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    Wikkibahi's Avatar
    Wikkibahi Posts: 42, Reputation: 1
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    #1

    Jan 12, 2012, 12:52 AM
    Question About Calculus..
    Derive the Standard Equation of Parabola with vertex at (h,k):
    (y-k)^2 = 4a(x-h)
    Please derive it with diagram.
    corrigan's Avatar
    corrigan Posts: 115, Reputation: 18
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    #2

    Jan 12, 2012, 10:57 AM
    First of all, gives a parabola, but it's sideways. We can still derive it, but the only way it's a function is if is the domain and is the range. It's a little weird, but it's more likely you copied the problem wrong. Deriving it isn't to hard, you just treat , , and , like you would if they were real numbers. Since I don't know exactly what the problem is, I'll derive:

    the first thing we'll do is expand it out, and solve for y





    I'm not sure what you mean by a diagram, but I think it's the chart that shows the inflection points, and the orientation of the derivative. The inflection points are the values of when the derivative is zero. That isn't to hard, is zero only when , (i'm assuming is nonzero). The orientation of on either side of depends on the orientation of . If is positive, then is positive for and negative for . If is negative, then is negative for and positive for .

    I hope this helps

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