Find Quadratic Equation From 3 Points

The ability to find quadratic equation from 3 points is a fundamental skill in algebra that bridges the gap between abstract functions and concrete data. Whether you're a student tackling homework, a mathematician modeling real-world phenomena, or simply curious about how curves are uniquely determined, understanding this process opens the door to deeper insights in physics, engineering, and computer graphics. Even so, a quadratic function, typically written in the form $y = ax^2 + bx + c$, is defined by three coefficients. Because of that, when you are given three distinct points $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ that lie on the same parabola, you can determine these coefficients uniquely by solving a system of three equations. This method not only reinforces algebraic manipulation but also illustrates the power of systems of equations in describing geometric shapes That's the part that actually makes a difference..

The mathematical foundation rests on the fact that any parabola that is not vertical can be expressed as a quadratic function. Given three points, each point satisfies the equation $y = ax^2 + bx + c$ when its coordinates are substituted. Which means this yields three linear equations in terms of the unknowns $a$, $b$, and $c$. Because three points (assuming no two share the same $x$-coordinate) determine a unique parabola, the system will have exactly one solution. This is a direct application of the principle that a polynomial of degree $n$ is uniquely determined by $n+1$ points. In practical terms, this means you can always find quadratic equation from 3 points as long as the $x$-coordinates are distinct.

The most straightforward approach to solving this involves setting up a system of equations and using either substitution, elimination, or matrix methods. Plus, let's walk through the logic. Suppose the three points are $(1, 2)$, $(2, 3)$, and $(3, 6)$.

• For $(1, 2)$: $a(1)^2 + b(1) + c = 2 \implies a + b + c = 2$
• For $(2, 3)$: $a(2)^2 + b(2) + c = 3 \implies 4a + 2b + c = 3$
• For $(3, 6)$: $a(3)^2 + b(3) + c = 6 \implies 9a + 3b + c = 6$

Now you have a system

of three linear equations with three unknowns:

$ \begin{cases} a + b + c = 2 \ 4a + 2b + c = 3 \ 9a + 3b + c = 6 \end{cases} $

Solving by Elimination

The elimination method is often the most intuitive for hand calculation. The goal is to eliminate one variable at a time to reduce the system to two equations with two unknowns, and finally to a single equation.

1. Eliminate $c$: Subtract the first equation from the second and the second from the third Simple, but easy to overlook..

   • $(4a + 2b + c) - (a + b + c) = 3 - 2 \implies \mathbf{3a + b = 1}$ (Equation 4)
   • $(9a + 3b + c) - (4a + 2b + c) = 6 - 3 \implies \mathbf{5a + b = 3}$ (Equation 5)

2. Eliminate $b$: Subtract Equation 4 from Equation 5 Worth keeping that in mind..

   • $(5a + b) - (3a + b) = 3 - 1 \implies \mathbf{2a = 2 \implies a = 1}$

3. Back-substitute to find $b$: Plug $a = 1$ into Equation 4.

   • $3(1) + b = 1 \implies \mathbf{b = -2}$

4. Back-substitute to find $c$: Plug $a = 1$ and $b = -2$ into the very first original equation That's the whole idea..

   • $1 + (-2) + c = 2 \implies -1 + c = 2 \implies \mathbf{c = 3}$

Thus, the quadratic equation passing through the points $(1, 2)$, $(2, 3)$, and $(3, 6)$ is $y = x^2 - 2x + 3$. A quick verification confirms this works: for $x=3$, $y = 9 - 6 + 3 = 6$ Small thing, real impact..

Alternative Method: Matrix Inversion (Gaussian Elimination)

For larger systems or computational implementations, representing the system as a matrix equation $A\vec{x} = \vec{b}$ is standard. For our example:

$ \begin{bmatrix} 1 & 1 & 1 \ 4 & 2 & 1 \ 9 & 3 & 1 \end{bmatrix} \begin{bmatrix} a \ b \ c \end{bmatrix}

\begin{bmatrix} 2 \ 3 \ 6 \end{bmatrix} $

The coefficient matrix here is a Vandermonde matrix, which is guaranteed to be invertible as long as the $x$-coordinates ($1, 2, 3$) are distinct. Solving via row reduction (Gaussian elimination) or calculating $A^{-1}\vec{b}$ yields the same vector $\vec{x} = [1, -2, 3]^T$. This matrix approach scales efficiently for computer algorithms used in curve fitting and spline interpolation It's one of those things that adds up. Which is the point..

Alternative Method: Lagrange Interpolation

There is a elegant formulaic approach that bypasses solving systems entirely: Lagrange Interpolation. It constructs the polynomial directly as a weighted sum of basis polynomials, each designed to be $1$ at one specific $x$-coordinate and $0$ at the others.

For points $(x_1, y_1), (x_2, y_2), (x_3, y_3)$, the quadratic is: $P(x) = y_1 \frac{(x-x_2)(x-x_3)}{(x_1-x_2)(x_1-x_3)} + y_2 \frac{(x-x_1)(x-x_3)}{(x_2-x_1)(x_2-x_3)} + y_3 \frac{(x-x_1)(x-x_2)}{(x_3-x_1)(x_3-x_2)}$

Applying this to our points $(1,2), (2,3), (3,6)$: $P(x) = 2\frac{(x-2)(x-3)}{(1-2)(1-3)} + 3\frac{(x-1)(x-3)}{(2-1)(2-3)} + 6\frac{(x-1)(x-2)}{(3-1)(3-2)}$ $P(x) = 2\frac{(x-2)(x-3)}{2} + 3\frac{(x-1)(x-3)}{-1} + 6\frac{(x-1)(x-2)}{2}$ $P(x) = (x^2 - 5x + 6) - 3(x^2 - 4x + 3) + 3(x^2 - 3x + 2)$ $P(x) = x^2 - 5x + 6 - 3x^2 + 12x - 9 + 3x^2 - 9x + 6$ $P(x) = x^2 - 2x + 3$

This method is particularly powerful in theoretical