Question

which of the following radical functions has a domain of $x \geq -2$ and a range of $y \leq 3$?\
a. $h(x) = -\sqrt{x - 2} + 3$\
b. $h(x) = \sqrt{x - 2} + 3$\
c. $h(x) = -\sqrt{x + 2} + 3$\
d. $h(x) = \sqrt{x + 2} - 3$

Explanation:

Step1: Analyze the domain of radical functions

For a square - root function \(y = \sqrt{u}\), the domain requires \(u\geq0\).

• For option A: \(h(x)=-\sqrt{x - 2}+3\), set \(x - 2\geq0\), then \(x\geq2\).
• For option B: \(h(x)=\sqrt{x - 2}+3\), set \(x - 2\geq0\), then \(x\geq2\).
• For option C: \(h(x)=-\sqrt{x + 2}+3\), set \(x + 2\geq0\), then \(x\geq - 2\).
• For option D: \(h(x)=\sqrt{x + 2}-3\), set \(x + 2\geq0\), then \(x\geq - 2\).

Step2: Analyze the range of radical functions

The range of \(y=\sqrt{u}\) is \(y\geq0\).

• For option A: \(y =-\sqrt{x - 2}+3\), since \(\sqrt{x - 2}\geq0\), then \(-\sqrt{x - 2}\leq0\), and \(y=-\sqrt{x - 2}+3\leq3\). But we already know from step 1 that its domain is \(x\geq2\), not \(x\geq - 2\), so A is incorrect.
• For option B: \(y=\sqrt{x - 2}+3\), since \(\sqrt{x - 2}\geq0\), then \(y=\sqrt{x - 2}+3\geq3\). And its domain is \(x\geq2\), so B is incorrect.
• For option C: \(y =-\sqrt{x + 2}+3\), since \(\sqrt{x + 2}\geq0\), then \(-\sqrt{x + 2}\leq0\), and \(y=-\sqrt{x + 2}+3\leq3\). And from step 1, its domain is \(x\geq - 2\), which matches the required domain \(x\geq - 2\) and range \(y\leq3\).
• For option D: \(y=\sqrt{x + 2}-3\), since \(\sqrt{x + 2}\geq0\), then \(y=\sqrt{x + 2}-3\geq - 3\), so the range is \(y\geq - 3\), not \(y\leq3\), so D is incorrect.

Answer:

C. \(h(x)=-\sqrt{x + 2}+3\)