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posted by Jeremy Saturday, February 23, 2013 at 11:51pm.
Find an equation of the tangent line to the curve at the given point y^2 = x^3 (2-x) (1,1)
y' = 3x^2 (2-x) - x^3 y'(1) = 3(1) - 1 = 2 so, now yu have a slope and a point: y-1 = 2(x-1)
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