The equation provided is:

$y = \frac{3}{2} x^3$

You are asked to plot five points on the graph of this function: one point for $x = 0$, two points for negative $x$-values, and two points for positive $x$-values.

Step-by-step process to find points:

1. For $x = 0$: $y = \frac{3}{2} (0)^3 = 0$ So, the point is $(0, 0)$.

2. For negative values of $x$:

   • Let $x = -2$: $y = \frac{3}{2} (-2)^3 = \frac{3}{2} (-8) = -12$ So, the point is $(-2, -12)$.

   • Let $x = -1$: $y = \frac{3}{2} (-1)^3 = \frac{3}{2} (-1) = -\frac{3}{2} \approx -1.5$ So, the point is $(-1, -1.5)$.

3. For positive values of $x$:

   • Let $x = 1$: $y = \frac{3}{2} (1)^3 = \frac{3}{2} = 1.5$ So, the point is $(1, 1.5)$.

   • Let $x = 2$: $y = \frac{3}{2} (2)^3 = \frac{3}{2} (8) = 12$ So, the point is $(2, 12)$.

Summary of the points:

• $(0, 0)$
• $(-2, -12)$
• $(-1, -1.5)$
• $(1, 1.5)$
• $(2, 12)$

Plot these points on the graph. The graph should resemble a cubic curve with increasing steepness as $|x|$ increases.

Would you like any additional explanations or have questions on how to interpret the graph?

Here are five related questions:

1. How does the coefficient $\frac{3}{2}$ affect the shape of the cubic graph?
2. What would happen to the graph if the exponent of $x$ was even instead of odd?
3. How can we find the slope at specific points on a cubic function?
4. How do negative values of $x$ affect the function $y = \frac{3}{2} x^3$?
5. How do you graph transformations of the form $y = a(x - h)^3 + k$?

Tip: Cubic functions like $y = \frac{3}{2}x^3$ always pass through the origin, and their graphs exhibit symmetry about the origin, called odd symmetry.