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

1 Answer
Truong-Son N.
Apr 27, 2018

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]}

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