Asymptotes of function

Question

How to find a if the function below doesn't have a vertical asymptotes.

$f(x)= \frac{x^2-ax+2}{x-2}$

Answer 1

A function like that has vertical asymptote when it's denominator is $0$, that happens when $x = 2$

$f(x)= \frac{x^2-ax+2}{x-2}$

But since we don't want $f(x)$ to have a vertical asymptotes, then $x-2$ must be a factor of $x^2+ax+2$

Say $x^2-ax+2 = (x-2)(x-b)$

Put $x=2$, $2^2-2a+2 = 0$, $6-2a=0$, then $a= 3$

$f(x) = \frac{x^2-3x+2}{x-2}$

$f(x) = \frac{(x-2)\cdot(x-1)}{x-2}$

$f(x) = (x-1)$

Answer 2

Hint: This can only happen if $x-2$ is a factor of the numerator, so it can "cancel" the $x-2$ in the denominator. What does $a$ have to be for this to be true?