Question

unit test
unit test complete
which represents the reflection of f(x) = √x over the y-axis?

x	f(x)
-1	undefined
0	0
1	1
4	2
x	f(x)
-1	1
0	0
1	-1
4	-2

mark and return
save and exit

Explanation:

Step1: Recall reflection rule over y-axis

Reflection over y-axis: replace $x$ with $-x$, so $g(x) = f(-x) = \sqrt{-x}$.

Step2: Check domain of $g(x)$

$\sqrt{-x}$ requires $-x \geq 0 \Rightarrow x \leq 0$. For $x=-1$: $\sqrt{-(-1)}=1$; $x=0$: $\sqrt{0}=0$; $x=1$: $\sqrt{-1}$ undefined (but second table shows -1, which is likely a typo in display, but first table has undefined for $x=-1$ which is wrong for reflection; second table uses negative x values with valid outputs matching $\sqrt{-x}$).

Step3: Compare tables

First table has undefined at $x=-1$ (wrong for reflection). Second table uses $x=-1,0$ (valid domain) with $g(-1)=1, g(0)=0$, matching reflection.

Wait, correction on Step2: The second table's positive x values (1,4) have negative outputs, which might be a misrepresentation, but the key is reflection over y-axis swaps x sign. The original function $f(x)=\sqrt{x}$ has domain $x\geq0$. Its reflection over y-axis has domain $x\leq0$, so only x values ≤0 are valid. The first table has undefined at x=-1 (correct for original, wrong for reflection). The second table has x=-1 with f(x)=1 (correct: $\sqrt{-(-1)}=1$), x=0 with 0 (correct). The positive x values in the second table are likely errors, but among the two options, the second table is the only one with valid reflection values at x=-1,0. So the answer is the second table.

Answer:

The second table (with x values -1, 0, 1, 4 and f(x) values 1, 0, -1, -2)