Question

what is the minimum value of the function $g(x) = x^2 - 6x - 12$? $3 - sqrt{21}$, $3 + sqrt{21}$, $3$, $-21$

Explanation:

Step1: Identify the function type

The function \( g(x) = x^2 - 6x - 12 \) is a quadratic function in the form \( ax^2 + bx + c \) where \( a = 1 \), \( b = -6 \), and \( c = -12 \). Since \( a > 0 \), the parabola opens upwards, and the vertex is the minimum point.

Step2: Find the x-coordinate of the vertex

The x-coordinate of the vertex of a quadratic function \( ax^2 + bx + c \) is given by \( x = -\frac{b}{2a} \). Substituting \( a = 1 \) and \( b = -6 \):

$$ x = -\frac{-6}{2(1)} = \frac{6}{2} = 3 $$

Step3: Find the minimum value (y-coordinate of the vertex)

Substitute \( x = 3 \) into the function \( g(x) \):

$$ g(3) = (3)^2 - 6(3) - 12 = 9 - 18 - 12 = -21 $$

Answer:

-21