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Review Exercises for Chapter 2

Review Exercises for Chapter 2 237

1. (a) (b)

Vertical stretch Vertical stretch and a reflection in the x-axis

(c) (d)

Vertical shift two units upward Horizontal shift two units to the left

(a) y

x−4 −3 −2

−4

−3

−2

−1

1

4

−1 1 2 3 4

y

x−4 −3 −2

−4

−3

−2

−1

1

3

4

−1 1 2 3 4

y 5 sx 1 2d2y 5 x2 1 2

y

x−4 −3 −2

−4

−3

2

1

3

4

−1 1 2 3 4

y

x−4 −3 −2

−4

−3

−2

−1

2

3

4

−1 1 2 3 4

y 5 22x2y 5 2x2

2. (a)

Vertical shift four units downwardy

x−4 −3 −1

−5

−2

−1

2

3

1 3 4

y 5 x2 2 4 (b)

Reflection in the x-axis and a vertical shift

four units upwardy

x−4 −3 −1

−3

−2

−1

1

2

3

5

1 3 4

y 5 4 2 x2

(c)

Horizontal shift three units to the righty

x−3 −2 −1

−3

−2

−1

1

2

3

4

5

21 3 4 5

y 5 sx 2 3d2(d)

Vertical shrink (each y-value is multiplied by

and a vertical shift one unit downward

y

x−4 −3 −2

−4

−3

−2

1

2

3

4

2 3 4

12d,

y 512 x2 2 1

238 Chapter 2 Polynomial and Rational Functions

3.

Vertex:

Axis of symmetry:

x-intercepts: s0, 0d, s2, 0d

0 5 x2 2 2x 5 xsx 2 2d

x 5 1

s1, 21d

5 sx 2 1d2 2 1

5 x2 2 2x 1 1 2 1

x1

−1

−2

−1−2−3

3

4

5

6

7

2 3 4 5 6

ygsxd 5 x2 2 2x 4.

Vertex:

Axis of symmetry:


0 5 6x 2 x2 5 xs6 2 xd

x 5 3

s3, 9d

5 2sx 2 3d2 1 9

5 2sx2 2 6x 1 9 2 9d

x4

−2

8−2

2

4

8

6

10

2 10

y fsxd 5 6x 2 x2

5.

Vertex:

Axis of symmetry:

x-intercepts: s24 ± !6, 0d x 5 24 ± !6

x 1 4 5 ±!6

sx 1 4d2 5 6

0 5 sx 1 4d2 2 6

x 5 24

s24, 26d

x

2

2

−8 −4

−2

−4

−6

y 5 sx 1 4d2 2 6

5 x2 1 8x 1 16 2 16 1 10

f sxd 5 x2 1 8x 1 10 6.

Vertex:

Axis of symmetry:

x-intercepts: s2 ± !7, 0d

54 ± !28

25 2 ± !7

x 52s24d ± !s24d2 2 4s1ds23d

2s1d

0 5 x2 2 4x 2 3

0 5 3 1 4x 2 x2

x 5 2

s2, 7d

5 2sx 2 2d2 1 7

x4−2

2

4

8

6

10

2 106 8

y

5 2fsx 2 2d2 2 7g

5 2sx2 2 4x 1 4 2 4 2 3d

5 2sx2 2 4x 2 3d

hsxd 5 3 1 4x 2 x2

7.

Vertex:

Axis of symmetry:

t-intercepts: 11 ±!6

2, 02

t 5 1 ±!6

2

t 2 1 5 ±!3

2

2st 2 1d2 5 3

0 5 22st 2 1d2 1 3

t 5 1

s1, 3d

t1 2−1−2−3 3 4 5 6

2

4

5

6

3

1

y 5 22st 2 1d2 1 3

5 22fst 2 1d2 2 1g 1 1

5 22st2 2 2t 1 1 2 1d 1 1

f std 5 22t2 1 4t 1 1 8.

Vertex:

Axis of symmetry:


0 5 sx 2 2dsx 2 6d

0 5 x2 2 8x 1 12

x 5 4

x4 8−2

2

6

4

8

10

−2

−4

ys4, 24d

5 sx 2 4d2 2 4

5 x2 2 8x 1 16 2 16 1 12

f sxd 5 x2 2 8x 1 12


9.

Vertex:

Axis of symmetry:

No real zeros

x-intercepts: none

1x 11

222

5 23

0 5 41x 11

222

1 12

x 5 21

2

121

2, 122

x

−1−2−3 1 2 3

5

10

15

20

y 5 41x 11

222

1 12

5 41x2 1 x 11

42 2 1 1 13

5 41x2 1 x 11

42

1

42 1 13

5 4sx2 1 xd 1 13

hsxd 5 4x2 1 4x 1 13 10.

Vertex:

Axis of symmetry:

x-intercepts: s3 ± 2!2, 0dx

4 8

2

102

−2

−2

−4

−6

−8

y 56 ± !32

25 3 ± 2!2

x 52s26d ± !s26d2 2 4s1ds1d

2s1d

0 5 x2 2 6x 1 1

x 5 3

s3, 28d

5 sx 2 3d2 2 8

5 x2 2 6x 1 9 2 9 1 1

f sxd 5 x2 2 6x 1 1

11.

Vertex:

Axis of symmetry:

By the Quadratic Formula,

x-intercepts: 125 ± !41

2, 02

x 525 ± !41

2.

0 5 x2 1 5x 2 4

x 5 25

2

125

2, 2

41

4 2 5 1x 1

5

222

241

4

x

−2

−2 2−4−6−8

−4

−10

y 5 1x 15

222

225

42

16

4

5 x2 1 5x 125

42

25

42 4

hsxd 5 x2 1 5x 2 4 12.

Vertex:

Axis of symmetry:


The equation has no real zeros.

intercepts: Nonex-

x 524 ± 8i

85 2

1

2± i.

0 5 4x2 1 4x 1 5

x 5 21

2

121

2, 42

5 41x 11

222

1 4

x

8

10

12

2

−2−2

2 4 6−4−6−8

y 5 431x 11

222

1 14

5 41x2 1 x 11

42

1

41

5

42 fsxd 5 4x2 1 4x 1 5

13.

Vertex:

Axis of symmetry: x 5 25

2

125

2, 2

41

122

51

31x 15

222

241

12

51

331x 15

222

241

4 4 5

1

31x2 1 5x 125

42

25

42 42

x

4

−4

−2 2−4−6−8

−6

2

y

f sxd 51

3sx2 1 5x 2 4d


x-intercepts: 125 ± !41

2, 02

x 525 ± !41

2.

0 5 x2 1 5x 2 4


14.

