How do you find the vertical, horizontal or slant asymptotes for #((x^2)+3)/((9x^2)-80x-9)#?

1 Answer
Dec 31, 2016

Answer:

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

Explanation:

We can factorise the denominator

#9x^2-80x-9=(9x+1)(x-9)#

Let, #f(x)=(x^2+3)/(9x^2-80x-9)=(x^2+3)/((9x+1)(x-9))#

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

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

The vertical asymptotes are #x=-1/9# and #x=9#

As the degree of the numerator is #=# to the degree of the denominator, there is no slant asymptote

Now, we calculate the limits of #f(x)# as #x->+-oo#

#lim_(x->+-oo)f(x)=lim_(x->+-oo)x^2/(9x^2)=1/9#

The horizontal asymptote is #y=1/9#

graph{(y-(x^2+3)/(9x^2-80x-9))(y-1/9)=0 [-52.1, 51.93, -26, 25.94]}