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Determine the slope of the tangent line to the curve y=2x^3-3x^2+1 at the point (1,0)
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y' = 6x^2-6x
at x = 1, y' = slope = 6-6 = 0
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original curve: 2y^3+6(x^2)y-12x^2+6y=1
dy/dx=(4x-2xy)/(x^2+y^2+1) a) write an equation of each horizontal tangent line to the