Question

2 which is the correct graph of $y = \frac{2x}{x + 1}$

Explanation:

To determine the correct graph of \( y = \frac{2x}{x + 1} \), we analyze its key features:

Step 1: Vertical Asymptote

The denominator \( x + 1 = 0 \) when \( x = -1 \). Thus, there is a vertical asymptote at \( x = -1 \).

Step 2: Horizontal Asymptote

For rational functions \( \frac{ax^n + \dots}{bx^m + \dots} \), if \( n = m \), the horizontal asymptote is \( y = \frac{a}{b} \). Here, \( a = 2 \), \( b = 1 \), so the horizontal asymptote is \( y = 2 \).

Step 3: Intercepts

• x-intercept: Set \( y = 0 \): \( \frac{2x}{x + 1} = 0 \implies 2x = 0 \implies x = 0 \). So the x-intercept is \( (0, 0) \).
• y-intercept: Set \( x = 0 \): \( y = \frac{0}{0 + 1} = 0 \). So the y-intercept is also \( (0, 0) \).

Step 4: Analyze the Graphs

• The first graph (left) lacks a vertical asymptote at \( x = -1 \) (no break in the curve).
• The second graph (right) has a vertical asymptote (the curve splits near \( x = -1 \)) and approaches \( y = 2 \) (horizontal asymptote) as \( x \to \pm\infty \). It also passes through \( (0, 0) \), matching the intercept.

Thus, the correct graph is the second one (with the vertical asymptote at \( x = -1 \), horizontal asymptote at \( y = 2 \), and passing through the origin).

Answer:

The second graph (the one with two branches, a vertical asymptote at \( x = -1 \), horizontal asymptote at \( y = 2 \), and passing through \( (0, 0) \))