Vertex:

Axis of symmetry:

x-intercepts: 12 ±!3

3, 02

512 ± !12

65 2 ±

!3

3 x 5

2s212d ± !s212d2 2 4s3ds11d2s3d

0 5 3x2 2 12x 1 11

x 5 2

s2, 21d

5 3sx 2 2d2 2 1

5 3sx 2 2d2 1 3s24d 1 11

5 3sx2 2 4x 1 4 2 4d 1 11

5 3x2 2 12x 1 11

–6 –4 –2 4 6 8 10

2

4

6

8

10

12

14

x

y

f sxd 51

2s6x2 2 24x 1 22d

15. Vertex:

Point:

Thus, f sxd 5 212sx 2 4d2 1 1.

212 5 a

22 5 4a

s2, 21d ⇒ 21 5 as2 2 4d2 1 1

s4, 1d ⇒ f sxd 5 asx 2 4d2 1 1 16. Vertex:

fsxd 514sx 2 2d2 1 2

14 5 a

1 5 4a

3 5 4a 1 2

Point: s0, 3d ⇒ 3 5 as0 2 2d2 1 2

s2, 2d ⇒ fsxd 5 asx 2 2d2 1 2

17. Vertex:

Point:

Thus, f sxd 5 sx 2 1d2 2 4.

1 5 a

s2, 23d ⇒ 23 5 as2 2 1d2 2 4

s1, 24d ⇒ f sxd 5 asx 2 1d2 2 4 18. Vertex:

fsxd 513sx 2 2d2 1 3

13 5 a

3 5 9a

6 5 9a 1 3

Point: s21, 6d ⇒ 6 5 as21 2 2d2 1 3

s2, 3d ⇒ fsxd 5 asx 2 2d2 1 3

19. (a)

x

y

(c)

The maximum area occurs at the vertex when

and The dimensions with the

maximum area are meters and meters.y 5 50x 5 50

y 5 100 2 50 5 50.

x 5 50

5 2sx 2 50d2 1 2500

5 2fsx 2 50d2 2 2500g

5 2sx2 2 100x 1 2500 2 2500d

Area 5 100x 2 x2(b)

5 100x 2 x2

5 xs100 2 xd

Area 5 xy

y 5 100 2 x

x 1 y 5 100

2x 1 2y 5 200

20.

(a)

Rs30d 5 $15,000

Rs25d 5 $13,750

Rs20d 5 $12,000

R 5 210p2 1 800p

(b) The maximum revenue occurs at the vertex of the parabola.

The revenue is maximum when the price is $40 per unit.

The maximum revenue is $16,000.

Rs40d 5 $16,000

2b

2a5

2800

2s210d 5 $40


21.

The minimum cost occurs at the vertex of the parabola.

Approximately 1091 units should be produced each day to

yield a minimum cost.

Vertex: 2b

2a5 2

2120

2s0.055d< 1091 units

C 5 70,000 2 120x 1 0.055x2 22.

The age of the bride is

approximately 24 years

when the age of the groom

is 26 years.

y

x20 21 22 23 24 25

22

23

24

25

26

27

Age of bride

Age

of

gro

om

x < 23.7, 29.4

x 525.68 ± !s5.68d2 2 4s20.107ds274.5d

2s20.107d

0 5 20.107x2 1 5.68x 2 74.5

26 5 20.107x2 1 5.68x 2 48.5

23.

Transformation: Reflection in

the x-axis and a horizontal shift

four units to the right

x

2

1

1 2 7−2

5

−3

−4

3

4

3 4 6

y

y 5 x3, f sxd 5 2sx 2 4d3 24.

is a reflection in the x-axis

and a vertical stretch of the graph

of y 5 x3.

f sxd

x−1−2−3

2

1

1 2 3

3

−1

−2

−3

y

f sxd 5 24x3y 5 x3, 25.

Transformation: Reflection in

the x-axis and a vertical shift

two units upward

x

1

1 2 3−2−3

3

−1

−2

−3

y

y 5 x4, f sxd 5 2 2 x4

26.

is a shift to the right two

units and a vertical stretch of the

graph of y 5 x4.

f sxd

x−1−2−3 1 2 3 4 5 6

2

−2

−3

4

5

6

3

1

y

f sxd 5 2sx 2 2d4y 5 x4, 27.

Transformation: Horizontal shift

three units to the right

x

2

1

1−2

5

−5

3

4

3 4 5 6 7

y

y 5 x5, f sxd 5 sx 2 3d5 28.

is a vertical shrink and a

vertical shift three units upward

of the graph of y 5 x5.

f sxd

x6−2−4−6 2 4

4

8

6

y

f sxd 512x5 1 3y 5 x5,

29.

The degree is even and the leading coefficient is negative.

The graph falls to the left and falls to the right.

fsxd 5 2x2 1 6x 1 9 30.

The degree is odd and the leading coefficient is positive.

The graph falls to the left and rises to the right.

fsxd 512x3 1 2x

31.

The degree is even and the leading coefficient is positive.

The graph rises to the left and rises to the right.

g sxd 534sx4 1 3x2 1 2d 32.

The degree is odd and the leading coefficient is negative.

The graph rises to the left and falls to the right.

hsxd 5 2x5 2 7x2 1 10x


33.

Zeros: all of

multiplicity 1 (odd multiplicity)

Turning points: 1

x 532, 27,

5 s2x 2 3dsx 1 7d

0 5 2x2 1 11x 2 21 −9

−40

9

20 f sxd 5 2x2 1 11x 2 21 34.

Zeros: of multiplicity 1

(odd multiplicity)

of multiplicity 2

(even multiplicity)

Turning points: 2

x 5 23

x 5 0

0 5 xsx 1 3d2−6

−5

6

3f sxd 5 xsx 1 3d 2

35.

Zeros: all of

multiplicity 1 (odd multiplicity)

Turning points: 2

t 5 0, ±!3

0 5 tst2 2 3d

0 5 t3 2 3t−5

−3

4

3 f std 5 t3 2 3t 36.


(even multiplicity)

of multiplicity 1

(odd multiplicity)

Turning points: 2

x 5 8

x 5 0

0 5 x2sx 2 8d

0 5 x3 2 8x2−10

−80

10

10 f sxd 5 x3 2 8x2

37.


(even multiplicity)

of multiplicity 1

(odd multiplicity)

Turning points: 2

x 5 53

x 5 0

0 5 24x2s3x 2 5d

0 5 212x3 1 20x2

−5

−5

5

10 f sxd 5 212x3 1 20x2 38.

Zeros: of multiplicity 2 (even multiplicity)

of multiplicity 1 (odd multiplicity)

of multiplicity 1 (odd multiplicity)

Turning points: 3

x 5 2

x 5 21

x 5 0

5 x2sx 1 1dsx 2 2d

0 5 x2sx2 2 x 2 2d

0 5 x4 2 x3 2 2x2

−4

−3

5

3gsxd 5 x4 2 x3 2 2x2

39.

