If f(x)= 3x^2-x^3 find f'(1) and use it to find the equation of the tangent line to the curve y=3x^2-x^3 at the point (1,2) find f'(a)

Hello : 
f(x)= 3x^2-x^3 find f'(1) ; 
 f'(x) = 6x - 3x²
f'(1) = 6(1) -3 (1)² = 3 ( the slope )
the tangent line to the curve y=3x^2-x^3 at the point (1,2)  is : 
y - 2 = 3(x - 1 ) 

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