Answer
The graph is shown below:
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.