(a) The degree is odd and the leading coefficient is

negative. The graph rises to the left and falls to

the right.

(b) Zero:

(c)

(d) y

x−4 −3 −2

−4

−3

1

2

3

4

1 2 3 4

(−1, 0)

x 5 21

f sxd 5 2x3 1 x2 2 2 40.

(a) The degree is odd and the leading coefficient, 2, is

positive. The graph rises to the right and falls to the left.

(b)

The zeros are 0 and

(c)

(d) y

x−4 −3 −1

−4

−3

−2

−1

2

3

4

21 3 4

(−2, 0) (0, 0)

22.

0 5 x2sx 1 2d

0 5 2x2sx 1 2d

0 5 2x3 1 4x2

gsxd 5 2x3 1 4x2

gsxd 5 2x3 1 4x2

x 0 1 2

34 10 0 262222f sxd

212223

x 0 1

0 2 0 6218gsxd

212223


48.

4x 1 7

3x 2 25

4

31

29

3s3x 2 2d

293

4x 283

3x 2 2 ) 4x 1 7

43

41.

(a) The degree is even and the leading coefficient is

positive. The graph rises to the left and rises to

the right.

(b) Zeros:

(c)

(d) y

x−4 −1−2

3

21 3 4

−15

−18

−21

(−3, 0)

(0, 0)

(1, 0)

x 5 0, 1, 23

f sxd 5 xsx3 1 x2 2 5x 1 3d

x 0 1 2 3

100 0 0 0 10 7228218f sxd

21222324

42.

(a) The degree is even and the leading coefficient, , is

negative. The graph falls to the left and falls to the right.

(b)

The zeros are 0, and

(c)

(d) y

x−4 −3 −1

−4

−3

−2

−1

2

3

4

1 3 4

( (3, 0 ( (3, 0−

(0, 0)

!3.2!3,

0 5 x2s3 2 x2d

0 5 3x2 2 x4

gsxd 5 3x2 2 x4

21

hsxd 5 3x2 2 x4

x 0 1 2

2 0 2 2424hsxd

2122

43. (a)

(b) The zero is in the interval

Zero: x < 20.900

f21, 0g.

f sxd 5 3x3 2 x2 1 3

x 0 1 2 3

3 5 23 7521225287f sxd

212223

44. (a)

(b) The only zero is in the interval

It is x < 24.479.

s25, 24d.

f sxd 5 0.25x3 2 3.65x 1 6.12

45. (a)

(b) There are two zeros, one in the interval and

one in the interval

Zeros: x < 20.200, x < 1.772

f1, 2gf21, 0g

f sxd 5 x4 2 5x 2 1

x 0 1 2 3

95 25 5 5 652521f sxd

212223

46. (a)

(b) There are zeros in the intervals and

They are and x < 20.509.x < 21.211

s21, 0d.s22, 21d

f sxd 5 7x4 1 3x3 2 8x2 1 2

x

4.72 10.32 11.4226.88225.98f sxd

2223242526

x 0 1 2 3 4

9.52 6.12 2.72 0.82 1.92 7.52f sxd

21

x 0 1 2

416 58 2 4 10622f sxd

212223

47.

Thus,24x2 2 x 2 8

3x 2 25 8x 1 5 1

2

3x 2 2.

2

15x 2 10

15x 2 8

24x2 2 16x

3x 2 2 ) 24x2 2 x 2 8

8x 1 5


49.

Thus,5x3 2 13x2 2 x 1 2

x2 2 3x 1 15 5x 1 2.

0

2x2 2 6x 1 2

2x2 2 6x 1 2

5x3 2 15x2 1 5x

x2 2 3x 1 1 ) 5x3 2 13x2 2 x 1 2

5x 1 2 50.

3x4

x2 2 15 3x2 1 3 1

3

x2 2 1

3

3x2 2 3

3x2 1 0

3x4 2 3x2

x2 2 1 ) 3x4 1 0x3 1 0x2 1 0x 1 0

3x2 1 3

51.

Thus,x4 2 3x3 1 4x2 2 6x 1 3

x2 1 25 x2 2 3x 1 2 2

1

x2 1 2.

21

2x2 1 0x 1 4

2x2 1 0x 1 3

23x3 1 0x2 2 6x

23x3 1 2x2 2 6x

x4 1 0x3 1 2x2

x2 1 0x 1 2 ) x4 2 3x3 1 4x2 2 6x 1 3

x2 2 3x 1 2 52.

6x4 1 10x3 1 13x2 2 5x 1 2

2x2 2 15 3x2 1 5x 1 8 1

10

2x2 2 1

10

16x2 1 0x 2 8

16x2 2 0x 1 2

10x3 1 0x2 2 5x

10x3 1 16x2 2 5x

6x4 1 0x3 2 3x2

2x2 1 0x 2 1 ) 6x4 1 10x3 1 13x2 2 5x 1 2

3x2 1 5x 1 8

53.

Thus,

6x4 2 4x3 2 27x2 1 18x

x 2 25 6x3 1 8x2 2 11x 2 4 2

8

x 2 2.

2 6

6

24

12

8

227

16

211

18

222

24

0

28

28

54.

0.1x3 1 0.3x2 2 0.5

x 2 55 0.1x2 1 0.8x 1 4 1

19.5

x 2 5

5 0.1

0.1

0.3

0.5

0.8

0

4

4

20.5

20

19.5

55.

Thus,2x3 2 19x2 1 38x 1 24

x 2 45 2x2 2 11x 2 6.

4 2

2

219

8

211

38

244

26

24

224

0

56.

3x3 1 20x2 1 29x 2 12

x 1 35 3x2 1 11x 2 4

23 3

3

20

29

11

29

233

24

212

12

0

57.

(a)

Yes, is a zero of

(c)

Yes, is a zero of f.x 5 0

0 20

20

9

0

9

214

0

214

23

0

23

0

0

0

f.x 5 21

21 20

20

9

220

211

214

11

23

23

3

0

0

0

0

f sxd 5 20x 4 1 9x3 2 14x2 2 3x

(b)

Yes, is a zero of

(d)

No, is not a zero of f.x 5 1

1 20

20

9

20

29

214

29

15

23

15

12

0

12

12

f.x 534

34 20

20

9

15

24

214

18

4

23

3

0

0

0

0


59.

(a)

Thus, f s23d 5 2421.

