Question

the function $f(x)$ is defined as $f(x) = \frac{1}{3}(6)^x$. which table of values could be used to graph $g(x)$, a reflection of $f(x)$ across the $x$-axis?

$x$	$f(x)$	$g(x)$
$-1$	$\frac{1}{18}$	$2$
$0$	$\frac{1}{3}$	$\frac{1}{3}$
$1$	$2$	$\frac{1}{18}$
$2$	$12$	$\frac{1}{108}$

$x$	$f(x)$	$g(x)$
$-1$	$\frac{1}{18}$	$2$
$0$	$0$	$0$
$1$	$2$	$\frac{1}{18}$
$2$	$12$	$\frac{1}{108}$

Explanation:

Step1: Recall reflection rule

Reflecting a function \( f(x) \) across the \( x \)-axis gives \( g(x)=-f(x) \). So we need to check which table has \( g(x)=-f(x) \) for each \( x \).

Step2: Analyze \( f(0) \)

First, calculate \( f(0) \) using \( f(x)=\frac{1}{3}(6)^x \). Substitute \( x = 0 \): \( f(0)=\frac{1}{3}(6)^0=\frac{1}{3}(1)=\frac{1}{3} \). Then \( g(0)=-f(0)=-\frac{1}{3} \)? Wait, no, wait the first table: for \( x = 0 \), \( f(x)=\frac{1}{3} \), \( g(x)=\frac{1}{3} \)? No, wait the second table has \( f(0)=0 \), which is wrong. Wait, no, let's recalculate \( f(0) \) again. \( f(x)=\frac{1}{3}(6)^x \), so \( x = 0 \): \( 6^0 = 1 \), so \( f(0)=\frac{1}{3}(1)=\frac{1}{3} \). So \( g(0)=-f(0)=-\frac{1}{3} \)? But in the first table, \( g(0)=\frac{1}{3} \), second table \( g(0)=0 \). Wait, maybe I misread the tables. Wait the first table:

Wait the first table:

\( x=-2 \): \( f(x)=\frac{1}{108} \), \( g(x)=12 \). Let's check \( g(x)=-f(x) \)? \( -f(-2)=-\frac{1}{108} \), no. Wait, maybe reflection across y-axis? No, the problem says x-axis. Wait, maybe I made a mistake. Wait, let's recalculate \( f(x) \) for each x.

For \( x=-2 \): \( f(-2)=\frac{1}{3}(6)^{-2}=\frac{1}{3}\times\frac{1}{36}=\frac{1}{108} \). Then reflection over x-axis: \( g(-2)=-f(-2)=-\frac{1}{108} \), but the first table has \( g(-2)=12 \), second table \( g(-2)=12 \). Wait, that's not negative. Wait, maybe the tables are wrong, or maybe I misread the reflection. Wait, no, maybe the function is \( f(x)=\frac{1}{3}(6)^x \), so reflection over x-axis is \( g(x)=-\frac{1}{3}(6)^x \). Let's calculate \( g(x) \) for each x:

For \( x=-2 \): \( g(-2)=-\frac{1}{3}(6)^{-2}=-\frac{1}{3}\times\frac{1}{36}=-\frac{1}{108} \), but the tables have 12. Wait, 12 is \( -f(-2) \) multiplied by -108? No. Wait, maybe the reflection is over y-axis? No, the problem says x-axis. Wait, maybe the tables are actually for \( g(x)=-f(x) \), but let's check \( x = 1 \): \( f(1)=\frac{1}{3}(6)^1=\frac{1}{3}\times6 = 2 \). Then \( g(1)=-f(1)=-2 \). But in the first table, \( g(1)=\frac{1}{18} \), second table \( g(1)=\frac{1}{18} \). Wait, this is confusing. Wait, maybe the tables are reversed? Wait, no, the problem says "a reflection of f(x) across the x-axis", so \( g(x) = -f(x) \). Let's check the first table:

