Question

Find the points on the curve y=2x^3+3x^2-12x+1 where the tangent is horizontal

Answer 1

( - 2 , 21 ) and (1 , -6 )

The tangent is horizontal when dy / dx = 0

dy / dx = 6x² +6x - 12

equate to zero to obtain x - coordinates

6x² + 6x - 12 = 0 ⇒ 6 ( x² + x - 2 ) = 0

6( x + 2 )( x - 1 ) = 0 ⇒ x = - 2 or x = 1

substitute these values into y to obtain corresponding y - coordinates

x = - 2 → y =2(-2)³ + 3(-2)² - 12(-2) + 1 = 21 ⇒ (- 2 , 21)

x = 1 → y = 2 + 3 - 12 + 1 = - 6 ⇒ (1 , - 6 )