c 5 17
3 (21)
x
Range: y > 2
9a 1 3b 1 (21) 5 14 l 9a 1 3b 5 15
9a 1 3b 1 c 5 8
21
29a 2 3b 2 c 5 28
16a 1 4b 1 c 5 17
7a 1 b
59
(1, 3)
(0, 1)
21
x
4. Using the graph, you can estimate that the number of
incidents was about 250,000 during 2003 (t ø 7).
5. In the exponential growth model y 5 527(1.39) x, the
initial amount is 527, the growth factor is 1.39, and the
percent increase is 1 2 1.39 5 0.39 or 39%.
Algebra 2
Worked-Out Solution Key
371
Chapter 7,
continued
6. When p 5 2000, r 5 0.04, t 5 3, and n 5 365:
14.
1
h(x)
r nt
A 5 P 1 1 1 }n 2
1
x
A 5 2000 1 1 1 }
365 2
0.04 (365)(3)
5 2000 1 }
365 2
365.04 1095
15.
ø 2254.98
y
1
y 5 23 ? 2x
(22, 23)
y 5 23 ?
amount is 2.4, the growth factor is 1.5, and the percent
increase is 1 2 1.5 5 0.5 or 50%.
(21, )
5
4
y
18.
y 5 2x 1 1 1 3
(21, 4)
4. A; When x 5 0, y 5 23 + 20 5 23;
(
22,
When x 5 1, y 5 23 + 21 5 26
7
2
)
(1, 3)
2
y 5 3x
y 5 2x
2
(21, )
21
1
y
1
x
(2, 0) x
Domain: all real numbers
Domain: all real numbers
Range: y > 3
Range: y > 21
y
19.
2
(3, 2)
x
1
1
2
When x 5 1, y 5 2 + 31 5 6
(0, 1)
y 5 3x 2 2 2 1
(0, 1)
5. B; When x 5 0, y 5 2 + 30 5 2;
x
Range: y > 2
y
When x 5 1, y 5 3 + 21 5 6
1
Domain: all real numbers
Range y < 0
3. C; When x 5 0, y 5 3 + 20 5 3;
1
y 5 5 ? 4x
17.
more closely.
Domain: all real numbers
(1, 6)
Range: y > 21
3
(0, 2)
y 5 2 ? 3x
8.
9.
f(x)
y
(2, 1)
x
21
20.
2
y 5 2 ? 3x 2 2 2 1
1
y 5 23 ? 4x
10.
x
(0, 23)
(1, 25)
x
y 5 23 ? 4x 2 1 2 2
11.
y
1
g(x)
1
x
21.
Domain: all real numbers
f(x)
f (x) 5 6 ?
2x 2 3 1
Range: y > 3
3
(3, 9)
2
(0, 6)
(2, 6)
x
21
(21, 3)
12.
13.
y
f (x) 5 6 ? 2x
y
1
21
1
21
372
Algebra 2
Worked-Out Solution Key
Range: y < 22
(0, 22.75)
1
21
Domain: all real numbers
1
(21, 20.75)
x
21
y
x
x
2
2
x
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
(3, 5)
x
21
y 5 5 ? 4x 1 2
(1, 26)
2x 1 2
Domain: all real number
2. An asymptote is a line that a graph approaches more and
(0, 5)
13
4
(21, 26)
1. In the exponential growth model y 5 2.4(1.5) x, the initial
7.
(21, )
(0, 23)
Skill Practice
y
x
1
7.1 Exercises (pp. 482–485)
y
(0, 7)
The balance after 3 years is $2254.98.
6.
16.
Chapter 7,
22.
continued
Domain: all real numbers
g(x)
g(x) 5 5 ? 3x
(0, 5)
(22, 1)
( 21, )
(
Range: y > 24
33. y 5 ab x 2 h 1 k l y 5 2 + 5 x 1 4 1 3
a. If a changes to 1, there will be a vertical shrink.
b. If b changes to 4, the graph will increase slower.
5
3
7
23, 2 3
21
)
c. If h changes to 3, there will be a horizontal
shift (of 7 units right)
x
1
d. If k changes to 21, there will be a vertical
shift (of 4 units down).
34. a. f(x) 5 abx; f (x 1 1) 5 ab x 1 1
g(x) 5 5 ? 3x 1 2 2 4
23.
