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Completing the Square

  1. Mar 2, 2006 #1
    Ok so can you please tell me where I go wrong here. I want to put
    y=3x^2+2x-1 into y=a(x-h)^2+K

    ax^2-2ah+ah^2+k=3x^2-2x+ah^2-1

    ax^2=3x^2 -----> a=3
    -2ahx=-2x
    -2(3)hx=-2x ----->h=1/3
    K=----->-1

    Therefore:

    y=3(x-1/3)^2-1

    But I think it's suppost to be: y=3(x+1/3)^2-4/3

    Is that right??? So what is wrong....
     
  2. jcsd
  3. Mar 2, 2006 #2
    There's nothing wrong with your method...except that I find it very confusing! :confused:

    I know you are trying to compare terms to find h and k here, but how did the 2x get a negative sign?

    I'd recommend looking at it like this:
    [tex]y=3x^2+2x-1[/tex]
    [tex]y=3x^2+2x+D-D-1[/tex]

    Now, a perfect square in the form [tex]a(x-h)^2[/tex] looks like:
    [tex]ax^2-2ahx+ah^2=3x^2+2x+D[/tex]
    We see that -2ah=2 and ah^2=D.
    We of course know what a is, right? I leave it to you to finish finding h and D.

    Then you've got:
    [tex]y=(3x^x+2x+D)-D-1[/tex]
    [tex]y=a(x-h)^2-D-1[/tex]
    with whatever D value you have.

    -Dan
     
  4. Mar 2, 2006 #3

    mathman

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    Gold Member

    In going from the first line to the second, it looks quite screwed up.
     
  5. Mar 2, 2006 #4
    Yea thanks Dan for your help I now clearly understand it..

    Btw ya that was a mistake with the -2 , what should have bee +2
     
  6. Mar 3, 2006 #5

    HallsofIvy

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    Science Advisor

    This is where you went wrong: you want
    [tex]ax^2- 2ah+ ah^2+ k= 3x^2- 3x-1[/tex]
    not "ah2- 1".


    No, ah2+ k= -1. Since a= 3 and h= 1/3, ah2= 1/3 and k= -1- 1/3= -4/3.

     
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