Question

move the graphs that represent f(x) and g(x) into the table
function\tgraph
f(x) = \log_{2}(x - 2)

g(x) = \log_{4}(x - 2)

four graphs are shown below the table, each with a coordinate plane and a logarithmic - like curve

Explanation:

Step1: Analyze \( f(x)=\log_2(x - 2) \)

The parent function is \( y = \log_2 x \), shifted right by 2 units (vertical asymptote at \( x = 2 \)). When \( x=3 \), \( f(3)=\log_2(1)=0 \); when \( x = 4 \), \( f(4)=\log_2(2)=1 \). The graph should pass through \( (3,0) \) and \( (4,1) \), with a steeper slope (since base 2 > 1, and smaller base? Wait, no: base 2 is larger than base \( \frac{1}{4} \), but for \( \log_b(x) \), larger \( b>1 \) means slower growth? Wait, no: \( \log_2 x \) grows faster than \( \log_4 x \) because \( \log_4 x=\frac{\log_2 x}{\log_2 4}=\frac{1}{2}\log_2 x \). Wait, but here \( f(x)=\log_2(x - 2) \), \( g(x)=\log_{\frac{1}{4}}(x - 2)=\frac{\log_2(x - 2)}{\log_2 \frac{1}{4}}=\frac{\log_2(x - 2)}{-2}=-\frac{1}{2}\log_2(x - 2) \). Wait, no: \( \log_{\frac{1}{4}}a=\frac{\ln a}{\ln \frac{1}{4}}=\frac{\ln a}{-2\ln 2}=-\frac{1}{2}\log_2 a \). So \( g(x)=-\frac{1}{2}\log_2(x - 2) \), which is a reflection over x-axis and vertical compression of \( \log_2(x - 2) \). Wait, maybe better to find key points.

For \( f(x)=\log_2(x - 2) \):
- Vertical asymptote: \( x = 2 \)
- When \( x - 2 = 1 \) (i.e., \( x = 3 \)), \( f(3)=0 \)
- When \( x - 2 = 2 \) (i.e., \( x = 4 \)), \( f(4)=1 \)
- When \( x - 2 = 4 \) (i.e., \( x = 6 \)), \( f(6)=2 \)

For \( g(x)=\log_{\frac{1}{4}}(x - 2) \):
- Vertical asymptote: \( x = 2 \)
- When \( x - 2 = 1 \) (i.e., \( x = 3 \)), \( g(3)=\log_{\frac{1}{4}}1 = 0 \)
- When \( x - 2 = \frac{1}{4} \)? No, better: \( \log_{\frac{1}{4}}a = y \) means \( (\frac{1}{4})^y = a \). So when \( a = 1 \), \( y = 0 \); when \( a = \frac{1}{4} \), \( y = 1 \) (since \( (\frac{1}{4})^1=\frac{1}{4} \)), so \( x - 2=\frac{1}{4} \Rightarrow x = 2.25 \), \( g(2.25)=1 \); when \( a = 4 \), \( (\frac{1}{4})^y = 4 \Rightarrow y = -1 \), so \( x - 2 = 4 \Rightarrow x = 6 \), \( g(6)=-1 \).

Now, looking at the graphs:
- The first two graphs (leftmost) have vertical asymptote at \( x = 2 \)? Wait, the x-axis labels: first graph: x=1,2,3,... Wait, maybe the asymptote is at \( x = 2 \) (since \( x - 2 > 0 \Rightarrow x > 2 \)). So the graphs should start at x > 2.

Wait, maybe the four graphs: let's check the key points.

For \( f(x)=\log_2(x - 2) \): increasing function (since base 2 > 1), passes through (3,0), (4,1), (6,2).

For \( g(x)=\log_{\frac{1}{4}}(x - 2) \): since base \( \frac{1}{4} < 1 \), it's a decreasing function, passes through (3,0), (2.25,1), (6,-1).

Now, looking at the four graphs:

First graph: starts at x=2 (asymptote), passes through (3,0), (4,1) – increasing, so this is \( f(x) \).

Second graph: starts at x=2, passes through (3,0), but maybe different slope? Wait, no, maybe the fourth graph? Wait, maybe the labels are:

Wait, the user's image: four graphs. Let's assume:

Graph 1 (leftmost): increasing, passes through (3,0), (4,1) – \( f(x) \)

Graph 4 (rightmost): decreasing, passes through (3,0), (6,-1) – \( g(x) \)

Wait, maybe the correct graphs are:

\( f(x)=\log_2(x - 2) \) corresponds to the graph that is increasing, with vertical asymptote at \( x = 2 \), passing through (3,0), (4,1) – let's say the second graph? Wait, maybe I made a mistake. Alternatively, recall that \( \log_b(x - h) \) has vertical asymptote at \( x = h \), and for \( b > 1 \), it's increasing; for \( 0 < b < 1 \), it's decreasing.

So \( f(x)=\log_2(x - 2) \): \( b = 2 > 1 \), increasing, asymptote \( x = 2 \).

\( g(x)=\log_{\frac{1}{4}}(x - 2) \): \( b = \frac{1}{4} < 1 \), decreasing, asymptote \( x = 2 \).

Now, looking at the four graphs:

- First graph: increasing, asymptote at x=2 (since x starts at 2), passes through (3,0) – this is \( f(x) \)

- Fourth graph: decreasing, asymptote at x=2, passes through (3,0) – this is \( g(x) \)

Wait, maybe the correct assignment is:

\( f(x)=\log_2(x - 2) \) → the graph that is increasing (second from left? Or first? Let's check the x-intercept: when x=3, y=0. So the graph with x-intercept at (3,0) and increasing is \( f(x) \), and the one with x-intercept at (3,0) and decreasing is \( g(x) \).

So, moving the increasing graph (with x-intercept (3,0), increasing) to \( f(x) \), and the decreasing graph (with x-intercept (3,0), decreasing) to \( g(x) \).

Step2: Assign Graphs

After analyzing the functions' behavior (increasing/decreasing, asymptote, key points), we determine:

- \( f(x)=\log_2(x - 2) \) (increasing, base > 1) corresponds to the graph that is increasing, with vertical asymptote at \( x = 2 \), passing through (3,0).

- \( g(x)=\log_{\frac{1}{4}}(x - 2) \) (decreasing, base < 1) corresponds to the graph that is decreasing, with vertical asymptote at \( x = 2 \), passing through (3,0).

Assuming the four graphs: let's say the second graph (from left) is \( f(x) \) and the fourth is \( g(x) \), but more accurately, based on the key points, the increasing graph (passing through (3,0), (4,1)) is \( f(x) \), and the decreasing graph (passing through (3,0), (6,-1)) is \( g(x) \).

Answer:

For \( f(x) = \log_2(x - 2) \), the graph is the increasing logarithmic graph with vertical asymptote \( x = 2 \), passing through \( (3, 0) \) and \( (4, 1) \) (e.g., the second graph from the left, depending on the exact plot). For \( g(x) = \log_{\frac{1}{4}}(x - 2) \), the graph is the decreasing logarithmic graph with vertical asymptote \( x = 2 \), passing through \( (3, 0) \) and \( (6, -1) \) (e.g., the fourth graph from the left, depending on the exact plot).

(Note: Since the exact graph labels are not fully clear, the key is to identify the increasing graph for \( f(x) \) (base > 1) and the decreasing graph for \( g(x) \) (base < 1), both with vertical asymptote \( x = 2 \) and passing through \( (3, 0) \).)