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Roots and Radical Expressions
ALGEBRA 2 LESSON 7-1
(For help, go to Lesson 5-4.)
Write each number as a square of a number.
1. 25
3. 4
2. 0.09
49
Write each expression as a square of an expression.
4. x10
5. x4y2
6. 169x6y12
7-1
Roots and Radical Expressions
ALGEBRA 2 LESSON 7-1
Solutions
4
22
2
2
1. 25 = 52
2. 0.09 = 0.32
3. 49 = 2 =
7
7
4. x10 = (x5)2
5. x4y2 = (x2y)2
6. 169x6y12 = (13x3y6)2
7-1
Roots and Radical Expressions
ALGEBRA 2 LESSON 7-1
Find all the real roots.
1
a. the cube root of 0.027, –125, and 64
Since 0.33 = 0.027, 0.3 is the cube root of 0.027.
Since (–5)3 = –125, –5 is the cube root of –125.
Since
1
4
3
=
1
1
1
,
is the cube root of
.
64 4
64
81
b. the fourth roots of 625, –0.0016, and 625
Since 54 = 625 and (–5)4 = 625, 5 and –5 are fourth roots of 625.
There is no real number with a fourth power of –0.0016.
3
Since 5
4
81
–3
= 625 and 5
4
81
3
3
81
= 625 , 5 and – 5 are fourth roots of 625 .
7-1
Roots and Radical Expressions
ALGEBRA 2 LESSON 7-1
Find each real-number root.
a.
3
–1000
=
3
(–10)3
= –10
b.
Rewrite –1000 as the third power of a number.
Definition of nth root when n = 3, there is only one
real cube root.
–81
There is no real number whose square is –81.
7-1
Roots and Radical Expressions
ALGEBRA 2 LESSON 7-1
Simplify each radical expression.
9x10
a.
9x10 =
32(x5)2 =
(3x5)2 = 3| x5|
Absolute value symbols ensure that the root is positive when x is negative.
b.
3
a3b3
3
(ab)3 = ab
Absolute value symbols must not be used here. If a or b is negative, then
the radicand is negative and the root must also be negative.
7-1
Roots and Radical Expressions
ALGEBRA 2 LESSON 7-1
(continued)
c.
4
x16y4
4 x16y14
=
4 (x4)4(y)4
= x4 |y|
Absolute value symbols ensure that root is positive when y is negative.
They are not needed for x because x2 is never negative.
7-1
Roots and Radical Expressions
ALGEBRA 2 LESSON 7-1
A cheese manufacturer wants to ship cheese balls that weigh
from 10 to 11 ounces in cartons that will have 3 layers of 3 cheese
balls by 4 cheese balls. The weight of a cheese ball is related to its
3
diameter by the formula w = d , where d is the diameter in inches and
5
w is the weight in ounces. What size cartons should be used? Assume
whole-inch dimensions.
Find the diameter of the cheese balls.
10 <
Write an inequality.
– 11
– w<
d3
10 <
– 5 <
– 11
Substitute for w in terms of d.
3
50 <
– d <
– 55
Multiply by 5.
d3 <
– 3 55
3.68 <
– d <
– 3.80
3
50 <
–
3
Take cube roots.
The diameters range from 3.68 in. to 3.80 in.
7-1
Roots and Radical Expressions
ALGEBRA 2 LESSON 7-1
(continued)
The length of a row of 4 of the largest cheese balls is 4(3.80 in.) = 15.2 in.
The length of a row of 3 of the largest cheese balls 3(3.80 in.) = 11.4 in.
The manufacturer should order cartons that are 16 in. long by 12 in. wide
by 12 in. high to accommodate three dozen of the largest cheese balls.
7-1
Roots and Radical Expressions
ALGEBRA 2 LESSON 7-1
12. 10 , – 10
24. 8b24
1. 15, –15
13. 6
25. –4a
2. 0.07, –0.07
14. –6
26. 3y2
3. none
15. no real-number root
27. x2|y3|
4. 8 , – 8
16. 0.6
28. 2y2
17. –4
29. 1.34 in.
6. 0.5
18. –4
30. 1.68 ft
7. – 1
19. –3
31. 0.48 cm
8. 0.07
20. no real-number root
32. 0.08 mm
9. 2, –2
21. 4|x|
33. 10, –10
10. none
22. 0.5|x3|
34. 1, –1
11. 0.3, –0.3
23. x4|y9|
35. 0.5, –0.5
pages 366–367 Exercises
13
5. –4
13
2
3
3
7-1
Roots and Radical Expressions
ALGEBRA 2 LESSON 7-1
36. 2 , – 2
3
3
37.
47. –y4
3
–64,
38. a.
b.
6
64, –
3
–64,
48. k3
64
49. –k3
35 ft
20 ft longer
50. |x + 3|
39. 0.5
51. (x + 1)2
40. 1
52. |x|
41. 0.2
53. x2
42. 1
54. |x3|
43. 2|c|
55. Answers may vary. Sample:
3
4
44. 3xy2
3
3
45. 12y2z2x
3
xy
–8x6, –
4
16x8,
5
–32x10
56. a. for all positive integers
b. for all odd positive integers
46. y4
7-1
Roots and Radical Expressions
ALGEBRA 2 LESSON 7-1
57. Yes, because 10 is really 101.
69. m4
58. All; x2 is always nonnegative.
70. B
59. Some; true only for x > 0.
71. I
60. Some; true only for x = –1, 0, 1.
72. B
61. Some; true only for x > 0.
73. H
62. |m|
74. [2]
63. m2
64. |m3|
65. m4
66. m
67. m2
68. m3
3
x2y4 equals x3y6 whenever
y = 0, regardless of the value of x.
This is true because x2y4 and x3y6
will be 0 whenever y = 0. x2y4 also
3
equals x3y6 when x > 0. This is
true because x2y4 = |x|y2 and
3 3 6
x y = xy2, and |x|y2 = xy2 when
x > 0.
[1] answer only, with no explanation
7-1
Roots and Radical Expressions
ALGEBRA 2 LESSON 7-1
75. x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + y5
76. 16 – 96y + 216y2 – 216y3 + 81y4
77. 729x6 – 7290x5 + 30,375x4 – 67,500x3 + 84,375x2 – 56,250x + 15,625
78. 128a7 – 448a6b + 672a5b2 – 560a4b3 + 280a3b4 – 84a2b5 + 14ab6 – b7
79. y = x(2x – 7)(2x + 7)
80. y = (9x + 2)2
81. y = 4x(x + 1)2
82. y = 2x(6x + 1)(x + 1)
83. y = 3(x – 0)2 – 7
84. y = –2 – x – 1
2
4
1
2
85. y = (x + 4) – 5
4
– 79
8
7-1
Roots and Radical Expressions
ALGEBRA 2 LESSON 7-1
1. Find all the real square roots of each number.
a. 121 ± 11
b. –49 none
d. – 1
c. 64 ± 8
25
none
2. Find all the real cube roots of each number.
a. –8000 –20
b.
1
216
1
6
3. Find each real-number root.
a. 0.49 0.7
b. 3 125 5
4. Simplify each radical expression.
a.
3
–8x3
–2x
b.
16y4
4y2
c. –
c.
81 –9
36x14
d.
4 –625
none
6 | x7 |
5. The formula for the volume of a cone with a base of radius r and height
r is V = 1
3
r 3. Find the radius to the nearest hundredth of a centimeter
if the volume is 40 cm3.
about 3.37 cm
7-1
Multiplying and Dividing Radical Expressions
ALGEBRA 2 LESSON 7-2
(For help, go to page 362.)
Find each missing factor.
1. 150 = 52(
3. 48 = 42(
5. 3a3b4 = (
)
)
)3(3b)
2. 54 = (
)3(2)
4. x5 = (
)2(x)
6. 75a7b8 = (
7-2
)2(3a)
Multiplying and Dividing Radical Expressions
ALGEBRA 2 LESSON 7-2
Solutions
1. 150 = 52(6)
2. 54 = (3)32
3. 48 = 42(3)
4. x5 = (x2)2(x)
5. 3a3b4 = (ab)3(3b)
6. 75a7b8 = (5a3b4)2(3a)
7-2
Multiplying and Dividing Radical Expressions
ALGEBRA 2 LESSON 7-2
Multiply. Simplify if possible.
a.
b.
c.
3•
12
3•
12 =
3
–16 •
3
4
3
–16 •
3
4=
–4 •
16
3 • 12 =
3
36 = 6
–16 • 4 =
3
64 = –4
The property for multiplying radicals does not apply.
–4 is not a real number.
7-2
Multiplying and Dividing Radical Expressions
ALGEBRA 2 LESSON 7-2
Simplify each expression. Assume all variables are positive.
50x5
a.
50x5 =
=
52 • 2 • (x2)2 • x
Factor into perfect squares.
52 • (x2)2 •
n
= 5x2
b.
2•x
2x
54n8
3
54n8 =
3
33 • 2 • (n2)3 • n2
=
3
33(n2)3 •
3
2n2
n
b =
n
ab
definition of nth root
3
= 3n2
a•
3
2n2
Factor into perfect cubes.
n
a•
n
b =
n
ab
definition of nth root
7-2
Multiplying and Dividing Radical Expressions
ALGEBRA 2 LESSON 7-2
Multiply and simplify
are positive.
3
25xy8 •
3
3
25xy8 • 5x4y3
=
3
53x3(y3)3 • x2y2
=
3
53x3(y3)3 •
= 5xy3
3
25xy8 •
n
5x4y3 =
3
3
a•
3
n
5x4y3 . Assume all variables
b =
n
ab
Factor into perfect cubes.
x2y2
x2y2
n
a•
n
b =
n
ab
definition of nth root
7-2
Multiplying and Dividing Radical Expressions
ALGEBRA 2 LESSON 7-2
Divide and simplify. Assume all variables are positive.
a.
3
=
=
–81
3
3
b.
3
3
=
–81
3
=
192x8
3
3
3x
192x8
3x
=
3
–27
=
3
64x7
=
3
(–3)3
=
3
43(x2)3 • x
= –3
= 4x2
7-2
3
x
Multiplying and Dividing Radical Expressions
ALGEBRA 2 LESSON 7-2
Rationalize the denominator of each expression. Assume that
all variables are positive.
