Question

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

Answer 1

take the derivative and set it equal to 0, find x then solve for y in the original equation

Answer 2

y'=6x^2+6x-12

Answer 3

6x^2+6x-12=0
x^2+x-2=0
(x+2)(x-1)=0
x=-2, 1

Answer 4

plug these values of x into y = 2x^3 + 3x^2 - 12x + 1 to find the y-coordinate

Answer 5

so (-2,0) and (1,0) are the answers?

Answer 6

no

Answer 7

you plug x=-2 and x=1 into the original equation, not the derivative

Answer 8

ah

Answer 9

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

Answer 10

right