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.
