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Math Help - Parabola

  1. #1
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    Post Parabola

    Hey guys,
    Can you help me answer this question?
    *Find the equation, in vertex form, of the quadratic realtion with zeors at -2 and 8 and y-intercept of 8.

    Thanks
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  2. #2
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    Quote Originally Posted by dgolverk View Post
    Hey guys,
    Can you help me answer this question?
    *Find the equation, in vertex form, of the quadratic realtion with zeors at -2 and 8 and y-intercept of 8.

    Thanks
    You want the form:
    y = a(x - h)^2 + k

    You have zeros at (-2, 0) and (8, 0) and a y-intecept of (0, 8).

    Plugging these points into the equation individually:
    0 = a(-2 - h)^2 + k
    0 = a(8 - h)^2 + k
    8 = a(0 - h)^2 + k

    Or:
    ah^2 + 4ah + 4a + k = 0
    ah^2 - 16ah + 64a + k = 0
    ah^2 + k = 8

    Probably the simplest thing to do here would be to take the last equation:
    ah^2 = 8 - k and sub it into the top two equations.

    (8 - k) + 4ah + 4a + k = 0 ==> 4ah + 4a + 8 = 0
    (8 - k) - 16ah + 64a + k = 0 ==> -16ah + 64a + 8 = 0

    These turn into:
    ah + a + 2 = 0
    -2ah + 8a + 1 = 0

    I'm going to multiply the top equation by 2 and add the bottom equation to it:
    2ah + 2a + 4 = 0
    -2ah + 8a + 1 = 0
    --------------------------------
    10a + 5 = 0

    Thus a = -\frac{5}{10} = -\frac{1}{2}

    So
    ah + a + 2 = 0
    Becomes
    -\frac{1}{2}h + -\frac{1}{2} + 2 = 0

    -h - 1 + 4 = 0

    h = 3

    Finally:
    ah^2 + k = 8

    -\frac{1}{2}(3)^2 + k = 8

    -\frac{9}{2} + k = 8

    -9 + 2k = 16

    2k = 25

    k = \frac{25}{2}

    So
    y = -\frac{1}{2}(x - 3)^2 + \frac{25}{2}

    -Dan
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