Answer
There will not be any vertical asymptotes, but there will be a hole at $x=-6$
Work Step by Step
Write $r\left( x \right)$ in the form of $\frac{p\left( x \right)}{q\left( x \right)}$.
$\begin{align}
& r\left( x \right)=\frac{{{x}^{2}}+2x-24}{x+6} \\
& =\frac{p\left( x \right)}{q\left( x \right)}
\end{align}$
So, $p\left( x \right)={{x}^{2}}+2x-24$ and $q\left( x \right)=x+6$.
Here, the degree of the polynomial in the denominator is less than the numerator.
Thus, no vertical asymptotes.
Now,
$\begin{align}
& r\left( x \right)=\frac{{{x}^{2}}+2x-24}{x+6} \\
& =\frac{\left( x+6 \right)\cdot \left( x-4 \right)}{\left( x+6 \right)} \\
& =x-4
\end{align}$
But, after rearranging, $x+6$ from the denominator will be eliminated.
Here, when x approaches $-6$ , the polynomial approaches to $-10$ , but at the point $x=-6$ the polynomial will be discontinuous. So, there will be a hole at $x=-6$
Therefore, there are no vertical asymptotes, but there will be a hole at $x=-6$.
