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: 83

Answer

a) The slant asymptote of the given rational function is $y=x$. b) Graph the function as:
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Work Step by Step

(a) Consider the rational function $f\left( x \right)=\frac{{{x}^{2}}+1}{x}$. 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 ${{x}^{2}}+1$ , since the graph of the function has a slant asymptote. Now, divide the function ${{x}^{2}}+1$ with $x$. So, $\frac{{{x}^{2}}+1}{x}=x+\frac{1}{x}$ Therefore, the equation has a slant asymptote at $y=x$. (b) Consider the rational function $f\left( x \right)=\frac{{{x}^{2}}+1}{x}$. 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 ${{x}^{2}}-4$ , since, the graph of the function has a slant asymptote. Step 1: Substitute $x=-x$ $\begin{align} & f\left( -x \right)=\frac{{{\left( -x \right)}^{2}}+1}{\left( -x \right)} \\ & =-\frac{{{x}^{2}}+1}{x} \\ & =-f\left( x \right) \end{align}$ The graph is symmetrical about the $y$ axis. Step 2: To calculate the x intercept equate $f\left( x \right)=0$. $\begin{align} & \frac{{{x}^{2}}+1}{x}=0 \\ & x=\pm i \end{align}$ There are no real x-intercepts. Step 3: To calculate the y intercept evaluate $f\left( 0 \right)$. $f\left( 0 \right)=\infty $ Step 4: Since the degree of the numerator is greater than the denominator, there is no horizontal asymptote. Step 5: For the vertical asymptote, equate the denominator to 0. $x=0$
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