How do you find the maximum or minimum value of y=2x^2 +32x - 4?

1 Answer
Aug 13, 2017

Find the derivative, dy/dx, and solve for dy/dx =0.
The x-coordinate of the turning point is at x=-8.

Explanation:

The derivative of any term x^n is given by nx^(n-1), so for this polynomial, 2x^2+32x-4, the derivative is:

2*2x + 1*32 + 0*-4.
:. dy/dx = 4x +32.

To calculate the turning point(s), let this equal 0.

4x + 32 = 0
:. x= -8.

By substituting this into the original equation, the turning point is at (-8, -137)

Since dy/dy > 0 for x = -7 and dy/dx < 0 for x = -9, this is a minimum turning point.