Answer
The function $ f\left( x \right)=\frac{\left( x+2 \right)}{\left( x+2 \right)\left( x-5 \right)}$ is discontinuous for the points $-2\text{ and 5}$.
Work Step by Step
Consider the rational function $ f\left( x \right)=\frac{\left( x+2 \right)}{\left( x+2 \right)\left( x-5 \right)}$,
Here, $ p\left( x \right)=x+2\text{ and }q\left( x \right)=\left( x+2 \right)\left( x-5 \right)$
Find the zeros of the function $ q\left( x \right)=\left( x+2 \right)\left( x-5 \right)$ by $ q\left( x \right)=0$,
$\left( x+2 \right)\left( x-5 \right)=0$
Solve for the x,
$\begin{align}
& \left( x+2 \right)=0 \\
& x=-2
\end{align}$
Or
$\begin{align}
& \left( x-5 \right)=0 \\
& x=5
\end{align}$
The zeros of the function $ q\left( x \right)=\left( x+2 \right)\left( x-5 \right)$ are $-2\text{ and 5}$.
Thus, the function $ f\left( x \right)=\frac{\left( x+2 \right)}{\left( x+2 \right)\left( x-5 \right)}$ is discontinuous for the points $-2\text{ and 5}$.