23 1

1

10

23

7

224

221

245

20

135

155

44

2465

2421

f sxd 5 x4 1 10x3 2 24x2 1 20x 1 44

(b)

f s21d 5 29

21 1

1

10

21

9

224

29

233

20

33

53

44

253

29

(b)

No, is not a zero of f.

(d)

No, is not a zero of f.x 5 21

21 3

3

28

23

211

220

11

29

16

9

25

x 5 24

24 3

3

28

212

220

220

80

60

16

2240

2224

58.

(a)

Yes, is a zero of f.

(c)

Yes, is a zero of f.x 523

23 3

3

28

2

26

220

24

224

16

216

0

x 5 4

4 3

3

28

12

4

220

16

24

16

216

0

f sxd 5 3x3 2 8x2 2 20x 1 16

60.

(a)

Thus,

(b)

Thus, gs!2d 5 0.

!2 2

2

25

2!2

25 1 2!2

0

25!2 1 4

25!2 1 4

0

210 1 4!2

210 1 4!2

28

210!2 1 8

210!2

20

220

0

gs24d 5 23276.

24 2

2

25

28

213

0

52

52

0

2208

2208

28

832

824

20

23296

23276

gstd 5 2t5 2 5t4 2 8t 1 20

61. ; Factor:

(a)

Yes, is a factor of

(b)

The remaining factors of are and

(c)

(d) Zeros:

(e)

−8 5

−60

80

27, 21, 4

5 sx 1 7dsx 1 1dsx 2 4d

f sxd 5 x3 1 4x2 2 25x 2 28

sx 1 1d.sx 1 7df

x2 1 8x 1 7 5 sx 1 7dsx 1 1d

f sxd.x 2 4

4 1

1

4

4

8

225

32

7

228

28

0

sx 2 4df sxd 5 x3 1 4x2 2 25x 2 28 62.

(a)

Yes, is a factor of

(b)

The remaining factors are and

(c)

(d) Zeros:

(e)

−7 5

−100

50

x 5 252, 3, 26

f sxd 5 s2x 1 5dsx 2 3dsx 1 6d

sx 2 3d.s2x 1 5d

2x2 2 x 2 15 5 s2x 1 5dsx 2 3d

f sxd.sx 1 6d

26 2

2

11

212

21

221

6

215

290

90

0

f sxd 5 2x3 1 11x2 2 21x 2 90


63.

Factors:

(a)

Both are factors since the remainders are zero.

(b)


(c)

(d) Zeros:

(e)

−3

−10

5

40

22, 21, 3, 4

f sxd 5 sx 1 1dsx 2 4dsx 1 2dsx 2 3d

sx 2 4d.sx 1 1d

x2 2 3x 2 4 5 sx 1 1dsx 2 4d

3 1

1

26

3

23

5

29

24

12

212

0

22 1

1

24

22

26

27

12

5

22

210

12

24

224

0

sx 1 2d, sx 2 3d

f sxd 5 x 4 2 4x3 2 7x2 1 22x 1 24 64.

(a)

Yes, and are both factors of

(b)


(c)

(d) Zeros:

(e)

−6 12

−8

4

x 5 1, 2, 3, 5

f sxd 5 sx 2 1dsx 2 3dsx 2 2dsx 2 5d

sx 2 3d.sx 2 1d

x2 2 4x 1 3 5 sx 2 1dsx 2 3d

f sxd.sx 2 5dsx 2 2d

5 1

1

29

5

24

23

220

3

215

15

0

2 1

1

211

2

29

41

218

23

261

46

215

30

230

0

f sxd 5 x4 2 11x3 1 41x2 2 61x 1 30

65. 6 1 !24 5 6 1 2i 66. 3 2 !225 5 3 2 5i 67. i2 1 3i 5 21 1 3i

68. 25i 1 i2 5 21 2 5i 69. s7 1 5id 1 s24 1 2id 5 s7 2 4d 1 s5i 1 2id 5 3 1 7i

70. 5 221!2

2i2 5 2!2 i1!2

22

!2

2i2 2 1!2

21

!2

2i2 5

!2

22

!2

2i 2

!2

22

!2

2i

71. 5is13 2 8id 5 65i 2 40i2 5 40 1 65i 72.

5 17 1 28i

5 5 1 28i 1 12

s1 1 6ids5 2 2id 5 5 2 2i 1 30i 2 12i2

73.

5 24 2 46i

s10 2 8ids2 2 3id 5 20 2 30i 2 16i 1 24i2 74.

5 9 1 20i

5 20i 2 9i2

5 is20 2 9id

is6 1 ids3 2 2id 5 is18 2 12i 1 3i 2 2i2d

75.

523

171

10

17i

523 1 10i

17

524 1 10i 1 i2

16 1 1

6 1 i

4 2 i5

6 1 i

4 2 i?

4 1 i

4 1 i76.

517

261

7i

26

517 1 7i

26

515 2 3i 1 10i 2 2i2

25 2 i2

3 1 2i

5 1 i5

3 1 2i

5 1 i?

5 2 i

5 2 i


77.

521

132

1

13i

5 1 8

131 12 1 112

13i 2 i2

58

131

12

13i 1 1 2 i

58 1 12i

4 1 91

2 2 2i

1 1 1

4

2 2 3i1

2

1 1 i5

4

2 2 3i?

2 1 3i

2 1 3i1

2

1 1 i?

1 2 i

1 2 i78.

59 1 83i

855

9

851

83i

85

518 1 81i 1 2i 1 9i2

4 2 81i2

529 2 i

22 1 9i?

s22 2 9ids22 2 9id

51 1 4i 2 10 2 5i

2 1 8i 1 i 1 4i2

1

2 1 i2

5

1 1 4i5

s1 1 4id 2 5s2 1 ids2 1 ids1 1 4id

79.

5 ±!1

3 i 5 ±

!3

3i

x 5 ±!21

3

x2 5 21

3

3x2 5 21

3x2 1 1 5 0 80.

x 5 ±1

2i

x2 5 21

4

8x2 5 22

2 1 8x2 5 0

81.

x 5 1 ± 3i

x 2 1 5 ±!29

sx 2 1d2 5 29

x2 2 2x 1 1 5 210 1 1

x2 2 2x 1 10 5 0 82.

523 ± 3i!71

125 2

1

4±!71

4i

523 ± !2639

12

523 ± !32 2 4s6ds27d

2s6d

x 52b ± !b2 2 4ac

2a

6x2 1 3x 1 27 5 0

83.

Zeros: x 5 0, x 5 2

f sxd 5 3xsx 2 2d2 84.

Zeros: x 5 29, 4

f sxd 5 sx 2 4dsx 1 9d2 85.

Zeros: x 5 1, x 5 8

5 sx 2 1dsx 2 8d

f sxd 5 x2 2 9x 1 8 86.

