Let's break down the function provided:

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

3. Name the real zeros of $f(x)$.

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

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

Solve for $x$ by setting each factor to zero:

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

So, the real zeros are: 0, 3, and -4.

4. State the end behavior of $f(x)$.

The end behavior of a polynomial is determined by the leading term, which comes from multiplying the highest degree terms in each factor. Here:

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

The highest degree term is found by multiplying:

$-2x \times x^2 \times x = -2x^4$

Since the degree is 4 (an even number) and the leading coefficient is negative, the end behavior is:

• As $x \to +\infty$, $f(x) \to -\infty$
• As $x \to -\infty$, $f(x) \to -\infty$

5. Use the coordinate plane to sketch $f(x)$.

Let's sketch the function considering the following:

• The zeros are at $x = -4$, $x = 0$, and $x = 3$.
• At $x = 3$, the factor $(x - 3)^2$ indicates a double root, so the graph will "bounce" off the x-axis at this point.
• Since the end behavior as $x \to \infty$ and $x \to -\infty$ is $-\infty$, the graph falls on both sides.

Would you like a visual sketch of the graph, or more detailed steps to plot it?

Here are 5 related questions:

1. How does the multiplicity of a root affect the graph's shape near the zero?
2. How do you determine the end behavior of a polynomial?
3. What is the effect of a negative leading coefficient on the graph?
4. How would you calculate the y-intercept of $f(x)$?
5. How does changing the coefficient of the polynomial affect the graph?

Tip: Always analyze the degree and leading coefficient to determine the end behavior of polynomials.