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Math Help - in "layman's terms"....

  1. #1
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    in "layman's terms"....

    I am beginning to work with Polynomial, Rational, and Quadratic functions...

    I can understand most of it but I am having issues trying to understand who it works if written like this,
    f(x)=x^2+2x

    I am told that the first thing to do is completing the Sqr, by adding one and subtracting one, thus;

    f(x)=x^2+2x+1-1

    x^2+2x+1=(x+1)^2-1

    x+1=0

    x=-1

    and so my vertex is (-1,-1)

    but now how about the X and the Y coordinates?


    Thank you and sorry for such a long post.

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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Morzilla View Post
    I am beginning to work with Polynomial, Rational, and Quadratic functions...

    I can understand most of it but I am having issues trying to understand who it works if written like this,
    f(x)=x^2+2x

    I am told that the first thing to do is completing the Sqr, by adding one and subtracting one, thus;

    f(x)=x^2+2x+1-1

    x^2+2x+1=(x+1)^2-1

    x+1=0

    x=-1

    and so my vertex is (-1,-1)

    but now how about the X and the Y coordinates?


    Thank you and sorry for such a long post.

    x and y coordinates for what?

    you didn't have to solve x + 1 = 0 explicitly.

    if a quadratic is expressed as y = a(x - h)^2 + k, then the vertex is given by (h,k)
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    x and y coordinates for what?

    you didn't have to solve x + 1 = 0 explicitly.

    if a quadratic is expressed as y = a(x - h)^2 + k, then the vertex is given by (h,k)
    soooo........how does x^2 and 2x fit into y = a(x - h)^2 + k

    Thanks
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Morzilla View Post
    soooo........how does x^2 and 2x fit into y = a(x - h)^2 + k

    Thanks
    i'm not sure i got your question. we completed the square for f(x) = x^2 + 2x to get it into the above form. in that form, a = 1, h = -1 and k = -1
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