Question

indicate which of the following four graphs is the correct graph of this function.
$h(x) = \log_{\frac{1}{3}} (x - 4)$

Explanation:

Step1: Analyze the function's domain

The function is \( h(x)=\log_{\frac{1}{3}}(x - 4) \). For a logarithmic function \( \log_b(u) \), the argument \( u>0 \). So \( x - 4>0\implies x > 4 \). This means the graph should be defined for \( x>4 \), so we can eliminate graphs where the domain starts at \( x < 4 \) (like the first two graphs with vertical asymptote at \( x=-4 \) or \( x = 4 \) but increasing left of 4, and the third graph with vertical asymptote at \( x = 4 \) but increasing from below, wait no—wait, let's check the vertical asymptote. The vertical asymptote of \( \log_b(x - a) \) is \( x=a \). Here \( a = 4 \), so vertical asymptote at \( x = 4 \), and domain \( x>4 \). So the graph should be to the right of \( x = 4 \). Now, the base of the logarithm is \( \frac{1}{3} \), which is between 0 and 1, so the logarithmic function is decreasing (since for \( 0 < b<1 \), \( \log_b(u) \) is decreasing in \( u \)).

Step2: Check the shape and asymptote

- The vertical asymptote is \( x = 4 \), so the graph approaches \( x = 4 \) from the right (since domain is \( x>4 \)).
- Since the base \( \frac{1}{3}<1 \), the function is decreasing as \( x \) increases (because as \( x \) increases, \( x - 4 \) increases, and \( \log_{\frac{1}{3}}(u) \) decreases as \( u \) increases).

Now let's check the graphs:
- Top left: vertical asymptote at \( x=-4 \), domain \( x>-4 \), decreasing? But our domain is \( x>4 \), so no.
- Top right: vertical asymptote at \( x = 4 \), but the graph is to the right of \( x = 4 \), but is it decreasing? Wait, the top right graph: as \( x \) increases (to the right), the graph goes down? Wait no, the top right graph: the curve is on the right of \( x = 4 \), and as \( x \) increases (towards 8), the graph goes down? Wait, no, the top right graph's curve: when \( x \) is 4, it's near the asymptote, and as \( x \) increases (to 8), the graph goes down? Wait, no, let's look at the bottom graphs. Bottom left: vertical asymptote at \( x = 4 \), but the graph is increasing (since it goes up as \( x \) increases), but our function is decreasing, so no. Bottom right: vertical asymptote at \( x = 4 \) (approaches from the right), and as \( x \) increases (to the right), the graph decreases (goes down), which matches the decreasing nature of \( \log_{\frac{1}{3}}(x - 4) \) (since base \( <1 \), decreasing in \( x \)). Wait, no, wait: bottom right graph: when \( x \) is 5, \( x - 4 = 1 \), \( h(5)=\log_{\frac{1}{3}}(1)=0 \), so the graph should pass through \( (5, 0) \). Let's check the bottom right graph: does it pass through \( (5, 0) \)? The bottom right graph: at \( x = 5 \), which is to the right of 4, the \( y \)-value is 0? Let's see, the bottom right graph: when \( x \) is 5 (one unit right of 4), \( y = 0 \), and as \( x \) increases (to 8), \( y \) decreases (goes down), which is correct. Wait, but let's recheck the base: \( \log_{\frac{1}{3}}(u) \) is decreasing, so as \( u = x - 4 \) increases ( \( x \) increases), \( h(x) \) decreases. So the graph should be decreasing, with vertical asymptote \( x = 4 \), domain \( x>4 \), and passing through \( (5, 0) \) (since \( h(5)=\log_{\frac{1}{3}}(1)=0 \)).

Looking at the bottom right graph: it's to the right of \( x = 4 \), has vertical asymptote at \( x = 4 \), and as \( x \) increases, it decreases, which matches. The top right graph: the curve is on the right of \( x = 4 \), but does it pass through \( (5, 0) \)? Wait, the top right graph: when \( x = 5 \), is \( y = 0 \)? The top right graph's curve: at \( x = 5 \), it's near the asymptote? No, the top right graph's curve is on the right of \( x = 4 \), but the direction: if the base is \( \frac{1}{3} \), the function is decreasing, so as \( x \) increases, \( y \) decreases. The bottom right graph: as \( x \) increases (from 4 to 8), \( y \) decreases, which is correct. The bottom left graph is increasing (goes up as \( x \) increases), which is wrong because our function is decreasing. The top left has wrong asymptote. The top right: wait, maybe I mixed up. Wait, let's recast: the function \( h(x)=\log_{\frac{1}{3}}(x - 4) \) can be rewritten as \( h(x)=\frac{\ln(x - 4)}{\ln(\frac{1}{3})} \), and since \( \ln(\frac{1}{3})<0 \), it's equivalent to \( -\frac{\ln(x - 4)}{\ln(3)} \), so it's a reflection of \( \ln(x - 4) \) over the x-axis, hence decreasing. So the graph should be decreasing, to the right of \( x = 4 \), with vertical asymptote \( x = 4 \).

So the correct graph is the bottom right one (the fourth graph, bottom right), which has vertical asymptote at \( x = 4 \), is to the right of \( x = 4 \), and is decreasing as \( x \) increases.

Answer:

The bottom right graph (the fourth graph, located at the bottom right among the four graphs)