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

Question

Cora Alexander · Accepted Answer

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#

Theodore Amidon · Answer

undefined To find the asymptotes for the function $ f(x) = \frac{9x}{\sqrt{16x^2 - 5}} $, you need to consider both vertical and horizontal asymptotes.

Vertical asymptotes occur when the denominator approaches zero but the numerator doesn't. So, set the denominator equal to zero and solve for $ x $. This gives you the equation of the vertical asymptote(s).

Horizontal asymptotes occur when $ x $ approaches positive or negative infinity. To find horizontal asymptotes, you can analyze the degree of the numerator and denominator. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is at $ y = 0 $. If the degree of the numerator is equal to the degree of the denominator, the horizontal asymptote is the ratio of the leading coefficients. If the degree of the numerator is greater than the degree of the denominator, there are no horizontal asymptotes.

In this case, $ f(x) $ has a vertical asymptote when $ 16x^2 - 5 = 0 $, and you can find the horizontal asymptotes by analyzing the degrees of the numerator and denominator.