# How do you find a horizontal asymptote (x^2 - 5x + 6)/(x - 3?

$\frac{\left(x - 2\right) \left(x - 3\right)}{\left(x - 3\right)}$ and cancel the $\left(x - 3\right)$'s
And this will turn the function into $x - 2$ which has no asymptotes at all.
$x = 3$ is not allowed , as this would make the numerator $= 0$
(Actually it would turn the fraction into $\frac{0}{0}$, which is not defined at all).