at what points does the curve have a horizontal tangent?

Question

At what points does the curve
y = 2x^3 + 3x^2 - 36x + 40
have a horizontal tangent?

Answer 1

A curve has an horizontal tangent when its derivative equals 0.
So all you have to do is calculate the derivative of your function (which is y'= 6x^2 +6x -36) and solve the equation y' = 0.
You'll find two values of x, which are the two points where your curve has an horizontal tangent.

Answer 2

y = 2x³ + 3x² + 3 The tangent is got here upon with the help of placing the by-product, y ', of y to 0 and fixing for x: y ' = 6x² + 6x 6x² + 6x = 0 6x(x + a million) = 0 If the manufactured from 2 aspects equals 0, then one or both aspects equivalent 0. If 6x = 0, x = 0 and y = 2(0)³ + 3(0)² + 3 y = 3 One element of Tangency (0, 3) ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ If x + a million = 0, x = - a million and y = 2(- a million)³ + 3(- a million)² + 3 y = - 2 + 3 + 3 y = 4 different element of Tangency (- a million, 4) ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯  

Answer 3

Horizontal tangent is when the gradient is 0.
y' = 6x^2 + 6x - 36 = 0
6(x^2 + x - 6) = 0
6(x+3)(x-2) = 0
x = 2 and x = -3.
You're asked for the points, so find the y values too:
(2,-4) and (-3, 121).

Answer 4

y = 2x^3 + 3x^2 - 36x + 40
y'=6x^2+6x-36=0
x^2+x-6=0
(x+3)(x-2)=0
x=-3
x=2

there are horizontal tangents at x=-3 & x=2
y = 2x^3 + 3x^2 - 36x + 40
y=2(-3)^2+3(-3)^2-36(-3)+40
y=-54+27+108+40=121
y = 2x^3 + 3x^2 - 36x + 40
y=2*2^3+3*2^2-36*2+40=16+12-72+40=-4

points are (-3, 121) and (2,-4)