# How do you find the vertical, horizontal and slant asymptotes of: y = (x)/(x^2+4)?

Nov 30, 2016

horizontal asymptote at y = 0

#### Explanation:

The denominator of y cannot be zero as this would make y 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

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

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

$y = \frac{\frac{x}{x} ^ 2}{{x}^{2} / {x}^{2} + \frac{4}{x} ^ 2} = \frac{\frac{1}{x}}{1 + \frac{4}{x} ^ 2}$

as $x \to \pm \infty , y \to \frac{0}{1 + 0}$

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

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