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Math Help - Further Transformation issues with y=af(k(x-d)+c

  1. #1
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    Further Transformation issues with y=af(k(x-d)+c

    I am given a base function of f=x2

    and given a transformed equation of y = 7f((-1/6)(x-1))+1

    I am being asked to describe the transformations that must be applied to the graph of the base function f(x) to obtain the transformed function, which I understand, and then I am ask to write the transformed equation in simplified form. This is where I'm hung up.

    I guess the next line would be

    y=7((-1/6)(x-1)2+1This is where I get stuck. How do I multiply -1/6 by x-1? What do I do with the square? Also, I don't know how to write fractions on the site yet, so -1/6 is negative one sixt to clear up any confusion. I've placed it in parenthesis to try to indicate that.
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  2. #2
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    Re: Further Transformation issues with y=af(k(x-d)+c

    Hi OverlyFrustrated, this seems to be what you are required to do.

    f\left(-\frac{1}{16}(x-1)\right) = \left(-\frac{1}{16}(x-1)\right)^2 = \frac{1}{256}(x-1)^2

    From here on you can expand using the rules of algebra and fractions and you'll find out the simplified form. The fact that you are asking about multiplying -1/16 with (x-1), and then saying that -1/6 is negative one sixth, suggests to me that you are unsure about the basics of fractions. For example

    -\frac{1}{16} = \frac{-1}{16} = \frac{1}{-16}

    Multiplying fractions works like this

    \frac{a}{b}\cdot\frac{c}{d} = \frac{a\cdot c}{b \cdot d}

    In your case you have
    -\frac{1}{16}\cdot(x-1) = - \frac{1}{16}\cdot\frac{x-1}{1} = - \frac{x-1}{16}

    Finally, the way to write math in this forum is to write [..TEX..] before the math expression and [../TEX..] after the math (remove the dots). Fractions are written as \frac{}{}, for example, \frac{x-1}{16}
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