The problem asks for the vertical asymptote of the function $f(x) = \frac{2x}{x+3}$ and the sign of infinity ($\infty$) as $x$ approaches the asymptote from the right.

1. Problem Description

The problem asks for the vertical asymptote of the function $f(x) = \frac{2x}{x+3}$ and the sign of infinity ($\infty$) 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. So we set the denominator equal to zero:

$x+3 = 0$

$x = -3$

The vertical asymptote is at $x = -3$.

Now we consider the sign of the function as $x$ approaches $-3$ from the right. This means $x$ is slightly greater than $-3$. Let $x = -3 + \epsilon$, where $\epsilon$ is a very small positive number.

Then $f(x) = \frac{2(-3+\epsilon)}{(-3+\epsilon)+3} = \frac{-6+2\epsilon}{\epsilon}$.

Since $\epsilon$ is a small positive number, $-6 + 2\epsilon$ will be close to $-6$, which is negative. The denominator $\epsilon$ is positive. Therefore, the fraction $\frac{-6+2\epsilon}{\epsilon}$ will be negative. Thus, as $x$ approaches $-3$ from the right, $f(x)$ approaches negative infinity.

3. Final Answer

The equation of the vertical asymptote is $x = -3$, and the sign of infinity as $x$ approaches $-3$ from the right is negative.