# How do you find the Vertical, Horizontal, and Oblique Asymptote given y = (8 x^2 + x - 2)/(x^2 + x - 72)?

Nov 28, 2016

The vertical asymptotes are $x = - 8$ and $x = 9$
No oblique asymptote
The horizontal asymptote is $y = 8$

#### Explanation:

Let's factorise the denominator

${x}^{2} + x - 72 = \left(x + 8\right) \left(x - 9\right)$

So,

$y = \frac{8 {x}^{2} + x - 2}{{x}^{2} + x - 72} = \frac{8 {x}^{2} + x - 2}{\left(x + 8\right) \left(x - 9\right)}$

The domain of $y$ is ${D}_{y} = \mathbb{R} - \left\{- 8 , + 9\right\}$

As we cannot divide by $0$, $x \ne - 8$ and $x \ne 9$

The vertical asymptotes are $x = - 8$ and $x = 9$

As the degree of the numerator is $=$ to the degree of the denominator, there are no oblique asymptotes.

To calculate the limits of $x \to \pm \infty$, we take the terms of highest degree in the numerator and the denominator.

${\lim}_{x \to \pm \infty} y = {\lim}_{x \to \pm \infty} \frac{8 {x}^{2}}{x} ^ 2 = 8$

So, the horizontal asymptote is $y = 8$

graph{(y-(8x^2+x-2)/(x^2+x-72))(y-8)=0 [-58.5, 58.55, -29.3, 29.23]}