The problem asks us to determine the equation of the rational function represented by the given graph. The possible equations are: a) $y = \frac{2x+1}{x-3}$ b) $y = \frac{-2x+1}{x+3}$ c) $y = \frac{-2x+1}{x-3}$ d) $y = \frac{2x+1}{x+3}$

1. Problem Description

The problem asks us to determine the equation of the rational function represented by the given graph. The possible equations are:

a) $y = \frac{2x+1}{x-3}$

b) $y = \frac{-2x+1}{x+3}$

c) $y = \frac{-2x+1}{x-3}$

d) $y = \frac{2x+1}{x+3}$

2. Solution Steps

We can analyze the graph to determine the equation.

First, we observe that the graph has a vertical asymptote at $x = 3$. This means that the denominator of the rational function must be zero when $x = 3$. This eliminates option b) and d) because their denominators are $x+3$.

The remaining options are:

a) $y = \frac{2x+1}{x-3}$

c) $y = \frac{-2x+1}{x-3}$

Next, we observe the horizontal asymptote. As $x$ approaches positive or negative infinity, the value of $y$ approaches $-2$.

For option a), $y = \frac{2x+1}{x-3}$, we can divide both the numerator and denominator by $x$ to get

$y = \frac{2 + \frac{1}{x}}{1 - \frac{3}{x}}$.

As $x$ approaches infinity, $\frac{1}{x}$ approaches 0, so $y$ approaches $\frac{2}{1} = 2$.

For option c), $y = \frac{-2x+1}{x-3}$, we can divide both the numerator and denominator by $x$ to get

$y = \frac{-2 + \frac{1}{x}}{1 - \frac{3}{x}}$.

As $x$ approaches infinity, $\frac{1}{x}$ approaches 0, so $y$ approaches $\frac{-2}{1} = -2$.

Since the graph has a horizontal asymptote at $y=-2$, the equation must be $y = \frac{-2x+1}{x-3}$.

3. Final Answer

$y = \frac{-2x+1}{x-3}$