Question

indicate which of the following four graphs is the correct graph of this function.
$q(x) = \log_{\frac{1}{4}} (x - 2)$
answer

Explanation:

Step1: Analyze the domain

The function is \( q(x)=\log_{\frac{1}{4}}(x - 2) \). The argument of a logarithm must be positive, so \( x-2>0\implies x > 2 \). So the vertical asymptote is \( x = 2 \), and the graph is defined for \( x>2 \).

Step2: Analyze the behavior of the logarithm

For a logarithmic function \( y=\log_{a}(x - h) \), when \( 0 < a<1 \), the function is decreasing. Here \( a=\frac{1}{4} \) (which is between 0 and 1), so \( q(x) \) is a decreasing function. Also, when \( x - 2=1\) (i.e., \( x = 3 \)), \( q(3)=\log_{\frac{1}{4}}(1) = 0 \), so the graph passes through \( (3,0) \).

Now, check the graphs:
- The first graph (top - left): Vertical asymptote around \( x = 3 \)? No, and the domain seems to start around \( x=3 \) but the function is decreasing? Wait, no, let's check the asymptote. The correct asymptote is \( x = 2 \). Wait, maybe the grids: Let's see the x - axis. The first graph (top - left) has the curve starting near \( x = 3 \), but let's check the direction. Wait, the function \( \log_{\frac{1}{4}}(x - 2) \) is decreasing (since base \( \frac{1}{4}<1 \)). So as \( x \) increases (moves to the right), \( y=\log_{\frac{1}{4}}(x - 2) \) decreases (moves down).

Wait, the top - left graph: when \( x \) increases (moves right), the \( y \) - value decreases (goes down), and the vertical asymptote is at \( x = 2 \) (since \( x-2>0\implies x > 2 \)). Let's check the x - intercept: when \( y = 0 \), \( \log_{\frac{1}{4}}(x - 2)=0\implies x - 2 = 1\implies x=3 \). So the graph should pass through \( (3,0) \).

Looking at the four graphs:
- Top - left: The curve is decreasing (as \( x \) increases, \( y \) decreases), has vertical asymptote at \( x = 2 \) (since the curve approaches a vertical line at \( x = 2 \) from the right), and passes through \( (3,0) \) (since at \( x = 3 \), \( y = 0 \)).
- Top - right: The curve is increasing (as \( x \) increases, \( y \) increases), which is wrong because the base is less than 1, so the log function should be decreasing.
- Bottom - left: The domain is \( x>-4 \) (since the curve is near \( x=-4 \)), which is wrong because the domain should be \( x > 2 \).
- Bottom - right: The curve is increasing (as \( x \) increases, \( y \) increases), and the asymptote is at \( x = 2 \), but the function is increasing, which is wrong.

Wait, maybe I made a mistake. Wait, the function \( \log_{a}(x - h) \): if \( a>1 \), it's increasing; if \( 0 < a<1 \), it's decreasing. So \( \log_{\frac{1}{4}}(x - 2) \) is decreasing. So as \( x \) increases, \( y \) decreases. So the top - left graph: when \( x \) moves to the right (increases), \( y \) moves down (decreases), has vertical asymptote at \( x = 2 \) (since the curve is approaching a vertical line at \( x = 2 \) from the right), and passes through \( (3,0) \). So the correct graph is the top - left graph (the first graph).

Wait, let's re - check:

The function \( q(x)=\log_{\frac{1}{4}}(x - 2) \):

1. Domain: \( x>2 \), so the graph exists only for \( x>2 \).
2. Vertical asymptote: \( x = 2 \).
3. Monotonicity: Since \( 0<\frac{1}{4}<1 \), \( q(x) \) is a decreasing function (as \( x \) increases, \( q(x) \) decreases).
4. X - intercept: \( q(x)=0\implies x - 2 = 1\implies x = 3 \), so the graph passes through \( (3,0) \).

Now, check each graph:

- Top - left graph:
- Domain: \( x>2 \) (the curve starts to the right of \( x = 2 \)).
- Vertical asymptote: At \( x = 2 \) (the curve approaches a vertical line at \( x = 2 \) from the right).
- Monotonicity: As \( x \) increases (moves right), \( y \) decreases (moves down) – correct, since the function is decreasing.
- X - intercept: Passes through \( (3,0) \) (since at \( x = 3 \), \( y = 0 \)) – correct.

- Top - right graph:
- Monotonicity: As \( x \) increases, \( y \) increases – wrong, since the function should be decreasing.

- Bottom - left graph:
- Domain: \( x>-4 \) – wrong, since the domain should be \( x>2 \).

- Bottom - right graph:
- Monotonicity: As \( x \) increases, \( y \) increases – wrong, and the vertical asymptote seems to be at \( x = 2 \), but the function is increasing, which is incorrect for \( \log_{\frac{1}{4}}(x - 2) \).

Answer:

The top - left graph (the first graph in the top - left corner)