Determine the absolute minimum and maximum values of the function $f\left(x\right)=\sqrt{16-x^2}$ on $[-1,4]$ and identify where they occur.

The absolute minimum is $\sqrt{15}$ at $x=1$, and the absolute maximum is $4$ at $x=0$.

The absolute minimum is $\sqrt{15}$ at $x=-1$, and the absolute maximum is $4$ at $x=0$.

The absolute minimum is $0$ at $x=4$, and the absolute maximum is $4$ at $x=0$.

The absolute minimum is $0$ at $x=4$, and the absolute maximum is $\sqrt{15}$ at $x=-1$.