# How do you find the asympotes for f (x)= (5x-1)/(x^2 + 9)?

Jul 10, 2015

This one has only a horizontal asymptote.

#### Explanation:

The vertical asymptote would happen if the numerator would go to $= 0$. Since ${x}^{2}$ is always non-negative, the numerator will always be at least $9$

The horizontal asymptote can be found by making $x$ larger and larger. The $+ 1$ and $- 9$ then make less and less of a difference and the whole thing will tend to look like:
$\frac{5 x}{x} ^ 2 \approx \frac{5}{x}$
As $x$ gets larger $\frac{5}{x}$ gets smaller, or:

${\lim}_{x \to \pm \infty} f \left(x\right) = 0$

In fact this is not a real asymptote, as the value of $f \left(x\right) = 0$ is also reached when $5 x - 1 = 0 \to x = \frac{1}{5}$ (see graph)
graph{(5x-1)/(x^2+9) [-10, 10, -5, 5]}