What are the coordinates of the hole in the graph of the function f(x)?

f(x)= x^2+4x-12x-2

Answer:

[ x^2 + 4x ] / [ -12x - 2 ]

This graph will have a vertical asymptote where -12x - 2 = 0 ⇒ at x = -1/6

We won't have any "holes"....the denominator would have to be a factor of the numerator for this to happen

Do you mean f(x)= x^2+4x/(-12x-2) ???

Yes, that is what I meant. I am sorry about the confusing. It has no holes?

Omi is right ...

has no holes.