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

1 Answer
Ryan
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.

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