Domain: all real numbers
h(x)
h(x) 5 22 ? 5x 2 1 1 1
(0, 0.6)
h(x) 5 22 ? 5x
(21, 20.4)
(0, 22)
Range: y < 1
x
2
(1, 21)
f (x 1 1)
f (x)
ab x 1 1
ab
bx 1 1
b
(x 1 1) 2 x
}5}
5b
x 5}
x 5b
b. Check to see if the ratios of consecutive pairs of
f (x 1 1)
f (x)
y-values satisfy the equation } 5 b, where b > 1.
f (1)
f (0)
24
f (3)
f (2)
When x 5 1: 2(1.5)1 1 1 5 3 1 1 5 4 l (1, 4)
25. D; Using the exponential growth model y 5 a(1 1 r)t
with r 5 0.1 and a 5 1310, the model is y 5 1310(1.1)t.
26. The y-intercept should be at (0, 2), not (0, 1).
y
8
4
}5}52
24
8
}5}53
}5}53
24. B; When x 5 0: 2(1.5)0 1 1 5 2(1) 1 1 5 3 l (0, 3)
f (2)
f (1)
4
4
}5}51
f (4)
f (3)
72
24
The ratios are not all equal and are not all greater than
1. So, there is no exponential function f (x) 5 abx whose
graph passes through the points.
Problem Solving
35. a. initial amount 5 0.42 million or $420,000;
growth factor 5 2.47; annual percent increase
5 2.47 2 1 5 1.47 or 147%
b. Using the graph, you can estimate that the number of
DVD players sold in 2001 was about 16 million.
x
x
27. The graph for y 5 2 should have been translated 3 units
right and 3 units up rather than 3 units left and 3 units up.
28. y 5 number of monk parakeets
t 5 number of years since 1992
y 5 a(1 1 r)t
y 5 1219(1 1 0.12)
t
y 5 1219(1.12)t
29. A 5 amount in account
t 5 years since deposit
r nt
A 5 p 1 1 1 }n 2
0.02 365t
A 5 800 1 1 1 }
365 2
30. y 5 value of table
t 5 years since purchase
y 5 a(1 1 r)t
y 5 450(1 1 0.06)t
y 5 450(1.06)t
31. a. The balance after 7 years is $1844.81.
b. You have to press enter 18 times, so it takes 18 years.
32. Sample answer: y 5 12 + 4 x 2 1 1 2
n
32
28
24
20
16
12
8
4
0
0 1 2 3 4 5 t
Years since 1997
36. a. Initial amount 5 2500; growth factor 5 1.50; annual
percent increase 5 1.5 2 1 5 0.5 or 50%
b. Domain: 0, 1, 2, 3, 4, 5
Range: 2500, 3750, 5625, 8438, 12,656, 18,984
Using the graph, you can estimate that the number of
referrals in March of 2002 was about 13,000.
Number of referrals
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
21
Number of DVD players sold
(millions)
2
y
16,000
14,000
12,000
10,000
8000
6000
4000
2000
0
0 1 2 3 4 5 t
Years since 1998
Algebra 2
Worked-Out Solution Key
373
Chapter 7,
continued
r n(4)
37. A 5 22001 1 1 }
; P 5 2200; t 5 4
n2
b. Domain: 0 a ta10
0.03 4 + 4
5 2200(1.0075)16 ø 2479.38
4
2
The balance after 4 years is $2479.38.
b. r 5 0.0225, n 5 12
0.0225 12 + 4
A 5 22001 1 1 }
12 2
5 2200(1.001875)48
ø 2406.98
Year (t 5 0; 1990)
c. Using the graph, you can estimate that the population
The balance after 4 years is $2406.98.
was about 590,000 during 1996 (t ø 6).
c. r 5 0.02, n 5 365
5 2200(1.00005479452)1460
ø 2383.23
l p 5 50(1.105)n
38. a. When A 5 3000, r 5 0.0225, n 5 4, and t 5 3:
0.0225 4 + 3
3000 5 p 1 1 1 }
4 2
3000 5 p (1.005625)12
41. a. Initial amount 5 48.28; percent increase 5 0.06
2804.71 ø p
b. When A 5 3000, r 5 0.035, n 5 12, and t 5 3:
r nt
A 5 p 1 1 1 }n 2
0.035 12 + 3
3000 5 p 1 1}
12
l p 5 48.28(1.06)t
ticket was $60 during 2004 (t ø 4).
2
2
2701.39 ø p
You should deposit $2701.39 to have $3000 in your
account after 3 years.
c. When A 5 3000, r 5 0.04, n 5 1, and t 5 3:
r nt
A 5 p 1 1 1 }n 2
p
80
70
60
50
40
30
20
10
0
0
1
2
3
4
t
Years (t 5 0; 2000)
0.04 1 + 3
3000 5 p 1 1 1 }
1 2
c. Over the domain tq0, the minimum value of t is
3000 5 p (1.04) 3
2666.99 ø p
You should deposit $2666.99 to have $3000 in your
account after 3 years.
