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Thread: 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:
    $\displaystyle y = a(x - h)^2 + k$

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

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

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

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

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

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

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

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

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

    $\displaystyle -h - 1 + 4 = 0$

    $\displaystyle h = 3$

    Finally:
    $\displaystyle ah^2 + k = 8$

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

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

    $\displaystyle -9 + 2k = 16$

    $\displaystyle 2k = 25$

    $\displaystyle k = \frac{25}{2}$

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

    -Dan
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