3 Unit Lesson 8 Section 3.2 3.
Practice Answers
f ( x ) = 2 x 2 5 x is a parabola with x-intercepts ( 0, 0 ) and ( 5 , 0 ) and opens 2 downward. Matches graph (h). f ( x ) = 1 x 4 + 3x 2 has intercepts ( 0, 0 ) and 2 3, 0 . Matches graph (a). 4
5.
(
)
9. y = x 3 (a) f ( x ) = ( x 2 )
3
Horizontal shift two units to the right
3 (b) f ( x ) = x 2
Vertical shift two units downward
1
Unit 3 Lesson 8
3 (c) f ( x ) = 1 x 2
Practice Answers
Reflection in the x-axis and a vertical shrink (d) f ( x ) = ( x 2 ) 2
3
Horizontal shift two units to the right and a vertical shift two units downward
3 13. f ( x ) = 1 x + 5 x 3 Degree: 3 Leading coefficient: 1 3 The degree is odd and the leading coefficient is positive. The graph falls to the left and rises to the right.
2
Unit 3 Lesson 8
Practice Answers
2 15. f ( x ) = 5 7 x 3 x 2 Degree: 2 Leading coefficient: 3 The degree is even and the leading coefficient is negative. The graph falls to the left and falls to the right. 2 3 19. f ( x ) = 6 2 x + 4 x 5 x Degree: 3 Leading coefficient: 5 The degree is odd and the leading coefficient is negative. The graph rises to the left and falls to the right. 3 3 23. f ( x ) = 3 x 9 x + 1; g ( x ) = 3 x
33. f ( x ) = 3x 3 12 x 2 + 3
0 = 3x x 2 4 x + 1 x = 0, x =
(
)
4 16 4 = 2 3 each of multiplicity 1 2
39. f ( x ) = 5 x 4 + 15 x 2 + 10
( 0 = 5( x
0 = 5 x 4 + 3x 2 + 2
2
+ 2 x2 + 2
)(
)
)
No real zeros
3
Unit 3 Lesson 8 41. g ( x ) = x 3 + 3x 2 4 x 12 0 = x 3 + 3x 2 4 x 12 0 = x 2 ( x + 3) 4 ( x + 3 ) 0 = x 2 4 ( x + 3) 0 = ( x + 2 ) ( x 2 ) ( x + 3) x = 2, 3 each of multiplicity 1 43. y = 4 x 3 20 x 2 + 25 x 0 = 4 x 3 20 x 2 + 25 x 0 = x ( 2 x 5) x = 0 or x =
5 2 2
Practice Answers
(
)
x-intercepts: ( 0, 0 ) ,
( 5 , 0) 2
53. f ( x ) = ( x 4 ) ( x + 3) ( x 3) ( x 0 ) = ( x 4) x2 9 x = x 4 4 x 3 9 x 2 + 36 x 4 3 2 Note: f ( x ) = a x 4 x 9 x + 36 x has these zeros for all real numbers a 0.
(
)
(
)
61. f ( x ) = ( x 0 ) x 3 x 3 = x x
(
(
)( ( 3) ( x + 3)
( )
))
3, 3.
= x3 3x 3 Note: f ( x ) = a x 3 x , x 0 has degree 3 and zeros x = 0,
4
Unit 3 Lesson 8
4 5 4 65. f ( x ) = x ( x + 4 ) = x + 4 x
Practice Answers
or f ( x ) = x 3 ( x + 4 ) = x 5 + 8 x 4 + 16 x 3
2
or f ( x ) = x 2 ( x + 4 ) = x 5 + 12 x 4 + 48 x 3 + 64 x 2
3
or f ( x ) = x ( x + 4 ) = x 5 + 16 x 4 + 96 x 3 + 256 x 2 + 256 x
4
Note: Any nonzero scalar multiple of these functions would also have degree 5 and zeros x = 0 and 4.
3 2 67. f ( x ) = x 9 x = x x 9 = x ( x + 3) ( x 3) (a) Falls to the left Rises to the right (b) Zeros: 0, 3, 3 (c)
(
)
x f(x)
-3 0
-2 10
-1 8
0 0
1 -8
2 -10
3 0
(d)
2 77. f ( x ) = x ( x 4 ) (a) Falls to the left Rises to the right (b) Zeros: 0, 4 (c)
x f(x)
-1 -5
0 0
1 -3
2 -8
3 -9
4 0
5 25
5
Unit 3 Lesson 8 (d)
Practice Answers
83. g ( x ) = 1 ( x + 1) 5
2
( x 3) ( 2 x 9 )
Zeros: 1 of multiplicity 2, 3 of multiplicity 1, and
9 2
of multiplicity 1
6
Unit 3 Lesson 8 Section 3.3 x2 4 1. y1 = and y2 = x 2 + x+2 x+2 x2 x + 2 x2 + 0x + 0 x2 + 2x 2x + 0 2x 4 4 2 x 4 Thus, = x2+ and y1 = y2 . x+2 x+2 5. 2x + 4 x + 3 2 x + 10 x + 12
2
Practice Answers
2 x2 + 6x 4x + 12 4 x + 12 0 2 2 x + 10 x + 12 = 2x + 4 x+3 11. 7 x + 2 7x + 3 7 x + 14 11 7x + 3 11 =7 x+2 x+2 13. x2 2 x + 0 x + 1 6 x + 10 x + x + 8
2 3 2
6 x3 + 0 x 2 + 3x 10 x 2 2 x + 8 10 x 2 + 0 x + 5 2x + 3
7
Unit 3 Lesson 8 6 x 3 + 10 x 2 + x + 8 2 x 2x = 3x + 5 + 2 = 3x + 5 2 2 2x +1 2x +1 2x +1 21. 2 4 4
3
Practice Answers
8 9 18 8 0 18 0 9
2
0
4 x + 8 x 9 x 18 = 4 x2 9 x+2 23. 10 1 0 10 1 10 75 250 100 25 250 0
x 3 + 75 x 250 = x 2 + 10 x 25 x + 10 25. 4 5 6 0 8 20 56 224 5 14 56 232 5 x3 6 x 2 + 8 232 = 5 x 2 + 14 x + 56 + x4 x4 49. 2 1 0 7 6 2 4 6 1 2 3
3
0
x 7 x + 6 = ( x 2) x2 + 2 x 3 Zeros: 3, 2 51.
