Solution Manual for Precalculus: Graphs and Models, A Right Triangle Approach, 6th Edition

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SOLUTIONS MANUAL
JUDITH A. PENNA
A LGEBRA & T RIGONOMETRY
G RAPHS AND M ODELS
S IXTH E DITION
P RECALCULUS
G RAPHS AND M ODELS ,
A R IGHT T RIANGLE A PPROACH
S IXTH E DITION
Marvin L. Bittinger
Indiana University Purdue University Indianapolis
Judith A. Beecher
David J. Ellenbogen
Community College of Vermont
Judith A. Penna

Contents
Just-in-Time Review . . . . . . . . . . . . . . . . . . . . . . . 1
Chapter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . 13
Chapter 2 . . . . . . . . . . . . . . . . . . . . . . . . . . 73
Chapter 3 . . . . . . . . . . . . . . . . . . . . . . . . . . 127
Chapter 4 . . . . . . . . . . . . . . . . . . . . . . . . . . 185
Chapter 5 . . . . . . . . . . . . . . . . . . . . . . . . . . 273
Chapter 6 . . . . . . . . . . . . . . . . . . . . . . . . . . 333
Chapter 7 . . . . . . . . . . . . . . . . . . . . . . . . . . 387
Chapter 8 . . . . . . . . . . . . . . . . . . . . . . . . . . 429
Chapter 9 . . . . . . . . . . . . . . . . . . . . . . . . . . 483
Chapter 10 . . . . . . . . . . . . . . . . . . . . . . . . . . 583
Chapter 11 . . . . . . . . . . . . . . . . . . . . . . . . . . 667
Chapter R . . . . . . . . . . . . . . . . . . . . . . . . . . 709

Just-in-Time Review
1. Real Numbers
1. Rational numbers: 2
3 , 6, −2.45, 18.4, −11, 3
√27, 5 1
6 , − 8
7 ,
0, √16
2. Rational numbers but not integers: 2
3 , −2.45, 18.4, 5 1
6 ,
− 8
7
3. Irrational numbers: √3, 6
√26, 7.151551555 . . . , −√35, 5
√3
(Although there is a pattern in 7.151551555 . . . , there is
no repeating block of digits.)
4. Integers: 6, −11, 3
√27, 0, √16
5. Whole numbers: 6, 3
√27, 0, √16
6. Real numbers: All of them
2. Properties of Real Numbers
1. −24 + 24 = 0 illustrates the additive inverse property.
2. 7(xy) = (7x)y illustrates the associative property of mul-
tiplication.
3. 9(r − s) = 9r − 9s illustrates a distributive property.
4. 11 + z = z + 11 illustrates the commutative property of
addition.
5. −20 · 1 = −20 illustrates the multiplicative identity prop-
erty.
6. 5(x + y) = (x + y)5 illustrates the commutative property
of multiplication.
7. q + 0 = q illustrates the additive identity property.
8. 75 · 1
75 = 1 illustrates the multiplicative inverse property.
9. (x+y)+w = x+(y+w) illustrates the associative property
of addition.
10. 8(a + b) = 8a + 8b illustrates a distributive property.
3. Order on the Number Line
1. 9 is to the right of −9 on the number line, so it is false
that 9 < −9.
2. −10 is to the left of −1 on the number line, so it is true
that −10 ≤ −1.
3. −5 = −√25, and −√26 is to the left of −√25, or −5, on
the number line. Thus it is true that −√26 < −5.
4. √6 = √6, so it is true that √6 ≤ √6.
5. −30 is to the left of −25 on the number line, so it is false
that −30 > −25.
6. − 4
5 = − 16
20 and − 5
4 = − 25
20 ; − 16
20 is to the right of − 25
20 ,
so it is true that − 4
5 > − 5
4 .
4. Absolute Value
1. | − 98| = 98 (|a| = −a, if a < 0.)
2. |0| = 0 (|a| = a, if a ≥ 0.)
3. |4.7| = 4.7 (|a| = a, if a ≥ 0.)
4.
∣
∣
∣
∣ − 2
3
∣
∣
∣
∣ = 2
3 (|a| = −a, if a < 0.)
5. | − 7 − 13| = | − 20| = 20, or
|13 − (−7)| = |13 + 7| = |20| = 20
6. |2 − 14.6| = | − 12.6| = 12.6, or
|14.6 − 2| = |12.6| = 12.6
7. | − 39 − (−28)| = | − 39 + 28| = | − 11| = 11, or
| − 28 − (−39)| = | − 28 + 39| = |11| = 11
8.
∣
∣
∣
∣ − 3
4 − 15
8
∣
∣
∣
∣ =
∣
∣
∣
∣ − 6
8 − 15
8
∣
∣
∣
∣ =
∣
∣
∣
∣ − 21
8
∣
∣
∣
∣ = 21
8 , or
∣
∣
∣
∣
15
8 −
(
− 3
4
)∣
∣
∣
∣ =
∣
∣
∣
∣
15
8 + 6
8
∣
∣
∣
∣ =
∣
∣
∣
∣
21
8
∣
∣
∣
∣ = 21
8
5. Operations with Real Numbers
1. 8 − (−11) = 8 + 11 = 19
2. − 3
10 ·
(
− 1
3
)
= 3 · 1
10 · 3 = 3
3 · 1
10 = 1 · 1
10 = 1
10
3. 15 ÷ (−3) = −5
4. −4 − (−1) = −4 + 1 = −3
5. 7 · (−50) = −350
6. −0.5 − 5 = −0.5 + (−5) = −5.5
7. −3 + 27 = 24
8. −400 ÷ −40 = 10
9. 4.2 · (−3) = −12.6

