# How do you find the Vertical, Horizontal, and Oblique Asymptote given Q(x) =( 2x^2) / (x^2 - 5x - 6)?

Jul 29, 2016

Vertical Asymptotes : x=6 and x=-1
Horizontal asymptotes:y=2

#### Explanation:

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

The first step is factor the numerator and denominator
if possible .
We can factor the denominator
(x^2-5x-6=(x-6)(x+1)#

$q \left(x\right) = \frac{2 {x}^{2}}{\left(x - 6\right) \left(x + 1\right)}$

Vertical asymptote is a point to which x approaches make the
function approaches + or - infinity .

So
Set denominator as 0, solve for x.
$\left(x - 6\right) \left(x + 1\right) = 0$
$\left(x - 6\right) = 0 \mathmr{and} \left(x + 1\right) = 0$
$x = 6 \mathmr{and} x = - 1$
So vertical asymtotes are x=6 and x=-1
Horizontal asymptote is the value of the function
when x approches infinity.

To find horizontal asymptote
we have to use the degree of numerator and denominator.
Here both have the same degree
y= ( leading coefficient of numerator) /(leading coefficient of denominator) is the horizontal asymptote.
$y = \frac{2}{1}$
$y = 2$
If the degree of numerator > degree of the denominator , then we would get the slant asymptote.
Here no slant asymptote are there.