##Introduction
Finding a quadratic equation that passes through three given points is a classic problem that blends algebraic manipulation with geometric intuition. Whether you are a high‑school student tackling homework, a college freshman reviewing coordinate geometry, or a data analyst fitting a parabolic trend to experimental data, the ability to derive the equation from three points is essential. In this guide we will demystify the process, explain why three points are sufficient, and walk you through a clear, step‑by‑step method that works every time. By the end, you will not only know the mechanics but also understand the underlying theory that makes the solution possible Most people skip this — try not to..
Detailed Explanation
A quadratic equation in standard form is written as
[ \boxed{y = ax^{2} + bx + c} ]
where a, b, and c are constants that determine the shape and position of the parabola. Given three distinct points ((x_{1},y_{1})), ((x_{2},y_{2})), and ((x_{3},y_{3})) with different (x)-coordinates, we can substitute each point into the equation to obtain a system of three linear equations in the unknowns (a), (b), and (c). Solving this system yields the unique quadratic that interpolates the three points—provided the points are not collinear in a way that would force the parabola to degenerate Worth knowing..
Not obvious, but once you see it — you'll see it everywhere Simple, but easy to overlook..
The reason three points are enough lies in the degrees of freedom of a quadratic: there are exactly three coefficients to determine, so three independent conditions (the points) are required to pin down a single solution. If the (x)-values repeat or the points lie on a straight line, the system either has no solution or infinitely many, indicating that a genuine quadratic does not exist for that set And that's really what it comes down to. Still holds up..
Step‑by‑Step or Concept Breakdown
To find the quadratic equation, follow these logical steps:
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Set up the system – Write three equations by plugging each point into (y = ax^{2}+bx+c).
[ \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} ] -
Eliminate one variable – Subtract the first equation from the second and the first from the third. This removes (c) and gives two equations in (a) and (b).
[ \begin{aligned} y_{2}-y_{1}&=a(x_{2}^{2}-x_{1}^{2})+b(x_{2}-x_{1})\ y_{3}-y_{1}&=a(x_{3}^{2}-x_{1}^{2})+b(x_{3}-x_{1}) \end{aligned} ] -
Solve the 2×2 linear system – Treat the two resulting equations as a system for (a) and (b). Use substitution or the determinant method (Cramer's rule) to find their values.
[ \begin{vmatrix} x_{2}^{2}-x_{1}^{2} & x_{2}-x_{1}\ x_{3}^{2}-x_{1}^{2} & x_{3}-x_{1} \end{vmatrix} \neq 0 ] ensures a unique solution Small thing, real impact.. -
Back‑substitute to find (c) – Once (a) and (b) are known, plug either of the original equations (or the simplified ones) to solve for (c).
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Write the final equation – Substitute (a), (b), and (c) back into (y = ax^{2}+bx+c).
Optional shortcut: If the three points are symmetric or include the origin, you can exploit those patterns to simplify calculations, but the general method above works for any valid set of points.
Real Examples
Example 1 – Simple integer points
Suppose the points are ((1,2)), ((2,3)), and ((3,6)) Easy to understand, harder to ignore..
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Set up the system:
[ \begin{cases} 2 = a(1)^{2}+b(1)+c\ 3 = a(2)^{2}+b(2)+c\ 6 = a(3)^{2}+b(3)+c \end{cases} ] which simplifies to
[ \begin{cases} 2 = a+b+c\ 3 = 4a+2b+c\ 6 = 9a+3b+c \end{cases} ] -
Subtract the first equation from the second and third:
[ \begin{aligned} 1 &= 3a+b\ 4 &= 8a+2b \end{aligned} ] -
Solve: from the first, (b = 1-3a). Plug into the second:
[ 4 = 8a + 2(1-3a) = 8a + 2 - 6a = 2a + 2 \implies a = 1. ]
Then (b = 1-3(1) = -2). Finally, use (2 = a+b+c) to get (c = 3) And that's really what it comes down to. That's the whole idea.. -
The quadratic is (y = x^{2} - 2x +
