Answer
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.