Zeros: x 5 0, ±!6i

5 xsx2 1 6d

f sxd 5 x3 1 6x

87.

Zeros: x 5 24, x 5 6, x 5 2i, x 5 22i

f sxd 5 sx 1 4dsx 2 6dsx 2 2idsx 1 2id 88.

Zeros: x 5 5, 8, 3 ± i

f sxd 5 sx 2 8dsx 2 5d2sx 2 3 1 idsx 2 3 2 id

89.

Possible rational zeros:

±14, ±

34, ±

54, ±

154

±1, ±3, ±5, ±15, ±12, ±

32, ±

52, ±

152 ,

f sxd 5 24x3 1 8x2 2 3x 1 15 90.

Possible rational zeros: ±13, ±

23, ±

43, ±

83±1, ±2, ±4, ±8,

fsxd 5 3x4 1 4x3 2 5x2 2 8


91.


The zeros of are and x 5 23.x 5 21, x 5 6,f sxd

5 sx 1 1dsx 2 6dsx 1 3d

x3 2 2x2 2 21x 2 18 5 sx 1 1dsx2 2 3x 2 18d

21 1

1

22

21

23

221

3

218

218

18

0

±1, ±2, ±3, ±6, ±9, ±18

f sxd 5 x3 2 2x2 2 21x 2 18 92.


Zeros: x 5 21, 53, 6

0 5 sx 1 1ds3x 2 5dsx 2 6d.

5 sx 1 1ds3x 2 5dsx 2 6d

5 sx 1 1ds3x2 2 23x 1 30d

So, f sxd 5 3x3 2 20x2 1 7x 1 30

21 3

3

220

23

223

7

23

30

30

230

0

±103±30, ±

13, ±

23, ±

53,

±15,±10,±1, ±2, ±3, ±5, ±6,

f sxd 5 3x3 2 20x2 1 7x 1 30

93.


The zeros of are and x 5 8.x 5 1f sxd

5 sx 2 1d2sx 2 8d

5 sx 2 1dsx 2 1dsx 2 8d

x3 2 10x2 1 17x 2 8 5 sx 2 1dsx2 2 9x 1 8d

1 1

1

210

1

29

17

29

8

28

8

0

±1, ±2, ±4, ±8

f sxd 5 x3 2 10x2 1 17x 2 8 94.


Zeros: x 5 25, 22

5 sx 1 5dsx 1 2d2.

5 sx 1 5dsx2 1 4x 1 4d

So, f sxd 5 x3 1 9x2 1 24x 1 20

25 1

1

9

25

4

24

220

4

20

220

0

±20±1, ±2, ±4, ±5, ±10,

f sxd 5 x3 1 9x2 1 24x 1 20

95.


The real zeros of are and x 5 24.x 5 3,f sxd

x4 1 x3 2 11x2 1 x 2 12 5 sx 2 3dsx 1 4dsx2 1 1d

24 1

1

4

24

0

1

0

1

4

24

0

3 1

1

1

3

4

211

12

1

1

3

4

212

12

0

±1, ±2, ±3, ±4, ±6, ±12

f sxd 5 x4 1 x3 2 11x2 1 x 2 12

96.


Zeros: x 5 23, 2, ±25

5 sx 1 3dsx 2 2ds5x 1 2ds5x 2 2d.

5 sx 1 3dsx 2 2ds25x2 2 4d

So, f sxd 5 25x 4 1 25x3 2 154x2 2 4x 1 24

2 25

25

250

50

0

24

0

24

8

28

0

23 25

25

25

275

250

2154

150

24

24

12

8

24

224

0

±625, ±

825, ±

1225, ±

2425±

425,±

125, ±

225, ±

325,±

245 ,

±24, ±15, ±

25, ±

35, ±

45, ±

65, ±

85, ±

125 ,±12,±1, ±2, ±3, ±4, ±6, ±8,

f sxd 5 25x 4 1 25x3 2 154x2 2 4x 1 24


97. Since is a zero, so is

Multiply by 3 to clear the fraction.

Note: where a is any real nonzero number, has zeros 23, 4, and ±!3 i.f sxd 5 as3x4 2 14x3 1 17x2 2 42x 1 24d,

5 3x4 2 14x3 1 17x2 2 42x 1 24

5 s3x2 2 14x 1 8dsx2 1 3d

5 s3x 2 2dsx 2 4dsx2 1 3d

2!3i.!3if sxd 5 3sx 223dsx 2 4dsx 2 !3 idsx 1 !3 id

98. Since is a zero and the coefficients are real,

must also be a zero.

5 x 4 2 x3 2 3x2 1 17x 2 30

5 sx2 1 x 2 6dsx2 2 2x 1 5d

5 sx2 1 x 2 6dfsx 2 1d2 1 4g

f sxd 5 sx 2 2dsx 1 3dsx 2 1 1 2idsx 2 1 2 2id

1 1 2i

1 2 2i 99. Zero: i

Since i is a zero, so is

Zeros: x 5 ± i, 4f sxd 5 sx 2 idsx 1 idsx 2 4d,

2i 1

1

24 1 i

2i

24

24i

4i

0

i 1

1

24

i

24 1 i

1

21 2 4i

24i

24

4

0

2i.

fsxd 5 x3 2 4x2 1 x 2 4,

100.

Since is a zero, so is

Zeros: x 5 ± 4i, 2

hsxd 5 sx 1 4idsx 2 4ids2x 1 2d

4i 21

21

2 1 4i

24i

2

28i

8i

0

24i 21

21

2

4i

2 1 4i

216

16 2 8i

28i

32

232

0

4i.24i

hsxd 5 2x3 1 2x2 2 16x 1 32 101. Zero:

Since is a zero, so is

Zeros: x 5 2 ± i, 12, 23

5 sx 2 2 2 idsx 2 2 1 ids2x 2 1dsx 1 3d

gsxd 5 fx 2 s2 1 idgfx 2 s2 2 idgs2x2 1 5x 2 3d

2 2 i 2

2

1 1 2i

4 2 2i

5

213 1 5i

10 2 5i

23

6 2 3i

26 1 3i

0

2 1 i 2

2

23

4 1 2i

1 1 2i

213

5i

213 1 5i

37

231 2 3i

6 2 3i

215

15

0

2 2 i2 1 i

2 1 igsxd 5 2x 4 2 3x3 2 13x2 1 37x 2 15,

102.

One zero is Since is a zero, so is

Zeros: 0, 34, 1 1 i, 1 2 i

5 xsx 2 1 1 idsx 2 1 2 ids4x 2 3d

f sxd 5 xfx 2 s1 2 idgfx 2 s1 1 idgs4x 2 3d

1 1 i 4

4

27 2 4i

4 1 4i

23

3 1 3i

23 2 3i

0

1 2 i 4

4

211

4 2 4i

27 2 4i

14

211 1 3i

3 1 3i

26

6

0

1 1 i.1 2 ix 5 0.

