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Thread: Equation of parabola question

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    Equation of parabola question

    Would anyone be able to help me with this question where I have to find the equation of a parabola in vertex form. The information given is that the vertex is (-4,2) and the y intercept is -16. Thank you

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    Junior Member Sp3ctre's Avatar
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    Re: Equation of parabola question

    By inspection, the parabola is concave down

    Therefore general equation is (x-h)^2 = -4a(y-k)
    Given that centre is (-4,2), h=-4, k=2
    (x+4)^2 = -4a(y-2)

    When x=0, y=-16

    16 = -4a(-18)
    16 = 72a
    a = 2/9

    Therefore the equation is (x+4)^2 = -8/9(y-2)
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    Re: Equation of parabola question

    y = a(x-h)^2 + k is the vertex form for a Parabola (note that h,k are the coordinates for the vertex).

    Therefore sub (-4,2) in the equation giving you: y = a(x+4)^2 + 2

    To get 'a', you must sub a mentioned point in the equation (which in this case is the y-intercept).

    After subbing (0,-16 - which is the y-intercept), you should get:

    -16 = a(0+4)^2+2
    -16 = 16a +2
    -18 = 16a
    -9/8 = a

    Now that you have 'a', you can write the equation as: y = -9/8(x+4)^2 +2
    Last edited by _Anonymous; 14 Oct 2017 at 9:26 PM.

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