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I know that $(x - h)^2 + (y - k)^2 = 1$ is a circle, but what would the graph look like for $|x-h| + |y-k| = 1$ and why would it look like that?

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"Related":… – Asaf Karagila May 12 '12 at 22:59
The equation $|x|+|y|=1$ can be split into four cases and then translated. – anon May 12 '12 at 23:02
Since you make the comparison to a (Euclidean) circle, I'll point out that $|x-h| + |y-k| = 1$ is the corresponding taxicab circle. See also – Isaac May 12 '12 at 23:22

It will have a graph very similar to the graph of $|x|+|y|=1$, except displaced so as to have "centre" $(h,k)$. Now you need to graph $|x|+|y|=1$. First deal with the first quadrant part. That's just $x+y=1$. Now deal with the second quadrant part. In the second quadrant, $|x|=-x$, and $|y|=y$, so we are looking at $-x+y=1$. Continue.

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