2 Just-in-Time Review
10. −13 − (−33) = −13 + 33 = 20
11. −60 + 45 = −15
12. 1
2 − 2
3 = 1
2 +
(
− 2
3
)
= 3
6 +
(
− 4
6
)
= − 1
6
13. −24 ÷ 3 = −8
14. −6 + (−16) = −22
15. − 1
2 ÷
(
− 5
8
)
= − 1
2 ·
(
− 8
5
)
= 1 · 8
2 · 5 = 1 · 2/ · 4
2/ · 5 = 4
5
6. Interval Notation
1. This is a closed interval, so we use brackets. Interval no-
tation is [−5, 5].
2. This is a half-open interval. We use a parenthesis on
the left and a bracket on the right. Interval notation is
(−3, −1].
3. This interval is of unlimited extent in the negative direc-
tion, and the endpoint −2 is included. Interval notation is
(−∞, −2].
4. This interval is of unlimited extent in the positive direc-
tion, and the endpoint 3.8 is not included. Interval nota-
tion is (3.8, ∞).
5. {x|7 < x}, or {x|x > 7}.
This interval is of unlimited extent in the positive direction
and the endpoint 7 is not included. Interval notation is
(7, ∞).
6. The endpoints −2 and 2 are not included in the interval,
so we use parentheses. Interval notation is (−2, 2).
7. The endpoints −4 and 5 are not included in the interval,
so we use parentheses. Interval notation is (−4, 5).
8. The interval is of unlimited extent in the positive direc-
tion, and the endpoint 1.7 is included. Internal notation
is [1.7, ∞).
9. The endpoint −5 is not included in the interval, so we use a
parenthesis before −5. The endpoint −2 is included in the
interval, so we use a bracket after −2. Interval notation is
(−5, −2].
10. This interval is of unlimited extent in the negative direc-
tion, and the endpoint √5 is not included. Interval nota-
tion is (−∞, √5).
7. Integers as Exponents
1. 3−6 = 1
36 Using a−m = 1
am
2. 1
(0.2)−5 = (0.2)5 Using a−m = 1
am
3. w−4
z−9 = z9
w4 Using a−m
b−n = bn
am
4.
( z
y
)2
= z2
y2 Raising a quotient to a power
5. 1000 = 1 Using a0 = 1, a = 0
6. a5
a−3 = a5−(−3) = a5+3 = a8 Using the quotient rule
7. (2xy3)(−3x−5y) = 2(−3)x · x−5 · y3 · y
= −6x1+(−5)y3+1
= −6x−4y4, or − 6y4
x4
8. x−4 · x−7 = x−4+(−7) = x−11, or 1
x11
9. (mn)−6 = m−6n−6, or 1
m6n6
10. (t−5)4 = t−5·4 = t−20, or 1
t20
8. Scientific Notation
1. Convert 18,500,000 to scientific notation.
We want the decimal point to be positioned between the
1 and the 8, so we move it 7 places to the left. Since
18,500,000 is greater than 10, the exponent must be posi-
tive.
18, 500, 000 = 1.85 × 107
2. Convert 0.000786 to scientific notation.
We want the decimal point to be positioned between the
7 and the 8, so we move it 4 places to the right. Since
0.000786 is between 0 and 1, the exponent must be nega-
tive.
0.000786 = 7.86 × 10−4
3. Convert 0.0000000023 to scientific notation.
We want the decimal point to be positioned between the
2 and the 3, so we move it 9 places to the right. Since
0.0000000023 is between 0 and 1, the exponent must be
negative.
0.0000000023 = 2.3 × 10−9
4. Convert 8,927,000,000 to scientific notation.
We want the decimal point to be positioned between the
8 and the 9, so we move it 9 places to the left. Since
8,927,000,000 is greater than 10, the exponent must be
positive.
8, 927, 000, 000 = 8.927 × 109
5. Convert 4.3 × 10−8 to decimal notation.
The exponent is negative, so the number is between 0 and
1. We move the decimal point 8 places to the left.
4.3 × 10−8 = 0.000000043

