To determine the minimum value of the function [tex]\( g(x) = x^2 - 6x - 12 \)[/tex]:
1. Identify and Define the Quadratic Function:
The function given is a quadratic function in the form [tex]\( g(x) = ax^2 + bx + c \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -6 \)[/tex], and [tex]\( c = -12 \)[/tex].
2. Find the Derivative:
To locate the critical points, we need to find the derivative of [tex]\( g(x) \)[/tex]. The first derivative [tex]\( g'(x) \)[/tex] is:
[tex]\[
g'(x) = \frac{d}{dx}(x^2 - 6x - 12) = 2x - 6
\][/tex]
3. Set the Derivative Equal to Zero:
To find the critical points, solve [tex]\( g'(x) = 0 \)[/tex]:
[tex]\[
2x - 6 = 0 \implies x = 3
\][/tex]
4. Determine the Type of Critical Point:
Since [tex]\( g(x) \)[/tex] is a quadratic function and [tex]\( a = 1 \)[/tex] (which is positive), the parabola opens upwards. Therefore, the critical point [tex]\( x = 3 \)[/tex] is actually the vertex of the parabola, representing a minimum point.
5. Evaluate the Function at the Critical Point:
Now, we substitute [tex]\( x = 3 \)[/tex] back into the original function to find the minimum value:
[tex]\[
g(3) = (3)^2 - 6(3) - 12 = 9 - 18 - 12 = -21
\][/tex]
Therefore, the minimum value of the function [tex]\( g(x) = x^2 - 6x - 12 \)[/tex] is [tex]\(\boxed{-21}\)[/tex].
