[tex]y = 4 + 4 x^{2}-2x^{3} Lets \ write \ it \ y = -2 x^{3} +4 x^{2} +4
[/tex]
The tangential line at a certain point is just the derivative so.
[tex]y ' = -6 x^{2} +8x[/tex]. At the point (1,6) we plug the x value in and get the slope at the point (y ' = 2)
The tangential line at that point is
y - 6 = 2(x - 1) (this is the answer)
