# How do you find the vertical asymptotes and holes of f(x)=1/(x^2+5x+6)?

Jun 17, 2018

$\text{vertical asymptotes at "x=-3" and } x = - 2$
$\text{horizontal asymptote at } y = 0$

#### Explanation:

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

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

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

$\text{horizontal asymptotes occur as}$

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

$\text{divide terms on numerator/denominator by the }$
$\text{highest power of x that is } {x}^{2}$

$f \left(x\right) = \frac{\frac{1}{x} ^ 2}{{x}^{2} + {x}^{2} + \frac{5 x}{x} ^ 2 + \frac{6}{x} ^ 2} = \frac{\frac{1}{x} ^ 2}{1 + \frac{5}{x} + \frac{6}{x} ^ 2}$

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

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

$\text{holes occur when a common factor is removed from}$
$\text{the numerator/denominator. This is not the case here}$
$\text{hence there are no holes}$
graph{1/(x^2+5x+6) [-10, 10, -5, 5]}