Question

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

Answer 1

x= `+- sqrt 5 / 4`

y=±9/4

Explanation:

Vertical asymptotes would be given by `16x^2-5 =0`

Thesy=
±
9
4
e would be x= `+- sqrt 5 / 4`

For horizontal asymptotes write f(x)= 9 `1/sqrt (16-5/x^2`. Now letting `lim x-> oo`, it would be y=`+-9/4`