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Math Help - completing the square help

  1. #1
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    completing the square help

    Find the axis of symmetry and coordinates of vertex of the graph of the following quadratic functions using the method of completing the square.

    A, y=-x^2+7x-4


    B, y=3x^2-8x+6


    C, y= -1/3x^2+2x-3/2
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Ok

    A. complete the square and get y=\frac{-(2x-7)^2}{4}+\frac{33}{4}...the axis of symmetry is x=\frac{7}{2}...and the vertex is at \bigg(\frac{7}{2},\frac{33}{4}\bigg)

    B.The secon one factors into y=3\bigg(x-\frac{4}{3}\bigg)^2+\frac{2}{3}


    and I am gettign tired of this so Ill come back and finish if someone else doesnt
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Ok

    I am back for

    C. the equation is y=\frac{-(x-3)^2}{3}+\frac{3}{2}..the axis of symmetry is x=3...and the vertex is \bigg(3,\frac{3}{2}\bigg)

    and for b part two the axis of symmetry is x=\frac{4}{3}...and the vertex is at \bigg(\frac{4}{3},\frac{2}{3}\bigg)
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  4. #4
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    thanks for the answers but if it isn't too much to ask can you please explain the steps b/c i'm lost.
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Here is what you do

    If I had x^2-4x+13 you could factor that if it was x^2-4x+4=(x-2)^2...but you dont have that 4...or do you...you can rewrite it it as x^2-4x+4+9 then factor the one part to get your final asnwer as (x-2)^2+9 now that you have it in that form to find the axis of symmetry it is just the value that makes the inside of the parenthases zero...so in this case x-2=0,x=2 now put it in line form x=2 is the axis of symmetry....to find the vertex it is jsut a(x-h)^2+k (h,k) so once you get it in that order its simple
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