How do you find the Vertical, Horizontal, and Oblique Asymptote given #y = (8 x^2 + x - 2)/(x^2 + x - 72)#?

1 Answer
Nov 28, 2016

Answer:

The vertical asymptotes are #x=-8# and #x=9#
No oblique asymptote
The horizontal asymptote is #y=8#

Explanation:

Let's factorise the denominator

#x^2+x-72=(x+8)(x-9)#

So,

#y=(8x^2+x-2)/(x^2+x-72)=(8x^2+x-2)/((x+8)(x-9))#

The domain of #y# is #D_y=RR-{-8,+9}#

As we cannot divide by #0#, #x!=-8# and #x!=9#

The vertical asymptotes are #x=-8# and #x=9#

As the degree of the numerator is #=# to the degree of the denominator, there are no oblique asymptotes.

To calculate the limits of #x->+-oo#, we take the terms of highest degree in the numerator and the denominator.

#lim_(x->+-oo)y=lim_(x->+-oo)(8x^2)/x^2=8#

So, the horizontal asymptote is #y=8#

graph{(y-(8x^2+x-2)/(x^2+x-72))(y-8)=0 [-58.5, 58.55, -29.3, 29.23]}