Graph of $f(x)$ given, find graph of $f(|x|)$ I know the graph of $f(x)=x^2-2x$.

Google calculator https://www.google.com/#q=graph+of+x%5E2-2x

But how can I find the graph of $f(|x|)=|x|^2-2|x|$? What is the best method to approach here? Also, what's the difference between $f(|x|)$ and $|f(x)|$? 
Thanks!
 A: take the graph that is an the right half-plane and reflect it about the $y$-axis to obtain the whole graph. 
Indeed if $x\ge 0$ then $f(|x|)=f(x)$ so for $x\ge0$ (the right half-plane) the graph does not change. 
If $x<0$ then $-x>0$ and $f(|x|)=f(-x)$, which makes the points $(x,f(-x))$ and $(-x,f(-x))$ to both belong to the graph, and these points are symmetric about the $y$-axis. 
Concerning $|f(x)|$ if the original graph of $f(x)$ is above (or on) the $x$-axis, then it remains unchanged. But those points of the original graph that are below the $x$-axis get reflected about the $x$-axis, so they end up in the upper half-plane, and the whole graph of $|f(x)|$ is in the upper half-plane (including possibly the $x$-axis itself). 
For $f(|x|)$ you start with $x$, apply $|x|$ and then apply the original $f$. As explained above (and in additional detail in the comments), the result is a graph symmetric about the $y$-axis. 
For $|f(x)|$ you start with $x$ and immediately apply the original $f$, but of the result is negative, then you "make it positive" and the point of the graph ends up above the $x$-axis. 
A: The short answer: the graph where $x \geq 0$ gets mirrored in the vertical axis.
As for the difference between $f(|x|)$ and $|f(x)|$. Suppose $f(x) = x-4$. Then $f(|0|) = f(0) = -4$ whereas $|f(0)| = |-4| = 4$.
