Find the absolute maximum and minimum values of the function over the indicated interval, and indicate the $\mathrm{x}$-values at which they occur.
$$f(x)=x^2-8x-4;[0,5]$$
The absolute maximum value is at $x=$ (Use a comma to separate answers as needed.)

Step 1 ：The function given is $f(x)=x^2-8x-4$ and we are asked to find the absolute maximum value over the interval [0,5].

Step 2 ：The maximum or minimum value of a function can occur either at the endpoints of the interval or at the critical points within the interval.

Step 3 ：The critical points are the points where the derivative of the function is zero or undefined. In this case, the derivative of the function is always defined, so we only need to find where it is zero.

Step 4 ：The derivative of the function $f(x)$ is $f^{\prime }(x)=2x-8$. Setting this equal to zero gives us the critical point $x=4$.

Step 5 ：We evaluate the function at the critical point and the endpoints of the interval to find the maximum and minimum values. The values are $(4,-20)$, $(0,-4)$, and $(5,-19)$.

Step 6 ：Comparing these values, we find that the maximum value of the function over the interval [0,5] is -4, which occurs at $x=0$. The minimum value is -20, which occurs at $x=4$.

Step 7 ：However, the question only asks for the maximum value and the x-value at which it occurs. So, the absolute maximum value of the function over the interval [0,5] is at $x=0$.

Step 8 ：$\boxed{{\displaystyle x=0}}$ is the final answer.