Question

Find the equation of the tangent line at the point (1, 6). for y = 4 + 4x2 - 2x3.

Answer 1

$y = 4 + 4 x^{2}-2x^{3} Lets \ write \ it \ y = -2 x^{3} +4 x^{2} +4$
The tangential line at a certain point is just the derivative so.
$y ' = -6 x^{2} +8x$. 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)