# How do you find the Vertical, Horizontal, and Oblique Asymptote given y = (2x^2 - 11)/( x^2 + 9)?

Nov 10, 2016

Only a horizontal asymptote $y = 2$.

#### Explanation:

As $y = \frac{2 {x}^{2} - 11}{{x}^{2} + 9} = \frac{2 - \frac{11}{x} ^ 2}{1 + \frac{9}{x} ^ 2}$

$L {t}_{x \to \pm \infty} \frac{2 - \frac{11}{x} ^ 2}{1 + \frac{9}{x} ^ 2} = 2$

Also observe that the denominator ${x}^{2} + 9 > 0$, as ${x}^{2}$ is always positive.

Hence we have only a horizontal asymptote $y = 2$.
graph{(y-(2x^2-11)/(x^2+9))(y-2)=0 [-10, 10, -5, 5]}