The problem asks us to find the vertical asymptote of the function $f(x) = \frac{2x}{x+3}$, and to determine the sign of infinity that $f(x)$ approaches as $x$ approaches the asymptote from the right.

1. Problem Description

The problem asks us to find the vertical asymptote of the function $f(x) = \frac{2x}{x+3}$, and to determine the sign of infinity that $f(x)$ approaches as $x$ approaches the asymptote from the right.

2. Solution Steps

First, we find the vertical asymptote. Vertical asymptotes occur where the denominator of a rational function is equal to zero.

$x + 3 = 0$

$x = -3$

So the vertical asymptote is $x = -3$.

Next, we need to determine the sign of infinity as $x$ approaches $-3$ from the right (i.e., $x \to -3^+$). We can choose a value slightly greater than $-3$, for example $x = -2.9$.

$f(-2.9) = \frac{2(-2.9)}{-2.9 + 3} = \frac{-5.8}{0.1} = -58$.

We can also consider the limit as $x$ approaches $-3$ from the right:

$\lim_{x \to -3^+} \frac{2x}{x+3}$.

As $x$ approaches $-3$ from the right, $x$ is slightly larger than $-3$, so $x+3$ approaches $0$ from the positive side (i.e., $x+3 \to 0^+$).

$2x$ approaches $2(-3) = -6$.

Therefore, $\lim_{x \to -3^+} \frac{2x}{x+3} = \frac{-6}{0^+} = -\infty$.

3. Final Answer

The equation of the vertical asymptote is $x = -3$. As $x$ approaches the asymptote from the right, the function approaches negative infinity.