Precalculus: Mathematics for Calculus - 6th Edition

6th Edition

ISBN: 9780840068071

Author: Stewart, James, Redlin, Lothar, Watson, Saleem

Publisher: Cengage Learning

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Question

Chapter 3.7, Problem 63E

To determine

To find: the intercepts and asymptotes, and then sketch a graph of the rational function. To state the domain and range.

Expert Solution & Answer

Answer to Problem 63E

The function has no y intercept.

The vertical asymptote occurs at x=0,x=3 .

The function has no horizontal asymptote.

The domain of the function is equal to {x|x≠0,x≠3} .

The range of the function is equal to {y|y∈ℝ} .

Explanation of Solution

Given information:

Given function,

r(x)=x2−2x+1x3−3x2

Calculation:

Consider the function,

r(x)=x2−2x+1x3−3x2

To solve for the y intercept, simplify need to plug zero for x .

Consider,

x2−2x+1x3−3x2=(0)2−2(0)+1(0)3−3(0)2=∞

Therefore, the function has no y intercept.

To solve for the x intercept, set the function equal to 0 and solve for x .

Consider,

x2−2x+1x3−3x2=0x2−2x+1=0(x−1)(x−1)=0x=1

Vertical asymptotes:

The vertical asymptote occurs at the point that makes the denominator of the function zero.

The denominator of the function is,

x3−3x2

Find the vertical asymptotes,

x3−3x2=0x2(x−3)=0x=0,x=3

Therefore, the vertical asymptote occurs at x=0,x=3 .

Horizontal asymptotes:

The degree of the numerator is lesser than denominator then the function has no horizontal asymptote.

In the function the degree of the numerator is lesser than the degree of denominator.

The function has no horizontal asymptote.

Sketch the graph of the function r(x)=x2−2x+1x3−3x2 .

Precalculus: Mathematics for Calculus - 6th Edition, Chapter 3.7, Problem 63E

From the graph,

The vertical asymptote at x=0,x=3 .

The domain of the function is equal to {x|x≠0,x≠3} .

The function has no horizontal asymptote.

The range of the function is equal to {y|y∈ℝ} .

Chapter 3.7, Problem 62E

Chapter 3.7, Problem 64E

Chapter 3 Solutions

Precalculus: Mathematics for Calculus - 6th Edition

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