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?
What kind of graph would $|x-h| + |y-k| = 1$ give?
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algebra-precalculus
graphing-functions
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1"hint": http://www.wolframalpha.com/input/?i=%7Cx-5%7C+%2B+%7Cy-4%7C+%3D+1 – 2012-05-12
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0"Related": http://math.stackexchange.com/questions/144211/relationship-between-two-absolute-value-curves – 2012-05-12
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1The equation $|x|+|y|=1$ can be split into four cases and then translated. – 2012-05-12
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1Since 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 http://en.wikipedia.org/wiki/Taxicab_geometry – 2012-05-12