To find the equation of the function $R(x)$ represented in this graph, let’s analyze its key features:

1. Vertical Asymptotes: The graph has vertical asymptotes at $x = -3$ and $x = 2$. This suggests the function has factors in the denominator that cause division by zero at these points, likely of the form $(x + 3)$ and $(x - 2)$.

2. Horizontal Asymptote: The graph appears to approach a horizontal line (likely $y = 0$) as $x$ moves far to the left or right, indicating a rational function where the degree of the numerator is less than or equal to the degree of the denominator.

3. Holes: There is a removable discontinuity (hole) at $x = 5$, represented by an open circle on the graph. This suggests that the function has a factor $(x - 5)$ in both the numerator and denominator, which cancels out.

4. Point on the Graph: The point $(-4, 1)$ is given, which allows us to check our function by plugging in $x = -4$ and seeing if it outputs $y = 1$.

Given these observations, a reasonable form for $R(x)$ is: $R(x) = \frac{A(x - 5)}{(x + 3)(x - 2)}$ where $A$ is a constant that can be adjusted to fit the known point $(-4, 1)$.

Determining the Constant $A$

Using the point $(-4, 1)$: $R(-4) = \frac{A(-4 - 5)}{(-4 + 3)(-4 - 2)} = 1$ $R(-4) = \frac{A \cdot (-9)}{(-1)(-6)} = \frac{-9A}{6} = 1$ Solving for $A$: $-9A = 6 \Rightarrow A = -\frac{2}{3}$

Final Equation

Thus, the equation of $R(x)$ is: $R(x) = \frac{-\frac{2}{3}(x - 5)}{(x + 3)(x - 2)}$ or simplified: $R(x) = \frac{-2(x - 5)}{3(x + 3)(x - 2)}$

Would you like a more detailed explanation or have any questions?

Here are some related questions to explore further:

1. What would happen to the graph if the value of $A$ were positive?
2. How does the hole at $x = 5$ affect the function’s behavior near that point?
3. How would we modify the function if there were additional vertical asymptotes?
4. What would the function look like if there were a horizontal asymptote at $y = 1$ instead of $y = 0$?
5. Can we have a similar function with a slant asymptote instead of a horizontal asymptote?

Tip: For rational functions, vertical asymptotes correspond to factors in the denominator that do not cancel with the numerator, while holes occur where factors cancel out.