Answer
There will not be any vertical asymptotes, but there will be a hole at $x=-7$
Work Step by Step
Write $r\left( x \right)$ in the form of $\frac{p\left( x \right)}{q\left( x \right)}$.
$\begin{align}
& r\left( x \right)=\frac{{{x}^{2}}+4x-21}{x+7} \\
& =\frac{p\left( x \right)}{q\left( x \right)}
\end{align}$
So, $p\left( x \right)={{x}^{2}}+4x-21$ and $q\left( x \right)=x+7$.
Here, the degree of the polynomial in the denominator is less than the numerator.
Therefore, there will not be any vertical asymptotes.
Now,
$\begin{align}
& r\left( x \right)=\frac{{{x}^{2}}+4x-21}{x+7} \\
& =\frac{\left( x+7 \right)\cdot \left( x-3 \right)}{\left( x+7 \right)} \\
& =x-3
\end{align}$
But, after rearranging, $x+7$ from the denominator will be eliminated.
Here, when x approaches $-7$ , the polynomial approaches to $4$ , but at the point $x=-10$ the polynomial will be discontinuous. So, there will be a hole at $x=-7$
Thus, there are no vertical asymptotes, but there will be a hole at $x=-7$.
