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Math Help - parabola HW help plz

  1. #1
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    parabola HW help plz

    A, write an equation whose graph is a parabola whose vertex is (3/4,-3)


    B, write an equation whose graph is a parabola that opens downward and has an axis of symmetry x=-4


    C, Write an equation whose graph is a parabola opens upward and has a y-intercept of -1/2

    D, Write an equation whose graph is a parabola whose vertex is (-3,4) with a y intercept of 1
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    OK here is what you do

    A. the equation for a parabola with vertex (h,k)...is y=a(x-h)^2+k...so for \frac{3}{4},-3)...is y=\bigg(x-\frac{3}{4}\bigg)^2-3

    B.the equation for a line facing down with its axis of symmetry at 4 is y=-a(x+4)^2+k

    C.one of the possibilities y=x^2-\frac{1}{2}

    D.to do this you need to set up your equation [tex]y=a(x+3)^2+4...I know...now all you have to do is say is put in your values 1=a(0+3)^2+4 and solve for a...so you get a=\frac{-1}{3}...so your equation is y=\frac{-(x+3)^2}{3}+4
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