SOLUTION: what is the equation of the quadratic function with roots 0 and 1 and a vertex at (1/2,-2/33)?

Question 391016: what is the equation of the quadratic function with roots 0 and 1 and a vertex at (1/2,-2/33)?
Found 3 solutions by stanbon, ewatrrr, solver91311:
Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
what is the equation of the quadratic function with roots 0 and 1 and a vertex at (1/2,-2/33)?
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Using the two roots.
Form: y = a(x-0)(x-1)
y = ax(x-1)
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Using (1/2,-2/33)
-2/33 = a(1/2)(-1/2)
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-2/33 = -a/4
a = 8/33
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Equation:
y = (8/33)x(x-1)
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Cheers,
Stan H.
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Answer by ewatrrr(24785)   (Show Source): You can put this solution on YOUR website!

Hi
parabola with vertex at (1/2,-2/33) and x-inercepts of (0,0) and (1,0)
Using the vertex form of a parabola, where(h,k) is the vertex
y = a(x - 1/2)^2 - 2/33 using Pt(0,0) to solve for a
0 = a(-1/2)^2 - 2/33
2/33 = (1/4)a
8/33 = a
y = 8/33(x-1/2)^2 - 2/33

Answer by solver91311(24713)   (Show Source): You can put this solution on YOUR website!

Begin with:

Since the roots are 0 and 1 we can say:

and

Since is a point on the graph, we can say:

Giving us the 3X3 system:

Solve the system to get your coefficients.

John

My calculator said it, I believe it, that settles it

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