Answer
a)
The slant asymptote of the given function is $y=x-9$.
b) The graph of the function is shown below as:
Work Step by Step
(a)
Consider the rational function $f\left( x \right)=\frac{{{x}^{3}}-1}{{{x}^{2}}-9}$.
Here, the degree of the numerator term is 2 and degree of the denominator term is 1. And $x=0$ does not constitute 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 ${{x}^{2}}-9$.
$\frac{{{x}^{3}}-1}{{{x}^{2}}-9}=x+\frac{13}{3\left( x-3 \right)}+\frac{14}{3\left( x+3 \right)}$
Therefore, the equation has a slant asymptote at $y=x-9$.
(b)
Consider the rational function $f\left( x \right)=\frac{{{x}^{3}}-1}{{{x}^{2}}-9}$.
Here, the degree of the numerator term is 2 and degree of the denominator term is 1. And $x=0$ does not constitute factor of the function ${{x}^{3}}-1$ , so, the graph of the function has a slant asymptote.
Determine the symmetry.
$f\left( x \right)=\frac{{{x}^{3}}-1}{{{x}^{2}}-9}$
Substitute $x=-x$ and then,
$\begin{align}
& f\left( -x \right)=\frac{{{\left( -x \right)}^{3}}-1}{{{\left( -x \right)}^{2}}-9} \\
& =\frac{-{{x}^{3}}-1}{{{x}^{2}}-9}
\end{align}$
The graph is not symmetrical about the $y$ axis.
Compute the $y$ intercept.
Put $x=0$
$\begin{align}
& f\left( x \right)=\frac{{{x}^{3}}-1}{{{x}^{2}}-9} \\
& f\left( 0 \right)=\frac{{{\left( 0 \right)}^{3}}-1}{{{\left( 0 \right)}^{2}}-9} \\
& =\frac{1}{9}
\end{align}$
The $y$ intercept is $y=\frac{1}{9}$.
Compute the $x$ intercept.
Put $p\left( x \right)=0$ and then,
$\begin{align}
& {{x}^{3}}-1=0 \\
& {{x}^{3}}=1 \\
& =\sqrt[3]{1} \\
& =1
\end{align}$
The $x$ intercept is $x=1$.
Now, compute the vertical asymptotes.
Set $q\left( x \right)=0$
$\begin{align}
& {{x}^{2}}-9=0 \\
& x=3,-3
\end{align}$
Thus, the graph has a vertical asymptote as $x=3$ and $x=-3$.
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.
