Finding a quadratic equation given three points is a fundamental algebraic skill that bridges coordinate geometry and polynomial functions. This is because a quadratic has three unknown coefficients, and each point provides one linear equation, resulting in a solvable system of three equations with three unknowns. When you have three distinct points ((x_1, y_1)), ((x_2, y_2)), and ((x_3, y_3)) with unique (x)-coordinates, there exists exactly one quadratic polynomial (y = ax^2 + bx + c) that passes through all three. Mastering this process not only strengthens your ability to work with parabolas but also lays the groundwork for more advanced topics like polynomial interpolation and curve fitting in calculus and data analysis.
The general form of a quadratic equation is [ y = ax^2 + bx + c, ] where (a), (b), and (c) are real numbers and (a \neq 0). That's why the coefficient (a) determines the direction and width of the parabola: if (a > 0), the parabola opens upward; if (a < 0), it opens downward. The coefficients (b) and (c) influence the position and orientation of the graph relative to the coordinate axes. When three points are given, substituting each ((x, y)) pair into the general form yields a system of equations that can be solved simultaneously to reveal the specific values of (a), (b), and (c) that define the unique quadratic passing through those points.
To begin the process, write the general form and substitute the coordinates of each point. You can solve it using standard algebraic techniques such as substitution, elimination, or matrix methods like Gaussian elimination. On the flip side, substitution gives: [ \begin{cases} y_1 = a x_1^2 + b x_1 + c \ y_2 = a x_2^2 + b x_2 + c \ y_3 = a x_3^2 + b x_3 + c \end{cases} ] This system is linear in (a), (b), and (c), even though the equation itself is quadratic in (x). On top of that, suppose the three points are ((x_1, y_1)), ((x_2, y_2)), and ((x_3, y_3)). The goal is to isolate (a), (b), and (c) so that the resulting equation precisely models the relationship captured by the three points.
Let’s walk through a concrete example to illustrate the method. Consider the points ((1, 2)), ((2, 5)), and ((3, 10)). Substituting into (y = ax^2 + bx + c) produces: [ \begin{aligned} 2 &= a(1)^2 + b(
