(a) Answer The domain of function is all the real numbers except x = ± 3 . Explanation Given information: The function is f ( x ) = ( x 2 − 2 x − 8 ) ( x 2 − 9 ) Calculation: Consider the function. . f ( x ) = ( x 2 − 2 x − 8 ) ( x 2 − 9 ) The function is valid for all the real numbers except x = ± 3 as denominator becomes zero for these two values which makes the function undefined Therefore, the domain of function is all the real numbers except x = ± 3 . (b) To determine To find: The intercepts of the equation. Answer The y- intercept is ( 0 , 8 9 ) and x- intercept is ( − 2 , 0 ) and ( 4 , 0 ) . Explanation Given information: The function is f ( x ) = ( x 2 − 2 x − 8 ) ( x 2 − 9 ) Calculation: Put x = 0 in function to find y-intercept. f ( 0 ) = ( 0 2 − 2 ( 0 ) − 8 ) ( 0 2 − 9 ) = 8 9 The y- intercept is ( 0 , 8 9 ) . Equate f ( 0 ) to zero to find x-intercept. f ( 0 ) = 0 ( x 2 − 2 x − 8 ) ( x 2 − 9 ) = 0 x 2 − 2 x − 8 = 0 x = − 2 , 4 The x- intercept is ( − 2 , 0 ) and ( 4 , 0 ) . Therefore, the y- intercept is ( 0 , 8 9 ) and x- intercept is ( − 2 , 0 ) and ( 4 , 0 ) . (c) To determine To find: The asymptotes of the function. Answer The vertical asymptote is x = ± 3 and horizontal asymptote is y = 1 . Explanation Given information: The function is f ( x ) = ( x 2 − 2 x − 8 ) ( x 2 − 9 ) Calculation: Calculate vertical asymptote by finding solution of denominator. x 2 − 9 = 0 x 2 = 9 x = ± 3 The degree of numerator is same as the degree of denominator so the horizontal asymptote will be ratio of leading coefficient of numerator to denominator which is unity for numerator as well as denominator. So horizontal asymptote is y = 1 . Therefore, the vertical asymptote is x = ± 3 and horizontal asymptote is y = 1 . (d) To determine To find: The sketch of graph. Answer The graph is shown in Figure-(1). Explanation Given information: The function is f ( x ) = ( x 2 − 2 x − 8 ) ( x 2 − 9 ) Calculation: The additional points are tabulated below. Test Interval Value of x Value of f Sign Point of f ( − ∞ , 3 ) -4 16 7 Positive ( − 4 , 16 7 ) ( − 3 , − 2 ) -2.5 -1.18 Negative ( − 2.5 , − 1.18 ) ( − 2 , 3 ) 1 9 8 Positive ( 1 , 9 8 ) ( 3 , 4 ) 3.5 − 0.846 Negative ( 03.5 , − 0.846 ) ( 4 , ∞ ) 5 7 16 ( 5 , 7 16 ) Draw the sketch for the function by using the equations of asymptotes. Figure-(1) Therefore, the graph is shown in Figure-(1).

3rd Edition

ISBN: 9781285607160

Author: Larson

Publisher: CENGAGE CO

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Question

Chapter 2.6, Problem 28E

(a)

To determine

To find:The domain of function.

(a)

Expert Solution

Answer to Problem 28E

The domain of function is all the real numbers except x=±3 .

Explanation of Solution

Given information:

The function is f(x)=(x2−2x−8)(x2−9)

Calculation:

Consider the function.

. f(x)=(x2−2x−8)(x2−9)

The function is valid for all the real numbers except x=±3 as denominator becomes zero for these two values which makes the function undefined

Therefore, the domain of function is all the real numbers except x=±3 .

(b)

To determine

To find:The intercepts of the equation.

(b)

Expert Solution

Answer to Problem 28E

The y- intercept is (0,89) and x- intercept is (−2,0) and (4,0) .

Explanation of Solution

Given information:

The function is f(x)=(x2−2x−8)(x2−9)

Calculation:

Put x=0 in function to find y-intercept.

f(0)=(02−2(0)−8)(02−9)=89

The y- intercept is (0,89) .

Equate f(0) to zero to find x-intercept.

f(0)=0(x2−2x−8)(x2−9)=0x2−2x−8=0x=−2,4

The x- intercept is (−2,0) and (4,0) .

Therefore, the y- intercept is (0,89) and x- intercept is (−2,0) and (4,0) .

(c)

To determine

To find:The asymptotes of the function.

(c)

Expert Solution

Answer to Problem 28E

The vertical asymptote is x=±3 and horizontal asymptote is y=1 .

