+0  
 
-1
819
4
avatar+589 

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

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

 

Answer: 

 Feb 3, 2018

Best Answer 

 #4
avatar+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

 

 

cool cool cool

 Feb 3, 2018
edited by CPhill  Feb 3, 2018
 #1
avatar+9922 
+1

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

 Feb 3, 2018
 #3
avatar+589 
+1

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

adore.nuk  Feb 3, 2018
 #2
avatar+97529 
+1

Omi is right ... 

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

has no holes.

 Feb 3, 2018
 #4
avatar+95951 
+1
Best 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

 

 

cool cool cool

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

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