I have to find some level curves for: $\displaystyle f(x,y)=1-|x|-|y|$

So, if we call $\displaystyle S$ at the surface given by the equation $\displaystyle z=f(x,y)$, then

$\displaystyle z=1\Rightarrow{-|x|-|y|=0}\Rightarrow{x=y=0} \therefore P(0,0,1)\in{S}$

Now, that particular case its simple, cause it gives just a point, but if I go downwards I get:

$\displaystyle z=0\Rightarrow{-|x|-|y|=-1\Rightarrow{|x|+|y|=1\Rightarrow{|y|=1-|x|}}}$

I'm not sure how to represent this. How does this look on the xy plane?

I know that:

$\displaystyle |y|=\begin{Bmatrix} 1-|x| & \mbox{ si }& y\geq{0}\\-1+|x| & \mbox{si}& y<0\end{matrix}$

And

$\displaystyle |x|=\begin{Bmatrix} x & \mbox{ si }& x\geq{0}\\-x & \mbox{si}& x<0\end{matrix}$

But it don't helps me to visualize the "curve". I know actually that it looks like a parallelogram, but thats because I've used mathematica to compute the surface :P I don't know how to deduce it analytically.