# How do you find the asymptotes for y = (2x^2 - 11)/( x^2 + 9)?

Feb 18, 2016

This function has only one horizontal asymptote $y = 2$

#### Explanation:

To find vertical asymptotes of such rational function you have to check if its denominator has any zeroes. Since ${x}^{2} + 9$ has no real roots (it only has 2 complex conjugate roots) the vertical asymptotes do not exist.

To find horizontal asymptotes you have to calculate

${\lim}_{x \to - \infty} f \left(x\right)$ and

${\lim}_{x \to + \infty} f \left(x\right)$

${\lim}_{x \to - \infty} f \left(x\right) =$

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

The same calculations can be repeated for ${\lim}_{x \to + \infty} f \left(x\right)$. So we can write that $y = 2$ is a horizontal asymptote for the function $f \left(x\right)$