Let's analyze the given polynomial function:

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

Step 1: Finding the Zeros

The zeros of the function occur when $f(x) = 0$. Setting the equation equal to zero:

$-3(x - 2)(x + 4)^2 = 0$

This equation will be zero when either factor equals zero:

• $x - 2 = 0$ gives $x = 2$.
• $(x + 4)^2 = 0$ gives $x = -4$.

Thus, the zeros are $x = 2$ and $x = -4$.

Step 2: Determining Multiplicities

• $x = 2$ comes from the factor $(x - 2)$, which is raised to the power of 1. Therefore, its multiplicity is 1.
• $x = -4$ comes from the factor $(x + 4)^2$, which is raised to the power of 2. Therefore, its multiplicity is 2.

Step 3: Behavior at the Zeros

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

Step 4: Answering the Multiple-Choice Question

In the multiple-choice options, the correct choice is:

• A: There are two zeros. The multiplicity of the largest zero (in absolute value, $-4$) is 2. The multiplicity of the smallest zero (in absolute value, $2$) is 1.

Do you need further clarification or details?

Here are 5 related questions for practice:

1. How would the graph change if the multiplicity of $(x + 4)^2$ was 3 instead of 2?
2. What are the end behaviors of the function $f(x) = -3(x - 2)(x + 4)^2$?
3. How does the negative sign in front of the function $f(x)$ affect the graph?
4. What happens to the graph if $(x + 4)^2$ becomes $(x + 4)^4$?
5. How do you find the y-intercept of this function?

Tip: For polynomial functions, zeros with odd multiplicities cross the x-axis, while zeros with even multiplicities touch and turn around at the x-axis.