# How do you find vertical, horizontal and oblique asymptotes for  (x^2 + 5x + 6)/(x+3)?

Aug 4, 2018

$\text{no asymptotes}$

#### Explanation:

$\text{factor the numerator}$

$= \frac{\left(x + 2\right) \cancel{\left(x + 3\right)}}{\cancel{\left(x + 3\right)}} = x + 2$

$\text{the removal of the factor "(x+3)" indicates a removable}$
$\text{discontinuity (hole) at } x = - 3$

$\text{The simplified version is a linear equation with }$
$\text{no asymptotes}$
graph{(x^2+5x+6)/(x+3) [-10, 10, -5, 5]}