If f(x)= 3x^2-x^3, find f'(x) and use it to find an equation of the tangent line to the curve y=3x^2-x^3 at the point (1,2)
f'(x) = 6x-3x^2
f'(1) = 6-3 = 3
So, now you have a point and a slope, and the tangent line is thus
y=3(x-1)+2
see
www.wolframalpha.com/input/?i=plot+y%3D3x%5E2-x%5E3,+y%3D3(x-1)%2B2