Just-in-Time Review 3
6. Convert 5.17 × 106 to decimal notation.
The exponent is positive, so the number is greater than
10. We move the decimal point 6 places to the right.
5.17 × 106 = 5, 170, 000
7. Convert 6.203 × 1011 to decimal notation.
The exponent is positive, so the number is greater than
10. We move the decimal point 11 places to the right.
6.203 × 1011 = 620, 300, 000, 000
8. Convert 2.94 × 10−5 to scientific notation.
The exponent is negative, so the number is between 0 and
1. We move the decimal point 5 places to the left.
2.94 × 10−5 = 0.0000294
9. Order of Operations
1. 3 + 18 ÷ 6 − 3 = 3 + 3 − 3 Dividing
= 6 − 3 = 3 Adding and subtracting
2. = 5 · 3 + 8 · 32 + 4(6 − 2)
= 5 · 3 + 8 · 32 + 4 · 4 Working inside parentheses
= 5 · 3 + 8 · 9 + 4 · 4 Evaluating 32
= 15 + 72 + 16 Multiplying
= 87 + 16 Adding in order
= 103 from left to right
3. 5[3 − 8 · 32 + 4 · 6 − 2]
= 5[3 − 8 · 9 + 4 · 6 − 2]
= 5[3 − 72 + 24 − 2]
= 5[−69 + 24 − 2]
= 5[−45 − 2]
= 5[−47]
= −235
4. 16 ÷ 4 · 4 ÷ 2 · 256
= 4 · 4 ÷ 2 · 256 Multiplying and dividing
in order from left to right
= 16 ÷ 2 · 256
= 8 · 256
= 2048
5. 26 · 2−3 ÷ 210 ÷ 2−8
= 23 ÷ 210 ÷ 2−8
= 2−7 ÷ 2−8
= 2
6. 4(8 − 6)2 − 4 · 3 + 2 · 8
31 + 190
= 4 · 22 − 4 · 3 + 2 · 8
3 + 1 Calculating in the
numerator and in
the denominator
= 4 · 4 − 4 · 3 + 2 · 8
4
= 16 − 12 + 16
4
= 4 + 16
4
= 20
4
= 5
7. 64 ÷ [(−4) ÷ (−2)] = 64 ÷ 2 = 32
8. 6[9 − (3 − 2)] + 4(2 − 3)
= 6[9 − 1] + 4(2 − 3)
= 6 · 8 + 4(−1)
= 48 − 4
= 44
10. Introduction to Polynomials
1. 5 − x6
The term of highest degree is −x6, so the degree of the
polynomial is 6.
2. x2y5 − x7y + 4
The degree of x2y5 is 2 + 5, or 7; the degree of −x7y is
7 + 1, or 8; the degree of 4 is 0 (4 = 4x0). Thus the degree
of the polynomial is 8.
3. 2a4 − 3 + a2
The term of highest degree is 2a4, so the degree of the
polynomial is 4.
4. −41 = −41x0, so the degree of the polynomial is 0.
5. 4x − x3 + 0.1x8 − 2x5
The term of highest degree is 0.1x8, so the degree of the
polynomial is 8.
6. x − 3 has two terms. It is a binomial.
7. 14y5

