horizontal tangents, again

[racuna]

I need to find the points on the curve

y=2x^3+3x^2-12x+1

where the tangent is horizontal.

I got the derivative, y'=6x^2+6x-12, but I don't know what to do next.
What does it mean when the tangent is horizontal? Is it when x=0?

Help!

 

[stapel]

For the tangent line to be horizontal, the line's slope must be zero. For a curve, the slope at a point is the derivative. So you are looking for the x-values at which dy/dx = 0.

A good first step would be to divide through by 6.

Eliz.