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

Sep 19, 2016

The minimum is located at $y = - \frac{675}{4}$ (this function opens up, so it has no maximum).

#### Explanation:

Start by multiplying out.

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

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

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

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.

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

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

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

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

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

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

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

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

Hopefully this helps!