which function has a domain of $xgeq5$ and a range of $yleq3$?\n$y = sqrt{x - 5}+3$\n$y=sqrt{x +…

Explanation:

Step1: Recall domain - condition for square - root functions

For a square - root function of the form $y = \sqrt{u}$, the expression inside the square - root $u\geq0$.

Step2: Analyze the domain of each option

For $y=\sqrt{x - 5}+3$, we set $x−5\geq0$, which gives $x\geq5$. For $y=\sqrt{x + 5}-3$, we set $x + 5\geq0$, which gives $x\geq - 5$. For $y=-\sqrt{x - 5}+3$, we set $x−5\geq0$, which gives $x\geq5$. For $y=-\sqrt{x + 5}-3$, we set $x + 5\geq0$, which gives $x\geq - 5$. So, options with domain $x\geq5$ are $y=\sqrt{x - 5}+3$ and $y=-\sqrt{x - 5}+3$.

Step3: Recall range - condition for square - root functions

The square - root function $\sqrt{u}\geq0$. For $y=\sqrt{x - 5}+3$, since $\sqrt{x - 5}\geq0$, then $y=\sqrt{x - 5}+3\geq3$. For $y=-\sqrt{x - 5}+3$, since $\sqrt{x - 5}\geq0$, then $-\sqrt{x - 5}\leq0$, and $y=-\sqrt{x - 5}+3\leq3$.

Answer:

$y =-\sqrt{x - 5}+3$