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Math Help - Transformations of quadratic equations

  1. #1
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    Transformations of quadratic equations

    Can you please explain the transformations involved when transforming y = x^2 to y = 4(3x -6)^2 + 1
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  2. #2
    Junior Member Barioth's Avatar
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    Re: Transformations of quadratic equations

    I dont understand what your question is sorry
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    Re: Transformations of quadratic equations

    Hello, elmidge!

    \text{Can you please explain the transformations involved}
    \text{when transforming }\,y \:=\: x^2\,\text{ to }\,y\:=\:4(3x -6)^2 + 1

    \text{We are going from }\,y \:=\:x^2\,\text{ to }\,y \:=\:36(x-2)^2 + 1

    \text{I assume you know the graph of: }\:y \,=\,x^2.

    \text{We have: }\:y \:=\:(x\,{\color{red}-\,2})^2
    . . \text{The graph is moved 2 units to the right.}

    \text{Then: }\:y \:=\:{\color{red}36}(x-2)^2
    . . The graph is stretched vertically by a factor of 36.

    \text{Finally: }\:y \:=\:36(x-2)^2\, {\color{red}+ 1}
    . . The graph is moved one unit upward.
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    Re: Transformations of quadratic equations

    Thank you can you also show me the transalations involved with transforming y = x^2 to y = 9x^2 -6x
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  5. #5
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    Re: Transformations of quadratic equations

    Quote Originally Posted by elmidge View Post
    Thank you can you also show me the transalations involved with transforming y = x^2 to y = 9x^2 -6x
    Start by Completing the Square, then the transformations will be easier to see.
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    Re: Transformations of quadratic equations

    Quote Originally Posted by elmidge View Post
    Thank you can you also show me the transalations involved with transforming y = x^2 to y = 9x^2 -6x
    The vertex of  y = x^2 is at (0,0)

    The vertex of  y = (x + p)^2 + q is at (- p, q). A translation p units in the negative x direction and q units in the positive y.

    If you complete the square for y = 9x^2 - 6x, you will have it in the above, vertex, form and will therefore know the transformations involved.

    Can you do that?
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