Precalculus (6th Edition) Blitzer

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

Chapter 7 - Cumulative Review Exercises - Page 880: 25

Answer

The graph is shown below:
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Work Step by Step

In order to plot the graph of $ f\left( x \right)=\frac{{{x}^{2}}-x-6}{x+1}$, substitute y in place of $ f\left( x \right)$ and evaluate the value of the y variable for each value of the x variable. The function $ f\left( x \right)=\frac{{{x}^{2}}-x-6}{x+1}$ is undefined at $ x=-1$. Therefore, $ x=-1$ is the vertical asymptote of the function. To find the value of the y-intercept, substitute $ x=0$ as given below: $\begin{align} & y=\frac{{{x}^{2}}-x-6}{x+1} \\ & =\frac{{{\left( 0 \right)}^{2}}-0-6}{0+1} \\ & =-\frac{6}{1} \\ & =-6 \end{align}$ To find the value of the y-intercept, substitute $ x=3$ as given below: $\begin{align} & y=\frac{{{x}^{2}}-x-6}{x+1} \\ & =\frac{{{\left( 3 \right)}^{2}}-3-6}{3+1} \\ & =0 \end{align}$ To find the value of the y-intercept, substitute $ x=-2$ as given below: $\begin{align} & y=\frac{{{x}^{2}}-x-6}{x+1} \\ & =\frac{{{\left( -2 \right)}^{2}}-\left( -2 \right)-6}{-2+1} \\ & =\frac{4+2-6}{-1} \\ & =0 \end{align}$ To find the value of the y-intercept, substitute $ x=-3$ as given below: $\begin{align} & y=\frac{{{x}^{2}}-x-6}{x+1} \\ & =\frac{{{\left( -3 \right)}^{2}}-\left( -3 \right)-6}{-3+1} \\ & =\frac{9+3-6}{-2} \\ & =-\frac{6}{2} \\ & =-3 \end{align}$ Therefore, plot the intercepts $\left( 0,-6 \right),\left( 3,0 \right)\text{,}\left( -2,0 \right)\text{, and }\left( -3,-3 \right)$ and join them with a free hand and draw a vertical asymptote line $ x=-1$ in order to get the graph of the equation $ f\left( x \right)=\frac{{{x}^{2}}-x-6}{x+1}$ as given below: Note that there is a break in the graph of the function at $ x=-1$; it changes its position at this point.
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