How do you find the vertical, horizontal or slant asymptotes for #(4x^2+11) /( x^2+8x−9)#?

1 Answer
Dec 27, 2016

Answer:

The vertical asymptotes are #x=-9# and #x=1#
No slant asymptote.
The horizontal asymptote is #y=4#

Explanation:

Let's factorise the denominator

#x^2+8x-9=(x-1)(x+9)#

Let, #f(x)=(4x^2+11)/(x^2+8x-9)#

The domain of #f(x)# is #D_f(x)=RR-{-9,1} #

As we cannot divide by #0#, #x!=-9# and #x!=1#

So, the vertical asymptotes are #x=-9# and #x=1#

The degree of the numerator is #=# to the degree of the denominator, we don't have a slant asymptote.

To find the horizontal asymptote, we compute the limits of #f(x)# as #x->+-oo#

#lim_(x->+-oo)f(x)=lim_(x-+-oo)4x^2/x^2=4#

The horizontal asymptote is #y=4#

graph{(y-(4x^2+11)/(x^2+8x-9))(y-4)(x-1)(x+9)=0 [-52, 52, -26, 26]}