Answer:
A: -21
Step-by-step explanation:
Vertex of a quadratic function:
Suppose we have a quadratic function in the following format:
[tex]f(x) = ax^{2} + bx + c[/tex]
It's vertex is the point [tex](x_{v}, y_{v})[/tex]
In which
[tex]x_{v} = -\frac{b}{2a}[/tex]
[tex]y_{v} = -\frac{\Delta}{4a}[/tex]
Where
[tex]\Delta = b^2-4ac[/tex]
If a<0, the vertex is a maximum point, that is, the maximum value happens at [tex]x_{v}[/tex], and it's value is [tex]y_{v}[/tex].
In this question:
Quadratic function:
[tex]g(x) = x^2 - 6x - 12[/tex]
So [tex]a = 1, b = -6, c = -12[/tex].
Minimum value:
This is the y-value of the vertex. So
[tex]\Delta = b^2-4ac = (-6)^2 - 4(1)(-12) = 36+48 = 84[/tex]
[tex]y_{v} = -\frac{\Delta}{4a} = -\frac{84}{4} = -21[/tex]
The minimum value is -21, and the correct answer is given by option A.