For \( x=1 \): \( f(1)=2 \), \( g(1)=\frac{1}{18} \). \( -f(1)=-2 \), not \( \frac{1}{18} \). For \( x=2 \): \( f(2)=\frac{1}{3}(6)^2=\frac{1}{3}\times36 = 12 \), so \( g(2)=-12 \), but first table has \( g(2)=\frac{1}{108} \), second table (cut off) but first table's \( x=2 \) is \( \frac{1}{108} \). Wait, maybe the tables are for \( g(x) = \frac{1}{f(x)} \)? For \( x=-2 \), \( \frac{1}{f(-2)} = 108 \), no, first table has 12. Wait, \( f(-2)=\frac{1}{108} \), \( 12 = \frac{1}{108}\times1296 \), no. Wait, maybe the function is \( f(x)=\frac{1}{3}(6)^x \), so \( f(1)=\frac{1}{3}\times6 = 2 \), \( f(2)=\frac{1}{3}\times36 = 12 \), \( f(-1)=\frac{1}{3}\times\frac{1}{6}=\frac{1}{18} \), \( f(-2)=\frac{1}{3}\times\frac{1}{36}=\frac{1}{108} \). Now, if we take \( g(x) = -f(x) \), then:

\( x=-2 \): \( g(-2)=-\frac{1}{108} \) (not 12 or 12)

\( x=-1 \): \( g(-1)=-\frac{1}{18} \) (not 2)

\( x=0 \): \( g(0)=-\frac{1}{3} \) (not \( \frac{1}{3} \) or 0)

\( x=1 \): \( g(1)=-2 \) (not \( \frac{1}{18} \))

\( x=2 \): \( g(2)=-12 \) (not \( \frac{1}{108} \))

Wait, this is confusing. Maybe the tables are actually for reflection over y-axis? \( g(x)=f(-x) \). Let's check:

For \( x=-2 \): \( f(2)=12 \), so \( g(-2)=f(2)=12 \). Yes! That matches the first table's \( g(-2)=12 \).

\( x=-1 \): \( f(1)=2 \), so \( g(-1)=f(1)=2 \). Matches first table's \( g(-1)=2 \).

\( x=0 \): \( f(0)=\frac{1}{3} \), so \( g(0)=f(0)=\frac{1}{3} \). Matches first table's \( g(0)=\frac{1}{3} \).

\( x=1 \): \( f(-1)=\frac{1}{18} \), so \( g(1)=f(-1)=\frac{1}{18} \). Matches first table's \( g(1)=\frac{1}{18} \).

\( x=2 \): \( f(-2)=\frac{1}{108} \), so \( g(2)=f(-2)=\frac{1}{108} \). Matches first table's \( g(2)=\frac{1}{108} \).

Ah! So maybe the problem had a typo, and it's reflection over y-axis, but the first table matches \( g(x)=f(-x) \) (reflection over y-axis). But the problem says x-axis. Wait, but according to the calculations, the first table's \( g(x) \) values are \( f(-x) \), which is reflection over y-axis. But the problem says x-axis. Wait, maybe I misread the problem. Let me check again: "a reflection of f(x) across the x-axis". But according to the table, the first table's \( g(x) \) is \( f(-x) \). But maybe the options are wrong, or maybe I made a mistake. Wait, let's check the second table: for \( x=0 \), \( f(0)=0 \), which is wrong because \( f(0)=\frac{1}{3} \). So the second table has an error in \( f(0) \). So the first table is correct for reflection over y-axis, but the problem says x-axis. But since the second table has a wrong \( f(0) \), the first table must be the answer, even though it's reflection over y-axis? No, wait, maybe the problem meant y-axis. But according to the calculations, the first table's \( g(x) \) is \( f(-x) \), which is reflection over y-axis. But the problem says x-axis. Wait, maybe the function is \( f(x)=\frac{1}{3}(6)^{-x} \)? Let's check: \( f(x)=\frac{1}{3}(6)^{-x} \), then \( f(-2)=\frac{1}{3}(6)^2=12 \), \( f(-1)=\frac{1}{3}(6)^1=2 \), \( f(0)=\frac{1}{3} \), \( f(1)=\frac{1}{3}(6)^{-1}=\frac{1}{18} \), \( f(2)=\frac{1}{3}(6)^{-2}=\frac{1}{108} \). Then reflection over x-axis: \( g(x)=-f(x) \), so \( g(-2)=-12 \), no. Wait, this is confusing. But the second table has \( f(0)=0 \), which is wrong, so the first table is the only one with correct \( f(x) \) values (except for the reflection type). So the first table must be the answer, even though the reflection is over y-axis, maybe a typo in the problem.

Answer:

The first table (the one with \( f(0)=\frac{1}{3} \) and \( g(0)=\frac{1}{3} \), \( f(1)=2 \), \( g(1)=\frac{1}{18} \), etc.)