Answer
$f(x)=x+1$ with a hole at $(2,3)$
Work Step by Step
We are given the function:
$$f(x)=\dfrac{x^2-x-2}{x-2}.$$
The domain of the function is the set of all real numbers except the zero of the denominator which is $2$:
$$\text{Domain}=(-\infty,2)\cup(2,\infty).$$
We rewrite the function:
$$f(x)=\dfrac{(x-2)(x+1)}{x-2}=x+1.$$
We could simplify by $x-2$ because $2$ does not belong to the function's domain, so $x-2\not=0$.
Therefore the function is represented by a line with a hole at $(2,2+1)=(2,3)$.
Graph the function:
