Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 2 - Section 2.6 - Rational Functions and Their Graphs - Exercise Set - Page 399: 87

Answer

a) The slant asymptote of the given function is $y=x-2$. b) The graph of the function is:
1569560359

Work Step by Step

(a) Consider the rational function $f\left( x \right)=\frac{{{x}^{3}}+1}{{{x}^{2}}+2x}$. Here, the degree of the numerator term is 2 and degree of the denominator term is 1. And $x=0$ is not a factor of the function ${{x}^{3}}+1$ , so, the graph of the function has a slant asymptote. Now, divide the function ${{x}^{3}}+1$ with ${{x}^{2}}+2x$. $\begin{align} & \frac{{{x}^{3}}+1}{{{x}^{2}}+2x}=\frac{{{x}^{3}}}{{{x}^{2}}+2x}+\frac{1}{{{x}^{2}}+2x} \\ & =\frac{{{x}^{2}}}{x+2}+\frac{1}{{{x}^{2}}+2x} \\ & =x-2+\frac{1}{{{x}^{2}}+2x} \end{align}$ Therefore, the equation has a slant asymptote at $y=x-2$. (b) Consider the rational function $f\left( x \right)=\frac{{{x}^{3}}+1}{{{x}^{2}}+2x}$. Here, the degree of the numerator term is 2 and degree of the denominator term is 1. And $x=0$ is not a factor of the function ${{x}^{3}}+1$ , so, the graph of the function has a slant asymptote. Substitute $x=-x$ $\begin{align} & f\left( -x \right)=\frac{{{\left( -x \right)}^{3}}+1}{{{\left( -x \right)}^{2}}+2\left( -x \right)} \\ & =\frac{-{{x}^{3}}+1}{{{x}^{2}}-2x} \end{align}$ The graph is not symmetrical about the $y$ axis. Compute the $y$ intercept. Put $x=0$ $\begin{align} & f\left( x \right)=\frac{{{x}^{3}}+1}{{{x}^{2}}+2x} \\ & f\left( 0 \right)=\frac{{{\left( 0 \right)}^{3}}+1}{{{\left( 0 \right)}^{2}}+2\left( 0 \right)} \end{align}$ This is not a defined condition. There is no y intercept. Compute the x intercept. Put $p\left( x \right)=0$ and then, $\begin{align} & {{x}^{3}}+1=0 \\ & {{x}^{3}}=-1 \\ & =\sqrt[3]{-1} \end{align}$ There is no $x$ intercept Now, compute the vertical asymptotes. Set $q\left( x \right)=0$ So, $\begin{align} & {{x}^{2}}+2x=0 \\ & x\left( x+2 \right)=0 \\ & x=0,-2 \end{align}$ Thus, the graph has a vertical asymptote as $x=0$ and $x=-2$. Determine the horizontal asymptotes. There is no horizontal asymptote because the degree of the numerator term is not equal to the degree of the denominator term.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.