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Math Help - Parabola

  1. #1
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    Unhappy Parabola

    Please someone help. I have summer school and I got a Take-Home Quiz. My teacher knows about the whole, google or yahoo the answer thing, so he does not care. Anyways one of the questions is this: Write an equation of a parabola with vertex (-2,3) and passing through (0,15). I have tried to get this equation, but failed. Please help.
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  2. #2
    Eater of Worlds
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    You could try y=a(x-h)^{2}+k

    (h,k) is the vertex and (x,y) are the coordinates it passes through.

    Use these to find a and you're all set.
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  3. #3
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    How would I use it? Would i enter the vertex and the other thing into it?

    Sorry ... me = slow
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by iorga View Post
    Please someone help. I have summer school and I got a Take-Home Quiz. My teacher knows about the whole, google or yahoo the answer thing, so he does not care. Anyways one of the questions is this: Write an equation of a parabola with vertex (-2,3) and passing through (0,15). I have tried to get this equation, but failed. Please help.
    as galactus was saying, start with
    y = a(x - h)^2 + k

    you want, y = 15, x = 0, h = -2 and k = 3

    plug these into the formula to find a

    Once you've found a, rewrite the original equation, this time, filling in all the values except x and y and you will have your parabola
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  5. #5
    MHF Contributor red_dog's Avatar
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    Given a parabola y=ax^2+bx+c, a\neq 0, its vertex has the coordinates \left(-\frac{b}{2a},-\frac{\triangle}{4a}\right), where \triangle =b^2-4ac.
    Now, we have
    \displaystyle \left\{\begin{array}{ll}\displaystyle -\frac{b}{2a}=-2\\\displaystyle -\frac{\triangle}{4a}=3\end{array}\right.
    Because the parabola is passing through the point (0,15), replacing x=0,y=15 in the equation, we have c=15.
    Now, the system becomes
    \left\{\begin{array}{ll}b=4a\\b^2-48a=0\end{array}\right.\Rightarrow a=3,b=12.
    So the parabola is y=3x^2+12x+15
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