Question

d)

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( \boldsymbol{f(x) = -log_2 (x + 5)} )

( \boldsymbol{f(x) = -5 + log_2 x} )

( \boldsymbol{f(x) = 3 + log_2 x} )

( \boldsymbol{f(x) = log_2 (x - 4)} )

Explanation:

Step1: Analyze the domain and vertical asymptote

The graph has a vertical asymptote at \( x = -5 \) (since the curve approaches \( x = -5 \) from the right). For a logarithmic function \( \log_b(u) \), the argument \( u \) must be positive, so the vertical asymptote occurs where \( u = 0 \). Let's check the arguments of each function:

• For \( f(x)=-\log_2(x + 5) \), the argument is \( x + 5 \), so vertical asymptote at \( x + 5 = 0 \implies x=-5 \), which matches the graph.
• For \( f(x)=-5+\log_2x \), the argument is \( x \), vertical asymptote at \( x = 0 \), which does not match.
• For \( f(x)=3+\log_2x \), argument is \( x \), vertical asymptote at \( x = 0 \), no match.
• For \( f(x)=\log_2(x - 4) \), argument is \( x - 4 \), vertical asymptote at \( x = 4 \), no match.

Step2: Analyze the sign and behavior

The graph is decreasing (since as \( x \) increases, \( y \) decreases) and the coefficient of the log is negative (since \( f(x)=-\log_2(x + 5) \) has a negative sign, which reflects the log function over the x - axis, resulting in a decreasing function), which matches the graph's behavior.

Answer:

\( f(x)=-\log_2(x + 5) \) (the first option)