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

1 Answer
Jul 29, 2016

Answer:

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.