+0  
 
-1
224
4
avatar+588 

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

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

 

Answer: 

adore.nuk  Feb 3, 2018

Best Answer 

 #4
avatar+85757 
+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

CPhill  Feb 3, 2018
edited by CPhill  Feb 3, 2018
Sort: 

4+0 Answers

 #1
avatar+9232 
+1

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

Omi67  Feb 3, 2018
 #3
avatar+588 
+1

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

adore.nuk  Feb 3, 2018
 #2
avatar+92217 
+1

Omi is right ... 

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

has no holes.

Melody  Feb 3, 2018
 #4
avatar+85757 
+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

28 Online Users

avatar
avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details