Math Help - Transformation about a line other than an axis

1. 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;

$
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,

$
y = (x+4)^3-4(x+4)^2+(x+4)+6
$

$y = (x+4)^3-4(x+4)^2+x+10
$

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

$
y=(\frac{1}{2}x+4)^3-4(\frac{1}{2}x+4)^2+\frac{1}{2}x +10
$

$
y=(\frac{1}{2}x+4)^3-2(x+4)^2+\frac{1}{2}x +10
$

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.

$
y=\frac{1}{2}x^3-2x^2+\frac{1}{2}x+6
$

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

$
y=\frac{1}{8}x^3+\frac{1}{2}x^2-\frac{3}{2}x
$

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!

2. Originally Posted by Kasper
From here I stretched by a factor of 2 by replacing all $x$ with $\frac{1}{2}x$

$
y=(\frac{1}{2}x+4)^3-4(\frac{1}{2}x+4)^2+\frac{1}{2}x +10
$

$
y=(\frac{1}{2}x+4)^3-2(x+4)^2+\frac{1}{2}x +10
$
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.