# How do you find the asymptotes for f(x) = (9x) / sqrt(16x^2 - 5)?

Jun 19, 2016

x= $\pm \frac{\sqrt{5}}{4}$

y=±9/4

#### Explanation:

Vertical asymptotes would be given by $16 {x}^{2} - 5 = 0$

Thesy=
±
9
4
e would be x= $\pm \frac{\sqrt{5}}{4}$

For horizontal asymptotes write f(x)= 9 1/sqrt (16-5/x^2. Now letting $\lim x \to \infty$, it would be y=$\pm \frac{9}{4}$