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Math Help - please help me

  1. #1
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    please help me

    Hello everybody, how are you doing

    sorry to bother
    I just have two questions, I would really appreciate if you are so kind to answer step by step, the problem says

    write the equation in the form y=a(x-h)^2+k or x=a(y-k)^2+h

    1- x^2+2x
    2- x=3y^2-12y-36

    thanks!
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by jhonwashington View Post
    Hello everybody, how are you doing

    sorry to bother
    I just have two questions, I would really appreciate if you are so kind to answer step by step, the problem says

    write the equation in the form y=a(x-h)^2+k or x=a(y-k)^2+h

    1- x^2+2x
    2- x=3y^2-12y-36

    thanks!
    I'm presuming the first is y = x^2 + 2x?

    This is done by "completing the square." Note that:
    (x + a)^2 = x^2 + 2ax + a^2

    We have: x^2 + 2x, so by comparing the two we see that 2a = 2, so a = 1, so a^2 = 1.

    But remember we can't simply add a 1, we need to add and subtract 1 so that we don't change the equation:

    y = x^2 + 2x

    y = x^2 + 2x + 1 - 1

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

    y = (x + 1)^2 - 1

    The other problem is done exactly the same way.

    -Dan
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  3. #3
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    earboth's Avatar
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    Quote Originally Posted by jhonwashington View Post
    ...
    write the equation in the form y=a(x-h)^2+k or x=a(y-k)^2+h

    1- x^2+2x
    2- x=3y^2-12y-36...
    Hello, Jhonwashington,

    to 1): I assume that it reads:
    y = x^2+2x\ \Longleftrightarrow \  y= x^2+2x+\left(\frac{2}{2} \right)^2-\left(\frac{2}{2} \right)^2. Thus:
    y = x^2 + 2x \ \Longleftrightarrow\ y = (x+1)^2-1

    to 2)
    x=3y^2-12y-36 = 3\left( \left(y^2-4y+\left(\frac{4}{2} \right)^2\right)-\left(\frac{4}{2} \right)^2 \right) -36 =
    3 \left(y-2 \right)^2-12 -36 = 3 \left(y-2 \right)^2-48

    EB
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  4. #4
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    Thank you so much for your help topsquark, earboth
    God bless you.
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