Method 1:
a.
3
3
5
5
=
3
5
Rewrite as a square root of
a fraction.
=
3•5
5•5
Then make the denominator
a perfect square.
=
15
52
=
15
5
7-2
Multiplying and Dividing Radical Expressions
ALGEBRA 2 LESSON 7-2
(continued)
Method 2:
a.
3
3
5
5
=
3•
5•
=
15
5
5
5
Multiply the numerator and
denominator by 5 so the denominator
becomes a whole number.
7-2
Multiplying and Dividing Radical Expressions
ALGEBRA 2 LESSON 7-2
(continued)
b.
c.
x5
3
5
4y
3
5
4y
3x2y
x5
3x2y
=
=
=
=
x5 •
3x2y
3x2y •
3x2y
3x7y
3x2y
x3
x
3xy
2
3x y
=
3
5 • 42y2
4y • 42y2
=
3
80y2
=
3xy
3y
=
7-2
43y3
2
3 10y2
4y
3 10y2
2y
Rewrite the fraction
so the denominator
is a perfect cube.
Multiplying and Dividing Radical Expressions
ALGEBRA 2 LESSON 7-2
The distance d in meters that an object will fall in t seconds is
given by d = 4.9t 2. Express t in terms of d and rationalize the
denominator.
d = 4.9t 2
d
t 2 = 4.9
t=
=
=
d
4.9
d • 10
49
10d
7
7-2
Multiplying and Dividing Radical Expressions
ALGEBRA 2 LESSON 7-2
pages 371–373 Exercises
3
12. 2a
3
3y3
y
1. 16
13.
2. 4
14. 10a3b3
3. –9
2y
25. 2x2y2
2b
3
15. –5x2y
4
24. 4x
4a2
16. 2y
5. not possible
17. 2
6. not possible
18. 8y3
7. –6
19. 7x3y4
6y
8. 6
20. 40xy
3
9. 2x
10. 3
3
11. 5x2
4x3y2
21.
3x2
22. –2x2y
2x
5y
3
30y2
27.
28.
12
5x
29.
2y
3
30x
30.
31.
7-2
2x
2
10x
4x
3
4x
2
3
45x2
3x
4
250
5
32. 5x2
33.
23. 10
x2y2
2y2
4. 4
3
26. 5x3
2
15y
5y
5
Multiplying and Dividing Radical Expressions
ALGEBRA 2 LESSON 7-2
34.
3
x 10
2y
35. r =
36. a.
b.
37. 10
Gm1m2F
F
38. 4
5
49.
39. 3x6y5
6 +3
15
6 +3
15
c. Answers may vary. Sample:
First simplify the denominator.
Since 98 = 2 • 49 = 7 2,
to rationalize the denominator,
2
.
2
2•2+ 3•2 =
7 2•2
2y
40. 20x2y3
y
41. 10 + 7
2
42. 15 + 3
21
43. 5 + 5
44. 2x
3
45.
This yields
46. x
3
x
10y
2y 2
47. 5 14x
21x
7-2
3x2
3x
3
2 25x
x
3
2xy 2
xy
51. – 33x
4x
50.
52. –4
53.
3
2
3x2 3
multiply the fraction by
2+ 6 .
14
3
48.
2
54.
3
4–6
3
2
5 +5
5
6 –2
4
55. 212 mi/h greater
56. 20
22 cm2
Multiplying and Dividing Radical Expressions
ALGEBRA 2 LESSON 7-2
3
x2y
xy
57. A product of two square roots
can be simplified in this way
only if the square roots are
real numbers. –2 and –8
are not.
64.
58. 288a5 ft
67. a = –2c, b = –6d
59. For some values; it is easy to see
that the equation is true if x = 0 or
x = 1. But when x < 0, x3 is not
3
a real number, although x2 is.
68. No changes need to be made;
since they are both odd roots,
there is no need for absolute
value symbols.
60. Check students’ work.
69. C
61. 2xy
70. H
62. 2xy2
71. A
63. 2
72. G
65.
66.
5
7-2
5
yx
x
6
x4y3
y
Multiplying and Dividing Radical Expressions
ALGEBRA 2 LESSON 7-2
73. D
x is a real number if x >
– 0 and –x is a real number if
x<
– 0. So the only value that makes x • –x a real
number is x = 0.
[1] answer only OR error describing value(s) of variables
74. [2]
75. [4] You should multiply
3
3 •
2x
3
3
4x2
4x2
=
3
3
3
3 •
2x
3
3 by
2x
3
3
4x2
4x2
3
4x2 because
4x2
3
=
3
12x2
8x3
3
=
12x2 ,
2x
which has a denominator without a radical.
[3] appropriate methods, but with one minor error
[2] major error, but subsequent steps consistent with that error
[1] correct final expression, but no work shown
7-2
Multiplying and Dividing Radical Expressions
ALGEBRA 2 LESSON 7-2
76. 11|a45|
88. 25
77. –9c24d32
89. 25
78. –4a27
90. 121
79.
4
91. 121
4
92. 1
36
2y5
80. 0.5|x3|
81. x2y5
82.
93. 0.0225
2|x9|y24
94. 9
64
95. 9
100
83. 0.08x20
84. y2 – 4y + 16, R –128, not a factor
85. x2 – 3x + 9, a factor
86. 3a2 – a, R 4, not a factor
87. 2x3 + x2 + 2x, R 10, not a factor
7-2
Multiplying and Dividing Radical Expressions
ALGEBRA 2 LESSON 7-2
Assume that all variables are positive.
1. Multiply. Simplify if possible.
a.
5•
45 15
b.
3
4•
b.
3
–243x3y10
b.
3
10x2y4 •
3
2000
20
2. Simplify.
a.
8x5
2x2
2x
–3xy3
3
9y
3. Multiply and simplify.
a.
18x3 •
2x2y3
6x2y
xy
3
4x2y
2xy
4. Divide and simplify.
a.
128x3
2xy
8x
y
y
3
b.
3
270x
10xy2
3
3
y
y
5. Rationalize the denominator of each expression.
a.
7x
3
21x
3
b.
3
3
7-2
x2
4
3
2x2
2
3
5xy2
Binomial Radical Expressions
ALGEBRA 2 LESSON 7-3
(For help, go to Lesson 5-1 or Skills Handbook page 853.)
Multiply.
1. (5x + 4)(3x – 2)
2. (–8x + 5)(3x – 7)
3. (x + 4)(x – 4)
4. (4x + 5)(4x – 5)
5. (x + 5)2
6. (2x – 9)2
7-3
Binomial Radical Expressions
ALGEBRA 2 LESSON 7-3
Solutions
1. (5x + 4)(3x – 2)
= (5x)(3x) + (5x)(–2) + (4)(3x) + (4)(–2)
= 15x2 – 10x + 12x – 8
= 15x2 + 2x – 8
2. (–8x + 5)(3x – 7)
= (–8x)(3x) + (–8x)(–7) + (5)(3x) + (5)(–7)
= –24x2 + 56x + 15x – 35
= –24x2 + 71x – 35
3. (x + 4)(x – 4) = x2 – 42 = x2 – 16
4. (4x + 5)(4x – 5) = (4x)2 – 52 = 16x2 – 25
5. (x + 5)2 = x2 + 2(5)x + 52 = x2 + 10x + 25
6. (2x – 9)2 = (2x)2 – 2(2x)(9) + 92 = 4x2 – 36x + 81
7-3
Binomial Radical Expressions
ALGEBRA 2 LESSON 7-3
Add or subtract if possible.
a. 7
xy + 3
xy
7
xy + 3
xy = (7 + 3)
= 10
b. 2
3
x–2
3
xy
xy
Distributive Property
Subtract.
5
The radicals are not like radicals. They cannot be combined.
7-3
Binomial Radical Expressions
ALGEBRA 2 LESSON 7-3
The rectangular window shown below is made up of three
equilateral triangles and two right triangles. The equilateral triangles
have sides of length 4 feet. What is the perimeter of the window?
The height of an equilateral triangle with side length 4 ft is 2
So the window’s height is 2
3 ft.
The length of the window is 8 ft.
The perimeter is 2(2
3 + 8) ft or 16 + 4
7-3
3 ft.
3 ft.
Binomial Radical Expressions
ALGEBRA 2 LESSON 7-3
3
20 –
45 + 4
Simplify 3
20 –
80 = 3
22 • 5 –
=3•2
=6
45 + 4
32 • 5 + 4
5–3
5–3
5+4•4
5 + 16
= (6 – 3 + 16)
= 19
80.
5
5
7-3
5
42 • 5
5
Factor each radicand.
Simplify each radical.
Multiply.
Use the Distributive Property.
Binomial Radical Expressions
ALGEBRA 2 LESSON 7-3
Multiply (2 + 4
(2 + 4
3)(1 – 5
3)(1 – 5
3) = 2 • 1 – 2 • 5
= 2 + (4 – 10)
= –58 – 6
3+4
3 – 60
3
7-3
3).
3•1–4
3•5
3
Use FOIL.
Distributive Property
Binomial Radical Expressions
ALGEBRA 2 LESSON 7-3
7)(3 –
Multiply (3 +
(3 +
7)(3 –
7) = 32 – (
7)2
7).
(a + b)(a – b) = a2 – b2
=9–7=2
7-3
Binomial Radical Expressions
ALGEBRA 2 LESSON 7-3
Rationalize the denominator of
2–
3
4+
3
=
=
=
=
2–
3
4+
3
•
2–
3
4+
3
8–2
4–
3
4–
3
•
3–4
42 – (
11 – 6
13
4–
4–
3
4–
3
3+(
3)2
Simplify.
3
7-3
3
4+
3
3 is the conjugate of 4 +
Multiply.
3)2
2–
3.
