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)
y^2 = x^3 (2-x) (1,1)
y'(1) = 3(1) - 1 = 2
so, now yu have a slope and a point:
y-1 = 2(x-1)