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

y=−9/2x+10

step-by-step explanation:

let's solve for y.

0−9x−2y=−20

step 1: add 9x to both sides.

−9x−2y+9x=−20+9x

−2y=9x−20

step 2: divide both sides by -2.

−2y

−2

=

9x−20

−2

y=

−9

2

x+10

i dont know sorry

step-by-step explanation: