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Math Help - Transformation about a line other than an axis

  1. #1
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    Transformation about a line other than an axis

    Hey, I'm sorry if this is in the wrong place, I wasn't too sure where to put a transformation question. No idea what I'm doing wrong, thanks for taking the time to look over my problem!

    The question I've got here;

    <br />
y=x^3-4x^2+x+6 is stretched horizontally by a factor of 2 about the line x = 4 What is the resulting equation?


    I started by moving the graph to the left by 4 units to line up x=4 with the y-axis. Which is what a reference problem in my textbook is doing,

    <br />
y = (x+4)^3-4(x+4)^2+(x+4)+6<br />
     y = (x+4)^3-4(x+4)^2+x+10<br />

    From here I stretched by a factor of 2 by replacing all x with \frac{1}{2}x

    <br />
y=(\frac{1}{2}x+4)^3-4(\frac{1}{2}x+4)^2+\frac{1}{2}x +10<br />
    <br />
y=(\frac{1}{2}x+4)^3-2(x+4)^2+\frac{1}{2}x +10<br />

    Lastly, I moved the graph back by 4 units to the right, to place it back with the line about which it was being stretched.

    <br />
y=\frac{1}{2}x^3-2x^2+\frac{1}{2}x+6<br />

    I must be doing something wrong, as I don't see why the intermediate steps of translating it 4 units left and then back had anything to do with it.

    The correct answer to this problem is

    <br />
y=\frac{1}{8}x^3+\frac{1}{2}x^2-\frac{3}{2}x<br />

    I have NO idea how they could manage to get this! Does anyone have any idea how I'm going wrong? Thanks in advance for any and all help!
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  2. #2
    Super Member craig's Avatar
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    Quote Originally Posted by Kasper View Post
    From here I stretched by a factor of 2 by replacing all x with \frac{1}{2}x

    <br />
y=(\frac{1}{2}x+4)^3-4(\frac{1}{2}x+4)^2+\frac{1}{2}x +10<br />
    <br />
y=(\frac{1}{2}x+4)^3-2(x+4)^2+\frac{1}{2}x +10<br />
    For a start you working is incorrect.

    -4(\frac{1}{2}x+4)^2 \neq -2(x+4)^2

    For a start you can't just take a factor of \frac{1}{2} from the x and not the 4, secondly you can't take it out of a bracket that is being squared.
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