39. a. Initial amount 5 494.29 thousand; percent increase
5 0.03; growth factor 5 1 1 0.03 5 1.03
t 5 0. The function is defined in the interval
0ata4. You can look at the graph between these
values to determine the minumum or maximum that
gives meaningful results.
42. a. Initial amount 5 41; percent increase 5 0.089
Exponential growth model: n 5 41(1 1 0.089) t
l n 5 41(1.089) t
t
Exponential growth model: P 5 494.29(1.03)
When t 5 10: P 5 494.29(1.03)10 ø 664.284
The population in 2000 was about 664,284 people.
Algebra 2
Worked-Out Solution Key
Exponential growth model: p 5 48.28(1 1 0.06)t
b. Using the graph, you can estimate that the price of a
12.035 36
3000 5 p }
12
374
b. When n 5 5: p 5 50(1.105)5 ø 82.37. The price
after 5 bids was $82.37. After 100 bids (n 5 100),
the price will be p 5 50(1.105)100 ø $1,084,420.72.
No. Sample answer: This amount is unreasonable
because the model is only defined for 6 bids and
100 is out of this domain.
r nt
A 5 p 1 1 1 }n 2
1
number of bids 5 n
Exponential growth model: p 5 50(1 1 0.105)n
The balance after 4 years is $2383.23.
1
40. a. Initial amount 5 50; percent increase 5 0.105;
Average price (dollars)
0.02 365 + 4
A 5 22001 1 1 }
365 2
P
700
600
500
400
300
200
100
0
0 1 2 3 4 5 6 7 8 9 t
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
1
A 5 2200 1 1 }
Population (thousands)
Range: 494.29aPa664,284
a. r 5 0.03, n 5 4
Chapter 7,
b.
t
continued
n
t
n
t
46. H;
n
0
41
9
88.315
17 174.686
mŽC
AB
C
AB 5
+ 2:r
3608
1
44.649
10
96.175
18 190.233
5 } + 2:(8)
2
48.623
11 104.735
19 207.164
}
1208
3608
ø 16.8
The approximate length of arc AB is 16.8 centimeters.
3
52.950
12 114.056
20 225.601
4
57.663
13 124.207
21 245.680
Lesson 7.2
5
62.795
14 135.262
22 267.545
7.2 Guided Practice (pp. 487–488)
6
68.383
15 147.230
23 291.357
7
74.470
16 160.410
24 317.288
8
81.097
Number of breeding pairs
c.
1.
2.
y
x
21
3.
y
4.
f(x)
(0, 5)
210
180
(21, 4)
150
120
1 x21
1
4
1
(1, 2)
x
21
90
60
()
y5
1
(0, 1)
1
21
0 3 6 9 12 15 18 21 24 t
there were about 315 breeding pairs of bald eagles.
43. Investing $3000 at 6% annual interest: A1 5 3000(1.06)
5.
Investing $6000 at 7% annual interest: A3 5 6000(1.07)t
1
2
3
( 1, )
10
3
y55
(21, 3)
(
4
A3
6420
6870
7352.18
0,
6869.40 7350.26
7868.90
4
3
)
44. a. Initial amount 5 5200; final amount 5 9000; number
of years 5 5
9000 5 5200(1 1 r)5
45
26
Î4526 5 1 1 r
}
5
x
()
2 x11
2
3
2
7864.78
No, A1 1 A2 and A3 are not the same. After the first year,
the money split between the 6% and 8% accounts grows
at a faster rate.
} 5 (1 1 r)5 l
()
2 x
3
1
y55
The average annual growth rate was about 11.6%
b. y 5 9000(1 1 0.116)5 5 9000(1.116)5 ø 15,579.86
The cost will be $15,579.86 in 5 more years.
Mixed Review for TAKS
g(x) 5 23
()
3 x25
1
4
(6, ) x
7
4
(5, 1)
21
4
2
(1, 2 )
9
4
(0, 23)
g(x) 5 23
()
3 x
4
Domain: all real numbers
Range: y > 22
Range: y < 4
7. Initial amount 5 4200; percent decrease 5 0.20
Exponential decay model:
y 5 4200(1 2 0.20)t l y 5 4200(0.80) t using the graph,
you can estimate that the value of the snowmobile will be
$2500 after about 2 years.
y
4000
}
0.116 ø r
g(x)
Domain: all real numbers
Value (dollars)
6420
6.
