how do i find the equation of a parabola given x intercepts x=0 and x=6 and a vertex at (3,4)?

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asked by Cody
  1. you know the equation must be of the form

    y = a(x-3)^2 + 4
    all we need is the value of a to stear it so it passes through (6,0)
    0 = a(6-3)^2+4
    0 = 9a + 4
    a = -4/9

    so: y = (-4/9)(x-3)^2 + 4

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    posted by Reiny
  2. *facepalm* wow i can't believe i forgot that. thanks for including the
    y=a(x-3)^2+4 i cant believe i forgot to use point-slope form.

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    posted by Cody

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