Let's compare the given function with the model for a quadratic equation:
[tex]\begin{gathered} f(x)=ax^2+bx+b \\ a=2,b=12,c=-6 \end{gathered}[/tex]Since the value of a is positive, the parabola has its concavity upwards, and the function has a minimum value.
The minimum value can be found calculating the y-coordinate of the vertex:
[tex]\begin{gathered} x_v=-\frac{b}{2a}=-\frac{12}{4}=-3 \\ \\ y_v=2\cdot(-3)^2+12\cdot(-3)-6 \\ y_v=2\cdot9-36-6^{} \\ y_v=-24 \end{gathered}[/tex]Therefore the minimum value is -24.
