The polynomial function given is:

$f(x) = -9(x+5)(x-8)^2$

Step 1: Finding the Zeros

To find the zeros, set $f(x) = 0$:

$-9(x+5)(x-8)^2 = 0$

This equation equals zero when either factor equals zero:

1. $x+5 = 0 \Rightarrow x = -5$
2. $(x-8)^2 = 0 \Rightarrow x = 8$

Thus, the zeros are $x = -5$ and $x = 8$.

Step 2: Multiplicity of Each Zero

• The zero $x = -5$ comes from the factor $(x + 5)$, which is raised to the first power. Therefore, its multiplicity is 1.
• The zero $x = 8$ comes from the factor $(x - 8)^2$, which is squared. Therefore, its multiplicity is 2.

Step 3: Behavior of the Graph at Each Zero

• At $x = -5$: Since the multiplicity is 1 (odd), the graph crosses the x-axis at this point.
• At $x = 8$: Since the multiplicity is 2 (even), the graph touches the x-axis and turns around at this point.

Conclusion:

• The zeros are $x = -5$ and $x = 8$.
• Multiplicity of $x = -5$ is 1, and the graph crosses the x-axis here.
• Multiplicity of $x = 8$ is 2, and the graph touches the x-axis and turns around here.

Would you like further details on any part of this?

Here are 5 related questions to expand on this:

1. How does the leading coefficient of a polynomial affect the graph's end behavior?
2. Can a polynomial have a zero with multiplicity greater than 2?
3. What is the significance of the degree of a polynomial in relation to its zeros?
4. How can you factor higher-degree polynomials to find their zeros?
5. What happens to the graph of a polynomial if the multiplicity of a zero is 3?

Tip: The behavior at zeros can often be predicted by their multiplicity: odd multiplicities lead to crossings, while even multiplicities result in touching and turning.