𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑔𝑒𝑛𝑒𝑟𝑎𝑙 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑎𝑟𝑎𝑏𝑜𝑙𝑎 𝑤ℎ𝑜𝑠𝑒 𝑎𝑥𝑖𝑠 𝑖𝑠 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑎𝑛𝑑 𝑝𝑎𝑠𝑠𝑒𝑠 𝑡ℎ𝑟𝑜𝑢𝑔ℎ 𝑡ℎ𝑒 𝑝𝑜𝑖𝑛𝑡𝑠 𝐴(−1, −3), 𝐵(1, −2), 𝑎𝑛𝑑 𝐶(2, 1).

To find the general equation of the parabola, we start by assuming the equation of a parabola with a vertical axis takes the standard form:

\[
y = ax^2 + bx + c
\]

We are given three points that lie on the parabola: \( A(-1, -3) \), \( B(1, -2) \), and \( C(2, 1) \). These points will allow us to create a system of equations by plugging the \(x\) and \(y\) values into the general form of the parabola equation.

### Step 1: Create the system of equations
For point \( A(-1, -3) \):
\[
-3 = a(-1)^2 + b(-1) + c \implies -3 = a - b + c
\]
\[
\text{(Equation 1:)} \quad a - b + c = -3
\]

For point \( B(1, -2) \):
\[
-2 = a(1)^2 + b(1) + c \implies -2 = a + b + c
\]
\[
\text{(Equation 2:)} \quad a + b + c = -2
\]

For point \( C(2, 1) \):
\[
1 = a(2)^2 + b(2) + c \implies 1 = 4a + 2b + c
\]
\[
\text{(Equation 3:)} \quad 4a + 2b + c = 1
\]

### Step 2: Solve the system of equations

We now solve the system:

\[
\begin{aligned}
a - b + c &= -3 \quad \text{(Equation 1)} \\
a + b + c &= -2 \quad \text{(Equation 2)} \\
4a + 2b + c &= 1 \quad \text{(Equation 3)}
\end{aligned}
\]

#### Subtract Equation 1 from Equation 2:
\[
(a + b + c) - (a - b + c) = -2 - (-3)
\]
\[
2b = 1 \implies b = \frac{1}{2}
\]

#### Substitute \(b = \frac{1}{2}\) into Equations 1 and 2:

Using Equation 1:
\[
a - \frac{1}{2} + c = -3 \implies a + c = -\frac{5}{2} \quad \text{(Equation 4)}
\]

Using Equation 3:
\[
4a + 2\left(\frac{1}{2}\right) + c = 1 \implies 4a + 1 + c = 1 \implies 4a + c = 0 \quad \text{(Equation 5)}
\]

#### Solve Equations 4 and 5:

From Equation 5:
\[
c = -4a
\]

Substitute \(c = -4a\) into Equation 4:
\[
a + (-4a) = -\frac{5}{2} \implies -3a = -\frac{5}{2} \implies a = \frac{5}{6}
\]

#### Find \(c\):
\[
c = -4a = -4\left(\frac{5}{6}\right) = -\frac{10}{3}
\]

### Step 3: Write the equation of the parabola

Now that we have \(a = \frac{5}{6}\), \(b = \frac{1}{2}\), and \(c = -\frac{10}{3}\), the general equation of the parabola is:

\[
y = \frac{5}{6}x^2 + \frac{1}{2}x - \frac{10}{3}
\]

Let me know if you want any further clarification or have other questions.

Here are five related questions you might explore:

1. How do we find the vertex of a parabola given its general equation?
2. What is the geometric significance of the axis of symmetry for a parabola?
3. How can you determine if a parabola opens upwards or downwards from its equation?
4. What is the process for converting the general form of a parabola to vertex form?
5. How would the equation change if the parabola had a horizontal axis instead?

**Tip**: For quadratic equations in the form \(y = ax^2 + bx + c\), the vertex can be found using \(x = -\frac{b}{2a}\).

Find the general equation of the parabola whose axis is vertical and passes through the points A(-1, -3), B(1, -2), and C(2, 1).

Learn how to find the general equation of a parabola with a vertical axis that passes through the points A(-1, -3), B(1, -2), and C(2, 1). This problem involves quadratic equations, systems of equations, and properties of parabolas. Suitable for students in Grades 9-12.

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