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

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Answer 1 · 2016-06-19T19:15:02

x= $+- sqrt 5 / 4$

y=±9/4

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= $+-9/4$

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