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

Jan 2, 2016

VA: $x = \pm \sqrt{6}$
HA: $x = 2$

#### Explanation:

Vertical Asymptotes

These will occur when the denominator equals $0$.

${x}^{2} - 6 = 0$
${x}^{2} = 6$
$x = \pm \sqrt{6}$

Thus, there are vertical asymptotes at $x = - \sqrt{6}$ and $x = \sqrt{6}$.

Horizontal Asymptotes

When the degree of the numerator and denominator are equal, like they are here, divide the terms with the greatest degree. In this case, those are $2 {x}^{2}$ and ${x}^{2}$.

$\frac{2 {x}^{2}}{x} ^ 2 = 2$

There is a horizontal asymptote at $x = 2$.

graph{(2x^2+3)/(x^2-6) [-19.62, 20.93, -8.83, 11.45]}