Question

what is the equation of the function graphed? \\(\bigcirc\\) equation a \\(\bigcirc\\) equation b \\(\bigcirc\\) equation c \\(\bigcirc\\) equation d \\(\text{a. } g(x)=\frac{1}{2x}-4\\) \\(\text{b. } g(x)=\frac{2}{x - 4}\\) \\(\text{c. } g(x)=\frac{1}{2x}+4\\) \\(\text{d. } g(x)=\frac{2}{x + 4}\\)

Explanation:

Step1: Identify Vertical Asymptote

The vertical asymptote of a rational function \( G(x)=\frac{f(x)}{h(x)} \) occurs where \( h(x) = 0 \). From the graph, the vertical asymptote is at \( x = 4 \) (since the graph has a break at \( x = 4 \)). So we need to find which function has a denominator that is zero when \( x = 4 \).

For option A: \( G(x)=\frac{1}{2x}-4 \) (rewriting, but the denominator is \( 2x \), vertical asymptote at \( x = 0 \), not 4.

For option B: \( G(x)=\frac{2}{x - 4} \). Denominator \( x - 4 = 0 \) when \( x = 4 \), which matches the vertical asymptote.

For option C: \( G(x)=\frac{1}{2x}+4 \), denominator \( 2x \), vertical asymptote at \( x = 0 \).

For option D: \( G(x)=\frac{2}{x + 4} \), denominator \( x + 4 = 0 \) when \( x=-4 \), not 4.

Also, we can check the horizontal behavior or a point. Let's check the sign. For \( x>4 \), the function \( G(x)=\frac{2}{x - 4} \) should be positive (since numerator 2 is positive and denominator \( x - 4 \) is positive when \( x>4 \)), which matches the graph (the right part of the asymptote is in positive y-region). For \( x<4 \), denominator \( x - 4 \) is negative, so \( G(x)=\frac{2}{x - 4} \) is negative, which also matches the left part (below x-axis).

Step2: Confirm with Analysis

We analyzed the vertical asymptote and the sign of the function for different intervals of \( x \) relative to the asymptote. Option B's function \( G(x)=\frac{2}{x - 4} \) has vertical asymptote at \( x = 4 \) and the correct sign behavior as per the graph.

Answer:

B. \( G(x)=\frac{2}{x - 4} \)