Question

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

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

Answer 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)