5 xs4x3 2 11x2 1 14x 2 6d

f sxd 5 4x4 2 11x3 1 14x2 2 6x 103.

Zeros: x 5 0, 25, 1

5 xsx 1 5dsx 2 1d

5 xsx2 1 4x 2 5d

f sxd 5 x3 1 4x2 2 5x


104.

The zeros of are The zeros of are

gsxd 5 sx 1 2dsx 2 3dsx 2 6d

22, 3, 6.gsxdx 5 3, 6.x2 2 9x 1 18 5 sx 2 3dsx 2 6d

22 1

1

27

22

29

0

18

18

36

236

0

gsxd 5 x3 2 7x2 1 36

105. Zero:

By the Quadratic Formula the zeros of are

The zeros of are of multiplicity

2, and

5 sx 1 4d2sx 2 2 2 3idsx 2 2 1 3id

gsxd 5 sx 1 4d2fx 2 s2 1 3idgfx 2 s2 2 3idg

x 5 2 ± 3i.

x 5 24gsxdx 5 2 ± 3i.

x2 2 4x 1 13

gsxd 5 sx 1 4d2sx2 2 4x 1 13d

24 1

1

0

24

24

23

16

13

52

252

0

24 1

1

4

24

0

23

0

23

40

12

52

208

2208

0

x 5 24gsxd 5 x 4 1 4x3 2 3x2 1 40x 1 208, 106.

By the Quadratic Formula, the zeros of are

The zeros of are

f sxd 5 sx 1 3dsx 2 3dsx 1 4 2 idsx 1 4 1 id

23, 3, 24 2 i, 24 1 i.f sxd

x 528 ± !s8d2 2 4s1ds17d

2s1d 528 ± !24

25 24 ± i.

x2 1 8x 1 17

23 1

1

11

23

8

41

224

17

51

251

3 1

1

8

3

11

8

33

41

272

123

51

2153

153

0

f sxd 5 x4 1 8x3 1 8x2 2 72x 2 153

107.

has two variations in sign, so g has either two or no

positive real zeros.

has one variation in sign, so g has one negative

real zero.

g s2xd

gs2xd 5 25x3 1 3x2 1 6x 1 9

g sxd

g sxd 5 5x3 1 3x2 2 6x 1 9 108.

has three variations in sign, so h has either three or

one positive real zeros.

has two variations in sign, so h has either two or

no negative real zeros.

hs2xd

5 2x5 2 4x3 2 2x2 1 5

hs2xd 5 22s2xd5 1 4s2xd3 2 2s2xd2 1 5

hsxd

hsxd 5 22x5 1 4x3 2 2x2 1 5

109.

(a)

Since the last row has all positive entries,

is an upper bound.

(b)

Since the last row entries alternate in sign,

is a lower bound.x 5 214

214 4

4

23

21

24

4

1

5

23

254

2174

x 5 1

1 4

4

23

4

1

4

1

5

23

5

2

fsxd 5 4x3 2 3x2 1 4x 2 3 110.

(a)

Since the last row has all positive entries, is an

upper bound.

(b)

Since the last row entries alternate in sign, is a

lower bound.

x 5 4

24 2

2

25

28

213

214

52

38

8

2152

2144

x 5 8

8 2

2

25

16

11

214

88

74

8

592

600

gsxd 5 2x3 2 5x2 2 14x 1 8


111.

Domain: all real numbers

except x 5 212

x

f sxd 55x

x 1 12112.


except x 5 213

x

x 5 21

3

3x 5 21

1 1 3x 5 0

f sxd 53x2

1 1 3x113.


except and x 5 6x 5 4

x

58

sx 2 4dsx 2 6d

f sxd 58

x2 2 10x 1 24

114.


f sxd 5x2 2 x 2 2

x2 1 4115.

Vertical asymptote:

Horizontal asymptote: y 5 0

x 5 23

f sxd 54

x 1 3116.

Vertical asymptote: none


f sxd 52x2 1 5x 2 3

x2 1 2

117.

Vertical asymptote:


x 5 23

52

x 1 3, x Þ 5

52sx 2 5d

sx 1 3dsx 2 5d

hsxd 52x 2 10

x2 2 2x 2 15118.

Vertical asymptotes:

Horizontal asymptotes: none

x 5 22, x 5 21

hsxd 5x3 2 4x2

x2 1 3x 1 25

x2sx 2 4dsx 1 2dsx 1 1d

119.

(a) Domain: all real numbers except

(b) No intercepts

(c) Vertical asymptote:

Horizontal asymptote:

(d)

−1 1 2

−3

−2

1

x

y

y 5 0

x 5 0

x 5 0x

f sxd 525

x2

x

y 252542

59

±1±2±3

120.


(b) No intercepts



(d)

−3 −2 −1 1 2 3 4

−3

−2

1

2

3

4

x

y

y 5 0

x 5 0

x 5 0x

fsxd 54

x

1 2 3

4 24324222

43

y

212223x


121.


(b) x-intercept:

y-intercept:



(d)

−8

−6

−4

−2

4

6

x2(−2, 0)

(0, 2)

y

y 5 21

x 5 1

s0, 2ds22, 0d

x 5 1x

gsxd 52 1 x

1 2 x5 2

x 1 2

x 2 1

x 0 2 3

y 2 25224

12

21

122.


(b) x-intercept:

y-intercept:



(d)

−2 −1 1 4 5 6

−3

−2

3

4

5

x

(3, 0)

( (0,32

y

y 5 1

x 5 2

10, 3

22s3, 0d

x 5 2x

hsxd 5x 2 3

x 2 2

x 0 1 3 4 5

y 2 023

12

32

43

21

123.

(a) Domain: all real numbers

(b) Intercept:

(c) Horizontal asymptote:

(d)

(0, 0)−3 −2 −1 2 3

−2

2

3

4

x

y

y 5 1

s0, 0d

x

psxd 5x2

x2 1 1

x 0

y 012

45

910

±1±2±3

124.


(b) Intercept:


(d)

−3

−2

−1

1

2

3

x1 2 3

(0, 0)

y

y 5 0

s0, 0d

x

f sxd 52x

x2 1 4

x 0 1 2

y 012

252

252

12

2122

125.

(a) Domain: all real numbers (d)

(b) Intercept:

(c) Horizontal asymptote: y 5 0

s0, 0d

1 2

−2

−1

1

2

x(0, 0)

yx

f sxd 5x

x2 1 1

x 0 1 2

y 025

122

122

25

2122


126.