Binomial Radical Expressions
ALGEBRA 2 LESSON 7-3
pages 376–378 Exercises
1. 6
2. 4
3
12. 2
4
2+2
6
13. 8 + 4
3
14. 23 + 7
3. cannot combine
15. 63 – 38
4. –2
16. 8 + 2
x
5. cannot combine
6. 5
3 x2
4
3
24. 12 3 + 8
23
25. 13 + 7
5
26.
7
2
11 + 8 2
–14
27. 13
2
28. 8
15
3
17. 49 + 12
13
29. 48
2x
18. 38 + 12
10
30. 5
3–4
7. 33
2
19. 14
31. 33y
8. 13
5
20. 4
32. –2
21. –40
33. –11 +
22. –2
34. 8 +
9. 7
2
10.
5
3
2
11.
9
3
3–6
3
2
3
23. –2 + 2
7-3
3
3
2
6
2
21
10
35. 17 + 31
2
Binomial Radical Expressions
ALGEBRA 2 LESSON 7-3
36. –36 – 15
37. x + 3
46. a must be twice a perfect square.
2
47. 14
3x + 6
38. 8y – 22
48. Answers may vary. Sample: Without
simplifying first, you must estimate three
separate square roots, and then add the
estimates. If they are first simplified, then
they can be combined as 13. Then only
one square root need be estimated.
2y + 30
39. 89 + 42 3
–239
3–
40. 2
2
3– 7
2
3
42. 2 + 3 4
2 4
3
43. x + 5 x
x
41.
44. 2
3
2–
3
7m
49. Answers may vary.
Samples: ( 7 + 2)( 7 – 2);
(2 2 + 5)(2 2 – 5)
50. 60 – 20 2 s, or about 4.53 s
12
45. The reciprocal is –1 +
2
which is one less than
5,
51. – 1
1+ 5 .
52. 4
2
7-3
7
2
3
Binomial Radical Expressions
ALGEBRA 2 LESSON 7-3
53. (a = 0 and b > 0) or (b = 0 and a > 0)
54. In the second step the exponent was incorrectly distributed: (a – b)x =/ ax – bx.
55. a. m and n can be any positive integers.
b. m must be even or n must be odd.
c. m must be even, and n can be any positive integer.
56. B
57. I
58. D
59. I
60. D
61. [2] (1 –
3
8)(1 +
3
8) = (1 – 2)(1 + 2) = 1 – 4 = –3, which is rational.
[1] the value –3 only OR yes only OR minor error in calculation
7-3
Binomial Radical Expressions
ALGEBRA 2 LESSON 7-3
62. [4]
2
5+2 2
2
5+2 2
3
5–2 2
10 – 4 2
25 – 4 • 2
10 – 4
17
10 – 4
3
5–2 2
• 5–2 2
5–2 2
• 5+2 2
5+2 2
– 15 + 6 2
25 – 4 • 2
2 – 15 + 6 2
17
2 – 15 – 6 2
17
–
=
63. 3
–
64. x
=
2
15
75. ±2 5
68. 7x2
=
–5 – 10
17
2
2
3n
n
3
70. 100x2
5x
69.
71. 2, –1 ± i
3
72. –10, 5 ± 5i
and gets
73. 1 , –1 ± i 3
5
[1] answer only, no work shown
7-3
7
5
76. ± i , ± 1
3
3
67. 2x
[2] correct method, but finds
2 + 15 + 6
17
2
74. ±
93
3
66.
[3] minor error, but appropriate method
10 – 4
17
25 + 2
17
2
65. 4
=
=
3
10
3
Binomial Radical Expressions
ALGEBRA 2 LESSON 7-3
Simplify.
1. 6
10 – 4
3. 2
27 +
10
48
5. Multiply (7 + 2
2
10
10
2. 8
3
4.
3) (5 –
3
3). 29 + 3
3
x+5
128 –
x
54
13
3
3
x
2
3
6. Rationalize the denominator in 3 – 2 7 .
10 – 7
7-3
3
3
16 – 17
93
7
Rational Exponents
ALGEBRA 2 LESSON 7-4
(For help, go to page 362.)
Simplify.
1. 2–4
2. (3x)–2
3. (5x2y)–3
4. 2–2 + 4–1
5. (2a–2b3)4
6. (4a3b–1)–2
7-4
Rational Exponents
ALGEBRA 2 LESSON 7-4
Solutions
1
1
=
24
16
1
1
(3x)–2 =
=
(3x)2
9x2
1
(5x2y)–3 = 12 3 =
125x6y3
(5x y)
2–2 + 4–1 = 12 + 11 = 1 + 1 = 2 = 1
2
4
4
4
4
2
12
(2a–2b3)4 = 24a–8b12 = 16b8
a
2
2
(4a3b–1)–2 = 4–2a–6b2 = b2 6 = b 6
4a
16a
1. 2–4 =
2.
3.
4.
5.
6.
7-4
Rational Exponents
ALGEBRA 2 LESSON 7-4
Simplify each expression.
a. 64
1
3
1
3
64 =
3
64
Rewrite as a radical.
=
3
43
Rewrite 64 as a cube.
=4
1
2
1
2
1
2
1
2
b. 7 • 7
7 •7 =
=7
Definition of cube root.
7•
7
Rewrite as radicals.
By definition,
square is 7.
7-4
7 is the number whose
Rational Exponents
ALGEBRA 2 LESSON 7-4
(continued)
1
3
1
3
1
3
1
3
c. 5 • 25
5 • 25 =
3
=
3
=5
5•
3
5 • 25
25
Rewrite as radicals.
property for multiplying radical expressions
By definition,
cube is 5.
7-4
3
5 is the number whose
Rational Exponents
ALGEBRA 2 LESSON 7-4
2
7
a. Write x and y –0.4 in radical form.
2
7
x =
7
x2
or
7
x
2
b. Write the radical expressions
4 c3
=c
3
4
y
–0.4
=y
4 c3 and
3
b
7-4
5
2
–5
3
=b
=
1
or
5 y2
1
5 y
2
b 5 in exponential form.
5
3
Rational Exponents
ALGEBRA 2 LESSON 7-4
The time t in hours needed to cook a pot roast that weighs p
pounds can be approximated by using the equation t = 0.89p0.6. To
the nearest hundredth of an hour, how long would it take to cook a
pot roast that weighs 13 pounds?
t = 0.89p0.6
= 0.89(13)0.6
4.15
Write the formula.
Substitute for p.
Use a calculator.
7-4
Rational Exponents
ALGEBRA 2 LESSON 7-4
Simplify each number.
a. (–27)
2
3
Method 1
2
3
Method 2
2
2
3
(–27) = (
3
–27)2
= (–3)2
=(
3
(–3)3)2
=9
= (–3)2 = 9
(–27) =
((–3)3) 3
7-4
Rational Exponents
ALGEBRA 2 LESSON 7-4
(continued)
b. 25–2.5
Method 1
25–2.5
Method 2
= 25
5
–2
5
2 –2
25–2.5 = 25
5
–2
= (5 )
1
5
–
2 2
=
=5
25
5
2
= 5–5
=
1
= 55
1
= 3125
=
7-4
1
25)5
(
1
55
=
1
3125
Rational Exponents
ALGEBRA 2 LESSON 7-4
Write
2
(243a–10) 5
2
(243a–10) 5
in simplest form.
2
=
(35a–10) 5
=
35 5
2
•
2
a(–10) 5
= 32a–4
32
= 4
a
=
9
a4
7-4
Rational Exponents
ALGEBRA 2 LESSON 7-4
pages 382–384 Exercises
12.
1. 6
13.
2. 3
14.
3. 7
15.
4. 10
7
x2 or (
7
x)2
22. a
2
3
5
y2 or (
5
y)2
23. a
2
3
24. c
1
2
1
1
or
8
8
y9
1
or
4
t3
4
x3 or (
5. –3
16.
6. 6
17.
7. 8
18. (–10) 2
5
y6 or (
y
1
9
t
3
x)3
5
y)6
25. 25x2y2
26.
72.8 m
27.
15.1 m
28.
7.9 m
29.
1.6 m
1
1
2
8. 3
19. 7 x
9. 3
10.
6
11.
5
20. (7x)
3
2
21. (7x)
3
2
x
x
3
2
30. 4
31. 16
32. 4
33. 64
7-4
Rational Exponents
ALGEBRA 2 LESSON 7-4
34. 1
44. 1
16
x
35. 8
45. x
36. 64
37. 1000
1
x2
39. 14
x
40. 1
54. 2,097,152
46.
55. 1,000,000,000 or 109
13
3
56. 1
y4
4
57. 1
8
58. 1
36
x3
47. y2
38.
x8
48. x3y9
49.
2
3
3x
41. 5 2
x3
42. – 3
x3
43. –2y3
59. 16
y5
60. – 1
x10
81
50. –7
61. 10
51. –3
62. 78%, 61%, 37%
52. 64
63. 635.87
53. 729
64. 768
7-4
Rational Exponents
ALGEBRA 2 LESSON 7-4
65. x
1
2
66. y
4
5
67. x
1
2
68.
1
y2
69.
1
x6
70.
74.
1
x
75. 1
76.
79. a. Answers may vary.
7
24
1
Sample: 4 – 5 2 ,
1
x3
1
2(4 –
b. no
1
(xy) 2
1
y4
1
1
4
x y
71. 4x7
9y9
72. 9y8
4x6
73. 1
5
6
1
52 ),
77. The cube root of –64 is –4,
1
–(64) 2 = –
64 = –8.
78. The exponent 1 applies
2
only to the 5, not to the 25.
13
b.
7-4
4
x2
x
81. 49
82. 9
83. x2
84. 1
x 36
x•
x•
x • x = x2, so
1
3
which equals –(64) .
The square root of –64 is
not a real number, but
x•
80. a.
4–5
2
=
1
2
(x ) 4
1
2
=
x =
x2 =
2
x4
x
1
2
=x =
Rational Exponents
ALGEBRA 2 LESSON 7-4
85. 3
2
96. 21
2
86. 9
97. 1 + 3
87. 33.13 mi/h
98. 15 – 4
88. B
99. 4 + 6 5 + 2 10 + 15 2
89. H
100. 10 – 8 2
14
–41
90. [2] x = 1
16
[1] minor error
91. B
7
101.
