We are tasked with finding a polynomial of the form:

$f(x) = ax^3 + bx^2 + cx + d$

such that the following conditions hold:

• $f(0) = -6$
• $f(1) = -6$
• $f(3) = -1$
• $f(4) = -4$

Step 1: Apply the given conditions to form a system of equations

1. From $f(0) = -6$: $a(0)^3 + b(0)^2 + c(0) + d = -6 \quad \Rightarrow \quad d = -6$

2. From $f(1) = -6$: $a(1)^3 + b(1)^2 + c(1) + d = -6 \quad \Rightarrow \quad a + b + c - 6 = -6 \quad \Rightarrow \quad a + b + c = 0$

3. From $f(3) = -1$: $a(3)^3 + b(3)^2 + c(3) + d = -1 \quad \Rightarrow \quad 27a + 9b + 3c - 6 = -1 \quad \Rightarrow \quad 27a + 9b + 3c = 5$

4. From $f(4) = -4$: $a(4)^3 + b(4)^2 + c(4) + d = -4 \quad \Rightarrow \quad 64a + 16b + 4c - 6 = -4 \quad \Rightarrow \quad 64a + 16b + 4c = 2$

Step 2: Solve the system of equations

We now have the following system of equations:

1. $a + b + c = 0$
2. $27a + 9b + 3c = 5$
3. $64a + 16b + 4c = 2$

Let's solve this system to find $a$, $b$, and $c$. I will compute this.The solution to the system of equations is:

$a = -\frac{2}{3}, \quad b = \frac{7}{2}, \quad c = -\frac{17}{6}$

Thus, the polynomial is:

$f(x) = -\frac{2}{3}x^3 + \frac{7}{2}x^2 - \frac{17}{6}x - 6$

Let me know if you need further details or explanations! Here are some related questions you might find useful:

1. How do we solve systems of linear equations in general?
2. What are the conditions for a unique solution in a system of equations?
3. How does the degree of a polynomial affect its graph's shape?
4. What happens if you change one of the constraints for the polynomial?
5. How can this method be generalized for higher-degree polynomials?

Tip: When solving systems of equations, check if simplifying the equations step-by-step helps identify any common factors.