(0, 5)
21
A1 1 A2
x
1 x
4
Range: y > 1
y
t
Investing $3000 at 8% annual interest: A2 5 3000(1.08)t
x
()
y5
Domain: all real numbers
Years since 1977
d. Using the graph, you can estimate that in 2001 (t 5 24),
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
x
21
2
n
330
300
270
240
30
0
y
1
3000
2000
1000
0
0 1 2 3 4 5 6 7 8 9 t
Time (yr)
45. A;
2
The graph of y 5 x 2 2 is translated 10 units up.
Algebra 2
Worked-Out Solution Key
375
continued
3 x
15. B; y 5 22 }
5
1 2
8. Final cost 5 3000; number of years 5 3; percent
decrease 5 0.07
3 0
When x 5 0: y 5 22 1 }5 2 5 22(1) 5 22 (0, 22)
3000 5 a(1 2 0.07)3
3000 5 a(0.93)3
3 1
16.
The original cost of the snowmobile was $3729.69.
y5
((
1 x
1
3
1
21
21
x
2. The function y 5 b represents exponential growth if
b > 1, and exponential decay if 0 < b < 1.
Because 0 < b < 1, this
1 2
((
1
3
y5
1 x
3
((
y52
x
19.
y
( )
2
3
1,
y52
(0, 2 )
7
3
Because 0 < b < 1,
this function represents exponential decay.
y
((
2 x
3
((
2
5
x
(0, 1)
1 x11
2
3
3
((
(1, )
9.
21.
y
y
y5
1
1
(1, )
21
1
3
x
12.
1
(0, 23)
f(x)
f(x) 5 23
23.
()
1 x
4
14.
1
5
x
(2, )
3
24
g(x) 5 6
(25, 4)
()
21
2
21
376
Algebra 2
Worked-Out Solution Key
x
x
1 x x
3
f(x) 5 23
()
1 x21
4
(0, 6)
(1, 3)
1
(24, 1)
h(x)
((
g(x)
1 x
2
(1, 23)
y
y5
Domain: all real numbers
1
13.
7
3
Range: y > 2
3
4
x
( 3, )
Range: y > 3
(1, 2 )
21
2
Domain: all real numbers
g(x)
x
1 x22
1
3
(0, 1)
y 5 3(0.25)x x
1
x
21
(1, )
3
4
((
(2, 3)
3
1
22.
11.
(1, )
(0, 3)
1
y
1
3
Domain: all real numbers
y
21
21
(5, 2 )
15
4
x
10.
f(x)
x
(4, 0)
Range: y > 21
20.
21
1
Range: y > 23
1
x
2 x24
2
3
Domain: all real numbers
(0, 6)
21
((
1
2
3
y 5 3(0.25)x 1 3
1
y5
1 x
3
y52
(21, 21)
represents exponential growth.
1 x21
2
y
y5
(0, 2)
1
8.
((
Domain: all real numbers
Because b > 1, this function
y
(1, 21)
4
3
(1, )
x
Range: y < 0
Because b > 1, this function
6. f (x) 5 25(0.25) x, b 5 0.25
1
2
Range: y > 1
represents exponential growth.
2
5. f (x) 5 } + 4x, b 5 4
7
(2, 2 )
1 x
2
Domain: all real numbers
18.
function represents exponential decay.
5
5 x
4. f (x) 5 4 } , b 5 }
2
2
1
2
(0, 1)
decrease 5 1 2 0.85 5 0.15 or 15%
1 2
(1, 2 )
(0, 21)
24
(1, )
(0, 2)
1. Initial amount 5 1250; decay factor 5 0.85; percent
y
1
y52
Skill Practice
7.
3
17.
y
7.2 Exercises (pp. 489–491)
3
3 x
3. f (x) 5 3 } , b 5 }
4
4
1 1, 2}65 2
6
When x 5 1: y 5 22 1 }5 2 5 22 1 }5 2 5 2}5
3729.69 ø a
1
g(x) 5 6
()
1 x15
2
2
2
Domain: all real numbers
Domain: all real numbers
Range: y < 0
Range: y > 22
x
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
Chapter 7,
Chapter 7,
24.
continued
b–c. Using the graph, you can estimate that the value of
h(x)
h(x) 5 4
()
the bike will be $100 after about 2.5 years.
1 x
2
(0, 4)
(21, 4)
Value (dollars)
(1, 2)
(0, 2)
1
21
h(x) 5 4
()
1 x 1 1x
2
Domain: all real numbers
Range: y > 0
Years
25. y 5 ab x 2 h 1 k l y 5 3(0.4) x 2 2 2 1
a. If a changes to 4, there will be a vertical stretch.
b. If b changes to 0.2, the graph will be steeper because
the decay factor is smaller.
c. If h changes to 5, there will be a horizontal translation
(of 3 units right)
d. If k changes to 3, there will be a vertical translation
(of 4 units up).
