# How do you find the vertical, horizontal or slant asymptotes for g(x)=(2x^2) / (x^2 - 5x - 6)?

Sep 20, 2016

vertical asymptotes at x = - 1 , x = 6
horizontal asymptote at y = 2

#### Explanation:

The denominator of g(x) cannot be zero as this would make g(x) 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: ${x}^{2} - 5 x - 6 = 0 \Rightarrow \left(x - 6\right) \left(x + 1\right) = 0$

$\Rightarrow x = - 1 \text{ and " x=6" are the asymptotes}$

Horizontal asymptotes occur as

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

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

$g \left(x\right) = \frac{\frac{2 {x}^{2}}{{x}^{2}}}{{x}^{2} / {x}^{2} - \frac{5 x}{x} ^ 2 - \frac{6}{x} ^ 2} = \frac{2}{1 - \frac{5}{x} - \frac{6}{x} ^ 2}$

as $x \to \pm \infty , g \left(x\right) \to \frac{2}{1 - 0 - 0}$

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

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here ( both of degree 2 ) Hence there are no slant asymptotes.
graph{(2x^2)/(x^2-5x-6) [-20, 20, -10, 10]}