1 2
= ( x 2 ) ( x + 3) ( x 1)
(
)
2 15 27 10 1 7 10 2 14 20 0
8
Unit 3 Lesson 8 2 x 3 15 x 2 + 27 x 10 = ( x 1 ) 2 x 2 14 x + 20 2 = ( 2 x 1) ( x 2 ) ( x 5 ) Zeros: , 2, 5
1 2
3 2 57. f ( x ) = 2 x + x 5 x + 2; Factors: ( x + 2 ) , ( x 1) (a) 2 2 1 5 2 4 6 2
Practice Answers
(
)
2 3
1
0
1 2 3 1 2 1 2 1 0 Both are factors of f ( x ) since the remainders zero.
(b) are The remaining factor of f ( x ) is ( 2 x 1) . (c) f ( x ) = ( 2 x 1) ( x + 2 ) ( x 1) 1 (d) Zeros: , 2,1 2 (e)
9
Unit 3 Lesson 8
3 2 65. f ( x ) = x 2 x 5 x + 10 (a) The zeros of f are 2 and 2.236 (b) 2 1 2 5 10 2 0 10
Practice Answers
1
0 5
f ( x ) = ( x 2) x 5
2
(
0
= ( x 2) x 5 71. 1 1 1
3 2
(
)
) ( x + 5)
3 1 3 1 2 3 2 3 0
x + 3x x 3 = x 2 + 2 x 3, x 1 x +1
10
Unit 3 Lesson 8 Section 3.4 1. f ( x ) = x ( x 6) The zeros are: x = 0, x = 6.
2 3
Practice Answers
3. g ( x ) = ( x 2 ) ( x + 4 ) The zeros are: x = 2, x = 4. 5. f ( x ) = ( x + 6) ( x + i ) ( x i ) The three zeros are: x = 6, x = i, x = i f ( x ) = x3 + 3x 2 x 3 Possible rational zeros: 1, 3 Zeros shown on graph: 3, 1, 1 13. g ( x ) = x 3 4 x 2 x + 4 = x 2 ( x 4 ) 1( x 4 ) = ( x 4) x2 1 = ( x 4 ) ( x 1) ( x + 1) Thus, the zeros of g(x) are 4 and 1.
3 2 15. h ( t ) = t + 12t + +21t + 10 Possible rational zeros: 1, 2, 5, 10 1 1 12 21 10 1 11 10
7.
(
)
1 11
3 2
10
0
t + 12t + 21t + 10 = ( t + 1) t 2 + 11t + 10 = ( t + 1) ( t + 10 ) Thus, the zeros are 1 and 10.
2
= ( t + 1) ( t + 1) ( t + 10 )
(
)
11
Unit 3 Lesson 8
4 2 33. f ( x ) = x 3 x + 2 (a) From the calculator, we have x = 1 and x 1.414 (b) 1 1 0 3 0 2 1 1 2 2
Practice Answers
1 1 2 2 1 1 1
0
1 2 2 1 0 2 0 2 0
f ( x ) = ( x 1) ( x + 1) x 2 2
(
= ( x 1) ( x + 1) x 2
(
)
) ( x + 2)
The exact roots are x = 1, 2. 37. f ( x ) = ( x 1) ( x 5i ) ( x + 5i ) = ( x 1) x 2 + 25
(
)
= x 3 x 2 + 25 x 25 3 2 Note: f ( x ) = a x x + 25 x 25 , where a is any nonzero real number, has the zeros 1 and 5i.
(
)
39. f ( x ) = ( x 6 ) x ( 5 + 2i ) x ( 5 2i ) = ( x 6 ) ( x + 5 ) 2i ( x + 5 ) + 2i
2 2 = ( x 6 ) ( x + 5 ) ( 2i )
( = ( x 6) ( x (
= ( x 6 ) x 2 + 10 x + 25 + 4
2
+ 10 x + 29
)
) )
= x 3 + 4 x 2 31x 174 3 2 Note: f ( x ) = a x + 4 x 31x 174 , where a is any nonzero real number, has zeros 6, and 5 2i.
12
Unit 3 Lesson 8
3 2 47. f ( x ) = 2 x + 3 x + 50 x + 75 Since 5i is a zero, so is 5i. 5i 2 3 50 75 10i 50 + 15i 75
Practice Answers
2 5i 2
3 + 10i 3 + 10i 10i
15i 15i 15i
0
The zeros of f ( x ) are x = 3 and x = 5i. 2 Alternate Solution 2 Since x = 5i are zeros of f ( x ) , ( x + 5i ) ( x 5i ) = x + 25 is a factor of f ( x ) . By long division, we have: 2x + 3 2 3 2 x + 0 x + 25 2 x + 3 x + 50x + 75 2 x 3 + 0 x 2 + 50 x 3x 2 + 0 x + 75 3 x 2 + 0 x + 75 0 2 Thus, f ( x ) = x + 25 ( 2 x + 3) and the zero of f are x = 5i an...