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

Jun 11, 2016

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

#### Explanation:

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

solve : ${x}^{2} + 5 x + 6 = 0 \Rightarrow \left(x + 2\right) \left(x + 3\right) = 0$

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

Horizontal asymptotes occur as

${\lim}_{x \to \pm \infty} , f \left(x\right) \to c \text{ (a constant)}$

divide terms on numerator/denominator by ${x}^{2}$

$\frac{\frac{x}{x} ^ 2}{{x}^{2} / {x}^{2} + \frac{5 x}{x} ^ 2 + \frac{6}{x} ^ 2} = \frac{\frac{1}{x}}{1 + \frac{5}{x} + \frac{6}{x} ^ 2}$

as $x \to \pm \infty , f \left(x\right) \to \frac{0}{1 + 0 + 0}$

$\Rightarrow y = 0 \text{ is the asymptote}$

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (numerator-degree 1 , denominator-degree 2 ) Hence there are no slant asymptotes.
graph{x/(x^2+5x+6) [-10, 10, -5, 5]}