To solve this problem, let's analyze the two functions and their corresponding graphs.

1. Function $g(x) = -2(x - 4)(x^2 - 1)$

This is a cubic function because when expanded, the highest degree term will be $x^3$. The behavior of this function will be determined by:

• Zeros: The roots can be found by solving $(x - 4) = 0$ and $(x^2 - 1) = 0$. The roots are $x = 4$, $x = 1$, and $x = -1$.
• End behavior: Since the leading term is negative $-2x^3$, as $x \to \infty$, the function will tend to $-\infty$, and as $x \to -\infty$, the function will tend to $\infty$.
• Turning points: There will be turning points between the roots.

Looking at the graphs:

• We need a cubic function with three x-intercepts at 4, 1, and -1, and the function should start from the positive y-values and end in the negative direction.

2. Function $h(x) = x^4 - x^3 - 6x^2$

This is a quartic function (degree 4). Let's analyze the behavior:

• Factoring: We can factor this as $h(x) = x^2(x^2 - x - 6) = x^2(x - 3)(x + 2)$, giving roots at $x = 0$, $x = 3$, and $x = -2$.
• End behavior: Since the leading term is positive $x^4$, as $x \to \infty$, the function will tend to $\infty$, and as $x \to -\infty$, it will also tend to $\infty$.

Looking at the graphs:

• We need a quartic function with three x-intercepts at 3, 0, and -2, and it should have the typical "W" shape of a quartic function.

Graph Selection:

• For $g(x) = -2(x - 4)(x^2 - 1)$, the correct graph is Graph C (cubic behavior, crosses x-axis at 4, 1, -1).
• For $h(x) = x^4 - x^3 - 6x^2$, the correct graph is Graph F (quartic behavior, crosses x-axis at 3, 0, -2).

Let me know if you'd like further details!

Here are five related questions to deepen your understanding:

1. How can we expand $g(x) = -2(x - 4)(x^2 - 1)$ to confirm the shape of the graph?
2. What is the difference in end behavior between cubic and quartic functions?
3. Can you identify turning points algebraically for each function?
4. How does the sign of the leading coefficient affect the shape of the graph?
5. What would happen if we changed the sign of $g(x)$ from negative to positive?

Tip: Always check the degree of the polynomial to understand the general shape of the graph and the number of turning points it may have.