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Just-in-Time Review 5
8. y4 − 9y3 + 14y2 = y2(y2 − 9y + 14)
Now factor y2 − 9y + 14. Look for two numbers with a
product of 14 and a sum of −9. The numbers are −2 and
−7. Then y2 − 9y + 14 = (y − 2)(y − 7). We must include
the common factor, y2, in order to have a factorization of
the original trinomial.
y4 − 9y3 + 14y2 = y2(y − 2)(y − 7)
9. 2n2 + 9n − 56
1. There is no common factor other than 1 or −1.
2. The factorization must be of the form
(2n+ )(n+ ).
3. Factor the constant term, −56. The possibilities
are −1 · 56, 1(−56), −2 · 28, 2(−28), −4 · 16, 4(−16),
−7 · 8, and 7(−8). The factors can be written in
the opposite order as well: 56(−1), −56 · 1, 28(−2),
−28 · 2, 16(−4), −16 · 4, 8(−7), and −8 · 7.
4. Find a pair of factors for which the sum of the outer
and the inner products is the middle term, 9n. By
trial, we determine that the factorization is (2n −
7)(n + 8).
10. 2y2 + y − 6
1. There is no common factor other than 1 or −1.
2. The factorization must be of the form
(2y+ )(y+ ).
3. Factor the constant term, −6. The possibilities are
−1 · 6, 1(−6), −2 · 3, and 2(−3). The factors can be
written in the opposite order as well: 6(−1), −6 · 1,
3(−2) and −3 · 2.
4. Find a pair of factors for which the sum of the outer
and the inner products is the middle term, y. By
trial, we determine that the factorization is
(2y − 3)(y + 2).
11. b2 − 6bt + 5t2
We look for two numbers with a product of 5 and a sum
of −6. By trial, we determine that they are −1 and −5.
b2 − 6bt + 5t2 = (b − t)(b − 5t)
12. x4 − 7x2 − 30 = (x2)2 − 7x2 − 30
We look for two numbers with a product of −30 and a sum
of −7. By trial, we determine that they are 3 and

