Consider the function $f(x)=2-6 x^{2}, \quad-4 \leq x \leq 2$. The absolute maximum value is and this occurs at $x=$ The absolute minimum value is and this occurs at $x=$

Step 1 ：Consider the function \(f(x)=2-6 x^{2}\), for \(-4 \leq x \leq 2\).

Step 2 ：The derivative of the function \(f(x) = 2 - 6x^2\) is \(f'(x) = -12x\), which is defined for all \(x\).

Step 3 ：The critical points are the points where the derivative of the function is zero. So, we solve \(f'(x) = 0\), which gives us \(x = 0\).

Step 4 ：We also need to evaluate the function at the endpoints of the interval, which are \(x = -4\) and \(x = 2\).

Step 5 ：By substituting these values into the function, we get the values \(2\), \(-94\), and \(-22\).

Step 6 ：The maximum value of these is \(2\) and it occurs at \(x=0\).

Step 7 ：The minimum value of these is \(-94\) and it occurs at \(x=-4\).

Step 8 ：Final Answer: The absolute maximum value of the function is \(\boxed{2}\) and this occurs at \(x=\boxed{0}\). The absolute minimum value of the function is \(\boxed{-94}\) and this occurs at \(x=\boxed{-4}\).