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Let's analyze the given problem step by step.

The problem describes a rational function \( f \) with a domain \( D_f = \mathbb{R} - \{-4, 3\} \), and it provides several conditions to help determine the properties of the function:

### Given Conditions:
1. \( f(0) = -6 \)
2. \( f'(-1) = 0 \) (This suggests a critical point at \( x = -1 \))
3. \( f(2) = 0 \) (This is an x-intercept)
4. \( \lim_{x \to 3} f(x) = \infty \) and \( \lim_{x \to 3^-} f(x) = -\infty \) (Vertical asymptote at \( x = 3 \))
5. \( \lim_{x \to -4^+} f(x) = 5 \) and \( \lim_{x \to -4^-} f(x) = \infty \) (Discontinuity at \( x = -4 \), with vertical asymptote)
6. \( \lim_{x \to -\infty} f(x) = 0 \) and \( \lim_{x \to \infty} f(x) = 0 \) (Horizontal asymptote at \( y = 0 \))

### Questions:

a. **Determine the equation of the vertical asymptote(s), if any.**

There are vertical asymptotes at:
- \( x = 3 \), as indicated by \( \lim_{x \to 3} f(x) = \infty \)
- \( x = -4 \), as indicated by \( \lim_{x \to -4^-} f(x) = \infty \)

So, the equations for the vertical asymptotes are:
\[
x = 3 \quad \text{and} \quad x = -4.
\]

b. **Determine the equation of the horizontal asymptote(s), if any.**

The horizontal asymptotes are determined by the behavior of the function as \( x \to \infty \) or \( x \to -\infty \). We have:
\[
\lim_{x \to \infty} f(x) = 0 \quad \text{and} \quad \lim_{x \to -\infty} f(x) = 0.
\]
Therefore, the horizontal asymptote is:
\[
y = 0.
\]

c. **Determine the intersections with the Cartesian axes.**

To find the intersections with the axes:

- **x-intercepts**: These occur when \( f(x) = 0 \). Given \( f(2) = 0 \), there is an x-intercept at \( x = 2 \).

- **y-intercept**: This occurs when \( x = 0 \). Given \( f(0) = -6 \), there is a y-intercept at \( y = -6 \).

d. **Sketch the graph of the function \( f \) clearly representing all the given conditions.**

To sketch the graph, we need to consider all the features identified:

- **Vertical Asymptotes**: At \( x = 3 \) and \( x = -4 \).
- **Horizontal Asymptote**: At \( y = 0 \).
- **Intercepts**: x-intercept at \( x = 2 \), y-intercept at \( y = -6 \).
- **Critical Point**: \( f'(-1) = 0 \) suggests that there is either a local maximum, minimum, or saddle point at \( x = -1 \).

### Conclusion:

To summarize the graph:

1. The function has vertical asymptotes at \( x = -4 \) and \( x = 3 \).
2. The horizontal asymptote is at \( y = 0 \).
3. The function crosses the x-axis at \( x = 2 \) and the y-axis at \( y = -6 \).
4. The behavior near the asymptotes should match the given limits (e.g., \( \lim_{x \to 3^-} f(x) = -\infty \) and \( \lim_{x \to 3^+} f(x) = \infty \)).
5. There is a critical point at \( x = -1 \) where the derivative is zero, indicating a possible local maximum or minimum.

Would you like me to generate a sketch of this function or need any additional clarifications? 

### Follow-Up Questions:

1. What is the general form of a rational function with given vertical and horizontal asymptotes?
2. How do critical points affect the shape of the graph of a function?
3. How do you determine whether a critical point is a local maximum, minimum, or saddle point?
4. What additional information would be needed to precisely determine the function \( f(x) \)?
5. How does the behavior around discontinuities help determine the function's graph?

**Tip:** When sketching a graph of a rational function, always mark the asymptotes and intercepts first to understand the function's general behavior.

Sea f una función racional con dominio Df = R - {-4, 3} que satisfaga las siguientes condiciones: f(0) = -6, f'(-1) = 0, f(2) = 0, lim(x→3) f(x) = ∞, lim(x→3⁻) f(x) = -∞, con discontinuidad en x = -4. Con lim(x→-4⁺) f(x) = 5, lim(x→-4⁻) f(x) = ∞, lim(x→∞) f(x) = 0, y lim(x→-∞) f(x) = 0. Determine las asíntotas y los interceptos.

Solve a rational function problem with domain R - {-4, 3}. This involves determining vertical and horizontal asymptotes, x and y intercepts, and analyzing limits and continuity at specific points. Key mathematical concepts include rational functions, asymptotes, and limits.

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