4x(x2
– 2x + 4)
102. (x + 2)2
103. (x – 9)2
92. A
104. (4a – 3b)(4a + 3b)
93. C
105. (5x – 4y)2
94. A
95. 4
5
106. (3x + 8)2
3
3
7-4
Rational Exponents
ALGEBRA 2 LESSON 7-4
1. Simplify each expression.
a. 100
1
2
b. (–64)
10
1
3
–4
2. Write each expression in radical form.
a. x
1
7
7
x
b. y
4
5
5
y4
or
5
4
y
c.
k1.8
5
k9
or
5
k
9
d. a
1
–6
3. A container with curved sides is 10 ft tall. When water is in the
container to a depth of h ft, the number of cubic feet of water can be
approximated by using the formula V = 0.95h2.9. Find the amount of
water in the container when the depth of the water is 7.5 ft. Round to
the nearest hundredth. 327.64 ft3
4. Simplify each number.
a. (–64)
4
3
2
3
5. Write x y
256
1
– 4 –4
b. 810.75
in simplest form.
y
8
x3
7-4
27
1
6a
Solving Radical Equations
ALGEBRA 2 LESSON 7-5
(For help, go to Lesson 5-4.)
Solve by factoring.
1. x2 = –x + 6
2. x2 = 5x + 14
3. 2x2 + x = 3
4. 3x2 – 2 = 5x
5. 4x2 = –8x + 5
6. 6x2 = 5x + 6
7-5
Solving Radical Equations
ALGEBRA 2 LESSON 7-5
Solutions
1. x2 = –x + 6; x2 + x – 6 = 0
Factors of –6 with a sum of 1:
3 and –2
(x + 3)(x – 2) = 0
x + 3 = 0 or
x–2=0
x = –3
or
x=2
2. x2 = 5x + 14; x2 – 5x – 14 = 0
Factors of –14 with a sum of –5:
–7 and 2
(x – 7)(x + 2) = 0
x – 7 = 0 or
x+2=0
x=7
or
x = –2
3. 2x2 + x = 3; 2x2 + x – 3 = 0
(2x + 3)(x – 1) = 0
2x + 3 = 0 or
x–1=0
x=–3
or
x=1
4. 3x2 – 2 = 5x; 3x2 – 5x – 2 = 0
(3x + 1)(x – 2) = 0
3x + 1 = 0 or
x–2=0
1
x=–
or
x=2
5. 4x2 = –8x + 5; 4x2 + 8x – 5 = 0
(2x – 1)(2x + 5) = 0
2x – 1 = 0 or
2x + 5 = 0
x= 1
or
x=–5
6. 6x2 = 5x + 6; 6x2 – 5x – 6 = 0
(3x + 2)(2x – 3) = 0
3x + 2 = 0 or
2x – 3 = 0
x=– 2
or
x= 3
3
2
2
2
7-5
3
2
Solving Radical Equations
ALGEBRA 2 LESSON 7-5
Solve –10 +
–10 +
(
2x + 1 = –5.
2x + 1 = –5
2x + 1 = 5
Isolate the radical.
2x + 1 )2 = 52
2x + 1 = 25
2x = 24
x = 12
Square both sides.
Check: –10 + 2x + 1 = –5
–10 + 2(12) + 1 –5
–10 + 25 –5
–10 + 5 –5
–5 = –5
7-5
Solving Radical Equations
ALGEBRA 2 LESSON 7-5
3
5
Solve 3(x + 1) = 24.
3
5
3(x + 1) = 24
3
5
(x + 1) = 8
3
5
5
3
5
3
1)1
5
3
((x + 1) ) = 8
(x +
Divide each side by 3.
=8
x + 1 = 32
x = 31
5
Raise both sides to the 3 power.
3
5
Multiply the exponents 5 and 3 .
Simplify.
3
5
3
5
3
5 5
Check: 3(x + 1) = 24
3(31 + 1) 24
3(2 ) 24
3(2)3 24
24 = 24
7-5
Solving Radical Equations
ALGEBRA 2 LESSON 7-5
An artist wants to make a plastic sphere for a sculpture. The
plastic weighs 0.8 ounce per cubic inch. The maximum weight of the
sphere is to be 80 pounds. The formula for the volume V of a sphere
4
is V = • r 3, where r is the radius of the sphere.
3
What is the maximum radius the sphere can have?
Relate: volume of sphere • density of plastic <
– maximum weight
Define: Let r = radius in inches.
4
Write: 3 •
r 3 • 0.8 <
– 80
7-5
Solving Radical Equations
ALGEBRA 2 LESSON 7-5
(continued)
r3
4
3
• 0.8 <
– 80
3 • 80
• 0.8
r3 <
– 4•
75
r3 <
–
r <
– 2.88
Use a calculator.
The maximum radius is about 2.88 inches.
7-5
Solving Radical Equations
ALGEBRA 2 LESSON 7-5
Solve
x + 2 – 3 = 2x. Check for extraneous solutions.
x + 2 – 3 = 2x
(
x + 2 = 2x + 3
Isolate the radical.
x + 2)2 = (2x + 3)2
Square both sides.
x + 2 = 4x2 + 12x + 9
Simplify.
0 = 4x2 + 11x + 7
Combine like terms.
0 = (x + 1)(4x + 7)
Factor.
x+1=0
or 4x + 7 = 0
x = –1 or
Factor Theorem
7
x=–4
7-5
Solving Radical Equations
ALGEBRA 2 LESSON 7-5
(continued)
Check:
x + 2 – 3 = 2x
–1 + 2 – 3
x + 2 – 3 = 2x
–7 +2–3
2(–1)
4
–2
1 –3
4
–2 = –2
1 –3
2
1–3
2 –7
4
–7
2
–7
2
– 5 =/ – 7
2
2
The only solution is –1.
7-5
Solving Radical Equations
ALGEBRA 2 LESSON 7-5
2
3
1
3
Solve (x + 1) – (9x + 1) = 0. Check for extraneous
solutions.
2
3
1
3
(x + 1) – (9x + 1) = 0
2
3
(x + 1) = (9x + 1)
2
3 3
1
3
1
3 3
((x + 1) ) = ((9x + 1) )
(x + 1)2 = 9x + 1
x2 + 2x + 1 = 9x + 1
x2 – 7x = 0
x(x – 7) = 0
x = 0 or x = 7
7-5
Solving Radical Equations
ALGEBRA 2 LESSON 7-5
(continued)
Check:
1
2
(x + 1) 3 – (9x + 1) 3 = 0
2
3
(0 + 1) – (9(0) + 1)
2
3
(1) – (1)
2
3
1
3
1
3
1
(1) –
(12) 3
2
3
2
3
1 –1
0
0
0
=0
Both 0 and 7 are solutions.
7-5
2
1
(x + 1) 3 – (9x + 1) 3 = 0
2
3
(7 + 1) – (9(7) + 1)
1
3
1
3
2
3
(8) – (64 )
2
3
(8) –
1
(82) 3
2
3
2
3
0
0
0
8 –8 =0
Solving Radical Equations
ALGEBRA 2 LESSON 7-5
pages 388–390 Exercises
12. 0
24. 6
1. 16
13. 30.6 ft
25. 2
2. 1
14. 4 in.
26. –2
3. 22
15. 3
27. 5
4. 4
16. 1
28. –3
5. 23
17. –3, –4
29. 5
6. 2
18. 9
30. –2
7. 3
19. 1
31. s =
8. 29, –27
20. 1
9. 18
21. 3
3
A; 4
2 m,
or about 5.7 m
32. a. s =
2
3A
3
10. 78
22. –2
b. about 8.8 in.
11. 8
23. 1
c. about 15.2 in.
7-5
Solving Radical Equations
ALGEBRA 2 LESSON 7-5
33. a.
b.
c.
40. 6.5
41. 9, –7
42. 81
16
43. 9
d. The graph of each pair consists of two straight lines,
one of which is horizontal. They intersect at different
points, but these points have the same x-value,
about 1.236.
34. 8
44. 2
45. –1, –6
46. 2
47. 7
35. 4
48. 25
36. 5
49. 10
37. 23
50. –1
38. 1
51. 5
4
39. 5.0625
7-5
Solving Radical Equations
ALGEBRA 2 LESSON 7-5
v2
52. d =
64
58. a. A counterexample is a = 3, b = –3.
b. A counterexample is a = –5, b = 3.
53. Answers may vary. Sample:
x – 3 = 3x + 5
59. 12
60. 79
54. 1
61. 5
55. 2
4
62. 1
25
56. 0
57. Plan 1: Use a calculator to evaluate
2 + 2 and store the result as x.
Evaluate x + 2 and store the result
as x. Continue this procedure about
seven times until it becomes clear that
the values are approaching 2. Plan 2:
The given equation is equivalent to
x = 2 + x. Solve this equation to find
that x = 2.
7-5
63. 27
64. 2
65. 16
66. 125
67. 6
2
68. 8
5
Solving Radical Equations
ALGEBRA 2 LESSON 7-5
69. 3
3
2
81. 21
70. 1
82. 1
71. 128
83. 6
72. 25
84. 7
73. 2
85. 3, 4
3
74.
1
106
86. 3, 5
75. 7
87. –5, –4
76. 210
88. –2, – 2
77. 60
78. 1680
79. 24
3
89. – 1 , – 4
3
3
90. –2, – 3
4
80. 10
7-5
Solving Radical Equations
ALGEBRA 2 LESSON 7-5
Solve each equation. Check for extraneous solutions.
1. 7 +
2x – 1 = 10
1
3
2. 4(x – 9) = 8
3.
2x – 1 = x – 8
2
3
5
17
13
4. (4x + 3) = (16x + 44)
1
3
–
7 5
4, 4
5. A circular table is to be made that will have a top covered with material
that costs $3.50 per square foot. The covering is to cost no more than
$60. What is the maximum radius for the top of the table?
about 2.34 ft
7-5
Function Operations
ALGEBRA 2 LESSON 7-6
(For help, go to Lesson 2-1.)
Find the domain and range of each function.