26. The decay factor was written incorrectly. It should be
1 2 percent decrease 5 1 2 0.02 5 0.98.
t
1 2
28. You need a base between 0.25 and 0.5 and a vertical
translation up between 0 and 3 units. Sample answer:
y 5 (0.3)x 1 1
1 2
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
1 x
29. Yes; (4)2x is just another way of saying } or
4
x
2x
(0.25) , so f (x) 5 5(4)
the same function.
1762
ø 0.96
Ratio from year 2–3: }
1832
1692
ø 0.96
Ratio from year 3–4: }
1762
1627
ø 0.96
Ratio from year 4–5: }
1692
The ratio of depreciation remains a constant 0.96 or 96%.
1906
d 5 a(0.96)1 5 1906 l a 5 }
ø 1985
0.96
An equation is d 5 1985(0.96)t.
1 x 22
27. D; The graph y 5 }
1 3 has an asymptote at the
2
line y 5 3.
1832
32. Ratio from year 1–2: } ø 0.96
1906
and g(x) 5 5(0.25)x represent
33. a.
Value (dollars)
y 5 (inital amount)(decay factor)
y 5 500(0.98) t
y
200
175
150
125
100
75
50
25
0
0 1 2 3 4 5 6 7 8 t
y
28,000
24,000
20,000
16,000
12,000
8000
4000
0
0 1 2 3 4 5 6 7 8 t
Years
Using the graph, you can estimate that the value of the
car will be $10,000 after about 5 years.
Problem Solving
30. a. When I 5 200 and t 5 1.5: A 5 I(0.71) t
A 5 200(0.71)15 ø 119.65
There is about 120 milligrams of ibuprofen remaining
in the bloodstream.
b. When I 5 325 and t 5 3.5: A 5 I(0.71)t
A 5 325(0.71)3.5 ø 98.01
There is about 98 milligrams of ibuprofen remaining
in the bloodstream.
t
c. When I 5 400 and t 5 5: A 5 I(0.71)
b. When t 5 50: y 5 24,000(0.845)50 ø 5.29
The value after 50 years is $5.29 according to
the model. Sample answer: This is too low to be
reasonable for the price of a car. In addition, car values
usually begin to increase once they become antiques.
1 t/5730
34. a. P 5 100 }
2
1 2
When t 5 2500: P 5 1001 }2 2
1 2500/5730
1 5000/5730
A 5 400(0.71)5 ø 72.17
When t 5 5000: P 5 1001 }2 2
There is about 72 milligrams of ibuprofen remaining
in the bloodstream.
When t 5 10,000: P 5 1001 }2 2
31. a. When r 5 0.25 and a 5 200: y 5 a(1 2 r)t
y 5 200(1 2 0.25)t l y 5 200(0.75)t
When t 5 3: y 5 200(0.75)3 ø 84.38
The value of the bike after 3 years is about $84.38.
ø 73.90
ø 54.62
1 10,000/5730
ø 29.83
After 2500 years there will be about 73.9% of the
original carbon-14 remaining, after 5000 years there
will be about 54.6% remaining, and after 10,000 years
there will be about 29.8% remaining.
Algebra 2
Worked-Out Solution Key
377
Chapter 7,
38. J;
This year’s price 5 (1 2 0.167)(960) 5 799.68
799.68x 5 960
x ø 1.20
x ø 120%
Lesson 7.3
00
0
0
t
16
,
0
12
,0
0
0
Last year’s price is approximately 120% of this year’s
price for the computer.
80
0
c. Using the graph, you can estimate the age of the bison
bone is about 8000 years old when 37% of the
carbon-14 is present.
35. a. The decay facter is 0.89 and the percent decrease is
1 2 0.89 5 0.11, or 11%.
Number of eggs
produced per year
b.
E
180
170
160
150
140
130
120
110
100
0
7.3 Guided Practice (pp. 493–495)
1. e7 + e4 5 e7 1 4 5 e11
2. 2e23 + 6e 5 5 12e23 1 5 5 12e 2
24e8
3. }
5 6e8 2 5 5 6e3
4e5
1000
4. (10e24x) 3 5 103(e24x)3 5 1000e212x 5 }
e12x
5. e 3/4 ø 2.117
6.
Domain: all real numbers
y
Range: y > 0
1
0
20
40
60
80
100
120
140 w
c. There are 52 weeks in a year, so when a chicken is
7.
2.5 years old, it is 2.5(52) 5 130 weeks old. Using
the graph, you can estimate that the number of eggs
produced by a 130 week old chicken is about 134
per year.
