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:

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}$$.