Find the points on the curve y = 2x3 + 3x2 − 12x + 9 where the tangent line is horizontal.

Y = 2x^3 + 3x^2 − 12x + 9
y' = 6x^2 + 6x − 12

tangent line is horizontal y'=0
6x^2 + 6x − 12=0
x^2+x-2=0
(x+2)(x-1)=0
x=-2 or x=1
when x=-2, y=2x^3 + 3x^2 − 12x + 9 = -16 + 12 + 24 + 9 =29
when x=1, y=2x^3 + 3x^2 − 12x + 9 = 2 +3 -12 +9 =2
two points (-2,29),(1,2) 

2).  y = 2x^3 + 3x^2 − 12x + 9

slope of horizontal-tangent-line, y' = 6x^2+6x-12 = 0
SO,
6(x+2)(x-1) = 0,
x=1, and x=-2,
SO, for, x= 1, y= 2+3-12+9 = 2,
Hence,
the  1st. point is : Answer (1, 2)  

and, for, x=-2,
y = 2(-2)^3 +3(-2)^2 -12(-2) +9 = -16+12+24+9 = 29
Hence,
the 2nd-point is : Answer (-2, 29) 

The decimal expansion would be .27 repeated