Question

Find the points on the curve y = 2x^3 + 3x^2 − 12x + 9 where the tangent line is horizontal?

Answer 1

Horizontal tangent lines just tell you that the derivative at that point is zero. So,

`(dy)/(dx) = d/(dx)[2x^3 + 3x^2 - 12x + 9]`

`= 6x^2 + 6x - 12 = 0`

There should then be two solutions. Divide by `6` to get:

`0 = x^2 + x - 2`

This factors as:

`color(blue)((x-1)(x+2) = 0)`

Therefore, two horizontal tangent lines can be found, one at `x = -2` and one at `x = 1`. Those are your turning points. Makes sense given the shape of a cubic function.

graph{2x^3 + 3x^2 - 12x + 9 [-5.5, 5.5, -5.04, 30.04]}