Question

which graph represents the function $f(x)=\frac{2}{x - 1}+4$?

Explanation:

Step1: Identify the vertical asymptote

For the function \( f(x)=\frac{2}{x - 1}+4 \), the vertical asymptote occurs where the denominator is zero. So, \( x-1 = 0\) gives \( x = 1 \). This means the graph should have a vertical asymptote at \( x = 1 \).

Step2: Identify the horizontal asymptote

For a rational function of the form \( \frac{a}{x - h}+k \), the horizontal asymptote is \( y=k \). Here, \( k = 4 \), so the horizontal asymptote is \( y = 4 \).

Step3: Analyze the behavior around the asymptotes

• For \( x>1 \) (right of the vertical asymptote \( x = 1 \)), as \( x \) approaches \( 1 \) from the right (\( x

ightarrow1^+ \)), \( x - 1
ightarrow0^+ \), so \( \frac{2}{x - 1}
ightarrow+\infty \), and \( f(x)=\frac{2}{x - 1}+4
ightarrow+\infty \). As \( x
ightarrow+\infty \), \( \frac{2}{x - 1}
ightarrow0 \), so \( f(x)
ightarrow4 \) (approaching the horizontal asymptote \( y = 4 \) from above? Wait, no: \( \frac{2}{x - 1} \) is positive when \( x>1 \), so \( f(x)=\frac{2}{x - 1}+4 \) is \( 4 + \) a positive number approaching 0, so it approaches 4 from above? Wait, no, if \( \frac{2}{x - 1} \) approaches 0 from the positive side, then \( f(x) \) approaches 4 from above. But when \( x<1 \) (left of \( x = 1 \)), \( x - 1<0 \), so \( \frac{2}{x - 1}
ightarrow-\infty \) as \( x
ightarrow1^- \), so \( f(x)=\frac{2}{x - 1}+4
ightarrow-\infty \). As \( x
ightarrow-\infty \), \( \frac{2}{x - 1}
ightarrow0 \) (since the denominator goes to \( -\infty \)), so \( f(x)
ightarrow4 \) from below (because \( \frac{2}{x - 1} \) is negative when \( x<1 \), so \( f(x)=4+\) a negative number approaching 0, so it approaches 4 from below).

Now, looking at the given graph (the first one shown), it has a vertical asymptote at \( x = 1 \) (since the graph approaches a vertical line at \( x = 1 \)) and the horizontal asymptote behavior (approaching \( y = 4 \) for large \( x \) values) matches the analysis. Also, the behavior around \( x = 1 \) (shooting up to \( +\infty \) on the right of \( x = 1 \) and down to \( -\infty \) on the left of \( x = 1 \)) is consistent with the function \( f(x)=\frac{2}{x - 1}+4 \).

Answer:

The graph with vertical asymptote \( x = 1 \), horizontal asymptote \( y = 4 \), and the behavior as analyzed (shooting to \( +\infty \) on the right of \( x = 1 \), to \( -\infty \) on the left of \( x = 1 \), and approaching \( y = 4 \) for large \( x \) values) is the correct one (the first graph shown in the problem with the vertical asymptote at \( x = 1 \) and horizontal asymptote \( y = 4 \) behavior).