The problem asks for the zeros of the polynomial function:

$f(x) = 6(x - 9)(x + 4)^2$

Steps to find the zeros and their multiplicities:

1. Set $f(x) = 0$ to find the zeros: $6(x - 9)(x + 4)^2 = 0$ This equation is satisfied if either factor equals zero.

2. Solve for each factor:

   • For $x - 9 = 0$, the solution is $x = 9$. This zero has multiplicity 1 (because the exponent on this factor is 1).
   • For $(x + 4)^2 = 0$, the solution is $x = -4$. This zero has multiplicity 2 (because the exponent is 2).

Zeros and their multiplicities:

• Zero at $x = 9$, with multiplicity 1.
• Zero at $x = -4$, with multiplicity 2.

Behavior at the x-axis:

• At $x = 9$, since the multiplicity is 1 (odd), the graph crosses the x-axis.
• At $x = -4$, since the multiplicity is 2 (even), the graph touches the x-axis and turns around.

Final answer:

The zeros are $x = 9$ and $x = -4$.

Would you like more details or explanations on any step? Here are some related questions:

1. What is the general relationship between multiplicity and the graph's behavior at zeros?
2. How do higher-order multiplicities (greater than 2) affect the graph at the x-axis?
3. Can you always factor polynomials like this easily?
4. What happens if a polynomial has complex roots?
5. How does the degree of the polynomial affect the number of zeros?

Tip: The graph crosses the x-axis at zeros with odd multiplicity and touches (but does not cross) at zeros with even multiplicity!