How To Find A Quadratic Equation From 3 Points

How to Find a Quadratic Equation from 3 Points

Finding the equation of a parabola that passes through three given points is a common problem in algebra and coordinate geometry. Because a quadratic function has the general form

[ y = ax^{2}+bx+c, ]

three non‑collinear points provide exactly enough information to solve for the three unknown coefficients (a), (b), and (c). Below you’ll find a step‑by‑step guide, the underlying mathematics, practical tips, and a FAQ section to help you master the technique Took long enough..

Introduction

When you are given three points ((x_{1},y_{1})), ((x_{2},y_{2})), and ((x_{3},y_{3})) that lie on a parabola, the goal is to determine the unique quadratic equation (y = ax^{2}+bx+c) that fits them. This process is essentially solving a system of three linear equations in the unknowns (a), (b), and (c). Mastering it not only helps with homework problems but also builds intuition for curve fitting, data modeling, and even physics applications such as projectile motion.

Step‑by‑Step Procedure

1. Write the General Form for Each Point

Substitute each point into the quadratic template (y = ax^{2}+bx+c). This yields three equations:

[ \begin{aligned} 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{aligned} ]

2. Arrange the System in Matrix Form (Optional)

If you prefer a compact representation, write the system as

[ \begin{bmatrix} x_{1}^{2} & x_{1} & 1\ x_{2}^{2} & x_{2} & 1\ x_{3}^{2} & x_{3} & 1 \end{bmatrix} \begin{bmatrix} a\ b\ c \end{bmatrix}

\begin{bmatrix} y_{1}\ y_{2}\ y_{3} \end{bmatrix}. ]

Solving this matrix equation gives the coefficients directly, but you can also use elimination or substitution And that's really what it comes down to. Simple as that..

3. Eliminate One Variable (Typically (c))

Subtract the first equation from the second and the second from the third to eliminate (c):

[ \begin{aligned} (y_{2}-y_{1}) &= a(x_{2}^{2}-x_{1}^{2}) + b(x_{2}-x_{1}) \ (y_{3}-y_{2}) &= a(x_{3}^{2}-x_{2}^{2}) + b(x_{3}-x_{2}) \end{aligned} ]

These two equations now involve only (a) and (b) Easy to understand, harder to ignore..

4. Solve for (a) and (b)

Treat the two new equations as a linear system in (a) and (b). You can solve it by:

• Substitution: isolate (b) from one equation and plug into the other.
• Elimination: multiply each equation to make the coefficients of (b) (or (a)) opposites, then add.

As an example, solve the first for (b):

[ b = \frac{(y_{2}-y_{1}) - a(x_{2}^{2}-x_{1}^{2})}{x_{2}-x_{1}}. ]

Insert this expression into the second equation and solve for (a). Once (a) is known, back‑substitute to find (b).

5. Find (c)

With (a) and (b) determined, return to any of the original three equations (usually the first) and solve for (c):

[ c = y_{1} - a x_{1}^{2} - b x_{1}. ]

6. Write the Final Quadratic Equation

Plug the obtained values into (y = ax^{2}+bx+c). Double‑check by substituting the three points; each should satisfy the equation exactly (within rounding error if you used decimals).

Scientific Explanation

Why Three Points Suffice

A quadratic polynomial has three degrees of freedom (the coefficients (a), (b), (c)). In real terms, each point provides one linear constraint on these coefficients. Three independent constraints therefore determine a unique solution, assuming the points are not collinear (which would force (a=0) and reduce the curve to a line) Nothing fancy..

Connection to Linear Algebra

The coefficient matrix

[ V = \begin{bmatrix} x_{1}^{2} & x_{1} & 1\ x_{2}^{2} & x_{2} & 1\ x_{3}^{2} & x_{3} & 1 \end{bmatrix} ]

is a Vandermonde matrix. Its determinant is

[ \det(V) = (x_{2}-x_{1})(x_{3}-x_{1})(x_{3}-x_{2}), ]

which is non‑zero whenever the (x)-coordinates are distinct. Hence the matrix is invertible, guaranteeing a unique solution for ((a,b,c)^{T}).

Alternative Forms

• Vertex Form: (y = a(x-h)^{2}+k). If you know the vertex ((h,k)) and one other point, you can solve for (a) directly.
• Factored Form: (y = a(x-r_{1})(x-r_{2})) when the roots are known.

These forms are useful shortcuts but rely on additional information beyond three arbitrary points Worth keeping that in mind..

Numerical Stability

When the (x)-values are very close together or span a large range, the Vandermonde matrix can become ill‑conditioned, leading to large rounding errors. In such cases, using Lagrange interpolation or least‑squares fitting (for more than three points) may be preferable. For typical classroom problems with modest integer coordinates, the elimination method described above is perfectly stable And it works..

Practical Tips and Common Pitfalls