+0

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

-1
819
4
+589

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

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

Feb 3, 2018

#4
+95951
+1

[ 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

Feb 3, 2018
edited by CPhill  Feb 3, 2018

#1
+9922
+1

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

Feb 3, 2018
#3
+589
+1

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

#2
+97529
+1

Omi is right ...

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

has no holes.

Feb 3, 2018
#4
+95951
+1

[ 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

CPhill Feb 3, 2018
edited by CPhill  Feb 3, 2018