# How do you find the vertical, horizontal or slant asymptotes for (x^2 - 5x + 6)/( x - 3)?

Nov 22, 2016

We have a hole when $x = 2$

#### Explanation:

Let's factorise the numerator.

${x}^{2} - 5 x + 6 = \left(x - 3\right) \left(x - 2\right)$

Therefore,

$\frac{{x}^{2} - 5 x + 6}{x - 3} = \frac{\cancel{x - 3} \left(x - 2\right)}{\cancel{x - 3}}$

$= \left(x - 2\right)$

We have a hole at $x = 2$

$y = \left(x - 2\right)$ is the equation of a straight line.

graph{x-2 [-10, 10, -5, 5]}