1. {(0, –5), (2, –3), (4, –1)}
2. {(–1, 0), (0, 0), (1, 0)}
3. ƒ(x) = 2x – 12
4. g(x) = x2
Evaluate each function for the given value of x.
5. Let ƒ(x) = 3x + 4. Find ƒ(2).
6. Let g(x) = 2x2 – 3x + 1. Find g(–3).
7-6
Function Operations
ALGEBRA 2 LESSON 7-6
Solutions
1. {(0, –5), (2, –3), (4, –1)}
The domain is {0, 2, 4}. The range is {–5, –3, –1}.
2. {(–1, 0), (0, 0), (1, 0)}
The domain is {–1, 0, 1}. The range is {0}.
3. ƒ(x) = 2x – 12
The domain is the set of all real numbers. The
range is the set of all real numbers.
4. g(x) = x2
The domain is the set of all real numbers. The
range is the set of all nonnegative real numbers.
5. ƒ(x) = 3x + 4; ƒ(2) = 3(2) + 4 = 6 + 4 = 10
6. g(x) = 2x2 – 3x + 1; g(–3) = 2(–3)2 – 3(–3) + 1 =
2(9) + 9 + 1 = 18 + 10 = 28
7-6
Function Operations
ALGEBRA 2 LESSON 7-6
Let ƒ(x) = –2x + 6 and g(x) = 5x – 7. Find ƒ + g and ƒ – g
and their domains.
(ƒ + g)(x) = ƒ(x) + g(x)
(ƒ – g)(x) = ƒ(x) – g(x)
= (–2x + 6) + (5x – 7)
= (–2x + 6) – (5x – 7)
= 3x – 1
= –7x + 13
The domains of ƒ + g and ƒ – g are the set of real numbers.
7-6
Function Operations
ALGEBRA 2 LESSON 7-6
ƒ
Let ƒ(x) = x2 + 1 and g(x) = x4 – 1. Find ƒ • g and
and
g
their domains.
ƒ(x)
ƒ
(x)
=
g(x)
g
(ƒ • g)(x) = ƒ(x) • g(x)
=
(x2
=
x6
+
+
1)(x4
x4
–
x2 + 1
= 4
x –1
– 1)
x2
x2 + 1
=
(x2 + 1)(x2 – 1)
–1
=
1
x2 – 1
The domains of ƒ and g are the set of real numbers, so the domain of ƒ • g
is also the set of real numbers.
ƒ
The domain of g does not include 1 and –1 because g(1) and g(–1) = 0.
7-6
Function Operations
ALGEBRA 2 LESSON 7-6
Let ƒ(x) = x3 and g(x) = x2 + 7. Find (g ° ƒ)(2).
Method 1:
Method 2:
(g ° ƒ)(x) = g(ƒ(x)) = g(x3) = x6 + 7
(g ° ƒ)(x) = g(ƒ(x))
(g ° ƒ)(2) = (2)6 + 7 = 64 + 7 = 71
g(ƒ(2)) = g(23)
= g(8)
= 82 + 7
= 71
7-6
Function Operations
ALGEBRA 2 LESSON 7-6
A store offers a 20% discount on all items. You have a coupon
worth $3.
a. Use functions to model discounting an item by 20% and to model applying
the coupon.
Let x = the original price.
ƒ(x) = x – 0.2x = 0.8x
Cost with 20% discount.
g(x) = x – 3
Cost with a coupon for $3.
b. Use a composition of your two functions to model how much you would
pay for an item if the clerk applies the discount first and then the coupon.
(g ° ƒ)(x) = g(ƒ(x))
Applying the discount first.
= g(0.8x)
= 0.8x – 3
7-6
Function Operations
ALGEBRA 2 LESSON 7-6
(continued)
c. Use a composition of your two functions to model how much you would
pay for an item if the clerk applies the coupon first and then the discount.
(ƒ ° g)(x) = ƒ(g(x))
Applying the coupon first.
= ƒ(x – 3)
= 0.8(x – 3)
= 0.8x – 2.4
d. How much more is any item if the clerk applies the discount first?
(ƒ ° g)(x) – (g ° ƒ)(x) = (0.8x – 2.4) – (0.8x – 3)
= 0.6
Any item will cost $.60 more.
7-6
Function Operations
ALGEBRA 2 LESSON 7-6
pages 394–398 Exercises
11.
3x + 5
x2
x2
3x + 5
1. x2 + 3x + 5
12.
2. x2 – 3x – 5
13. 2x2 + 2x – 4; domain: all real numbers
3. –x2 + 3x + 5
14. –2x2 + 2; domain: all real numbers
4. 3x3 + 5x2
15. 2x2 – 2; domain: all real numbers
5. 3x +2 5
16. 2x3 – x2 – 4x + 3; domain: all real numbers
6.
17. 2x + 3; domain: all real numbers except 1
x
x2
3x + 5
1 ; domain: all real numbers except and 1
2x + 3
7. x2 + 3x + 5
18.
8. –x2 + 3x + 5
9. x2 – 3x – 5
19. 27x2, domain: all real numbers; 3,
domain: all real numbers except 0
10. 3x3 + 5x2
20. 2x + 3; 9, –1
21. x2 + 5; 14, 9
7-6
Function Operations
ALGEBRA 2 LESSON 7-6
22. 8
34. 9
23. 104
35. 9.25
24. 20
36. 0.25
25. 16
37. 6.25
26. 8
38. –2.75
27. 10
39. c2 – 6c + 9
28. 12
40. c2 – 3
29. 68
41. a2 + 6a + 9
30. 404
42. a2 – 3
31. 1
43. a.
b.
c.
d.
32. 25
33. –3
44. a. (g ° ƒ)(x) = 1.0968x
b. 16.45 pesos
45. x2 – x + 7
46. 6x + 13
47. x2 – 5x – 3
48. –2x2 + 8x + 1
49. –x2 + 5x + 13
50. 2x2 + 2x + 24
51. –3x2 + 2x + 16, domain:
all real numbers
ƒ(x) = 0.9x
g(x) = x – 2000
$14,200
$14,400
7-6
52. 3x2 – 12, domain:
all real numbers
53. 3x3 + 8x2 – 4x – 16, domain:
all real numbers
Function Operations
ALGEBRA 2 LESSON 7-6
54. –9x3 – 24x2 + 12x + 48, domain:
all real numbers
55. 3x – 4, domain: all real numbers
except –2
62. a.
1963; the area after 2 seconds
is about 1963 in.2.
b. 7854 in.2
63. 3x2, 9x2
56. 15x – 20, domain: all real numbers
except –2
64. x – 2, x – 2
57. 7; answers may vary. Sample:
First evaluate ƒ(3) since the
expression is (g ° ƒ)(3), and that
means g(ƒ(3)). Then evaluate g(6).
66. x – 3, x – 6
58. 1
59. –4
60. 0
61. 2
65. 12x2 + 2, 6x2 + 4
67. –4x – 7, –4x – 28
68.
x2 + 5 , x2 + 10x + 25
2
4
69. Answers may vary. Sample:
a. g(x) = 0.12x
b. ƒ(x) = 9.50x
c. (g °ƒ)(x) = 1.14x; your savings will
be $1.14 for each hour you work.
7-6
Function Operations
ALGEBRA 2 LESSON 7-6
70. a.
b.
c.
d.
e.
f.
ƒ(x) and g(x)
0, 15, 30; 3, 28, 103
3x2 + 9
3*A1^2 + 9, 9, 84, 309
9x2 + 3
9*A1^2 + 3, 3, 228, 903
71. a. P(x) = 5295x – 1000
b. $157,850
72. a. g(x) is the bonus earned when
x is the amount of sales over
$5000. h(x) is the excess of x
sales over $5000.
b. (g ° h)(x) because you first need
to find the excess sales over
$5000 to calculate the bonus.
73. (ƒ+ g)(x) = ƒ(x) + g(x) Def. of Function
Add.
= 3x – 2 + (x2 + 1)
Substitution
= x2 + 3x – 2 + 1
Comm. Prop.
= x2 + 3x – 1
arithmetic
74. (ƒ – g)(x) = ƒ(x) – g(x) def. of function
subtraction
= 3x – 2 – (x2 + 1)
substitution
= 3x – 2 – x2 – 1
Opp. of Sum
Prop.
= –x2 + 3x – 2 – 1
Comm. Prop.
= –x2 + 3x – 3
arithmetic
75. (ƒ ° g)(x) = ƒ(g(x))
def. of comp.
functions
= ƒ(x2 + 1)
substitution
= 3(x2 + 1) – 2
substitution
= 3x2 + 3 – 2
Dist. Prop.
= 3x2 + 1
arithmetic
7-6
Function Operations
ALGEBRA 2 LESSON 7-6
76. a. ƒ(x) = x + 10; g(x) = 1.09x
b. Each grade is increased 9%
before adding the 10-point
bonus; 91.75.
c. Add the 10-point bonus and
then increase the sum by
9%; 92.65.
d. no
80. x
1
x
82. 6 – x
8
81.
83. 2
84. 4
77. x7 – x6 – 16x5 + 10x4 + 85x3 – 25x2 – 150x;
domain: all real numbers
85. B
86. F
87. [2] (ƒ • g)(x) means ƒ(x) times
x–3
g(x) or in this case (3x – 4)
5, and – 5
(x + 3). The value of (ƒ • g)
(x) is 3x2 + 5x – 12.
