If a graph has a degree of 3 and points (-2,0), (0,-2), and (2,0), what is the formula for this function?

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.

Find the formula of a cubic polynomial passing through the points (-2, 0), (0, -2), and (2, 0). This problem uses algebraic concepts such as polynomial interpolation and systems of equations, suitable for students in Grades 10-12.

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