(b) intercept:



(d)

−3 −2 −1 2 3 4 5

1

3

5

6

7

x

(0, 4)

y

y 5 0

x 5 1

s0, 4dy-

x 5 1x

hsxd 54

sx 2 1d2

x 0 2 3 4

y 1 4 4 149

49

2122

127.


(b) Intercept:


(d)

x

−2−4−6

2

6

−8

4

2 4

(0, 0)

y

y 5 26

s0, 0d

x

f sxd 526x2

x2 1 1

x 0

y 02322452

275

±1±2±3

128.


(b) Intercept:

(c) Vertical asymptotes:


(d)

−6 −4 4 6

4

6

x(0, 0)

y

y 5 2

x 5 2, x 5 22

s0, 0d

x 5 ±2x

y 52x2

x2 2 4

x 0

y 0223

185

83

5021

±1±3±4±5

129.

(a) Domain: all real numbers except and

(b) intercept:

intercept: none



(d)

y

x−2−4−6−8 4 6 8

−2

−4

−6

−8

, 03

2( (

2

y 5 2

x 5 0

y-

13

2, 02x-

x 51

3x 5 0x

5s3x 2 1ds2x 2 3d

xs3x 2 1d 52x 2 3

x, x Þ

1

3

f sxd 56x2 2 11x 1 3

3x2 2 x

1 2 3 4

5 154

1221

72

y

2122x


130.

(a) Domain: all real numbers x except

(b) intercept:

intercept:



(d)

y

x

, 02

3( (−1−2−3 2 3

2

y 53

2

x 5 21

2

12

3, 02x-

s0, 22dy-

x Þ ±1

2

5s2x 2 1ds3x 2 2ds2x 2 1ds2x 1 1d 5

3x 2 2

2x 1 1, x Þ

1

2

f sxd 56x2 2 7x 1 2

4x2 2 1

0 1 2

5 045

1322

83

115

y

23212223x

131.

(a) Domain: all real numbers x

(b) Intercept:

(c) Slant asymptote:

(d)

1

x

−2

2

3

3

−3

1 2−1−2−3

y

(0, 0)

y 5 2x

s0, 0d

f sxd 52x3

x2 1 15 2x 2

2x

x2 1 1

x 0 1 2

y 0 1165212

165

2122

132.

(a) Domain: all real numbers x except

(b) y-intercept:


Using long division,

Slant asymptote:

(d)

x−2−4−6 62

4

4

y

(0, 1)

y 5 x 2 1

f sxd 5x2 1 1

x 1 15 x 2 1 1

2

x 1 1.

x 5 21

s0, 1d

x 5 21

f sxd 5x2 1 1

x 1 1

x 0 4

y 1175

522

132252

375

2122

322226


133.


(b) intercepts: and

intercept:


Slant asymptote:

(d)

y

x(1, 0)

, 0

−1−2 2 3 4

−2

1

2

3

4

2

3( (0, − 1

2( (

y 5 x 21

3

x 54

3

10, 21

22y-

12

3, 02s1, 0dx-

x 5 21, x 54

3x

5 x 21

31

2y3

3x 2 4, x Þ 21

5s3x 2 2dsx 2 1d

3x 2 4

5s3x 2 2dsx 1 1dsx 2 1d

s3x 2 4dsx 1 1d

f sxd 53x3 2 2x2 2 3x 1 2

3x2 2 x 2 4

0 1 2 3

0 21452

122

1252

4413

y

2223x

134.

(a) Domain: all real except or

(b) intercept:

intercepts:


Using long division,

Slant asymptote:

(d)

y

x

, 04

3( (

−2−4−6 4 6

−2

−6

2

4

(2, 0)

(0, −8)

y 5 x 2 3

f sxd 53x2 2 10x 1 8

3x 2 15 x 2 3 1

5

3x 2 1.

x 51

3

14

3, 02, s2, 0dx-

s0, 28dy-

x 51

3x 5 22x

5sx 2 2ds3x 2 4d

3x 2 1, x Þ 22

5sx 2 2dsx 1 2ds3x 2 4d

sx 1 2ds3x 2 1d

fsxd 53x3 2 4x2 2 12x 1 16

3x2 1 5x 2 2

0 1 2 4

01611

12282

2142

9613

y

2124x

135.


As x increases, the average cost per unit approaches the

horizontal asymptote, C 5 0.5 5 $0.50.

C 50.5

15 0.5

C 5C

x5

0.5x 1 500

x, 0 < x 136.

(a)

(b) When million.

When million.

When million.

(c) As No, it is not possible.p → 100, C → `.

p 5 75, C 5528s75d

100 2 755 $1584

p 5 50, C 5528s50d

100 2 505 $528

C 5528s25d

100 2 255 $176p 5 25,

0

0

100

4000

C 5528p

100 2 p, 0 ≤ p < 100


137. (a)

(b) The area of print is which is

30 square inches.

Total area 5 xy 5 x32s2x 1 7dx 2 4 4 5

2xs2x 1 7dx 2 4

y 52s2x 1 7d

x 2 4

y 54x 1 14

x 2 4

y 530 1 4sx 2 4d

x 2 4

y 530

x 2 41 4

y 2 4 530

x 2 4

sx 2 4dsy 2 4d 5 30

sx 2 4dsy 2 4d,

y

x

2 in.

2 in.

2 in. 2 in.

(c) Because the horizontal margins total 4 inches, x must be

greater than 4 inches. The domain is

(d)

The minimum area occurs when inches, so

The least amount of paper used is for a page size of about

9.48 inches by 9.48 inches.

y <2s2 ? 9.477 1 7d

9.477 2 4< 9.477 inches.

x < 9.477

0

4 32

200

x > 4.

138.

The limiting amount of uptake is determined

by the horizontal asymptote,

00

100

90

y 518.47

0.23< 80.3 mgydm2yhr.

CO2

y 518.47x 2 2.96

0.23x 1 1, 0 < x 139.

Critical numbers:

Test intervals:

Test: Is

By testing an x-value in each test interval in the

inequality, we see that the solution set is: s243,

12d

s3x 1 4ds2x 2 1d < 0?

s2`, 243, d, s24

3, 12d, s1

2, `dx 5 2

43, x 5

12

s3x 1 4ds2x 2 1d < 0

6x2 1 5x 2 4 < 0

6x2 1 5x < 4

140.

Critical numbers:

Test intervals:

Solution interval: s2`, 23g < f52, `d

s52, `d ⇒ s2x 2 5dsx 1 3d > 0

s23, 52d ⇒ s2x 2 5dsx 1 3d < 0

s2`, 23d ⇒ s2x 2 5dsx 1 3d > 0

x 552, x 5 23

s2x 2 5dsx 1 3d ≥ 0

2x2 1 x 2 15 ≥ 0

2x2 1 x ≥ 15 141.

