Question

a)

graph of a logarithmic function on a coordinate plane with x-axis and y-axis, and four function options below:

• $f(x) = 3 + \log_{2} x$
• $f(x) = \log_{2} (x - 4)$
• $f(x) = -\log_{2} (x + 5)$
• $f(x) = -5 + \log_{2} x$

Explanation:

Step1: Analyze the domain and vertical shift

The graph has a vertical asymptote near \(x = 0\) (since it approaches the y - axis from the right), so the argument of the log function should be \(x\) (domain \(x>0\)). Now check the vertical shift. When \(x = 1\), \(\log_2(1)=0\). From the graph, when \(x = 1\), let's see the y - value. The function \(f(x)=- 5+\log_2x\) when \(x = 1\), \(f(1)=-5 + 0=-5\)? Wait, no, wait the graph: wait, maybe I made a mistake. Wait, let's check the options.

First, for \(f(x)=-5+\log_2x\): domain \(x>0\), vertical asymptote \(x = 0\). When \(x = 1\), \(f(1)=-5+\log_2(1)=-5 + 0=-5\). When \(x = 2\), \(f(2)=-5+\log_2(2)=-5 + 1=-4\). When \(x = 4\), \(f(4)=-5+\log_2(4)=-5 + 2=-3\). Let's check the graph: at \(x = 2\), the y - value is around - 4, at \(x = 4\), around - 3, and as \(x\) increases, it approaches a higher value (less negative), which matches the graph.

Check other options:

• \(f(x)=3+\log_2x\): when \(x = 1\), \(f(1)=3\), which is positive, but the graph is negative, so eliminate.
• \(f(x)=\log_2(x - 4)\): domain \(x>4\), but the graph is near \(x = 0\), so eliminate.
• \(f(x)=-\log_2(x + 5)\): domain \(x>-5\), and when \(x = 0\), \(f(0)=-\log_2(5)\approx - 2.3\), but the graph at \(x = 0\) is more negative, and the shape is increasing, but this function is decreasing (because of the negative sign), so eliminate.
• \(f(x)=-5+\log_2x\): domain \(x>0\), when \(x = 1\), \(f(1)=-5\), \(x = 2\), \(f(2)=-4\), \(x = 4\), \(f(4)=-3\), which matches the increasing, negative - valued graph with vertical asymptote \(x = 0\).

Step2: Confirm the function

By analyzing the domain (vertical asymptote at \(x = 0\), so argument is \(x\)), the vertical shift (the y - intercept behavior and the values at \(x = 1,2,4\)), the function \(f(x)=-5+\log_2x\) matches the graph.

Answer:

\(f(x)=-5+\log_2x\) (the option: \(\boldsymbol{f(x)=-5+\log_2x}\))