3
22
f(x) 5
(1, 0.18)
8.
( )e
1
2
2x x
Domain: all real numbers
y
(1, 1.93)
y 5 1.5e 0.25x
275 5 1300(1 2 r)4
Range: y > 1
11
(1, 1.18)
1
2
t
e2x
3
2
number of years 5 4
Range: y > 22
(0, )
3
2
1
21
0 + 2115 ø (1 2 r)4
(1, 2 )
x
(2, 20.07)
1
2
0.678 ø 1 2 r
The decay factor is 0.678, so an equation giving the
stereo’s resale value as a function of time is
V 5 1300(0.678)t.
()
1
2
(0, )
(0, )
E 5 179.2(0.89)t.
V 5 a(1 2 r)
Domain: all real numbers
f(x)
f(x) 5
d. Let t represent the chicken’s age in years.
w
, you can rewrite the equation as
Because t 5 }
52
36. Initial amount 5 1300; resale value after t years 5 275;
x
21
Age of chicken (weeks)
y 5 1.5e 0.25(x 2 1) 2 2
9.
Mixed Review for TAKS
37. C;
The image of point Q will be at (21, 21), which is in
Quadrant III.
X=5
Y=223.65901
Use the trace feature to determine that * ø 224
when t 5 5.
The length of a 5-year-old tiger shark is about
224 centimeters.
378
Algebra 2
Worked-Out Solution Key
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
0
P
100
90
80
70
60
50
40
30
20
10
0
40
0
Percent
b.
continued
Chapter 7,
continued
Î(
}
10. A 5 Pe rt where p 5 2500 and r 5 0.05
16. D;
a. When t 5 2: A 5 2500e0.05(2) 5 2500e0.1 ø 2762.93
The balance after 2 years is $2762.93.
b. When t 5 5: A 5 2500e0.05(5) 5 2500e (0.25) ø 3210.06
The balance after 5 years is $3210.06
c. When t 5 7.5: A 5 2500e0.05(7.5)
5 2500e0.375 ø 3637.48
The balance after 7.5 years is $3637.48.
11. Amount of interest earned 5 balance 2 principle
a. The interest earned after 2 years is
2762.93 2 2500 5 $262.93.
b. The interest earned after 5 years is
3210.06 2 2500 5 $710.06.
c. The interest earned after 7.5 years is
3637.48 2 2500 5 $1137.48.
7.3 Exercises (pp. 495–498)
Î
5
}}
}
5 Ï36e6x4
}
}
}
5 Ï36 Ïe6 Ïx4
5 6e3x2
17. The power 2 should have been applied to the 3 also.
(3e5x)2 5 32(e5x)2 5 9e10x
18. The term 22x rather than 2x should have been subtracted
from 6x.
e6x
e
}
5 e6x 2 (22x) 5 e8x
22x
19. e3 ø 20.086
20. e23/4 ø 0.472
21. e2.2 ø 9.025
22. e1/2 ø 1.649
24. e4.3 ø 73.700
ø 0.670
25. e7 ø 1096.633
27. 2e
equal to 2.71828.
108e13x
3e x
}
7 23
5 Ï36e13 2 7x1 2 (23)
23. e
1. The number e is an irrational number approximately
3e x
}
22/5
Skill Practice
20.3
26. e24 ø 0.018
28. 5e 2/3 ø 9.739
ø 1.482
1
2. The function f (x) 5 } e4x is an example of exponential
3
1
}
growth because 3 > 0 and 4 > 0.
29. 26e2.4 ø 266.139
3. e3 + e4 5 e3 1 4 5 e7
1
32. The function f (x) 5 } e4x is an example of exponential
3
4. e22 + e6 5 e22 1 6 5 e4
5. (2e
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
4 27e13x)
}
7 23
)
3x 3
5 2 (e
3
)
3x 3
5 8e
9x
30. 0.4e 4.1 ø 24.136
2x
31. The function f (x) 5 3e
is an example of exponential
decay.
growth.
33. The function f (x) 5 e24x is an example of exponential
decay.
e8
1
6. (2e22)24 5 224(e22)24 5 } e8 5 }
16
16
3
34. The function f (x) 5 } ex is an example of exponential
5
1
1
7. (3e 5x)21 5 321(e 5x)21 5 } e25x 5 }
3
3 e5x
1
35. The function f (x) 5 } e25x is an example of exponential
4
8. e x + e23x + e4 5 e x 2 3x 1 4 5 e22x 1 4
9.
