To find the formula of a polynomial function of degree 3 that passes through the points $(-2, 0)$, $(0, -2)$, and $(2, 0)$, we can follow these steps:

Step 1: General form of the cubic function

The general form of a degree 3 polynomial is: $f(x) = ax^3 + bx^2 + cx + d$ where $a$, $b$, $c$, and $d$ are constants to be determined.

Step 2: Using the points to find the constants

Since the function passes through the given points, we substitute each point into the cubic function and get a system of equations.

• For $(-2, 0)$: $0 = a(-2)^3 + b(-2)^2 + c(-2) + d = -8a + 4b - 2c + d$ $-8a + 4b - 2c + d = 0$

• For $(0, -2)$: $-2 = a(0)^3 + b(0)^2 + c(0) + d$ $d = -2$

• For $(2, 0)$: $0 = a(2)^3 + b(2)^2 + c(2) + d = 8a + 4b + 2c + d$ $8a + 4b + 2c - 2 = 0$ $8a + 4b + 2c = 2$

Step 3: Solve the system of equations

From the above, we have the following system of equations:

1. $-8a + 4b - 2c + d = 0$
2. $d = -2$
3. $8a + 4b + 2c - 2 = 0$

Substitute $d = -2$ into equations 1 and 3:

1. $-8a + 4b - 2c - 2 = 0$
2. $8a + 4b + 2c = 2$

Now we solve these two equations:

1. $-8a + 4b - 2c = 2$
2. $8a + 4b + 2c = 2$

Step 4: Add the two equations

$(-8a + 4b - 2c) + (8a + 4b + 2c) = 2 + 2$ $8b = 4$ $b = \frac{1}{2}$

Step 5: Substitute $b = \frac{1}{2}$ into one of the equations

Substitute into $8a + 4b + 2c = 2$: $8a + 4\left(\frac{1}{2}\right) + 2c = 2$ $8a + 2 + 2c = 2$ $8a + 2c = 0$ $2c = -8a$ $c = -4a$

Step 6: Substitute $c = -4a$ into $-8a + 4b - 2c = 2$

$-8a + 4\left(\frac{1}{2}\right) - 2(-4a) = 2$ $-8a + 2 + 8a = 2$ $2 = 2$ This is a true statement, so no further information is obtained from this equation. This means $a$ can be any value. To simplify, we typically set $a = 1$.

Step 7: Final formula

Since $a = 1$, $b = \frac{1}{2}$, and $c = -4a = -4$, the equation of the cubic function is: $f(x) = x^3 + \frac{1}{2}x^2 - 4x - 2$

Would you like any more details or have questions? Here are some related questions:

1. How can we confirm that the points satisfy this cubic function?
2. What would change if we chose a different value for $a$?
3. How do we find the roots of a cubic function like this one?
4. What is the behavior of this cubic function at large $x$ values?
5. How can we use this function to find the turning points?

Tip: The degree of a polynomial dictates the maximum number of turning points it can have. A cubic function can have up to 2 turning points.