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