}
}
}
Ï9e6 5 Ï9 Ïe6 5 3e3
10. e x + 5e x 1 3 5 5e x 1 x 1 3 5 5e 2x 1 3
3e
11. }x 5 3e1 2 x
e
3}
3}
decay.
36. The function f (x) 5 e 3x is an example of exponential
growth.
37. The function f (x) 5 2e4x is an example of exponential
growth.
4
4ex
12. }
5 4e x 2 4x 5 4e23x 5 }
e3x
e4x
13.
growth.
3}
Ï8e9x 5 Ï8 Ïe9x 5 2e3x
38. The function f (x) 5 4e22x is an example of exponential
decay.
39. B; When x 5 0, y 5 0.5e 0.5(0) 5 0.5; When x 5 1,
y 5 0.5e0.5(1) ø 0.82
3
6e4x
14. } 5 } e4x 2 1
4
8e
40. C; When x 5 0, y 5 2e 0.5(0) 5 2; when x 5 1,
15. C; (4e2x)3 5 43(e2x) 3 5 64e 6x
41. A; When x 5 0, y 5 e 0.5(0) 1 2 5 3; When x 5 22,
y 5 2e0.5(1) ø 3.30
y 5 e0.5(22) 1 2 ø 2.37
Algebra 2
Worked-Out Solution Key
379
Chapter 7,
42.
continued
43.
y
51. Using the table feature, you can notice that incrementing
y
small values of n increases the function very slowly.
Incrementing n by powers of 10 gives values that are one
digit closer to the actual value of e each time. When
n 5 1010, the value of the function is 2.718281828,
which is the value of e correct to 9 decimal places.
1
x
21
52. No, e cannot be expressed as a ratio of two integers
1
x
21
Domain: all real numbers
Domain: all real numbers
Range: y > 0
Range: y > 0
44.
45.
y
because it is an irrational number; it is a decimal that
neither terminates nor repeats.
53. Choose a, b, r, and q, such that a > 0, b > 0, r < 0, q < 0,
and r 2 q > 0.
y
Sample answer: f (x) 5 2e23x, g (x) 5 e24x
f (x)
g (x)
2e23x
e
n
1
r
54. Let m 5 }, so n 5 rm and } 5 } .
r
m
n
23x 2 (24x)
}5}
5 2ex
24x 5 2e
(0, 2)
(0, 1)
x
21
21
y 5 2e23x
x
2
y 5 2e23x 2 1
(1, 20.90)
Domain: all real numbers
Range: y > 0
Range: y > 21
47.
y
y 5 2.5e20.5x 1 2
(0, )
( )
0,
5
2
(3, 1.63)
(2, )
21
21
3
5
x
y 5 0.6e x 2 2
Domain: all real numbers
Domain: all real numbers
Range: y > 2
Range: y > 0
f(x) 5
1 x13
e
2
f(x) 5
1 x
e
2
Domain: all real numbers
f(x)
22
Range: y > 22
2
1
2
(23, 2 )
3
2
(2, 4.62)
(1, 3.62)
4 x21
e
3
g(x) 5
50.
5
4 x
e
3
21
(1, )
7
3
(0, )
4
3
3 x
Domain: all real numbers
h(x)
Range: y > 23
(1, 0.14)
h(x) 5 e 22x
5
(0, 22.86)
h(x) 5 e 22(x 1 1) 2 3
380
Algebra 2
Worked-Out Solution Key
About 23,247 termites were collected in 1999.
57. When P 5 2000, r 5 0.04, and t 5 5: A 5 Pert
The balance after 12.5 years is $1114.17.
59. a. When k 5 20.02: L(x) 5 100ekx l L(x) 5 100e20.02x
L(x)
100
90
80
70
60
50
40
30
20
10
0
0
20
40
60
80
x
Depth below water surface (m)
(0, 1)
(21, 22)
y 5 738e0.345(10) 5 738e3.45 ø 23,247
A 5 800e(0.0265)(12.5) 5 800e0.33125 ø 1114.17
Range: y > 1
11
56. When t 5 1999 2 1989 5 10: y 5 738e0.345t
Percent of light
g(x) 5
About 895 million camera phones were shipped globally
in 2002.
58. When P 5 800, r 5 0.0265 and t 5 12.5: A 5 Pert
Domain: all real numbers
g(x)
y 5 1.28e1.31(5) 5 1.28e6.55 ø 895
The balance after 5 years is $2442.81.
3 x
(22, 20.64)
49.
55. When x 5 2002 2 1997 5 5: y 5 1.28e1.31x
A 5 2000e(0.04 + 5) 5 2000e0.2 ø 2442.81
(1, 1.36)
(0, )
approximates e.
