# How do you find the vertical, horizontal or slant asymptotes for (x^2+4)/(6x-5x^2)?

Nov 30, 2017

Horizontal asymptote of $- \frac{1}{5}$
No slant asymptotes
Vertical asymptotes are $x = 1.2$ and $x = 0$

#### Explanation:

HA of $- \frac{1}{5}$ because the degrees are the same in the denominator and numerator, so you divide the coefficients.

No slant asymptotes because the numerator degree must be one higher than the denominator, and it is not.

Vertical asymptotes are $x = 1$ and $x = 5$ because $6 x - 5 {x}^{2}$ factored is $\left(x - 1.2\right)$ and $\left(x\right)$