Question

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

Answer 1

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

Explanation:

`Q(x)=(2x^2)/(x^2-5x-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(x)=(2x^2)/((x-6)(x+1))`

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

So
Set denominator as 0, solve for x.
`(x-6)(x+1)=0`
`(x-6)=0 or (x+1)=0`
`x=6 or 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=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.