Find the maximum or minimum value of the function: y = 6x^2 + 4x + 2

The easiest way to solve this problem is by using differentiation.

dy/dx = (6x^2 + 4x + 2)/dx

dy/dx = 62x^1 + 41x^0 + 0

dy/dx = 12x + 4

When you set the derivative equal to zero you find a point on the curve where there is no change in gradient (where the curve is momentarily horizontal).

12x + 4 = 0

Solving this gives the x value for when the function has a maximum or a minimum.

x = -1/3

You can then use this x value to find the value of y at that maximum or minimum. This is done by substituting x back into the formula for y.

y = 6(-1/3)^2 + 4(-1/3) + 2 = 4/3

You could go a step further and find whether that value was a maximum or a minimum. To do this you would differentiate again.

d^2y/dx^2 = (12x + 4)/dx

d^2y/dx^2 = 121x^0 + 0

d^2y/dx^2 = 12

This is greater than 0 and so the value is a minimum as the rate of change of the gradient is positive.