# How do you find the Vertical, Horizontal, and Oblique Asymptote given f(x)= (2x)/(x^2+16)?

Aug 26, 2016

horizontal asymptote at y = 0

#### 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: ${x}^{2} + 16 = 0 \Rightarrow {x}^{2} = - 16$

This has no real solutions hence 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 ${x}^{2}$

$f \left(x\right) = \frac{\frac{2 x}{x} ^ 2}{{x}^{2} / {x}^{2} + \frac{16}{x} ^ 2} = \frac{\frac{2}{x}}{1 + \frac{16}{x} ^ 2}$

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

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

Oblique 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 oblique asymptotes.
graph{(2x)/(x^2+16) [-10, 10, -5, 5]}