Solution:

Part (a): The degree is the highest power of $x$ in the polynomial.

The highest power is $x^5$, so the degree is 5.

Part (b): The leading coefficient is the coefficient of the term with the highest power.

The coefficient of $x^5$ is $-2$, so the leading coefficient is -2.

Part (c): For end behavior, we look at the degree and leading coefficient:

• Degree: 5 (odd)
• Leading coefficient: -2 (negative)

For a polynomial with odd degree and negative leading coefficient:

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

In words: The graph rises to the left and falls to the right.

Solution:

Step 1: Find the zeros by setting each factor equal to zero.

• $x^2 = 0 \Rightarrow x = 0$
• $(x - 3)^3 = 0 \Rightarrow x = 3$
• $(x + 1) = 0 \Rightarrow x = -1$

Step 2: Determine the multiplicity of each zero.

• $x = 0$: multiplicity 2 (even)
• $x = 3$: multiplicity 3 (odd)
• $x = -1$: multiplicity 1 (odd)

Step 3: Describe behavior at each zero.

• At $x = 0$: The graph touches the x-axis and turns around (even multiplicity)
• At $x = 3$: The graph crosses the x-axis (odd multiplicity)
• At $x = -1$: The graph crosses the x-axis (odd multiplicity)

Step 4: Determine overall end behavior.

• Degree: $2 + 3 + 1 = 6$ (even)
• Leading coefficient: positive (from expanding, the leading term is $x^6$)
• End behavior: Both ends go to $\infty$

Answer: Zeros at $x = -1$ (crosses), $x = 0$ (touches), and $x = 3$ (crosses)

Solution:

Part (a): With zeros at $x = -2, 1, 3$, the function has factors $(x + 2)$, $(x - 1)$, and $(x - 3)$.

Since the degree is 4 and we only have 3 zeros (each multiplicity 1), we need one more factor. Let's call it $(x - r)$ where $r$ is unknown.

General form: $g(x) = a(x + 2)(x - 1)(x - 3)(x - r)$

where $a$ is a positive constant.

Part (b): We use the point $(0, -12)$ to find $a$ and $r$.

However, with two unknowns and one equation, we need more information. Let's assume the fourth zero is actually a repeated zero. If $x = -2$ has multiplicity 2:

$g(x) = a(x + 2)^2(x - 1)(x - 3)$

Substituting $(0, -12)$: $-12 = a(2)^2(-1)(-3)$ $$

But we need $a$ positive! So let's try $x = 1$ with multiplicity 2:

$g(x) = a(x + 2)(x - 1)^2(x - 3)$

$-12 = a(2)(1)^2(-3)$ $-12 = a(2)(-3) = -6a$

Therefore: $g(x) = 2(x + 2)(x - 1)^2(x - 3)$

Part (c): Degree 4 (even), positive leading coefficient:

• As $x \to \infty$, $g(x) \to \infty$
• As $x \to -\infty$, $g(x$

The graph rises on both ends.

Solution:

Step 1: Write the polynomial in factored form using the zeros.

With zeros at $x = -2$ (mult. 2), $x = 1$ (mult. 1), and $x = 4$ (mult. 1): $f(x) = a(x + 2)^2(x - 1)(x - 4)$

Step 2: Determine the sign of the leading coefficient.

Total degree: $2 + 1 + 1 = 4$ (even)

For both ends to go to $-\infty$ with an even degree, we need a negative leading coefficient.

So $a < 0$. We can choose $a = -1$.

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

Step 3: Verify (optional - expand to check).

If we expand: $f(x) = -(x + 2)^2(x^2 - 5x + 4)$

The leading term will be $-x^4$, confirming our negative leading coefficient.

Answer: $f(x) = -(x + 2)^2(x - 1)(x - 4)$ or any negative constant multiple