Question

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a ( f(x) = \frac{x + 2}{x - 2} )
b ( f(x) = \frac{x - 2}{x + 2} )
c ( f(x) = \frac{3x - 6}{x^2 + 2x} )
d ( f(x) = \frac{3x - 6}{x + 2} )
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Explanation:

Step1: Analyze Vertical Asymptotes

For a rational function \( f(x)=\frac{N(x)}{D(x)} \), vertical asymptotes occur where \( D(x) = 0 \) (and \( N(x)
eq0 \) at those points). From the graph (implied vertical asymptotes, likely at \( x = 2 \) and \( x=- 2 \) or similar, but let's check each option:

• Option A: \( D(x)=x - 2 \), vertical asymptote at \( x = 2 \).
• Option B: \( D(x)=x + 2 \), vertical asymptote at \( x=-2 \).
• Option C: \( D(x)=x^{2}+2x=x(x + 2) \), vertical asymptotes at \( x = 0 \) and \( x=-2 \).
• Option D: \( D(x)=x + 2 \), vertical asymptote at \( x=-2 \), and \( N(x)=3x - 6=3(x - 2) \), no common factors with \( D(x) \). Wait, maybe we also check horizontal asymptotes or holes, but let's re - evaluate. Wait, maybe the graph has vertical asymptotes at \( x = 2 \) and \( x=-2 \)? No, let's check the denominators again. Wait, maybe the original graph (not fully shown) has vertical asymptotes that match the denominator roots. Wait, let's check the function in Option B: \( f(x)=\frac{x - 2}{x + 2} \). Denominator \( x+2 = 0\Rightarrow x=-2 \), numerator \( x - 2

eq0 \) when \( x=-2 \). Wait, maybe the graph has a vertical asymptote at \( x=-2 \) and a hole? No, let's check the first option again. Wait, maybe the key is to check for holes (common factors) or vertical asymptotes. Wait, Option C: \( f(x)=\frac{3x - 6}{x^{2}+2x}=\frac{3(x - 2)}{x(x + 2)} \), no common factors, vertical asymptotes at \( x = 0 \) and \( x=-2 \). Option D: \( f(x)=\frac{3x - 6}{x + 2}=\frac{3(x - 2)}{x + 2} \), denominator \( x + 2=0\Rightarrow x=-2 \), numerator \( 3(x - 2)
eq0 \) at \( x=-2 \). Wait, maybe the graph has a vertical asymptote at \( x=-2 \) and the function in Option B: \( f(x)=\frac{x - 2}{x + 2} \), let's check the behavior. Wait, maybe the correct approach is to see which function's vertical asymptote and numerator/denominator match. Wait, maybe the intended answer is Option B? Wait, no, let's re - check. Wait, maybe the graph has a vertical asymptote at \( x=-2 \) and the function in Option B: \( f(x)=\frac{x - 2}{x + 2} \) has vertical asymptote at \( x=-2 \), and when we check the x - intercept (where \( N(x)=0 \)), \( x - 2=0\Rightarrow x = 2 \), which is a valid x - intercept. Let's check other options:

• Option A: \( f(x)=\frac{x + 2}{x - 2} \), vertical asymptote at \( x = 2 \), x - intercept at \( x=-2 \).
• Option B: \( f(x)=\frac{x - 2}{x + 2} \), vertical asymptote at \( x=-2 \), x - intercept at \( x = 2 \).
• Option C: \( f(x)=\frac{3x - 6}{x^{2}+2x}=\frac{3(x - 2)}{x(x + 2)} \), x - intercept at \( x = 2 \), vertical asymptotes at \( x = 0 \) and \( x=-2 \).
• Option D: \( f(x)=\frac{3x - 6}{x + 2}=\frac{3(x - 2)}{x + 2} \), x - intercept at \( x = 2 \), vertical asymptote at \( x=-2 \).

Wait, maybe the graph has a vertical asymptote at \( x=-2 \) and the function in Option B is the correct one. Wait, maybe I made a mistake earlier. Let's assume that the vertical asymptote is at \( x=-2 \) and the x - intercept is at \( x = 2 \), which matches Option B: \( f(x)=\frac{x - 2}{x + 2} \) (x - intercept when \( x - 2=0\Rightarrow x = 2 \), vertical asymptote when \( x + 2=0\Rightarrow x=-2 \)).

Answer:

B. \( f(x)=\frac{x - 2}{x + 2} \)