SOLUTION: I am trying to find the vertex for : f(x)=-x^2+10x+5. I am stuck...
(-x^2+10x+25-25)+5
(-x^2+10x+25)+(25+5)
(x+5)(x+5)
I think I am doing this wrong. The negative in front of t
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Quadratic Equations and Parabolas
-> SOLUTION: I am trying to find the vertex for : f(x)=-x^2+10x+5. I am stuck...
(-x^2+10x+25-25)+5
(-x^2+10x+25)+(25+5)
(x+5)(x+5)
I think I am doing this wrong. The negative in front of t
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Question 293618: I am trying to find the vertex for : f(x)=-x^2+10x+5. I am stuck...
(-x^2+10x+25-25)+5
(-x^2+10x+25)+(25+5)
(x+5)(x+5)
I think I am doing this wrong. The negative in front of the x^2 is throwing me off... any help will be appreciated! Answer by nerdybill(7384) (Show Source):
You can put this solution on YOUR website! Standard "vertex form" is
y= a(x-h)^2+k
where (h,k) is the vertex
.
-x^2+10x+5
First group the x terms:
-(x^2-10x)+5
Complete square:
-(x^2-10x+__)+5
-(x^2-10x+25)+5+25
-(x-5)^2+30
So, the vertex is at:
(5,30)
