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

Feb 19, 2016

Three asymptotes are $x = \sqrt{6}$, $x = - \sqrt{6}$ and $y = 2$.

#### Explanation:

Vertical asymptotes can be found by factorizing function in the denominator i.e. here ${x}^{2} - 6$. As such here $x + \sqrt{6} = 0$ and $x - \sqrt{6} = 0$ or $x = - \sqrt{6} =$ and $x = \sqrt{6}$ are two such asymptotes.

Location of the horizontal asymptote is determined by looking at the degrees of the numerator and denominator.

If degree of numerator is less (than denominator), $y = 0$ is the asymptote. If degree of numerator is greater (than denominator), there is no horizontal asymptote (but if degree of numerator is just one degree higher there is a slanting asymptote).

In case the degrees are same, as in the instant case, let $a$ be the ratios of the coefficient of highest degrees in numerator and denominator, then asymptote is at $y = a$.

Here as the ratio is $\frac{2 {x}^{2}}{x} ^ 2$ or $2$, asymptote is at $y = 2$.