Problem Solving
3
5
1 y 5 2.5e20.5x x
21
m
1 1 1 }m1 2
G
1 m rt
approaches 1`.
(0, )
(1, 1.52)
F
5 P 11 1 }
m2
r nt
(1, 1.63)
(1, 3.52)
1 rmt
So, A 5 P1 1 1 }n 2 approximates A 5 Pe rt as n
y
y 5 0.6e x
9
2
r nt
A 5 P 1 1 1 }n 2 5 P 1 1 1 }
m2
As n approaches 1`, m approaches 1`, and
Domain: all real numbers
46.
48.
(1, 0.10)
x
b. Using the graph, you can estimate that the percent of
surface light is about 45% at a depth of 40 meters.
c. Using the graph, you can estimate that the submarine
can descend about 35 meters before only 50% of
surface light is available.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
1
Chapter 7,
continued
60. a. P(t) 5 P0e0.116t; when P0 5 30: P(t) 5 30e0.116t
Population
b.
P(t)
90
80
70
60
50
40
30
20
10
0
0
3.
Domain: all real numbers
f(x)
( 21, )
f (x) 5 ( )
14
3
Range: y > 2
3 x
1
8
( 21, )
8
3
(0, 3)
1
(0, 1)
1
2
4
6
f (x) 5
()
3 x x
8
4. 3e4 + e 3 5 3e4 1 3 5 3e7
t
8
2
5. (25e 3x) 3 5 253(e 3x) 3 5 2125e9x
Hours after 1:00 P.M.
c. Using the graph, you can estimate that the bacteria
population is about 48 at 5:00 p.m.
d. P(2.75) 5 30e0.116(2.75) 5 30e0.319 ø 41
1
e4x
6. } 5 } e4x 2 1
5
5e
4e 5x 2 2x
4e3x
8e 5x
7. }
5}
5}
3
3
6e2x
8.
9.
y
y
There are 2.75 hours between 1:00 p.m. and 3:45 p.m.
so, let t 5 2.75.
There are about 41 bacteria at 3:45 p.m.
1
1
61. A 5 A0e20.05t where A0 5 4 and t 5 14
x
21
A 5 4e20.05(14) 5 4e20.7 ø 1.99
21
Domain: all real numbers
The area after 14 days is about 2 square centimeters.
Domain: all real numbers
Range: y > 0
62. a.
x
Range: y > 0
10.
11.
y
g(x)
(0, 5)
2
X=0
y 5 ex
Y=630
(1, 2.72)
(0, 1)
(0, 0.72)
(21, 21)
The arch is 630 feet tall at its highest point.
(0, 4) g(x) 5 4e 23x 1 1
(1, 1.20)
x
1
1
y 5 ex 1 1 2 2
21
ends of the arch are about 315 2 (2315) 5 630 feet
apart.
x
(1, 0.20)
g(x) 5 4e 23x
Domain: all real numbers
Domain: all real numbers
Range: y > 22
Range: y > 1
12. Decay factor 5 0.85;
Mixed Review for TAKS
percent decrease 5 1 2 0.85 5 0.15
63. A;
}
Conjecture: Ï x < x
TVs sold (millions)
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
b. The x-intercepts are approximately x 5 6315, so the
Î
}
1 1 1
If x 5 }4 , }4 < }4
1
2
1
4
}ñ}
20
10
0
64. J;
2
2.
y
Lesson 7.4
y
7.4 Guided Practice (pp. 499 – 503)
(3, 6)
1. log3 81 5 4
3
3 4 5 81
1
(2, 2)
1
t
The balance after 5 years is $1502.79.
Quiz 7.1–7.3 (p. 498)
y 5 2 ? 3x 2 2
6
A 5 1200e(0.045)(5) 5 1200e(0.225) ø 1502.79
mŽQ 5 mŽN 5 1258
(0, 2)
y 5 2 ? 3x
4
Range: 14.0 a n a 26.8;
using the graph you can
estimate that the number
of black-and-white TVs
sold in 1999 (t 5 2) was
about 19 million.
13. When P 5 1200, r 5 0.045, and t 5 5: A 5 Pe rt
mŽN 5 1808 2 mŽP 5 1808 2 558 5 1258
(1, 6)
0
Years since 1997
mŽM 5 mŽP 5 558
1.
Domain: 0 a t a 4
n
30
21
x
x
3. log14 1 5 0
140 5 1
Domain: all real numbers
Domain: all real numbers
Range: y > 0
Range: y > 0
2. log7 7 5 1
71 5 7
4. log1/2 32 5 25
1 25
}
5 32
2
1 2
Algebra 2
Worked-Out Solution Key
381