# How do you find the vertical, horizontal or slant asymptotes for 1/(x^2+4)?

Apr 19, 2017

horizontal asymptote : y=0

vertical asymptote : none

slant asymptote : none

#### Explanation:

because the numerator is of lower degree than the denominator
then there are no slant asymptotes and the horizontal asymptote is :

y=0

to find the vertical asymptote you put the denominator= 0

${x}^{2} + 4 = 0$

${x}^{2} = - 4$

$x = \sqrt{- 4}$

you can see that it has no real solutions so there are no vertical asymptotes of this rational function

Apr 19, 2017

$\text{horizontal asymptote at } y = 0$

#### Explanation:

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

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 there values then they are vertical asymptotes.

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

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 ${x}^{2}$

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

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

$\Rightarrow y = 0 \text{ is the asymptote}$

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here, hence there are no slant asymptotes.
graph{1/(x^2+4) [-10, 10, -5, 5]}