Question

matk311_mathematics iii semester a_2026-080220207-integ math 3a
polynomials, rational, and radical relationships: part three
which function has a domain of ( x geq 5 ) and a range of ( y leq 3 )?
( y = -sqrt{x + 5} - 3 )
( y = sqrt{x + 5} - 3 )
( y = -sqrt{x - 5} + 3 )
( y = sqrt{x - 5} + 3 )

Explanation:

Step1: Analyze Domain

For a square - root function \(y = \sqrt{u}\) (or \(y=-\sqrt{u}\)), the domain requires \(u\geq0\). We want the domain \(x\geq5\). So we need to find the function where the expression inside the square root is \(x - 5\) (because when \(x-5\geq0\), \(x\geq5\)). Let's check each function:

• For \(y =-\sqrt{x + 5}-3\), the domain is \(x+5\geq0\Rightarrow x\geq - 5\), not \(x\geq5\).
• For \(y=\sqrt{x + 5}-3\), the domain is \(x + 5\geq0\Rightarrow x\geq - 5\), not \(x\geq5\).
• For \(y=-\sqrt{x - 5}+3\), the domain is \(x - 5\geq0\Rightarrow x\geq5\).
• For \(y=\sqrt{x - 5}+3\), the domain is \(x - 5\geq0\Rightarrow x\geq5\).

Step2: Analyze Range

Now, let's analyze the range. The range of \(y = \sqrt{u}\) is \(y\geq0\), and the range of \(y=-\sqrt{u}\) is \(y\leq0\).

• For \(y=-\sqrt{x - 5}+3\): Let \(u=x - 5\), \(y =-\sqrt{u}+3\). Since \(\sqrt{u}\geq0\), then \(-\sqrt{u}\leq0\), and \(y=-\sqrt{u}+3\leq3\). So the range is \(y\leq3\).
• For \(y=\sqrt{x - 5}+3\): Since \(\sqrt{u}\geq0\), \(y=\sqrt{u}+3\geq3\), the range is \(y\geq3\), which does not match the required range \(y\leq3\).

Answer:

\(y =-\sqrt{x - 5}+3\)