For f(x)=(2x^3)-(15x^2)+24x find the following:

a. determine the equation of the tangent line at x=4
b.find the point(s) where the tangent is horizontal.

can someone explain how to do this??

f(x) = 2x³ - 15x² + 24x
f(4) = -16

slope at x = f'(x) = 6x² - 30x + 24
f'(4) = 0
equation of tangent line at (4, -16):
y + 16 = 0(x - 4)
y = -16
:::::
f'(x) = 0 where tangent is horizontal
6x² - 30x + 24 = 0
Quadratic equation:
x = [30 ± √(30² - 4·6·24)] / (2·6)
   = [30 ± √(324)] / 12
   = [30 ± 18] / 12
   = 1, 4

f(1) = 11
The tangent is horizontal at (1, 11) and (4, -16)