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Math Help - graphing this parabola

  1. #1
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    graphing this parabola

    4y^2-4y-4x+5=0

    this is what i did:

    4y^2-4y=4x-5
    4y^2-4y=4x-5+4
    4^2-4y+4=4x-5+4
    4(y^2-2y+2)=4x-1
    4(y-2)^2=4x-1

    i don't really know how to move to the next step in order to get something like (y-k)^2=4p(x-h) or (x-h)^2=4p(y-k)

    it doesn't make sense cause i have the 4 outside which seems like its part of the 4p but (y-2) is squared.... and i can't complete the square on the right side since no x^2 value, only x.
    Last edited by Frenchie; August 16th 2010 at 05:34 PM.
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  2. #2
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    Quote Originally Posted by Frenchie View Post
    4y^2-4y-4x+5=0

    this is what i did:

    4y^2-4y=4x-5
    4y^2-4y=4x-5+4
    4^2-4y+4=4x-5+4
    4(y^2-2y+2)=4x-1
    4(y-2)^2=4x-1

    i don't really know how to move to the next step in order to get something like (y-k)^2=4p(x-h) or (x-h)^2=4p(y-k)

    it doesn't make sense cause i have the 4 outside which seems like its part of the 4p but (y-2) is squared.... and i can't complete the square on the right side since no x^2 value, only x.
    Sorry, this is wrong. If you are going to complete the square, the coefficient of the square term must be 1.

    4y^2-4y-4x+5=0

    4y^2 - 4y + 5 = 4x

    4\left(y^2 - y + \frac{5}{4}\right) = 4x

    4\left[y^2 - y + \left(-\frac{1}{2}\right)^2 - \left(-\frac{1}{2}\right)^2 + \frac{5}{4}\right] = 4x

    4\left[\left(y-\frac{1}{2}\right)^2 + 1\right] = 4x

    \left(y - \frac{1}{2}\right)^2 + 1 = x.


    You should be able to graph this now...
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  3. #3
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    thank you! I have another question, is that equivalent to this:
    (y-1/2)^2=(x-1)

    see it doesn't make sense because whats P? there is nothing infront of (x-1) such as 8(x-1) where in that case we could say 4p(x-1), p=2

    it doesn't look like this general form (y-k)^2=4p(x-h) =/
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  4. #4
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    ah nvm i think its (y-1/2)^2=1/4(x-1)... or is it (y-1/2^2)=4(x-1)... eh w.e good enough

    thank you
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  5. #5
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    Quote Originally Posted by Frenchie View Post
    thank you! I have another question, is that equivalent to this:
    (y-1/2)^2=(x-1)

    see it doesn't make sense because whats P? there is nothing infront of (x-1) such as 8(x-1) where in that case we could say 4p(x-1), p=2
    "no" coefficient means the coefficient is 1. If 4p= 1, then p= 1/4.

    it doesn't look like this general form (y-k)^2=4p(x-h) =/
    But it is.
    (y- 1/2)^2= x- 1 is the same as (y- 1/2)^2= 4(1/4)(x- 1). p= 1/4, k= 1/2, and h= 1.
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  6. #6
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    ah thank you all this is helpful
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