Graphing absolute value inside absolute value equation How would I graph an equation with an absolute value inside an absolute value?
For example, $\left|-\left|x\right|+1\right|+\left|y-2\right|=3$ 
I tried graphing this on Desmos and it gave me some weird closed polygon. https://www.desmos.com/calculator/ysnqhkiuuk
 A: You graph this equation manually the same way you graph any equation: try to find enough solution points to give you a feel for how the equation behaves, and interpolate the rest. When possible, use what you know about the functions to reduce the number of points needed.
In this case, you should know what $y=|x|$ looks like, which tells you that your equation is going to consist of line segments with vertices at places where the interior of the absolute values is $0$.
It is helpful to solve it for $y$:
$$|y - 2| = 3 - | 1 - |x|\,|\\y = 2\pm(3 - | 1 - |x|\,|)$$
Which also tells you that the graph will be symmetric about the line $y = 2$, and since $x$ only appears inside the absolute value, the graph will also be symmetric about $x = 0$.
If we examine just the quadrant where $x \ge 0$ and $y \ge 2$, we have
$$y = 5 - |1 - x|$$
For $x \le 1, y = 5 - (1 -x) = 4 + x$. For $x \ge 1, y = 5 + (1 - x) = 6 - x$ Note that since $y \ge 2$ in this quadrant, this only holds when $6 - x \ge 2$, thus $x \le 4$. So in this quadrant, this is the graph of $$y = \begin{cases} 4 + x& 0\le x \le 1\\6 -x & 1 \le x \le 4\end{cases}$$
Reflect that graph through $x = 0$, and then reflect again though $y = 2$ to get the full graph.
