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Answer:Holes and Rational Functions
A hole on a graph looks like a hollow circle. It represents the fact that the function approaches the point, but is not actually defined on that precise \begin{align*}x\end{align*} value.
Take a look at the graph of the following equation:
\begin{align*}f(x)=(2x+2) \cdot \frac{\left(x+\frac{1}{2}\right)}{\left(x+\frac{1}{2}\right)}\end{align*}
The reason why this function is not defined at \begin{align*}-\frac{1}{2}\end{align*} is because \begin{align*}-\frac{1}{2}\end{align*} is not in the domain of the function. As you can see, \begin{align*}f\left(-\frac{1}{2}\right)\end{align*} is undefined because it makes the denominator of the rational part of the function zero which makes the whole function undefined. Also notice that once the factors are canceled/removed then you are left with a normal function which in this case is \begin{align*}2x+2\end{align*}. The hole in this situation is at \begin{align*}(-\frac{1}{2},1)\end{align*} because after removing the factors that cancel, \begin{align*}f(-\frac{1}{2})=1\end{align*}.
This is the essence of dealing with holes in rational functions. You should cancel what you can and graph the function like normal making sure to note what \begin{align*}x\end{align*} values make the function undefined. Once the function is graphed without holes go back and insert the hollow circles indicating what \begin{align*}x\end{align*} values are removed from the domain. This is why holes are called removable discontinuities.