Correct Order of Applying Graphical Transformation with Absolute Value I was going through this website, reading about transformations of graph when $| |$ is applied to various parts of a given function, $y=f(x)$. Going through the fourth example of the page, I came across the graph $y=1/(|x|+1)$. I understood the way they solved the problem. But I tried to do it in another way, and later realized, it's of course a wrong graph. What I did is, I first drew the graph of $1/x$ then transformed it to $1/|x|$ and then again transformed it to $1/(|x| + 1)$. I ended up with something like this. I think this will be correct for $y=(1/|x+1|)$, not for the given problem. As I understood my mistake another thing struck my mind: for composite mod functions like $||y|-3|=|3|x|+2|$ what would be the correct way of applying various transformations to draw graphs of such complex functions involving mod. Please give me ideas or steps to be followed to draw graphs of such functions correctly.
Edit: I actually understood the graphing of $1/(|x|+1)$. I am looking for a systematic way to draw the graph of $||y|-3|=|3|x|+2|$ so that I can apply the ideas in drawing other similar graphs.
 A: One way to use transformations to graph $y=\dfrac{1}{\vert x\vert +1}$ is, because of the absolute value, to break it into two parts.
Case I: $x\ge0$
$$ y=\dfrac{1}{x+1} $$
which is the graph of $y=\dfrac{1}{x}$ shifted left by $1$ but we only want the portion of that shifted graph for which $x\ge0$.
Case II: $x<0$
$$ y=\dfrac{1}{-x+1}=-\dfrac{1}{x-1}$$
which is the graph of $y=-\dfrac{1}{x}$ shifted right by $1$ but we only want the portion of that shifted graph for which $x>0$.
Put the two parts together and you have your graph.
This approach becomes quite involved for a graph such as 
$$ \vert\,\vert y\vert-3\,\vert=\vert\,3\vert x\vert+2\,\vert$$
because one has to divide the coordinate plane into twelve regions according to whether $y<3,3\le y<0,0\le y<3, y\ge3,x<-\frac{2}{3},-\frac{2}{3}\le x<0,x\ge0 $.
Then one must rewrite the equation without absolute values for each of those sections and graph only the portions of the resulting graphs which fall within those particular sections.
Actually, I see that for this example there are only eight cases because the $x$ expression on the right side of the equation equals $3x+2$ for $x\ge0$ and equals $-3x+2$ for $x<0$.
Here is a desmos.com graph (just the black lines). The orange lines are the graphs of the eight different equations but only the black portions fall within the 'correct' section of the coordinate plane.
Graph of | |y| - 3 | = | 3|x| + 2 |
