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Find an equation of the tangent line to the curve at the given point

y^2 = x^3 (2-x) (1,1)

  • Calculus -

    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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