# How do you find the vertical, horizontal or slant asymptotes for f(x) = ((x-3)(9x+4))/(x^2-4)?

Nov 8, 2016

The vertical asymptotes are $x = 2$ and $x = - 2$
The horizontal asymptote is $y = 9$

#### Explanation:

The denominator, ${x}^{2} - 4 = \left(x + 2\right) \left(x - 2\right)$
As we cannot divide by zero, the vertical asymptotes are $x = 2$ and $x = - 2$
As the degree of the numerator is the same asthe degree of the denominator, there is no slant asymptote:
${\lim}_{x \to \pm \infty} f \left(x\right) = {\lim}_{x \to \pm \infty} \frac{9 {x}^{2}}{x} ^ 2 = 9$

$\therefore y = 9$ is a horizontal asymptote
graph{(y-((x-3)(9x+4))/((x+2)(x-2)))(y-9)=0 [-41.13, 41.15, -20.55, 20.57]}