Question 139356: if i have a parabola with a vertex of (4,10) and x-intercepts of 0 and 8, how do i establish the quadratic equation. i came up with y=(x-4)squared + 10.
Ho do you establish the "a" of ax squared + bx + c
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! if i have a parabola with a vertex of (4,10) and x-intercepts of 0 and 8, how do i establish the quadratic equation. i came up with y=(x-4)squared + 10.
Ho do you establish the "a" of ax squared + bx + c
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y = a(x-h)^2 + k
Vertex = (h,k) = (4,10)
so h=4 and k=10
Substitute to get:
y = a(x-4)^2 + 10
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Now use either the point (0,0) or the point (8,0) to find "a":
Using 0,0 you get 0 = a*16 + 10
a = -10/16 = -5/8
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EQUATION:
y = (-5/8)(x-4)^2 + 10
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Cheers,
Stan H.
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