79. 2x – 3 ; domain: all real numbers except 0,
x + 2x
[1] only 3x2 + 5x – 12 OR
–2, 5, and – 5
minor error in explanation
2
78. x + 2x ; domain: all real numbers except 3,
88. A
7-6
Function Operations
ALGEBRA 2 LESSON 7-6
89. C
90. B
99. x6 + 6x5y + 15x4y2 + 20x3y3 + 15x2y4 + 6xy5 + y6
100. 16x4 – 32x3y + 24x2y2 – 8xy3 + y4
91. D
101. 128x7 – 1344x6y + 6048x5y2 – 15,120x4y3 +
92. 1
93. –3
94. 4
22,680x3y4 – 20,412x2y5 + 10,206xy6 – 2187y7
102. 59,049 – 65,610x + 29,160x2 – 6480x3 +
720x4 – 32x5
103. 1024x5 – 1280x4y + 640x3y2 – 160x2y3 + 20xy4 – y5
95. 3
104. x8 + 4x7 + 6x6 + 4x5 + x4
96. 2
105. x12 + 12x10y3 + 60x8y6 + 160x6y9 + 240x4y12 +
192x2y15 + 64y18
106. 6 + 2i
97. 3
98. x8 + 32x7 + 448x6 +
107. –2 + 19i 2
3584x5 + 17,920x4 +
57,344x3 + 114,688x2 + 108. 13 + 5i 5
131,072x + 65,536
109. –8 – 36i
7-6
Function Operations
ALGEBRA 2 LESSON 7-6
Let ƒ(x) = 2x2 and g(x) = 3x – 1. Perform each function operation. Then
find the domain.
1. ƒ(x) + g(x) ƒ(x) + g(x) = 2x2 + 3x – 1; all reals
2. ƒ(x) – g(x)
ƒ(x) – g(x) = 2x2 – 3x + 1; all reals
3. ƒ(x) • g(x)
ƒ(x) • g(x) = 6x3 – 2x2; all reals
4. ƒ(x)
g(x)
ƒ(x)
2x2
1
= 3x – 1 all reals except 3
g(x)
5. (ƒ ° g)(x)
(ƒ ° g)(x) = 18x2 – 12x + 2; all reals
6. A store is offering a 15% discount on all items. You have a coupon
worth $2 off any item. Let x be the original cost of an item. Use a
composition of functions to find a function c(x) that gives the final cost
of an item if the discount is applied first and then the coupon. Then use
this function to find the final cost of an item originally priced at $10.
c(x) = 0.85x – 2; $6.50
7-6
Inverse Relations and Functions
ALGEBRA 2 LESSON 7-7
(For help, go to Lesson 2-1.)
Graph each pair of functions on a single coordinate plane.
1. y = x – 6
y=x+6
4. y = 0.5x + 1
y = 2x – 2
2. y = x – 7
2
3. y = 3x – 1
y = 2x + 7
y= x+1
5. y = –x + 4
6. y = x + 4
y = –x + 4
–1
7-7
3
5
y = 5x – 4
Inverse Relations and Functions
ALGEBRA 2 LESSON 7-7
Solutions
2. y = x – 7 , or
1. y = x – 6
2
y = 1x – 7
y=x+6
2
2
y = 2x + 7
3. y = 3x – 1
4. y = 0.5x + 1
y = 2x – 2
y = x + 1, or
3
y= 1x+ 1
3
3
7-7
Inverse Relations and Functions
ALGEBRA 2 LESSON 7-7
Solutions (continued)
6. y = x + 4, or
5. y = –x + 4
y=
5
–x + 4
, or
–1
y=1x+4
5
y=x–4
5
y = 5x – 4
7-7
Inverse Relations and Functions
ALGEBRA 2 LESSON 7-7
a. Find the inverse of relation m.
Relation m
x
y
–1
–2
0
–1
1
–1
2
–2
Interchange the x and y columns.
Inverse of Relation m
x
y
–2
–1
–1
0
–1
1
–2
2
7-7
Inverse Relations and Functions
ALGEBRA 2 LESSON 7-7
(continued)
b. Graph m and its inverse on the same graph.
Relation m
Reversing the
Ordered Pairs
7-7
Inverse of m
Inverse Relations and Functions
ALGEBRA 2 LESSON 7-7
Find the inverse of y = x2 – 2.
y = x2 – 2
x = y2 – 2
±
Interchange x and y.
x + 2 = y2
Solve for y.
x+2=y
Find the square root of each side.
7-7
Inverse Relations and Functions
ALGEBRA 2 LESSON 7-7
Graph y = –x2 – 2 and its inverse.
The graph of y = –x2 – 2 is a parabola
that opens downward with vertex (0, –2).
The reflection of the parabola in the line
x = y is the graph of the inverse.
You can also find points on the graph of
the inverse by reversing the coordinates
of points on y = –x2 – 2.
7-7
Inverse Relations and Functions
ALGEBRA 2 LESSON 7-7
Consider the function ƒ(x) =
2x + 2 .
a. Find the domain and range of ƒ.
Since the radicand cannot be negative, the domain is the set of numbers
greater than or equal to –1.
Since the principal square root is nonnegative, the range is the set of
nonnegative numbers.
b. Find ƒ –1
ƒ(x) =
2x + 2
y=
2x + 2
Rewrite the equation using y.
x=
2y + 2
Interchange x and y.
x2 = 2y + 2
y=
x2 – 2
2
Square both sides.
So, ƒ
Solve for y.
7-7
–1(x)
x2 – 2
= 2 .
Inverse Relations and Functions
ALGEBRA 2 LESSON 7-7
(continued)
c. Find the domain and range of ƒ –1.
The domain of ƒ –1 equals the range of ƒ, which is the set of nonnegative
numbers.
x – 2 –1. Thus the range of ƒ–1 is the set of numbers
Since x2 >
0,
>
–
2 –
greater than or equal to –1.
2
Note that the range of ƒ–1 is the same as the domain of ƒ.
d. Is ƒ –1 a function? Explain.
For each x in the domain of ƒ–1, there is only one value of ƒ –1(x). So ƒ –1 is
a function.
7-7
Inverse Relations and Functions
ALGEBRA 2 LESSON 7-7
The function d = 16t 2 models the distance d in feet that an
object falls in t seconds. Find the inverse function. Use the inverse to
estimate the time it takes an object to fall 50 feet.
d = 16t 2
t2 =
d
16
t=
d
4
1
t= 4
Solve for t. Do not interchange variables.
Quantity of time must be positive.
50
1.77
The time the object falls is 1.77 seconds.
7-7
Inverse Relations and Functions
ALGEBRA 2 LESSON 7-7
For the function ƒ(x) = 1 x + 5, find (ƒ–1 ° ƒ)(652) and
2
–1
(ƒ ° ƒ )(– 86).
Since ƒ is a linear function, so is ƒ –1.
Therefore ƒ –1 is a function.
So (ƒ –1 ° ƒ)(652) = 652
and (ƒ ° ƒ –1)(–
86) = –
86.
7-7
Inverse Relations and Functions
ALGEBRA 2 LESSON 7-7
pages 404–406 Exercises
1.
5. y = 1 x – 1 ; yes
3.
3
3
6. y = 1 x + 1 ; yes
2
2
7. y = – 1 x + 4 ; yes
3
3
5 – x no
8. y = y = ±
2
9. y = ±
2.
10. y = ±
4.
11. y = ±
12. y = ±
13. y = ±
7-7
x – 4; no
x + 5 no
3
x – 1; no
x + 4 no
2
x – 5 – 1 ; no
2
Inverse Relations and Functions
ALGEBRA 2 LESSON 7-7
14.
15.
17.
20.
21.
18.
22.
16.
19.
7-7
Inverse Relations and Functions
ALGEBRA 2 LESSON 7-7
23. ƒ–1(x) = x – 4 , and the domain
3
and range for both ƒ and ƒ–1 are
all real numbers; ƒ–1 is a function.
24. ƒ–1(x) = x2 + 5, x >
– 0, domain
ƒ: {x|x > 5}, range ƒ: {y|y >
– 0},
domain ƒ–1: {x|x >
– 0}, and
range ƒ–1: {y|y >
– 5}; ƒ–1 is
a function.
25. ƒ–1(x) = x2 – 7, x >
– 0, domain
ƒ: {x|x >
– –7}, range ƒ: {y|y >
– 0},
domain ƒ–1: {x|x >
– 0}, and range
–1
ƒ–1: {y|y >
– –7}; ƒ is a function.
26. ƒ–1(x) =
3 – x2
3
, x>
0, domain ƒ: {x|x <
},
–
–
3
2
–1
range ƒ: {y|y >
– 0}, domain ƒ : {x|x >
– 0},
3
and range ƒ–1: {y|y <
– 2 };
ƒ–1 is a function.
27. ƒ–1(x) = ±
x – 2 , x 2, domain ƒ: all reals,
>
–
2
range ƒ: {y|y > 2}, domain ƒ–1: {x|x >
– 2},
and range ƒ–1: all reals;
ƒ–1 is not a function.
28. ƒ–1(x) = ± 1 – x , x <
– 1, domain ƒ: all reals,
–1
range ƒ: {y|y <
– 1}, domain ƒ : {x|x <
– 1},
and range ƒ–1: all reals;
ƒ–1 is not a function.
7-7
Inverse Relations and Functions
ALGEBRA 2 LESSON 7-7
29. a. F = (C – 32); yes
38. ƒ–1(x) = ±
b. –3.89°F
30. a. r =
3
x – 1; no
39. ƒ–1(x) = 1± x ; no
2
3V ; yes
4
40. ƒ–1(x) = –1 ±
41. ƒ–1(x) =
b. 20.29 ft
3
x + 1; no
x; yes
31. 10
42. ƒ–1(x) = ±
32. –10
43. ƒ–1(x) = ±
33. 0.2
2
44. x = V ; 25 ft, 6.25 ft
4
x; no
5x – 5 ; no
2
64
34. d
45. The range of the inverse is the
domain of ƒ, which is x >
– 1.
2x + 8 ; no
3
36. ƒ–1(x) = ± 2 x ; no
3
37. ƒ–1(x) = ± x2 – 6x + 10 ; yes
2
35. ƒ–1(x) = ±
46. 2 and 5
7-7
Inverse Relations and Functions
ALGEBRA 2 LESSON 7-7
47. ƒ–1(x) = x2, x <
– 0, domain of
ƒ: {x|x >
– 0}, range of ƒ: {y|y >
– 0},
domain of ƒ–1: {x|x >
– 0},
range of ƒ–1: {y|y >
– 0},
and ƒ–1 is a function.
50. ƒ–1(x) = x2 – 2, x >
– 0, domain of
ƒ: {x|x >
– 0},
– –2}, range of ƒ: {y|y >
domain of ƒ–1: {x|x >
– 0},
range of ƒ–1: {y|y >
– –2},
and ƒ–1 is a function.
