I have the problem "Write an equation for a parabola that has x-intercepts equal to (4,0) and (8,0) and minimum point at (6,-1)" but my problem is that I'm not sure how to get an equation that has a minimum point of (6,-1). I'm supposed to turn it into standard form (ax^2+bx+c). Is there some method I could follow that could help me achieve this?

Question

I have the problem "Write an equation for a parabola that has x-intercepts equal to (4,0) and (8,0) and minimum point at (6,-1)" but my problem is that I'm not sure how to get an equation that has a minimum point of (6,-1).  I'm supposed to turn it into standard form (ax^2+bx+c).  Is there some method I could follow that could help me achieve this?

agent0smith · Answer

You might want to start with this form $$\Large f(x) = a(x-h)^2 + k$$
(h, k) is the vertex - this is the minimum point they gave you, (6, -1).

anonymous · Answer

How would that look in this equation when plugging everything needed into the formula? I'm just a bit confused on that part.

agent0smith · Answer

You have the vertex, so plug those numbers into that equation first.

agent0smith · Answer

Then you can use either of the x-intercepts, to find the value of a.