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Find the center, vertices, foci, and slope of the asymptotes for the hyperbola. (x-4)^2/16 - (y+4)^2/16 = 1
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u can use wolfram and check he graph out
http://www.wolframalpha.com/input/?i=%28x-4%29^2%2F16++-++%28y%2B4%29^2%2F16+%3D+1
center(4,-4) verts are a distance of 4 from teh center to get: (8,-4) (0,-4)
your bottoms ar ethe same whih means we get a distance of 4sqrt(2) for the foci foci = (4+4sqrt(2),-4) and (4-4sqrt(2),-4)
and slope of asymps are 1 and -1
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jeezzz thanks again lol!!!
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