Can anyone explain what (1-x)^2 - y^2 > 1 should look like when graphed?
First note that this is the same as
$\displaystyle [-(x-1)]^2-y^2>1 \iff (x-1)^2-y^2>1$
This is just the standard form of a hyperbola that intersects the x-axis shifted one unit to the right.
Now just check some test points in each region.
You will get a graph that looks like this...
Notice that TheEmptySet is using the fact that "f(x)= g(x)" forms the boundary between "f(x)< g(x)" and "f(x)> g(x)".
To graph $\displaystyle (1- x)^2- y^2> 1$, first graph $\displaystyle (1-x)^2- y^2= 1$ which is the hyperbola TheEmptySet shows. Then notice that x= 0,y= 0 ((0,0), the point between the two branchs of the hyperbola) does NOT satisfy the inequality you want. Therefore, the graph you want is the region "inside" each branch of the hyperbola that is shaded in TheEmptySet's graph.