Question

which graph represents the function $f(x)=\frac{3x - 2}{x - 2}$?

Explanation:

Step1: Analyze Vertical Asymptote

The function is \( f(x)=\frac{3x - 2}{x - 2} \). The vertical asymptote occurs where the denominator is zero, so \( x - 2=0\Rightarrow x = 2 \).

Step2: Analyze Horizontal Asymptote

For rational functions \( \frac{ax + b}{cx + d} \), the horizontal asymptote is \( y=\frac{a}{c} \) when the degrees of numerator and denominator are equal. Here, \( a = 3 \), \( c = 1 \), so horizontal asymptote is \( y = 3 \)? Wait, no, wait: Wait, the function can be rewritten by long division or manipulation: \( f(x)=\frac{3x-6 + 4}{x - 2}=\frac{3(x - 2)+4}{x - 2}=3+\frac{4}{x - 2} \). So the horizontal asymptote is \( y = 3 \)? Wait, no, when we rewrite it as \( 3+\frac{4}{x - 2} \), the horizontal asymptote is \( y=3 \) (since \( \frac{4}{x - 2} \) approaches 0 as \( x\to\pm\infty \)). Wait, but looking at the graph, the upper part of the hyperbola should approach \( y = 3 \)? Wait, maybe I made a mistake. Wait, original function: \( f(x)=\frac{3x-2}{x - 2} \). Let's check the horizontal asymptote: degree of numerator (1) equals degree of denominator (1), so horizontal asymptote is \( y=\frac{3}{1}=3 \). Now, check the vertical asymptote at \( x = 2 \). Also, check the y - intercept: when \( x = 0 \), \( f(0)=\frac{-2}{-2}=1 \). Wait, but in the given graph, let's see: the graph has a vertical asymptote at \( x = 2 \) (since the graph splits at \( x = 2 \)). The upper branch: as \( x\to2^+ \), \( f(x)=\frac{3x - 2}{x - 2} \), numerator at \( x = 2 \) is \( 3(2)-2 = 4 \), denominator approaches \( 0^+ \), so \( f(x)\to+\infty \). As \( x\to+\infty \), \( f(x)\to3 \) (since \( \frac{3x-2}{x - 2}\approx\frac{3x}{x}=3 \)). The lower branch: as \( x\to2^- \), denominator approaches \( 0^- \), numerator is 4, so \( f(x)\to-\infty \). As \( x\to-\infty \), \( f(x)\to3 \). Also, y - intercept: when \( x = 0 \), \( f(0)=1 \), so the graph should cross the y - axis at (0,1). Wait, but in the given graph, the lower branch at \( x = 0 \) is at \( y=-1 \)? Wait, maybe I miscalculated. Wait, \( f(0)=\frac{3(0)-2}{0 - 2}=\frac{-2}{-2}=1 \). So the y - intercept is (0,1). Now, looking at the graph: the upper part is above \( x = 2 \), going from \( +\infty \) (near \( x = 2^+ \)) to 3 as \( x\to+\infty \), and the lower part is below \( x = 2 \), going from \( -\infty \) (near \( x = 2^- \)) to 3 as \( x\to-\infty \), and passes through (0,1). Wait, the given graph (the one shown) has a vertical asymptote at \( x = 2 \), and the upper branch approaches a horizontal line (maybe \( y = 3 \))? Wait, maybe the graph shown is the correct one because it has vertical asymptote at \( x = 2 \), and the behavior around \( x = 2 \) and the horizontal asymptote (approaching \( y = 3 \) as \( x\to\pm\infty \)) matches the function \( f(x)=\frac{3x - 2}{x - 2} \) (after rewriting as \( 3+\frac{4}{x - 2} \), the horizontal asymptote is \( y = 3 \), and the vertical asymptote at \( x = 2 \)). So the graph with vertical asymptote at \( x = 2 \), horizontal asymptote at \( y = 3 \), and passing through (0,1) (wait, maybe my y - intercept calculation was wrong? Wait, no: \( f(0)=\frac{3(0)-2}{0 - 2}=\frac{-2}{-2}=1 \), so (0,1) is on the graph. The given graph: the lower branch at \( x = 0 \) is at \( y=-1 \)? Wait, maybe I messed up. Wait, no, let's re - express the function: \( f(x)=\frac{3x - 2}{x - 2}=3+\frac{4}{x - 2} \). So when \( x = 0 \), \( f(0)=3+\frac{4}{-2}=3 - 2 = 1 \), correct. So the graph should have a point (0,1). Now, looking at the graph, the upper branch is above \( x = 2 \), and as \( x\) increases, it approaches \( y = 3 \), and the lower branch is below \( x = 2 \), as \( x\) decreases, it approaches \( y = 3 \). The vertical asymptote is at \( x = 2 \). So the graph shown (the one with vertical asymptote at \( x = 2 \), two branches, one above \( x = 2 \) and one below \( x = 2 \), approaching \( y = 3 \) as \( x\to\pm\infty \)) is the correct graph for \( f(x)=\frac{3x - 2}{x - 2} \).

Answer:

The graph shown (the one with vertical asymptote at \( x = 2 \), two hyperbolic branches, one above \( x = 2 \) and one below \( x = 2 \), approaching \( y = 3 \) as \( x\to\pm\infty \))