Question

c)

o $f(x) = -5 + \log_{2} x$

o $f(x) = \log_{2} (x - 4)$

o $f(x) = -\log_{2} (x + 5)$

o $f(x) = 3 + \log_{2} x$

Explanation:

Step1: Identify Vertical Asymptote

The graph has a vertical asymptote at \( x = 4 \). For a logarithmic function \( \log_b(x - h) \), the vertical asymptote is \( x = h \). So we check which function has vertical asymptote \( x = 4 \).

Step2: Analyze Each Option

• Option 1: \( f(x)=-5+\log_2 x \) has vertical asymptote \( x = 0 \). Eliminate.
• Option 2: \( f(x)=\log_2(x - 4) \) has vertical asymptote \( x = 4 \). Check the graph's behavior: as \( x \to 4^+ \), \( \log_2(x - 4) \to -\infty \), and as \( x \) increases, \( \log_2(x - 4) \) increases, matching the graph.
• Option 3: \( f(x)=-\log_2(x + 5) \) has vertical asymptote \( x = -5 \). Eliminate.
• Option 4: \( f(x)=3+\log_2 x \) has vertical asymptote \( x = 0 \). Eliminate.

Answer:

\( \boldsymbol{f(x) = \log_2 (x - 4)} \) (corresponding to the option " \( f(x) = \log_2 (x - 4) \) ")