Explanation of Solution

Given information:

The function is f(x)=(x2−2x−8)(x2−9)

Calculation:

Calculate vertical asymptote by finding solution of denominator.

x2−9=0x2=9x=±3

The degree of numerator is same as the degree of denominator so the horizontal asymptote will be ratio of leading coefficient of numerator to denominator which is unity for numerator as well as denominator.

So horizontal asymptote is y=1 .

Therefore, the vertical asymptote is x=±3 and horizontal asymptote is y=1 .

(d)

To determine

To find:The sketch of graph.

(d)

Expert Solution

Answer to Problem 28E

The graph is shown in Figure-(1).

Explanation of Solution

Given information:

The function is f(x)=(x2−2x−8)(x2−9)

Calculation:

The additional points are tabulated below.

Test Interval	Value of x	Value of f	Sign	Point of f
(−∞,3)	-4	167	Positive	(−4,167)
(−3,−2)	-2.5	-1.18	Negative	(−2.5,−1.18)
(−2,3)	1	98	Positive	(1,98)
(3,4)	3.5	−0.846	Negative	(03.5,−0.846)
(4,∞)	5	716		(5,716)

Draw the sketch for the function by using the equations of asymptotes.

EBK PRECALCULUS W/LIMITS, Chapter 2.6, Problem 28E

Figure-(1)

Therefore, the graph is shown in Figure-(1).

