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Math Help - Complete the square to find the vertex

  1. #1
    Newbie duBxStep's Avatar
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    Complete the square to find the vertex

    Given: F(x)= x^2-6x-8 , complete the square to find the vertex.

    TY for your help it is much appreciated!
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  2. #2
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    Re: Complete the square to find the vertex

    h
    Quote Originally Posted by duBxStep View Post
    Given: F(x)= x^2-6x-8 , complete the square to find the vertex.

    TY for your help it is much appreciated!
    Have you tried completing the square?
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  3. #3
    Newbie duBxStep's Avatar
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    Re: Complete the square to find the vertex

    Well, I looked up how to do it on google and this is where I got:

    y=a(x-h)^2+K
    Y=1(x-h)^2+k
    Y+8=x^2-6x
    y+8-3=x^2-6x-3

    but I have no idea where to go from there, and im not even sure if i'm on the right track.
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  4. #4
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    Re: Complete the square to find the vertex

    Quote Originally Posted by duBxStep View Post
    Given: F(x)= x^2-6x-8 , complete the square to find the vertex.

    TY for your help it is much appreciated!
    Have you been taught how to complete the square? What do you class notes and textbook have to say about it?
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  5. #5
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    Re: Complete the square to find the vertex

    Think about this:
    f(x)=x^2-ax+b
    f(x)=x^2-ax+b+c-c is the same thing.
    f(x)=(x^2-ax+b+c)-c is also the same thing.

    So what you need to figure out is what number added to itself makes -6. That is: x+x=-6 and if you square it what number do you get? x^2=?.

    What ever that ? is you need to make the -8 in your equation equal that by adding and subtracting the some number to it. and splitting it up like above.
    Last edited by takatok; November 14th 2011 at 06:29 PM.
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  6. #6
    Newbie duBxStep's Avatar
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    Re: Complete the square to find the vertex

    Quote Originally Posted by takatok View Post
    Think about this:
    f(x)=x^2-ax+b
    f(x)=x^2-ax+b+c-c is the same thing.
    f(x)=(x^2-ax+b+c)-c is also the same thing.

    So what you need to figure out is what number added to itself makes -6( x+x=-6 ) and if you square it what number do you get? x^2=?.

    What ever that ? is you need to make the -8 in your equation equal that by adding and subtracting the some number to it. and splitting it up like above.
    I thought that the formula was f(x)=(ax^2+bx+c)? How come the A in the given formula has shifted over to the right one place?

    Okay, so -6(x+x=-6) x=1/2 for -6 to = -6 right? square it and you get 1/4.

    So I need to make 1/4 = to -8? Man im confused. I know I've done this before I just forgot how to do it!
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  7. #7
    Newbie duBxStep's Avatar
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    Re: Complete the square to find the vertex

    So here I am now, let me know if I'm on the right track:

    Y=x^2-6x-8

    Y+8+9=x^2-6x+9

    Y+17=x^2-6x+9

    Y=(x-3)^2-17
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  8. #8
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    Re: Complete the square to find the vertex

    You've got it exactly. My original post was spaced badly. I corrected. Your method you just did is essentially the same as what I described.

    So now from your earlier formula
    y=a(x-h)^2+K

    a=1, h=3, and k=-17. Fill that in with whatever you were working on.
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  9. #9
    Junior Member BobBali's Avatar
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    Re: Complete the square to find the vertex

    f(x) = x^2-6x-8
    (x-3)^2 = 8   *Add the square of 3 to both sides; obtained by halving middle term coefficient -6x = -3

    (x-3)^2 = 8 + (-3^2)
    (x-3)^2 = 17
    x = 3 \pm\sqrt{17}

    Hence, coordinates of vertex is (3, -17)
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