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6 Just-in-Time Review
16. Special Factorizations
1. z2 − 81 = z2 − 92 = (z + 9)(z − 9)
2. 16x2 − 9 = (4x)2 − 32 = (4x + 3)(4x − 3)
3. 7pq4 − 7py4 = 7p(q4 − y4)
= 7p[(q2)2 − (y2)2]
= 7p(q2 + y2)(q2 − y2)
= 7p(q2 + y2)(q + y)(q − y)
4. x2 + 12x + 36 = x2 + 2 · x · 6 + 62
= (x + 6)2
5. 9z2 − 12z + 4 = (3z)2 − 2 · 3z · 2 + 22 = (3z − 2)2
6. a3 + 24a2 + 144a
= a(a2 + 24a + 144)
= a(a2 + 2 · a · 12 + 122)
= a(a + 12)2
7. x3 + 64 = x3 + 43
= (x + 4)(x2 − 4x + 16)
8. m3 − 216 = m3 − 63
= (m − 6)(m2 + 6m + 36)
9. 3a5 − 24a2 = 3a2(a3 − 8)
= 3a2(a3 − 23)
= 3a2(a − 2)(a2 + 2a + 4)
10. t6 + 1 = (t2)3 + 13
= (t2 + 1)(t4 − t2 + 1)
17. Equation-Solving Principles
1. 7t = 70
t = 10 Dividing by 7
The solution is 10.
2. x − 5 = 7
x = 12 Adding 5
The solution is 12.
3. 3x + 4 = −8
3x = −12 Subtracting 4
x = −4 Dividing by 3
The solution is −4.
4. 6x − 15 = 45
6x = 60 Adding 15
x = 10 Dividing by 6
The solution is 10.
5. 7y − 1 = 23 − 5y
12y − 1 = 23 Adding 5y
12y = 24 Adding 1
y = 2 Dividing by 12
The solution is 2.
6. 3m − 7 = −13 + m
2m − 7 = −13 Subtracting m
2m = −6 Adding 7
m = −3 Dividing by 2
The solution is −3.
7. 2(x + 7) = 5x + 14
2x + 14 = 5x + 14
−3x + 14 = 14 Subtracting 5x
−3x = 0 Subtracting 14
x = 0
The solution is 0.
8. 5y − (2y − 10) = 25
5

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Just-in-Time Review 7
5. 1
3 y − 6 < 3
1
3 y < 9 Adding 6
y < 27 Multiplying by 3
The solution set is (−∞, 27).
6. 8 − 2w ≤ −14
−2w ≤ −22 Subtracting 8
w ≥ 11 Dividing by −2 and
reversing the inequality symbol
The solution set is [11, ∞).
19. The Principle of Zero Products
1. 2y2 + 42y = 0
2y(y + 21) = 0
2y = 0 or y + 21 = 0
y = 0 or y = −21
The solutions are 0 and −21.
2. (a + 7)(a − 1) = 0
a + 7 = 0 or a − 1 = 0
a = −7 or a = 1
The solutions are −7 and 1.
3. (5y + 3)(y − 4) = 0
5y + 3 = 0 or y − 4 = 0
5y = −3 or y = 4
y = − 3
5 or y = 4
The solutions are − 3
5 and 4.
4. 6x2 + 7x − 5 = 0
(3x + 5)(2x − 1) = 0
3x + 5 = 0 or 2x − 1 = 0
3x = −5 or 2x = 1
x = − 5
3 or x = 1
2
The solutions are − 5
3 and 1
2 .
5. t(t − 8) = 0
t = 0 or t − 8 = 0
t = 0 or t = 8
The solutions are 0 and 8.
6. x2 − 8x − 33 = 0
(x + 3)(x − 11) = 0
x + 3 = 0 or x − 11 = 0
x = −3 or x = 11
The solutions are −3 and 11.
7. x2 + 13x = 30
x2 + 13x − 30 = 0
(x + 15)(x − 2) = 0
x + 15 = 0 or x − 2 = 0
x = −15 or x = 2
The solutions are −15 and 2.
8. 12x2 − 7x − 12 = 0
(4x + 3)(3x − 4) = 0
4x + 3 = 0 or 3x − 4 = 0
4x = −3 or 3x = 4
x = − 3
4 or x = 4
3
The solutions are − 3
4 and 4
3 .
20. The Principle of Square Roots
1. x2 − 36 = 0
x2 = 36
x = √36 or x = −√36
x = 6 or x = −6
The solutions are 6 and −6, or ±6.
2. 2y2 − 20 = 0
2y2 = 20
y2 = 10
y = √10 or y = −√10
The solutions are √10 and −√10, or ±√10.
3. 6z2 = 18

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Just-in-Time Review 11
5. 3
√30 − 4 = 3
√30 − 4 ·
√30 + 4
√30 + 4
= 3√30 + 12
(√30)2 − 42
= 3√30 + 12
30 − 16
= 3√30 + 12
14
6. 4
√7 − √3 = 4
√7 − √3 ·
√7 + √3
√7 + √3
= 4√7 + 4√3
(√7)2 − (√3)2
= 4√7 + 4√3
7 − 3
= 4√7 + 4√3
4 = 4(√7 + √3)
4
= √7 + √3
7. 6
√m − √n = 6
√m − √n ·
√m + √n
√m + √n
= 6(√m + √n)
(√m)2 − (√n)2
= 6√

