# How do you find the asymptote and holes for y=(x-6)/(x^2+5x+6)?

May 28, 2016

vertical asymptotes x = -3 , x = -2
horizontal asymptote y = 0

#### Explanation:

First step is to factorise the function.

$\Rightarrow \frac{x - 6}{\left(x + 2\right) \left(x + 3\right)}$

Since there are no common factors on numerator/denominator this function has no holes.

Vertical asymptotes occur as the denominator of a rational tends to zero. To find the equation/s set the denominator equal to zero.

solve: (x+2)(x+3) = 0 → x = -3 , x= -2

$\Rightarrow x = - 3 \text{ and "x=-2" are the asymptotes}$

Horizontal asymptotes occur as ${\lim}_{x \to \pm \infty} , y \to 0$

When the degree of the numerator < degree of the denominator as is the case here (numerator-degree 1 , denominator-degree 2 ) then the equation of the asymptote is always y = 0
graph{(x-6)/(x^2+5x+6) [-10, 10, -5, 5]}