# How do you find the Vertical, Horizontal, and Oblique Asymptote given ((x-3)(9x+4))/(x^2-4)?

Jan 26, 2017

The vertical asymptotes are $x = 2$ and $x = - 2$
The horizontal asymptote is $y = 9$
No slant asymptote

#### Explanation:

Let $f \left(x\right) = \frac{\left(x - 3\right) \left(9 x + 4\right)}{{x}^{2} - 4}$

Let's factorise the denominator

${x}^{2} - 4 = \left(x + 2\right) \left(x - 2\right)$

As you cannot divide by $0$, $x \ne 2$ and $x \ne - 2$

The vertical asymptotes are $x = 2$ and $x = - 2$

As the degree of the numerator is $=$ the degree of the denominator, there is no slant asymptote

${\lim}_{x \to \pm \infty} f \left(x\right) = {\lim}_{x \to \pm \infty} \frac{9 {x}^{2}}{x} ^ 2 = 9$

The horizontal asymptote is $y = 9$

graph{(y-((x-3)(9x+4)/(x^2-4)))(y-9)=0 [-41.1, 41.1, -20.55, 20.57]}