which function has the following end behavior?\n- as ( x \to 2 ), ( y \to -1 )\n- as ( x \to infty ), ( y…

which function has the following end behavior?\n- as ( x \to 2 ), ( y \to -1 )\n- as ( x \to infty ), ( y \to infty )\nhint: left endpoint occurs when the radicand is 0.\nskill sheet: key features: domain/range and end behavior, p. 2\n\na ( y = sqrt{x - 2} - 1 )\nb ( y = -sqrt{x - 2} - 1 )\nc ( y = sqrt{2 - x} - 1 )\nd ( y = sqrt{x + 2} + 1 )

which function has the following end behavior?\n- as ( x \to 2 ), ( y \to -1 )\n- as ( x \to infty ), ( y \to infty )\nhint: left endpoint occurs when the radicand is 0.\nskill sheet: key features: domain/range and end behavior, p. 2\n\na ( y = sqrt{x - 2} - 1 )\nb ( y = -sqrt{x - 2} - 1 )\nc ( y = sqrt{2 - x} - 1 )\nd ( y = sqrt{x + 2} + 1 )

Answer

Brief Explanations:

  1. Analyze the end - behavior as (x\rightarrow2):
    • For option A: (y = \sqrt{x - 2}-1). When (x\rightarrow2), (\sqrt{x - 2}\rightarrow0), so (y\rightarrow0 - 1=-1). When (x\rightarrow\infty), (\sqrt{x - 2}\rightarrow\infty), so (y=\sqrt{x - 2}-1\rightarrow\infty).
    • For option B: (y=-\sqrt{x - 2}-1). When (x\rightarrow\infty), (\sqrt{x - 2}\rightarrow\infty), then (y =-\sqrt{x - 2}-1\rightarrow-\infty), which does not match the given end - behavior ((y\rightarrow\infty) as (x\rightarrow\infty)).
    • For option C: (y=\sqrt{2 - x}-1). The domain of this function is (2 - x\geq0) or (x\leq2). As (x\rightarrow\infty), the function is not defined (since (x>2) is not in the domain), so it does not match the end - behavior ((x\rightarrow\infty) is part of the end - behavior we need to consider).
    • For option D: (y=\sqrt{x + 2}+1). When (x\rightarrow2), (y=\sqrt{2 + 2}+1=\sqrt{4}+1 = 2 + 1=3\neq - 1), so it does not match the end - behavior as (x\rightarrow2).

Answer:

A. (y=\sqrt{x - 2}-1)