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

Sep 9, 2016

horizontal asymptote at y = 1

#### Explanation:

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

solve: ${x}^{2} + 4 = 0 \Rightarrow {x}^{2} = - 4$

This has no real solutions hence there are no vertical asymptotes.

Horizontal asymptotes occur as

${\lim}_{x \to \pm \infty} , f \left(x\right) \to c \text{ (a constant)}$

divide terms on numerator/denominator by the highest power of x that is ${x}^{2}$

$f \left(x\right) = \frac{{x}^{2} / {x}^{2} - \frac{1}{x} ^ 2}{{x}^{2} / {x}^{2} + \frac{4}{x} ^ 2} = \frac{1 - \frac{1}{x} ^ 2}{1 + \frac{4}{x} ^ 2}$

as $x \to \pm \infty , f \left(x\right) \to \frac{1 - 0}{1 + 0}$

$\Rightarrow y = 1 \text{ is the asymptote}$
graph{(x^2-1)/(x^2+4) [-10, 10, -5, 5]}