https://file.aiquickdraw.com/mathbot/file/1730252418660rjzt7wr1.jpg

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?

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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.

Learn how to determine the equation of a rational function with given vertical asymptotes, a removable discontinuity, and a specified point on the graph. This example covers key concepts in rational functions, including asymptotes, holes, and using points to find constants. Suitable for Grades 10-12.

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