Question

what are the domain and range of the function $f(x)=\sqrt{x - 7}+9$?
\bigcirc domain: $x\geq - 7$
\quad\\; range: $y\geq 9$
\bigcirc domain: $x\geq 7$
\quad\\; range: $y\geq - 9$
\bigcirc domain: $x\geq 7$
\quad\\; range: $y\geq 9$
\bigcirc domain: $x\geq 9$
\quad\\; range: $y\geq 7$

Explanation:

Step1: Find the domain

For the square root function $\sqrt{x - 7}$, the expression inside the square root (the radicand) must be non - negative. So we set up the inequality $x - 7\geq0$.
Solving for $x$, we add 7 to both sides of the inequality: $x\geq7$. So the domain of the function $f(x)=\sqrt{x - 7}+9$ is $x\geq7$.

Step2: Find the range

The square root function $\sqrt{x - 7}$ has a range of $y\geq0$ (because the square root of a non - negative number is always non - negative).
For the function $f(x)=\sqrt{x - 7}+9$, we are adding 9 to the square root part. If $\sqrt{x - 7}\geq0$, then $\sqrt{x - 7}+9\geq0 + 9$, which simplifies to $y\geq9$. So the range of the function is $y\geq9$.

Answer:

domain: $x\geq7$
range: $y\geq9$ (corresponding to the option "domain: $x\geq7$; range: $y\geq9$")