Rip brilliant community problems part 3

If we complete the square on $f(x),$ we end up with:

$f(x) = a(x + \frac{b}{2a})^2 + (18 - \frac{b^2}{4a})$

If $f(2) = 6$ is the global minimum attained by this parabola, then three necessary & sufficient conditions are:

$2 + \frac{b}{2a} = 0$ (i)

$18 - \frac{b^2}{4a} = 6$ (ii)

$a > 0$ (iii)

which this system of equations yields $a = 3, b = -12 \Rightarrow a+b = 3-12 = \boxed{-9}.$