# How do you find the asymptotes for y=(2x^2+3)/(x^2-6)?

Aug 7, 2016

vertical asymptotes at x=±sqrt6
horizontal asymptote at y = 2

#### Explanation:

The denominator of y cannot be zero as this is 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.

solve: x^2-6=0rArrx^2=6rArrx=±sqrt6

$\Rightarrow x = - \sqrt{6} \text{ and " x=sqrt6" are the asymptotes}$

Horizontal asymptotes occur as

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

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

$\frac{\frac{2 {x}^{2}}{x} ^ 2 + \frac{3}{x} ^ 2}{{x}^{2} / {x}^{2} - \frac{6}{x} ^ 2} = \frac{2 + \frac{3}{x} ^ 2}{1 - \frac{6}{x} ^ 2}$

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

$\Rightarrow y = 2 \text{ is the asymptote}$
graph{(2x^2+3)/(x^2-6) [-10, 10, -5, 5]}