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4
2
2
4
4224 x
y
(1, 4)
(3, 5)
(0, 2)(4, 0)
(2, 2)
22 42
2
4
22
24
24
y
x
(25, 0)
(2, 24)
(1, 4)
(0, 3)
(24, 22)
4
2
2
4
424 x
y
(5, 1)
(2, 3)
(0, 1) (5, 1)
(2, 1)
22 42
2
4
22
24
24
y
x
(25, 2)
(4, 23)
(0, 21)
(2, 0)
(21, 25)
Chapter 1
Graphs, Functions, and Models
Exercise Set 1.1
1. Point A is located 5 units to the left of the y-axis and
4 units up from the x-axis, so its coordinates are (−5, 4).
Point B is located 2 units to the right of the y-axis and
2 units down from the x-axis, so its coordinates are (2, −2).
Point Cis located 0 units to the right or left of the y-axis
and 5 units down from the x-axis, so its coordinates are
(0, −5).
Point D is located 3 units to the right of the y-axis and
5 units up from the x-axis, so its coordinates are (3, 5).
Point E is located 5 units to the left of the y-axis and
4 units down from the x-axis, so its coordinates are
(−5, −4).
Point F is located 3 units to the right of the y-axis and
0 units up or down from the x-axis, so its coordinates are
(3, 0).
2. G: (2, 1); H: (0, 0); I: (4, −3); J: (−4, 0); K: (−2, 3);
L: (0, 5)
3. To graph (4, 0) we move from the origin 4 units to the right
of the y-axis. Since the second coordinate is 0, we do not
move up or down from the x-axis.
To graph (−3, −5) we move from the origin 3 units to the
left of the y-axis. Then we move 5 units down from the
x-axis.
To graph (−1, 4) we move from the origin 1 unit to the left
of the y-axis. Then we move 4 units up from the x-axis.
To graph (0, 2) we do not move to the right or the left of
the y-axis since the first coordinate is 0. From the origin
we move 2 units up.
To graph (2, −2) we move from the origin 2 units to the
right of the y-axis. Then we move 2 units down from the
x-axis.
4.
5. To graph (−5, 1) we move from the origin 5 units to the
left of the y-axis. Then we move 1 unit up from the x-axis.
To graph (5, 1) we move from the origin 5 units to the right
of the y-axis. Then we move 1 unit up from the x-axis.
To graph (2, 3) we move from the origin 2 units to the right
of the y-axis. Then we move 3 units up from the x-axis.
To graph (2, −1) we move from the origin 2 units to the
right of the y-axis. Then we move 1 unit down from the
x-axis.
To graph (0, 1) we do not move to the right or the left of
the y-axis since the first coordinate is 0. From the origin
we move 1 unit up.
6.
7. The first coordinate represents the year and the corre-
sponding second coordinate represents the number of cities
served by Southwest Airlines. The ordered pairs are
(1971, 3), (1981, 15), (1991, 32), (2001, 59), (2011, 72),
and (2014, 96).
8. The first coordinate represents the year and the second
coordinate represents the percent of Marines who are
women. The ordered pairs are (1960, 1%), (1970, 0.9%),
(1980, 3.6%), (1990, 4.9%), (2000, 6.1%), (2011, 6.8%),
and (2014, 7.6%).
9. To determine whether (−1, −9) is a solution, substitute
−1 for x and −9 for y.
y = 7x − 2
−9 ? 7(−1) − 2∣
∣
∣ −7 − 2
−9 ∣
∣ −9 TRUE
The equation −9 = −9 is true, so (−1, −

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Solution Manual for Precalculus: Graphs and Models, A Right Triangle Approach, 6th Edition

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