Titik belok dari fungsi y = x^3 + 6x^2 + 9x + 7 tolong sama caranya yah need fast answer :<

turunan = 0
3x² + 12x + 9 = 0
----------------------- ÷ 3
x² + 4x + 3 = 0
(x + 3)(x + 1) = 0
x = -3
y = (-3)³ + 6(-3)² + 9(-3) + 7
y = -27+54-27+7 = 7
x = -1
y = (-1)³ + 6(-1)² + 9(-1) + 7
y = -1+6-9+7 = 3
titik belok (-3,7) dan (-1,3)