HELP ASAP PLEASE I WILL MARK BRAINLEST Show all work to identify the asymptotes and zero of the function f of x equals 6 x over quantity x squared minus 36.

vertical asymptotes

x=6, x=-6

horizontal asymptotes

y=0

zeros (0,0)

Step-by-step explanation:

f(x) = 6x / ( x^2 - 36)

First factor

f(x) = 6x / ( x-6)(x+6)

Since nothing cancels

The vertical asymptotes are when the denominator goes to zero

x-6 = 0   x+6=0

x=6          x= -6

Since the numerator has a smaller power than the denominator (1 < 2), there is an asymptote at y = 0

To find the zeros, we find where the numerator = 0

6x=0

x=0

i dont have an answer

step-by-step explanation: