The problem states:

Write a polynomial equation for a function with a graph that bounces off the x-axis at $(-1, 0)$, crosses it at $(4, 0)$, and passes through the point $(-2, -18)$.

Solution:

1. Understanding the conditions:

   • At $(-1, 0)$, the graph "bounces" off the x-axis. This means the factor corresponding to this root will have even multiplicity (typically 2). Thus, the factor is $(x + 1)^2$.
   • At $(4, 0)$, the graph crosses the x-axis. This means the factor corresponding to this root will have odd multiplicity (typically 1). Thus, the factor is $(x - 4)$.
   • The polynomial passes through the point $(-2, -18)$, which allows us to determine the leading coefficient.

2. Constructing the polynomial: Based on the given roots, the polynomial will take the form: $f(x) = a(x + 1)^2(x - 4),$ where $a$ is the leading coefficient to be determined.

3. Using the point $(-2, -18)$ to find $a$: Substitute $x = -2$ and $f(x) = -18$ into the equation: $-18 = a((-2 + 1)^2)((-2 - 4)).$ Simplify: $-18 = a((-1)^2)(-6),$ $-18 = a(1)(-6),$ $-18 = -6a.$ Solve for $a$: $a = \frac{-18}{-6} = 3.$

4. Final polynomial: Substituting $a = 3$ back into the equation, we get: $f(x) = 3(x + 1)^2(x - 4).$

Final Answer:

$f(x) = 3(x + 1)^2(x - 4).$

Do you need any clarification or further steps explained?

Here are some related questions to expand your understanding:

1. What happens to the graph of a polynomial when the root has an even multiplicity versus an odd multiplicity?
2. How can you determine the degree of a polynomial from its factors?
3. How would the polynomial change if it bounced at $(4, 0)$ instead of crossing?
4. Can this method be extended to higher-degree polynomials with more roots and conditions?
5. How do the signs of the leading coefficient affect the shape of the graph?

Tip:

To confirm the solution, always verify the given points by substituting them back into the polynomial!