Determine whether $f(x)=6 x^{2}+12 x-8$ has a minimum or maximum value, and find the value of the minimum or maximum. Also, find the axis of symmetry.

The value of the ? is help (numbers)

The axis of symmetry for $f(x)$ is given by ? help (numbers)

Step 1 ：The function $f(x)=6 x^{2}+12 x-8$ is a quadratic function. The general form of a quadratic function is $f(x) = ax^2 + bx + c$. If the coefficient $a$ is positive, the function opens upwards and has a minimum value. If $a$ is negative, the function opens downwards and has a maximum value. In this case, $a = 6$ which is positive, so the function has a minimum value.

Step 2 ：The minimum or maximum value of a quadratic function $f(x) = ax^2 + bx + c$ is given by $f(-\frac{b}{2a})$. So, we can substitute $-\frac{b}{2a}$ into the function to find the minimum value.

Step 3 ：The axis of symmetry of a quadratic function $f(x) = ax^2 + bx + c$ is given by $x = -\frac{b}{2a}$. So, we can calculate $-\frac{b}{2a}$ to find the axis of symmetry.

Step 4 ：The minimum value of the function $f(x)=6 x^{2}+12 x-8$ is -14 and the axis of symmetry is $x = -1$.

Step 5 ：Final Answer: The function $f(x)=6 x^{2}+12 x-8$ has a minimum value of \(\boxed{-14}\) at $x = \(\boxed{-1}\)$, which is the axis of symmetry.