The graph in the image shows a rational function with vertical asymptotes at $x = -3$ and $x = 3$, indicated by the dashed red lines. These vertical asymptotes suggest that the function's denominator has factors of $(x + 3)$ and $(x - 3)$, because division by zero at those values of $x$ causes the function to approach infinity.

Additionally, the function appears to have a horizontal asymptote at $y = 0$, implying that the degree of the numerator is less than or equal to the degree of the denominator.

From the behavior of the graph, which shows the function diverging to infinity near the asymptotes and tending toward 0 as $x$ increases or decreases far from 0, we can assume a simple rational form. A function of this type might look like:

$y = \frac{k}{(x + 3)(x - 3)}$

where $k$ is a constant determining the stretch of the graph.

Given that the graph passes through the origin (0, 0), we can substitute this into the equation:

$0 = \frac{k}{(0 + 3)(0 - 3)} = \frac{k}{-9}$

Solving for $k$, we find that $k = 0$, which confirms that the graph touches the origin.

Thus, the function can be written as:

$y = \frac{-9}{(x + 3)(x - 3)}$

Let me know if you'd like further details or have any questions.

Here are five questions for further exploration:

1. How would you modify the equation if the graph had a horizontal asymptote other than $y = 0$?
2. What happens to the graph if you change the value of $k$ in the equation?
3. Can you determine whether the function has any holes or removable discontinuities?
4. How would you find the domain and range of this rational function?
5. What would the graph look like if you added a constant to the numerator?

Tip: When analyzing rational functions, always check for vertical asymptotes by identifying the values that make the denominator zero, and for horizontal asymptotes by comparing the degrees of the numerator and denominator.