Question

what are the vertical and horizontal asymptotes of $f(x) = \frac{2x}{x - 1}$?
horizontal asymptote at $y = 1$, vertical asymptote at $x = 2$
horizontal asymptote at $y = 2$, vertical asymptote at $x = 1$
horizontal asymptote at $y = 0$, vertical asymptote at $x = 1$
horizontal asymptote at $y = 1$, vertical asymptote at $x = 0$

Explanation:

Step1: Find Vertical Asymptote

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). Here, \( D(x)=x - 1 \), so set \( x-1 = 0 \), which gives \( x = 1 \).

Step2: Find Horizontal Asymptote

For rational functions where the degrees of \( N(x) \) and \( D(x) \) are equal (both degree 1 here, \( N(x)=2x \), \( D(x)=x - 1 \)), the horizontal asymptote is the ratio of the leading coefficients. The leading coefficient of \( N(x) \) is 2, and of \( D(x) \) is 1, so \( y=\frac{2}{1}=2 \)? Wait, no, wait. Wait, \( f(x)=\frac{2x}{x - 1}=\frac{2x-2 + 2}{x - 1}=2+\frac{2}{x - 1} \). As \( x\to\pm\infty \), \( \frac{2}{x - 1}\to0 \), so \( f(x)\to2 \)? Wait, no, earlier mistake. Wait, degree of numerator and denominator are both 1, so horizontal asymptote is \( y=\frac{\text{leading coefficient of numerator}}{\text{leading coefficient of denominator}}=\frac{2}{1}=2 \). And vertical asymptote at \( x = 1 \). So the option with horizontal asymptote \( y = 2 \) and vertical asymptote \( x = 1 \) is the correct one.

Answer:

The option with "horizontal asymptote at \( y = 2 \), vertical asymptote at \( x = 1 \)" (the middle option among the first three, or the second boxed option as per the initial display)