Question

which function has the following end behavior?

• as ( x \to 2 ), ( y \to -1 )
• as ( x \to infty ), ( y \to infty )

hint: left endpoint occurs when the radicand is 0.
skill sheet: key features: domain/range and end behavior, p. 2

a ( y = sqrt{x - 2} - 1 )
b ( y = -sqrt{x - 2} - 1 )
c ( y = sqrt{2 - x} - 1 )
d ( y = sqrt{x + 2} + 1 )

Explanation:

Step1: Analyze the end behavior as \( x \to 2 \)

For a square - root function \( y = \sqrt{u}+k \) (or with a negative sign), when the radicand \( u = 0 \), we get the left - most (or starting) point. Let's check each option:

• Option A: \( y=\sqrt{x - 2}-1 \). When \( x = 2 \), the radicand \( x - 2=0 \), then \( y=\sqrt{0}-1=- 1 \), so as \( x\to2 \), \( y\to - 1 \).
• Option B: \( y =-\sqrt{x - 2}-1 \). When \( x = 2 \), \( y=-\sqrt{0}-1=-1 \), but let's check the behavior as \( x\to\infty \). The square - root function \( \sqrt{x - 2} \) as \( x\to\infty \), \( \sqrt{x - 2}\to\infty \), so \( y=-\sqrt{x - 2}-1\to-\infty \), which does not match the given end - behavior \( y\to\infty \) as \( x\to\infty \).
• Option C: \( y=\sqrt{2 - x}-1 \). When \( x = 2 \), \( y=\sqrt{0}-1=-1 \), but as \( x\to\infty \), the radicand \( 2 - x\to-\infty \), and the domain of \( y=\sqrt{2 - x}-1 \) is \( 2 - x\geq0\) or \( x\leq2 \). So as \( x\to\infty \), the function is not defined (since \( x>2 \) is outside the domain), so it does not match the end - behavior \( x\to\infty \), \( y\to\infty \).
• Option D: \( y=\sqrt{x + 2}+1 \). When \( x\to2 \), \( y=\sqrt{2 + 2}+1=\sqrt{4}+1 = 2 + 1=3

eq - 1 \), so it does not match the behavior as \( x\to2 \), \( y\to - 1 \).

Step2: Confirm the correct option

For option A, the domain of \( y=\sqrt{x - 2}-1 \) is \( x\geq2 \) (since \( x - 2\geq0\)). As \( x\to\infty \), \( \sqrt{x - 2}\to\infty \), so \( y=\sqrt{x - 2}-1\to\infty \). And as \( x\to2 \), \( \sqrt{x - 2}\to0 \), so \( y = 0-1=-1 \), which matches the given end - behaviors.

Answer:

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