Find the turning points of y = 3 x 3 + 6 x 2 + 3 x – 1 and determine their nature, also, find the equations of the tangents at these points.

Question

please help me

Answer 1

where is the slope zero?

dy/dx = 9 x^2 + 12 x + 3
so
9 x^2 + 12 x + 3 = 0 for turning point

(9x+3)(x+1) = 0

x = -1/3 and x = -1

I will start the x = -1 point. You do the x = -1/3

the slope is 0 at these points so just find y
if x = -1, y = 3(-1)+6(1)+3(-1)-1
= -1
so tangent is y = -1