what is the domain and range for the function $f(x)=\\sqrt{x + 2}-3$?

Answer:

Domain: $x\geq - 2$, Range: $y\geq - 3$

Explanation:

Step1: Find the domain

For the square root function $\sqrt{x + 2}$, the expression inside the square root must be non - negative. $$x+2\geq0$$ $$x\geq - 2$$

Step2: Find the range

The square root function $\sqrt{x + 2}$ has a minimum value of $0$ (when $x=-2$). Since $y = \sqrt{x + 2}-3$, when $\sqrt{x + 2}=0$, $y=-3$. As $\sqrt{x + 2}$ increases (as $x$ increases beyond $-2$), $y=\sqrt{x + 2}-3$ also increases. So the range is $y\geq - 3$.