Critical numbers:

Test intervals:

Test: Is

By testing an value in each test interval in the inequality,

we see that the solution set is: f24, 0g < f4, `d.x-

xsx 1 4dsx 2 4d ≥ 0?

s4, `ds0, 4d,s24, 0d,s2`, 24d,

x 5 0, x 5 ±4

xsx 1 4dsx 2 4d ≥ 0

x3 2 16x ≥ 0


142.

Critical numbers:

Test intervals:

Solution interval: s2`, 0d < s0, 53d

s53, `d ⇒ 12x3 2 20x2

> 0

s0, 53d ⇒ 12x3 2 20x2

< 0

s2`, 0d ⇒ 12x3 2 20x2< 0

x 5 0, x 553

4x2s3x 2 5d < 0

12x3 2 20x2< 0 143.

Critical numbers:

Test intervals:

Test: Is

By testing an x-value in each test interval in the inequality,

we see that the solution set is: f25, 21d < s1, `d

2sx 1 5dsx 1 1dsx 2 1d ≤ 0?

s2`, 25d, s25, 21d, s21, 1d, s1, `d

x 5 25, x 5 ±1

2sx 1 5d

sx 1 1dsx 2 1d≤ 0

2x 2 2 2 3x 2 3

sx 1 1dsx 2 1)≤ 0

2sx 2 1d 2 3sx 1 1d

sx 1 1dsx 2 1d≤ 0

2

x 1 1≤

3

x 2 1

144.

Critical numbers:

Test intervals:

Solution intervals: s2`, 3d < s5, `d

s5, `d ⇒ x 2 5

3 2 x< 0

s3, 5d ⇒ x 2 5

3 2 x> 0

s2`, 3d ⇒ x 2 5

3 2 x< 0

x 5 5, x 5 3

x 2 5

3 2 x< 0 145.

Critical numbers:

Test intervals:

Test: Is

By testing an x-value in each test interval in the inequality,

we see that the solution set is: f24, 23g < s0, `d

sx 1 4dsx 1 3dx

≥ 0?

s2`, 24d, s24, 23d, s23, 0d, s0, `d

x 5 24, x 5 23, x 5 0

sx 1 4dsx 1 3dx

≥ 0

x2 1 7x 1 12

x≥ 0

146.

Critical numbers:

Test intervals:

Solution interval: s2`, 0d < s2, `d

s2, `d ⇒ 1

x 2 22

1

x> 0

s0, 2d ⇒ 1

x 2 22

1

x< 0

s2`, 0d ⇒ 1

x 2 22

1

x> 0

x 5 2, x 5 0

1

x 2 22

1

x> 0

1

x 2 2>

1

x147.

r > 4.9%

r > 0.0488

1 1 r > 1.0488

s1 1 rd2> 1.1

5000s1 1 rd2> 5500

148.

t ≥ 9 days

21000t ≤ 29000

10,000 1 2000t ≤ 1000 1 3000t

2000s5 1 td ≤ 1000s1 1 3td

2000 ≤1000s1 1 3td

5 1 t

P 51000s1 1 3td

5 1 t149. False. A fourth-degree polynomial

can have at most four zeros and

complex zeros occur in conjugate

pairs.

150. False. (See Exercise 123.)

The domain of

is the set of all real numbers x.

f sxd 51

x2 1 1


1.

Thus, and

Since the remainder f skd 5 r.r 5 ak31 bk2

1 ck 1 d,f skd 5 ak31 bk2

1 ck 1 d.

f sxd 5 ax31 bx2

1 cx 1 d 5 sx 2 kdfax21 sak 1 bdx 1 sak2

1 bx 1 cdg 1 ak31 bk2

1 ck 1 d

sak31 bk2

1 ck 1 dd

sak21 bk 1 cdx 2 sak3

1 bk21 ckd

sak21 bk 1 cdx 1 d

sak 1 bdx22 sak2

1 bkdx

sak 1 bdx21 cx

ax32 akx2

x 2 k) ax31 bx2

1 cx 1 d

ax21 sak 1 bdx 1 sak2

1 bk 1 cd

f sxd 5 ax31 bx2

1 cx 1 d

2. (a)

(b)

(c)

1x

223

1 1x

222

5 36 ⇒ x

25 3 ⇒ x 5 6

1

8x3

11

8s2x2d 5

1

8s288d

x31 2x2

5 288; a 5 1, b 5 2 ⇒ a2

b35

1

8

x31 x2

5 252 ⇒ x 5 6

(d)

(e)

(f)

(g)

110x

3 23

1 110x

3 22

5 1100 ⇒ 10x

35 10 ⇒ x 5 3

100

27s10x3d 1

100

27s3x2d 5

100

27s297d

10x31 3x2

5 297; a 5 10, b 5 3 ⇒ a2

b35

100

27

17x

6 23

1 17x

6 22

5 392 ⇒ 7x

65 7 ⇒ x 5 6

49

216s7x3d1

49

216s6x2d 5

49

216s1728d

7x31 6x2

5 1728; a 5 7, b 5 6 ⇒ a2

b35

49

216

12x

5 23

1 12x

5 22

5 80 ⇒ 2x

55 4 ⇒ x 5 10

4

125s2x3d1

4

125s5x2d 5

4

125s2500d

2x31 5x2

5 2500; a 5 2, b 5 5 ⇒ a2

b35

4

125

s3xd31 s3xd2

5 810 ⇒ 3x 5 9 ⇒ x 5 3

9s3x3d19x25 9s90d

3x31 x2

5 90; a 5 3, b 5 1 ⇒ a2

b35 9

y

1 2

2 12

3 36

4 80

5 150

6 252

7 392

8 576

9 810

10 1100

y31 y2

Problem Solving for Chapter 2

153. An asymptote of a graph is a line to which the graph becomes arbitrarily close as increases or decreases without bound.x

151. The maximum (or minimum) value of a quadratic

function is located at its graph’s vertex. To find the

vertex, either write the equation in standard form or

use the formula

If the leading coefficient is positive, the vertex is a

minimum. If the leading coefficient is negative, the

vertex is a maximum.

12b

2a, f 12

b

2a22.

152. Answers will vary. Sample answer:

Polynomials of degree with real coefficients can

be written as the product of linear and quadratic factors

with real coefficients, where the quadratic factors have

no real zeros.

Setting the factors equal to zero and solving for the

variable can find the zeros of a polynomial function.

To solve an equation is to find all the values of the

variable for which the equation is true.

n > 0

chapter 2 review solution key - montville township public ...€¦ · review exercises for chapter...

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