# How to find the asymptotes of f(x)= (x^2+4)/(6x-5x^2)?

Jan 9, 2016

Vertical asymptotes at $x = 0 , \frac{6}{5}$, horizontal asymptote at $y = - \frac{1}{5}$

#### Explanation:

Vertical Asymptotes

These will occur when the denominator equals $0$.

$6 x - 5 {x}^{2} = 0$

$x \left(6 - 5 x\right) = 0$

Split this into two equations.

$x = 0$

and

$6 - 5 x = 0$

$x = \frac{6}{5}$

The vertical asymptotes occur at $x = 0 , \frac{6}{5}$.

Horizontal Asymptotes

When the numerator and denominator have the same degree, the horizontal asymptote will be the terms with the largest degree divided.

${x}^{2} / \left(- 5 {x}^{2}\right) = - \frac{1}{5}$

There is a horizontal asymptote at $y = - \frac{1}{5}$.

The function graphed:

graph{(x^2+4)/(6x-5x^2) [-10.35, 12.15, -4.37, 6.88]}