Answer
a)
The slant asymptote of the given function is $y=x+4$.
b) Graph the function as:
Work Step by Step
(a)
Consider the rational function $f\left( x \right)=\frac{{{x}^{2}}+x-6}{x-3}$.
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 ${{x}^{2}}+x-6$ , since the graph of the function has a slant asymptote.
Now, divide the function ${{x}^{2}}+x-6$ with $x-3$.
So,
$\frac{{{x}^{2}}+x-6}{x-3}=x+\frac{6}{x-3}+4$
Therefore, the equation has a slant asymptote at $y=x+4$.
(b)
Consider the rational function $f\left( x \right)=\frac{{{x}^{2}}+x-6}{x-3}$. 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 ${{x}^{2}}+x-6$ , since, the graph of the function has a slant asymptote.
Step 1:
Substitute $x=-x$
$\begin{align}
& f\left( -x \right)=\frac{{{\left( -x \right)}^{2}}+\left( -x \right)-6}{\left( -x \right)-3} \\
& =\frac{{{x}^{2}}-x-6}{-x-3}
\end{align}$
The graph is not symmetrical about the $y$ axis.
Step 2: To calculate the x intercepts equate $f\left( x \right)=0$.
$\begin{align}
& \frac{{{x}^{2}}+x-6}{x-3}=0 \\
& x=2
\end{align}$
There are no real x-intercepts.
Step 3: To calculate the y intercepts evaluate $f\left( 0 \right)$.
$f\left( 0 \right)=2$
Step 4: Since the degree of the numerator is greater than the denominator, there is no horizontal asymptote.
Step 5: For the vertical asymptote, equate the denominator to 0.
$x=3$
