Answer:
-21 for x = 3
Step-by-step explanation:
It's a quadratic function. The graph is a parabola.
The coefficient of x² is equal 1 > 0. Therefore the parabola is open up.
Conclusion: The minimum is in a vertex.
[tex]f(x)=ax^2+bx+c\\\\(h,\ k)-vertex\\\\h=\dfrac{-b}{2a},\ k=f(h)[/tex]
We have
[tex]g(x)=x^2-6x-12\to a=1,\ b=-6,\ c=-12\\\\h=\dfrac{-(-6)}{2(1)}=\dfrac{6}{2}=3\\\\k=g(3)=3^2-6(3)-12=9-18-12=-21\\\\(3,\ -21)-vertex[/tex]
