How do you find the asymptotes for $f (x) = \frac{1}{x^{2} + 4}$ $f(x)=1/(x^2+4)$ ?

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Answer 1 · 2016-03-23T23:27:54

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

Notice that $f (x) \neq 0$ $f(x) != 0$ 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]}

How do you find the asymptotes for f(x)=1/(x^2+4)f(x)=1x2+4f(x)=1/(x^2+4)?