## Precalculus (6th Edition) Blitzer

There will not be any vertical asymptotes, but there will be a hole at $x=-7$
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 $x=-7$.