Question

f of x equals 4 x over quantity x squared minus 16. Show all work to identify the asymptotes and zero of the function.

Answer 1

Given:

The function is

$f(x)=\dfrac{4x}{x^2-16}$

To find:

The asymptotes and zero of the function.

Solution:

We have,

$f(x)=\dfrac{4x}{x^2-16}$

For zeroes, f(x)=0.

$\dfrac{4x}{x^2-16}=0$

$4x=0$

$x=0$

Therefore, zero of the function is 0.

For vertical asymptote equate the denominator of the function equal to 0.

$x^2-16=0$

$x^2=16$

Taking square root on both sides, we get

$x=\pm \sqrt{16}$

$x=\pm 4$

So, vertical asymptotes are x=-4 and x=4.

Since degree of denominator is greater than degree of numerator, therefore, the horizontal asymptote is y=0.