48. ƒ–1(x) = (x – 3)2, x >
– 3, domain of
ƒ: {x|x >
– 0}, range of ƒ: {y|y >
– 3},
domain of ƒ–1: {x|x >
– 3},
range of ƒ–1: {y|y >
– 0},
and ƒ–1 is a function.
51. ƒ–1(x) = ± 2x , x >
– 0, domain of
ƒ: all reals, range of ƒ: {y|y >
– 0},
domain of ƒ–1: {x|x >
– 0},
range of ƒ–1: all reals,
and ƒ–1 is not a function.
49. ƒ–1(x) = 3 – x2, x >
– 0, domain of
ƒ: {x|x <
– 3}, range of ƒ: {y|y >
– 0},
domain of ƒ–1: {x|x >
– 0},
range of ƒ–1: {y|y <
– 3},
and ƒ–1 is a function.
52. ƒ–1(x) = ±
1 , x > 0, domain of
x
ƒ:{x|x =/ 0}, range of ƒ:{y|y > 0},
domain of ƒ–1: {x|x > 0},
range of ƒ–1: {y|y =/ 0},
and ƒ–1 is not a function.
7-7
Inverse Relations and Functions
ALGEBRA 2 LESSON 7-7
53. ƒ–1(x) = ± x + 4 x <
– 0, domain of
ƒ: all reals, range of ƒ: {y|y <
– 0},
domain of ƒ–1: {x|x <
– 0},
range of ƒ–1: all reals,
and ƒ–1 is not a function.
56. ƒ–1(x) = – x – 4
54. ƒ–1(x) = 7 ± x x >
– 0, domain of
ƒ: all reals, range of ƒ: {y|y >
– 0},
domain of ƒ–1: {x|x >
– 0},
range of ƒ–1: all reals,
and ƒ–1 is not a function.
57. ƒ–1(x) = 3
55. ƒ–1(x) = ±
2
2
x<
– 4, domain of
ƒ: {x|x >
– 0}, range of ƒ: {y|y <
– 4},
domain of ƒ–1: {x <
– 4},
range of ƒ–1: { y|y >
– 0},
and ƒ–1 is a function.
x
2
x>
– 0, domain of
ƒ: {x|x > 0}, range of ƒ: {y|y > 0},
domain of ƒ–1: {x|x > 0},
range of ƒ–1: {y|y > 0},
and ƒ–1 is a function.
1 – 1 x > 0, domain of
x
58. ƒ–1(x) = – 1 1
ƒ: {x|x =/ –1}, range of ƒ: {y|y > 0},
domain of ƒ–1: {x|x > 0},
range of ƒ–1: {y|y =/ –1},
and ƒ–1 is not a function.
2 x
2x
> 0, domain of
ƒ: {x|x < 0}, range of ƒ: {y|y > 0},
domain of ƒ–1: {x|x > 0},
range of ƒ–1: {y|y < 0},
and ƒ–1 is a function.
7-7
Inverse Relations and Functions
ALGEBRA 2 LESSON 7-7
65. ƒ–1(x) = 27x3; yes
59. a-b. Answers may vary. Sample:
a.
66. ƒ–1(x) = 2 +
3
x; yes
67. ƒ–1(x) = x4; yes
68. ƒ–1(x) = ±
b.
4
5x no
6
69. B
70. G
60. r is not a function because there
are two y-values for one x-value.
r –1 is a function because each
of its x-values has one y-value.
61. h = s 2; 3 2 in. 4.2 in.
71. C
72. [2] x = 4y2 + 5, so 4y2 = x – 5,
y2 = x – 5 , and y = ±
4
The inverse has values that
are real numbers when x >
– 5.
62. Check students’ work.
63. ƒ–1(x) =
3
x – 5.
2
5x; yes
[1] y = ± x – 5 OR x >
– 5 OR
2
minor error
64. ƒ–1(x) = x3 + 5; yes
7-7
Inverse Relations and Functions
ALGEBRA 2 LESSON 7-7
73. [4] x = y2 – 2y + 1 or x = (y – 1)2.
Then y – 1 = ± x or y = ± x + 1.
It is not a function because each
positive value of x gives two
values of y.
[3] minor error in finding inverse
[2] attempt to find inverse and a
statement that the inverse is
not a function
[1] attempt to find inverse OR a
statement that the inverse is
not a function
78. |–4x – 10|
74. 2x + 28
86. 0.4
75. 2x + 7
87. 30
76. |–x – 10|
88. ±1, ±2, ±3, ±4, ±6, ±12, ± 1 , ± 3;
79. 2x + 28 + |–2x + 4|
80. 2
81. –2
82. not a real number
83. 3
84. –3
85. –3
roots are – 3 , ±2.
2
77. x + 14
7-7
2
2
Inverse Relations and Functions
ALGEBRA 2 LESSON 7-7
89. ±1, ±2, ±4, ± 1 , ± 2 , ± 4 ; roots are 2 , 2, –1.
3
3
3
3
90. ±1, ±2, ±3, ±4, ±6, ±12, ± 1 , ± 2 , ± 4 ; roots are 1, – 4 , -3.
3
3
3
3
91. ±1, ±2, ±3, ±5, ±6, ±10, ±15, ±30, ± 1 , ± 3 , ± 5 , ± 15 ; roots are 5, – 3 , 2.
2
2
2
2
2
92. ±1, ±2, ±3, ±6; roots are 1, 2, 3.
93. ±1, ±2, ±3, ±4, ±6, ±12; roots are –2, 2, –3.
7-7
Inverse Relations and Functions
ALGEBRA 2 LESSON 7-7
1. Find the inverse of the function y = 4x – 7. Is the inverse a function?
2. Find the inverse of y =
x + 9. Is the inverse a function?
3. Graph the relation y = 2x2 – 2 and its inverse.
4. For the function ƒ(x) = 2x2 – 1, find ƒ–1, and the domain and range of ƒ
and ƒ–1. Determine whether ƒ–1 is a function.
5. If ƒ(x) = 5x + 8, find (ƒ–1 ° ƒ)(59) and (ƒ ° ƒ–1)(3001).
6. A right triangle has a leg of length x and a hypotenuse of length 2x.
Write an equation to find the length s of the other leg, and use it to
estimate s when the hypotenuse has a length of 5 cm.
7-7
Inverse Relations and Functions
ALGEBRA 2 LESSON 7-7
1. Find the inverse of the function y = 4x – 7. Is the inverse a function?
y=
x+7
; yes
4
2. Find the inverse of y =
x + 9. Is the inverse a function?
y = x2 – 9; yes
3. Graph the relation y = 2x2 – 2 and its inverse.
7-7
Inverse Relations and Functions
ALGEBRA 2 LESSON 7-7
4. For the function ƒ(x) = 2x2 – 1, find ƒ–1, and the domain and range of ƒ
and ƒ–1. Determine whether ƒ–1 is a function.
1
ƒ–1(x) = ± 2 2x + 2; domain and range of ƒ: all numbers, range of ƒ:
all numbers greater than or equal to –1, domain of ƒ–1: all numbers
greater than or equal to –1, range of ƒ–1: all numbers; not a function
5. If ƒ(x) = 5x + 8, find (ƒ–1 ° ƒ)(59) and (ƒ ° ƒ–1)(3001).
59, 3001
6. A right triangle has a leg of length x and a hypotenuse of length 2x.
Write an equation to find the length s of the other leg, and use it to
estimate s when the hypotenuse has a length of 5 cm.
s=x
3; about 4.33 cm
7-7
Graphing Radical Functions
ALGEBRA 2 LESSON 7-8
(For help, go to Lesson 5-3.)
Graph each equation.
1. y = (x + 2)2
2. y = (x – 3)2
3. y = –(x + 4)2
4. y = –x2 – 1
5. y = –(x + 1)2 + 1
6. y = 3x2 + 3
7-8
Graphing Radical Functions
ALGEBRA 2 LESSON 7-8
Solutions
1. y = (x + 2)2
parent function: y = x2
translate 2 units left
2. y = (x – 3)2
parent function: y = x2
translate 3 units right
3. y = –(x + 4)2
parent function: y = –x2
translate 4 units left
7-8
Graphing Radical Functions
ALGEBRA 2 LESSON 7-8
Solutions (continued)
4. y = –x2 – 1
parent function: y = –x2
translate 1 unit down
5. y = –(x + 1)2 + 1
parent function: y = –x2
translate 1 unit left
and 1 unit up
6. y = 3x2 + 3
parent function: y = 3x2
translate 3 units up
7-8
Graphing Radical Functions
ALGEBRA 2 LESSON 7-8
Graph y =
x + 5 and y =
x – 7.
The graph of y = x + 5 is the graph of
y = x shifted up 5 units.
The graph y = x – 7 is the graph of y =
shifted down 7 units.
The domains of both functions are the set of nonnegative numbers, but
their ranges differ.
7-8
x
Graphing Radical Functions
ALGEBRA 2 LESSON 7-8
Graph y =
x + 7 and y =
x – 5.
The graph of y = x + 7 is the graph of
y = x shifted left 7 units.
The graph y = x – 5 is the graph of y =
shifted right 5 units.
x
The ranges of both functions are the set of nonnegative numbers, but their
domains differ.
7-8
Graphing Radical Functions
ALGEBRA 2 LESSON 7-8
Graph y = –
x, y = –2
x, and y = –1.5
x.
1
Each y-value of y = –2 x is 3 times the
corresponding y-value of y = –1.5 x.
The domains of the three functions are the
set of nonnegative numbers, and the ranges
are the set of negative numbers.
7-8
Graphing Radical Functions
ALGEBRA 2 LESSON 7-8
Graph y = 3
x + 2 – 1.
y=3
x – (–2) + (–1),
so translate the graph of y = 3
and down 1 unit.
7-8
x left 2 units
Graphing Radical Functions
ALGEBRA 2 LESSON 7-8
The function h(x) = 0.5 3 x models the height h in meters of
one group of male giraffes with a body mass of x kilograms.
