# How do you find the vertical, horizontal and slant asymptotes of: y = (2x^2 - 11)/( x^2 + 9)?

Jul 7, 2017

$\text{horizontal asymptote at } y = 2$

#### Explanation:

The denominator of y cannot be zero as this would make y undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

$\text{solve } {x}^{2} + 9 = 0 \Rightarrow {x}^{2} = - 9$

$\text{there are no real solutions hence no vertical asymptotes}$

$\text{horizontal asymptotes occur as}$

${\lim}_{x \to \pm \infty} , y \to c \text{ ( a constant )}$

divide terms on numerator/denominator by the highest power of x, that is ${x}^{2}$

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

as $x \to \pm \infty , y \to \frac{2 - 0}{1 + 0}$

$\Rightarrow y = 2 \text{ is the asymptote}$

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here ( both degree 2 ) hence there are no slant asymptotes.
graph{(2x^2-11)/(x^2+9) [-10, 10, -5, 5]}