Answer
1) Domain: $[-12, \infty)$
2) Range: $[0,\infty )$
Work Step by Step
Given \begin{equation}
f(x)=\sqrt{x+12}.
\end{equation} a)This is an even root function because the index, $n=2$, is even. The radicand must be positive. So, we require $$x+12\geq 0\implies x\geq -12.$$ 1) The domain is $[-12, \infty)$.
2) The range is $[0,\infty )$.
See the graph for proof.
