# How do I find the asymptotes of f(x)= (- 4x)/(2 + x^2)?

Oct 5, 2016

There is not one.

#### Explanation:

In order for there to be an asymptote, the denominator has to equal zero and have real roots. However, the denominator of this function has no real roots.

This is what the graph looks like:
graph{-4x/(2+x^2) [-10, 10, -5, 5]}

As you can, see there are no asymptotes on the $x$-axis. However, the graph seems to be bounded between $y = 1.5$, and $y = - 1.5$, but these are not asymptotes.

Oct 5, 2016

horizontal asymptote at y = 0

#### Explanation:

The denominator of f(x) cannot equal 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: $2 + {x}^{2} = 0 \Rightarrow {x}^{2} = - 2$

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 ${x}^{2}$

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

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

$\Rightarrow y = 0 \text{ is the asymptote}$
graph{(-4x)/(2+x^2) [-10, 10, -5, 5]}