Graph the model with a graphing calculator.
Use the graph to estimate the body mass of a young
male giraffe with a height of 3 meters.
Graph y = 0.5
3
x and y = 3. Adjust the window so the graphs intersect.
Use the Intersect feature to find that x
The giraffe has a mass of about 216 kg.
7-8
216 when y = 3.
Graphing Radical Functions
ALGEBRA 2 LESSON 7-8
Graph y = 2
3
x + 2 – 2.
3
The graph of y = 2 x + 2 – 2 is the graph of
y = 2 3 x translated 2 units left and 2 units down.
7-8
Graphing Radical Functions
ALGEBRA 2 LESSON 7-8
Rewrite y = 9x + 18 to make it easy to graph using a
translation. Describe the graph.
y=
9x + 18
=
9(x + 2)
=3
x+2
=3
x – (–2)
The graph of y =
9x + 18 is the graph of y = 3
7-8
x translated 2 units left.
Graphing Radical Functions
ALGEBRA 2 LESSON 7-8
pages 411–413 Exercises
4.
1.
7.
8.
2.
5.
9.
6.
3.
7-8
Graphing Radical Functions
ALGEBRA 2 LESSON 7-8
10.
13.
16.
11.
14.
17.
18.
12.
15.
7-8
Graphing Radical Functions
ALGEBRA 2 LESSON 7-8
19.
22.
25.
26.
20.
23. a.
27.
21.
b.
745 ft,
24.
7-8
1053 ft,
1343 ft
Graphing Radical Functions
ALGEBRA 2 LESSON 7-8
28.
32. y = –14 x + 1; the graph is the graph
of y = –14 x translated 1 unit
to the left.
33. y = 4 3 x + 2; the graph is the graph
of y = 4 3 x translated 2 units
to the left.
29.
34. y = 8 x – 2 – 3; the graph is the
graph of y = 8 x translated 2 units
to the right and 3 units down.
35. y = 3 3 x – 2 + 1; the graph is the
graph of y = 3 3 x translated 2 units
30. y = 3 x – 1; the graph is the graph
to the right and 1 unit up.
of y = 3 x translated 1 unit to the right.
31. y = –4 x + 2; the graph is the graph
of y = –4 x translated 2 units to the left.
7-8
Graphing Radical Functions
ALGEBRA 2 LESSON 7-8
36.
38.
40.
D: x >
– 6, R: y >
–0
D: x >
– 0, R: y >
–7
D: x >
–0
– 0, R: y <
39.
41.
37.
D: x >
– –6
– 0, R: y >
D: x >
–2
– 0, R: y<
7-8
1
D: x >
–7
– 2, R: y <
Graphing Radical Functions
ALGEBRA 2 LESSON 7-8
42.
46.
44.
D: all real numbers,
R: all real numbers
D: all real numbers,
R: all real numbers
D: all real numbers,
R: all real numbers
47.
45.
43.
D: x >
–3
– 1, R: y >
1
D: x >
– – , R: y <
–0
2
7-8
D: all real numbers,
R: all real numbers
Graphing Radical Functions
ALGEBRA 2 LESSON 7-8
48.
50.
D: x >
– –5, R: y <
– –1
49.
D: all real numbers,
R: all real numbers
52. a.
3
D: x >
–7
– 4, R: y<
51. a.
b.
4.3 s;
7-8
6.1 s
b. D: x >
– 2, R: y >
– –2
c. (2, –2)
d. The domain is based on
the x-coordinate of that
point, and the range is
based on the
y-coordinate.
Graphing Radical Functions
ALGEBRA 2 LESSON 7-8
53. a. y =
b. y =
x–5–2
x–1–5
56. y = 6 x + 3 + 4; the graph is the
graph of y = 6 x translated 3 units
to the left and 4 up.
54. a.
3
57. y = – 2 x – 1 ; the graph is the graph of
4
y = –2 3 x translated 1 unit to the right.
4
58. y = 1
2
x – 1 – 2; the graph is the same
as y = 1
x translated 1 unit right and
2
b. Both domains are x >
– 2.
The range of y = x – 2 + 1
is y >
– 1. The range of
y = – x – 2 + 1 is y <
– 1.
55. y = 5 x – 4 – 1; the graph is
the same as y = 5
2 down.
59. y = 10 – 1
3
3
x + 3; the graph is the
same as y = – 1
3
3
x translated 3 units
to the left and 10 up.
x, translated
4 units to the right and 1 down.
7-8
Graphing Radical Functions
ALGEBRA 2 LESSON 7-8
60. y = 1 x + 9 + 5; the graph is the same 64. y = – 2 x + 4; the graph is the
3 1
graph of y = – 2x translated 4
as y =
x, translated 9 units to the
3
units to the left; domain: x >
– –4,
left and 5 up.
range: y <
– 0.
61. Answers may vary. Sample:
65. y = – 8 x – 3 ; the graph is the
3
y= x–2+4
4
graph of y = – 8x translated 3
62. a.
4
3
units to the right; domain: x >
– 4,
range: y <
– 0.
66. y =
b. 20 in.
63. If a > 0, the graph is stretched vertically
by a factor of a. If a < 0, the graph is
reflected over the x-axis and stretched
vertically by a factor of |a|.
7-8
3•
x – 5 + 6; the graph is
3
the graph of y = 3x translated 5
3
units to the right and 6 units up;
5
domain: x >
– , range: y >
– 6.
3
Graphing Radical Functions
ALGEBRA 2 LESSON 7-8
67. y = – 12 • x – 3 – 3; the graph
2
is the graph of y = – 12x translated
3
units to the left and 3 units down;
2
3
domain: x >
– – , range: y <
– –3.
72. B
73. H
74. [2] Both graphs have the same
shape as the graph of y = x.
2
68. a.
The graph of ƒ(x) = x – 1 is
the graph of y = x moved 1
unit to the right, and the graph
of g(x) = x – 1 is the graph
of y = x moved 1 unit down.
[1] minor error in either direction
b. The graph of y = h – x is a reflection
of the graph of y = x – h in the line
x = h.
69. for all odd positive integers
70. A
71. I
7-8
Graphing Radical Functions
ALGEBRA 2 LESSON 7-8
75. [4] For ƒ(x) = x – 1 the domain
is x >
– 0.
– 1 and the range is y >
For g(x) = x – 1 the domain
is x >
– –1.
– 0 and the range is y >
[3] minor error in one of the descriptions
of domains or ranges
[2] correct domains or ranges
[1] some attempt to describe the domains
76. ƒ–1(x) = x + 1 ; yes
4
77. ƒ–1(x) = 3(x + 3) ; yes
2
x – 1 ; no
78. ƒ–1(x) = ±
2.4
81. ƒ–1(x) =
82. x
83.
84.
3
5
12xy2
2y
3
9xy2
3y
3
48x3y4
2y
86. –1 ± 61
10
87. –3 ± 3 5
2
88. 9 ± 21
2
85.
89. –5 ±
79. ƒ–1(x) = (x + 4)2 – 3; yes
3
14
90. 6 ± 11
91. 5 , – 3
80. ƒ–1(x) = –1 ± x ; no
2
4
7-8
2
x ; yes
2
Graphing Radical Functions
ALGEBRA 2 LESSON 7-8
Graph each function on the same graph.
1. y =
2
3. y = 3
x–6
2. y =
x
4. y =
x+8
3
x–2+5
L
5. The formula t = 2
9.8 can be used to estimate the number of
seconds t it takes a pendulum of length L meters to make one
complete swing. Graph the equation on a graphing calculator. Then
use the graph to estimate the values of t for
pendulums of lengths 1.5 meters and 2.5 meters.
About 2.46 s, about 3.17 s
6. Rewrite y = 2
y=6
9x – 27 to make it easy to graph using a translation.
x–3
7-8
Radical Functions and Radical Exponents
ALGEBRA CHAPTER 7
Page 418
1. –0.3
2. 3|x|y2
12. 36 + 38
13. – 1 – 2
3
6xy
3. –2x2y4
5
14. 3 +
2x4
7. x
8.
16. x
10. 12
3+
27. 18x2 + 9x – 6, –6x2 – 3x + 20
1
2
28. The sixth power of a real
number is always
nonnegative.
18. 7
29. a.
b.
c.
d.
19. 2
2
20. –43
3
21. 8
2–
11. –17 – 4
5
5
6
25. –x3 + 5x2 – 8x + 4,
domain: all reals
26. 16x2 + 8x – 1, 4x2 – 7
4 3 2
17. 4x 6x
9y
2y
xy
3y
6x – 3x
6x
9. 24
2+
25
2
6. 6x3y4
6
15. 1
4. (x – 2)2
5. 7x2
24. x – 1, domain {x|x =/ 2}
3
22. x2 – 4x + 4; domain: all reals
23. –2x2 + 7x – 6; domain: all reals
7-A
ƒ(x) = 0.5x
g(x) = 0.75x
g(g(x)) = 0.5625x
The cashier’s solution is
too high by 6.25% of the
original price.
Radical Functions and Radical Exponents
ALGEBRA CHAPTER 7
30. y = 4 x + 5 – 1; y = 4 x
translated 5 units left and
1 unit down
31. y = 3
x + 1; y = 3
3
translated 1 unit left
3
33.
35.
x
3
D: {x|x >
– – },
2
R: {y|y >
– 0}
32.
34.
3 },
D: {x|x >
–
–
2
R: {y|y >
– –4}
36. –5831
37. 192
38. –63
D: {x|x >
– 0},
R: {y|y >
– 3}
D: {x|x >
– 4},
R: {y|y <
– 0}
7-A
Radical Functions and Radical Exponents
ALGEBRA CHAPTER 7
39.
ƒ–1(x)
3
=
x + 2 ; yes
3
40. g–1(x) = (x + 1)2 – 3; yes
2
41. g–1(x) = x – 1 ; yes
2
42. ƒ–1(x) = ±
4
4x; no
43. a. V = 256
268.08
3
b. r = 3V
4
c. r
1
3
2.88 in.
44. Check students’ work.
45.
0.6 seconds;
0.9 seconds
7-A