1 Answer

y = 2x³ + 3x² − 12x + 3
The tangent line is horizontal when dy/dx = 0 (y’ = 0)
y’ = 6x² + 6x – 12 = 0 , div by 6
y’ = x² + x – 2 = 0
(x – 1)(x + 2) = 0
x=1 => y = 2*1³ + 3*1² – 12*1 + 3 = 4
x=2 => y = 2(2)³ + 3(2)² – 12(2) + 3 = 23
The tangent points are: (1, 4) and (2, 23)
================================