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

Question

Ariana Alvarez · Accepted Answer

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

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]}

James Allen · Answer

undefined To find the asymptotes of the function $ f(x) = \frac{1}{x^2 + 4} $:

1. Vertical asymptotes occur where the denominator of the rational function equals zero. Set $ x^2 + 4 = 0 $ and solve for $ x $. Since $ x^2 + 4 = 0 $ has no real solutions, there are no vertical asymptotes.

2. Horizontal asymptotes occur as $ x $ approaches positive or negative infinity. To find them, compare the degrees of the numerator and denominator of the rational function. Since the degree of the numerator (which is 0) is less than the degree of the denominator (which is 2), there is a horizontal asymptote at $ y = 0 $.

So, the only asymptote for $ f(x) = \frac{1}{x^2 + 4} $ is the horizontal asymptote $ y = 0 $.