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 $y = - \frac{675}{4}$ $y = -675/4$ (this function opens up, so it has no maximum).

Start by multiplying out.

$g (x) = 3 (x^{2} - 12 x + 3 x - 36)$ $g(x) = 3(x^2 - 12x + 3x - 36)$

$g (x) = 3 (x^{2} - 9 x - 36)$ $g(x) = 3(x^2 - 9x - 36)$

$g (x) = 3 x^{2} - 27 x - 108$ $g(x) = 3x^2- 27x - 108$

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.

$g (x) = 3 (x^{2} - 9 x + m - m) - 108$ $g(x) = 3(x^2 -9x + m - m) - 108$

$m = {(\frac{b}{2})}^{2} = {(- \frac{9}{2})}^{2} = \frac{81}{4}$ $m = (b/2)^2 = (-9/2)^2 = 81/4$

$g (x) = 3 (x^{2} - 9 x + \frac{81}{4} - \frac{81}{4}) - 108$ $g(x) = 3(x^2 - 9x + 81/4 - 81/4) - 108$

$g (x) = 3 (x^{2} - 9 x + \frac{81}{4}) - \frac{243}{4} - 108$ $g(x) = 3(x^2 - 9x + 81/4) - 243/4 - 108$

$g (x) = 3 {(x - \frac{9}{2})}^{2} - \frac{675}{4}$ $g(x) = 3(x - 9/2)^2 - 675/4$

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

We need the y-coordinate of the vertex, which is $- \frac{675}{4}$ $-675/4$ .

Hence, the minimum of the function $y = 3 (x - 12) (x + 3)$ $y = 3(x - 12)(x + 3)$ is $y = - \frac{675}{4}$ $y = -675/4$

Hopefully this helps!