How do you find the asymptotes for #f(x)=1/(x^2+4)#?

1 Answer
Mar 23, 2016

Answer:

This function only has a horizontal asymptote: #y = 0#

Explanation:

Notice that #f(x) != 0# for all #x in RR#, since if #f(x) = 0# then #1 = 0(x^2+4) = 0#, which is false.

As #x->+-oo# we find #1/(x^2+4) -> 0#

So we have a horizontal asymptote #y=0#

On the other hand #x^2+4 >= 4 > 0# for all #x in RR#, so the denominator is non-zero for all Real values of #x#.

So there is no vertical asymptote.

graph{1/(x^2+4) [-2.5, 2.5, -1.25, 1.25]}