Answer
1) Domain: $(-\infty,8]$
2) Range: $[0,\infty )$
Work Step by Step
Given \begin{equation}
f{(x)}=\sqrt{-x+8}.
\end{equation} a) This is an even root function because the index, $n=2$, is even. The radicand must be positive. So, we require $$-x+8\geq 0\implies x\leq 8.$$ 1) The domain is $(-\infty,8]$.
2) The range is $[0,\infty )$.
See the graph for proof.
