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Math Help - Conics: Parabola

  1. #1
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    Conics: Parabola

    Hello! The question is asking me to find the vertex, focus and directrix of the parabola, and sketch its graph.

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

    I tried putting aside the 1/4 for the time being and tried completing the square

    y-5 = (x^2)-2x
    y-3 = (x^2)-2x+2

    y-3 = (x-1)^2


    Plugging back in the 1/4 I got
    y-3 = (1/4)(x-1)^2

    With that, my vertex is (1, 3)

    The book's answer is (1,1)


    Please help me solve this problem, thanks!
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  2. #2
    Super Member TheChaz's Avatar
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    You can't "put aside" the 1/4!!!!!!!!!
    Multiply by 4...
    4y = x^2 - 2x + 5
    4y - 5 = x^2 - 2x
    4y - 4 = x^2 - 2x + 1
    4(y - 1) = (x - 1)^2
    y - 1 = (1/4)(x - 1)^2
    y = (1/4)(x - 1)^2 + 1
    So (h, k) = (1, 1)
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  3. #3
    Behold, the power of SARDINES!
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    Quote Originally Posted by NeoSonata View Post
    Hello! The question is asking me to find the vertex, focus and directrix of the parabola, and sketch its graph.

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

    I tried putting aside the 1/4 for the time being and tried completing the square

    y-5 = (x^2)-2x
    y-3 = (x^2)-2x+2

    y-3 = (x-1)^2


    Plugging back in the 1/4 I got
    y-3 = (1/4)(x-1)^2

    With that, my vertex is (1, 3)

    The book's answer is (1,1)


    Please help me solve this problem, thanks!
    y= \left(1/4 \right) (x^2 -2x+5) multiply both sides by 4

    4y=x^2-2x+5 subract 5 from both sides


    4y-5=x^2-2x Now complete the square on the x

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

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

    Now that this is is standard from we can identify

    4p(y-k)=(x-h)^2

    So p=1,k=1,h=1

    Vertex (1,1) and the equation of the directrix is y=-1

    To slow
    Last edited by TheEmptySet; April 5th 2011 at 07:37 PM. Reason: To slow
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