Answer
Define $f(2) = 3$.
Work Step by Step
$f(x)=\dfrac{x^2-x-2}{x-2}$
We can not directly evaluate the function when $x=2$ because the denominator would be zero and division by zero is undefined.
Let's make a function that is identical to f(x) everywhere except at $x=2$
We can factor the numerator and then cancel.
$f(x)=\dfrac{x^2-x-2}{x-2} = \dfrac{(x-2)(x+1)}{(x-2)}$
$f(x)=x+1$ for $x\ne2$
$f(2)=2+1 = 3$
If we graph $f(2)=3$ the hole at $x=2$ is filled.
*Previous answer*
The graph of $f(x)$ has a removable discontinuity at $x=2$, so give the function a value at that point. Choosing $f(2) = 3$ fills in the hole in the graph. (Graphing the function will make this choice more apparent. )