Chapter 2.6, Problem 27E

Chapter 2.6, Problem 29E

Chapter 2 Solutions

EBK PRECALCULUS W/LIMITS

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Prob. 73E Ch. 2.6 - Prob. 74E Ch. 2.6 - Prob. 75E Ch. 2.6 - Prob. 76E Ch. 2.6 - Prob. 77E Ch. 2.6 - Prob. 78E Ch. 2.6 - Prob. 79E Ch. 2.6 - Prob. 80E Ch. 2.6 - Prob. 81E Ch. 2.6 - Prob. 82E Ch. 2.7 - Prob. 1E Ch. 2.7 - Prob. 2E Ch. 2.7 - Prob. 3E Ch. 2.7 - Prob. 4E Ch. 2.7 - Prob. 5E Ch. 2.7 - Prob. 6E Ch. 2.7 - Prob. 7E Ch. 2.7 - Prob. 8E Ch. 2.7 - Prob. 9E Ch. 2.7 - Prob. 10E Ch. 2.7 - Prob. 11E Ch. 2.7 - Prob. 12E Ch. 2.7 - Prob. 13E Ch. 2.7 - Prob. 14E Ch. 2.7 - Prob. 15E Ch. 2.7 - Prob. 16E Ch. 2.7 - Prob. 17E Ch. 2.7 - Prob. 18E Ch. 2.7 - Prob. 19E Ch. 2.7 - Prob. 20E Ch. 2.7 - Prob. 21E Ch. 2.7 - Prob. 22E Ch. 2.7 - Prob. 23E Ch. 2.7 - Prob. 24E Ch. 2.7 - Prob. 25E Ch. 2.7 - Prob. 26E Ch. 2.7 - Prob. 27E Ch. 2.7 - Prob. 28E Ch. 2.7 - Prob. 29E Ch. 2.7 - Prob. 30E Ch. 2.7 - Prob. 31E Ch. 2.7 - Prob. 32E Ch. 2.7 - Prob. 33E Ch. 2.7 - Prob. 34E Ch. 2.7 - Prob. 35E Ch. 2.7 - Prob. 36E Ch. 2.7 - Prob. 37E Ch. 2.7 - Prob. 38E Ch. 2.7 - Prob. 39E Ch. 2.7 - Prob. 40E Ch. 2.7 - Prob. 41E Ch. 2.7 - Prob. 42E Ch. 2.7 - Prob. 43E Ch. 2.7 - Prob. 44E Ch. 2.7 - Prob. 45E Ch. 2.7 - Prob. 46E Ch. 2.7 - Prob. 47E Ch. 2.7 - Prob. 48E Ch. 2.7 - Prob. 49E Ch. 2.7 - Prob. 50E Ch. 2.7 - Prob. 51E Ch. 2.7 - Prob. 52E Ch. 2.7 - Prob. 53E Ch. 2.7 - Prob. 54E Ch. 2.7 - Prob. 55E Ch. 2.7 - Prob. 56E Ch. 2.7 - Prob. 57E Ch. 2.7 - Prob. 58E Ch. 2.7 - Prob. 59E Ch. 2.7 - Prob. 60E Ch. 2.7 - Prob. 61E Ch. 2.7 - Prob. 62E Ch. 2.7 - Prob. 63E Ch. 2.7 - Prob. 64E Ch. 2.7 - Prob. 65E Ch. 2.7 - Prob. 66E Ch. 2.7 - Prob. 67E Ch. 2.7 - Prob. 68E Ch. 2.7 - Prob. 69E Ch. 2.7 - Prob. 70E Ch. 2.7 - Prob. 71E Ch. 2.7 - Prob. 72E Ch. 2.7 - Prob. 73E Ch. 2.7 - Prob. 74E Ch. 2.7 - Prob. 75E Ch. 2.7 - Prob. 76E Ch. 2.7 - Prob. 77E Ch. 2.7 - Prob. 78E Ch. 2.7 - Prob. 79E Ch. 2.7 - Prob. 80E Ch. 2.7 - Prob. 81E Ch. 2.7 - Prob. 82E Ch. 2.7 - Prob. 83E Ch. 2.7 - Prob. 84E Ch. 2.7 - Prob. 85E Ch. 2.7 - Prob. 86E Ch. 2.7 - Prob. 87E Ch. 2.7 - Prob. 88E Ch. 2.7 - Prob. 89E Ch. 2.7 - Prob. 90E Ch. 2 - Prob. 1RE Ch. 2 - Prob. 2RE Ch. 2 - Prob. 3RE Ch. 2 - Prob. 4RE Ch. 2 - Prob. 5RE Ch. 2 - Prob. 6RE Ch. 2 - Prob. 7RE Ch. 2 - Prob. 8RE Ch. 2 - Prob. 9RE Ch. 2 - Prob. 10RE Ch. 2 - Prob. 11RE Ch. 2 - Prob. 12RE Ch. 2 - Prob. 13RE Ch. 2 - Prob. 14RE Ch. 2 - Prob. 15RE Ch. 2 - Prob. 16RE Ch. 2 - Prob. 17RE Ch. 2 - Prob. 18RE Ch. 2 - Prob. 19RE Ch. 2 - Prob. 20RE Ch. 2 - Prob. 21RE Ch. 2 - Prob. 22RE Ch. 2 - Prob. 23RE Ch. 2 - Prob. 24RE Ch. 2 - Prob. 25RE Ch. 2 - Prob. 26RE Ch. 2 - Prob. 27RE Ch. 2 - Prob. 28RE Ch. 2 - Prob. 29RE Ch. 2 - Prob. 30RE Ch. 2 - Prob. 31RE Ch. 2 - Prob. 32RE Ch. 2 - Prob. 33RE Ch. 2 - Prob. 34RE Ch. 2 - Prob. 35RE Ch. 2 - Prob. 36RE Ch. 2 - Prob. 37RE Ch. 2 - Prob. 38RE Ch. 2 - Prob. 39RE Ch. 2 - Prob. 40RE Ch. 2 - Prob. 41RE Ch. 2 - Prob. 42RE Ch. 2 - Prob. 43RE Ch. 2 - Prob. 44RE Ch. 2 - Prob. 45RE Ch. 2 - Prob. 46RE Ch. 2 - Prob. 47RE Ch. 2 - Prob. 48RE Ch. 2 - Prob. 49RE Ch. 2 - Prob. 50RE Ch. 2 - Prob. 51RE Ch. 2 - Prob. 52RE Ch. 2 - Prob. 53RE Ch. 2 - Prob. 54RE Ch. 2 - Prob. 55RE Ch. 2 - Prob. 56RE Ch. 2 - Prob. 57RE Ch. 2 - Prob. 58RE Ch. 2 - Prob. 59RE Ch. 2 - Prob. 60RE Ch. 2 - Prob. 61RE Ch. 2 - Prob. 62RE Ch. 2 - Prob. 63RE Ch. 2 - Prob. 64RE Ch. 2 - Prob. 65RE Ch. 2 - Prob. 66RE Ch. 2 - Prob. 1CT Ch. 2 - Prob. 2CT Ch. 2 - Prob. 3CT Ch. 2 - Prob. 4CT Ch. 2 - Prob. 5CT Ch. 2 - Prob. 6CT Ch. 2 - Prob. 7CT Ch. 2 - Prob. 8CT Ch. 2 - Prob. 9CT Ch. 2 - Prob. 10CT Ch. 2 - Prob. 11CT Ch. 2 - Prob. 12CT Ch. 2 - Prob. 13CT Ch. 2 - Prob. 14CT Ch. 2 - Prob. 15CT Ch. 2 - Prob. 16CT Ch. 2 - Prob. 17CT Ch. 2 - Prob. 18CT Ch. 2 - Prob. 1PS Ch. 2 - Prob. 2PS Ch. 2 - Prob. 3PS Ch. 2 - Prob. 4PS Ch. 2 - Prob. 5PS Ch. 2 - Prob. 6PS Ch. 2 - Prob. 7PS Ch. 2 - Prob. 8PS Ch. 2 - Prob. 9PS Ch. 2 - Prob. 10PS Ch. 2 - Prob. 11PS Ch. 2 - Prob. 12PS Ch. 2 - Prob. 13PS

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