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 402: 117


The graph of the function will be obtained as below:

Work Step by Step

Rational functions are the proportions of the polynomial expressions. The function can be expressed as: $f\left( x \right)=\frac{p\left( x \right)}{q\left( x \right)}$ Where $p\left( x \right)$ and $q\left( x \right)$ are the polynomial functions in terms of $x$ , and $q\left( x \right)\ne 0$. Suppose a rational function is $f\left( x \right)=\frac{2x}{{{x}^{2}}-4}$. Then, its graph will be plotted using the following steps: First, find the symmetry of the function by substituting –x in the place of x. $\begin{align} & f\left( x \right)=\frac{2x}{{{x}^{2}}-4} \\ & f\left( -x \right)=\frac{2\left( -x \right)}{{{\left( -x \right)}^{2}}-4} \\ & =\frac{-2x}{{{x}^{2}}-4} \\ & =-f\left( x \right) \end{align}$ Since, the function $f\left( -x \right)$ is equal to $-f\left( x \right)$ , the graph of the function will be symmetrical about the $y$ -axis. Second, compute the $y$ -intercept by determining $f\left( 0 \right)$ as, $\begin{align} & f\left( x \right)=\frac{2x}{{{x}^{2}}-4} \\ & f\left( 0 \right)=\frac{2\left( 0 \right)}{{{\left( 0 \right)}^{2}}-4} \\ & =0 \end{align}$ Therefore, the $y$ -intercept is 0 and that implies that the graph passes through the origin. Third, compute the $x$ -intercept by solving $p\left( x \right)=0$ where the function $p\left( x \right)$ is the numerator of the provided function. So, $\begin{align} & 2x=0 \\ & x=0 \end{align}$ Therefore, the $x$ -intercept is 0 which implies that the graph passes through the origin. Fourth, find the vertical asymptote. For that, equate the denominator of the function to zero to find the zeros of the rational function as, $\begin{align} & {{x}^{2}}-4=0 \\ & {{x}^{2}}=4 \\ & x=2,-2 \end{align}$ Thus, the equation of the vertical asymptote is $x=2$ and $x=-2$. Fifth, determine the horizontal asymptote. The degree of the numerator is 1 and denominator term degree is 2. So, the horizontal asymptote is y = 0. Then, the last step is to compute the point between and beyond the $x$ -intercept $x=0$ and vertical asymptote $x=1$.
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