Question

move the slider on the graph on the right to graph each function: for the function: $y = \sqrt{x - 7} + 1$ the domain is : $x \geq \square$ the range is: $y \geq \square$

Explanation:

Step1: Find the domain of the square root function

For the square root function \( y = \sqrt{u} \), the expression inside the square root (the radicand) must be non - negative, i.e., \( u\geq0 \). In the function \( y=\sqrt{x - 7}+1 \), the radicand is \( x - 7 \). So we set up the inequality \( x-7\geq0 \).
Solving for \( x \), we add 7 to both sides of the inequality: \( x\geq7 \).

Step2: Find the range of the square root function

The square root function \( y=\sqrt{x - 7} \) has a range of \( y\geq0 \) because the square root of a non - negative number is always non - negative. Now, for the function \( y=\sqrt{x - 7}+1 \), we are adding 1 to the square root function. If \( \sqrt{x - 7}\geq0 \), then when we add 1 to both sides of the inequality, we get \( \sqrt{x - 7}+1\geq0 + 1 \), which simplifies to \( y\geq1 \).

Answer:

The domain is \( x\geq\boxed{7} \) and the range is \( y\geq\boxed{1} \)