How do you find the value that will produce the maximum value of the function g(x)=3(x-12)(x+3)?

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Answer 1 · 2016-09-19T02:42:06

The minimum is located at #y = -675/4# (this function opens up, so it has no maximum).

Start by multiplying out.

#g(x) = 3(x^2 - 12x + 3x - 36)#

#g(x) = 3(x^2 - 9x - 36)#

#g(x) = 3x^2- 27x - 108#

We now must complete the square to find the minimum, which will be the y-coordinate of the vertex, or the lowest point of the function.

#g(x) = 3(x^2 -9x + m - m) - 108#

#m = (b/2)^2 = (-9/2)^2 = 81/4#

#g(x) = 3(x^2 - 9x + 81/4 - 81/4) - 108#

#g(x) = 3(x^2 - 9x + 81/4) - 243/4 - 108#

#g(x) = 3(x - 9/2)^2 - 675/4#

The vertex in the form #y = a(x- p)^2 + q# is at the point #(p, q)#.

We need the y-coordinate of the vertex, which is #-675/4#.

Hence, the minimum of the function #y = 3(x - 12)(x + 3)# is #y = -675/4#

Hopefully this helps!