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

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

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.

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