We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
-1
1570
4
avatar+594 

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+101369 
+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+10409 
+1

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

 Feb 3, 2018
 #3
avatar+594 
+1

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

adore.nuk  Feb 3, 2018
 #2
avatar+101768 
+1

Omi is right ... 

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

has no holes.

 Feb 3, 2018
 #4
avatar+101369 
+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

10 Online Users

avatar
avatar