Problem 33 - Essential Calculus

Question

Prove that you can find a polynomial \(p(x)=A x^{2}+B x+C\) that passes through any three points \(\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right),\) and \(\left(x_{3}, y_{3}\right),\) where the \(x_{i}^{\prime} \mathrm{s}\) are distinct.

Step-by-Step Solution

Answer

The quadratic polynomial that passes through any three points \(\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right),\left(x_{3}, y_{3}\right)\) with distinct \(x\) values can be found by substituting these points into the polynomial form \(p(x)=A x^{2}+B x+C\), setting up a system of three linear equations, solving for the coefficients \(A\), \(B\), and \(C\), and then forming the quadratic polynomial.

Step 1: Setting up the System of Equations

Plug the given points \(\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right),\left(x_{3}, y_{3}\right)\) into the polynomial \(p(x)=A x^{2}+B x+C\), thus obtaining a system of three equations: \[y_1 = Ax_1^2 + Bx_1 + C\], \[y_2 = Ax_2^2 + Bx_2 + C\] and \[y_3 = Ax_3^2 + Bx_3 + C\]. These equations represent the fact that \(p(x)\) passes through each of these points.

Step 2: Solving the System of Equations

Solve the system of equations, which is linear in the parameters \(A\), \(B\), and \(C\). There are various methods to solve a system of linear equations like substitution, elimination or using matrix operations.

Step 3: Forming the Quadratic Polynomial

Once the values of \(A\), \(B\), and \(C\) are found, they are substituted back into the polynomial \(p(x)=A x^{2}+B x+C\) to get the equation of the quadratic polynomial that passes through the three given points.

Key Concepts

Quadratic Polynomial

A quadratic polynomial is a simple yet powerful mathematical expression. It is of the form \(p(x) = Ax^{2} + Bx + C\), where \(A\), \(B\), and \(C\) are coefficients, and \(x\) is the variable. This polynomial can describe a wide range of curves, all parabolic in nature, providing elegant solutions in various real-world applications. For instance, they can model projectile motion, or the path of a thrown ball. The quadratic polynomial is uniquely defined by its coefficient values, and these dictate the shape and direction of its parabola. What makes quadratic polynomials especially interesting is their capability to perfectly fit any set of three distinct points in a coordinate plane. By determining the specific values of \(A\), \(B\), and \(C\), we can create a curve that passes exactly through these points. This process, part of a broader mathematical field called interpolation, provides a seamless way to predict or estimate intermediate values within the range of data points.

System of Equations

When working to define a quadratic polynomial that goes through specific points, we set up a system of equations. These equations emanate from substituting each point into the polynomial expression \(p(x) = Ax^{2} + Bx + C\).
To illustrate, suppose you have three points: \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\). Substituting each pair into the polynomial yields three separate equations:

• \(y_1 = Ax_1^2 + Bx_1 + C\)
• \(y_2 = Ax_2^2 + Bx_2 + C\)
• \(y_3 = Ax_3^2 + Bx_3 + C\)

These equations are linear in terms of the coefficients \(A\), \(B\), and \(C\), meaning the highest power of these variables is one. Solving this system gives us the coefficients we need for our quadratic polynomial.
This problem exemplifies how a system of linear equations can arise in many fields, requiring methods to find their solutions. By doing so, we harness the power to model complex behaviors and phenomena accurately.

Matrix Operations

Solving systems of equations, particularly when dealing with multiple equations and unknowns, often employs matrix operations for efficiency and clarity. Matrix methods streamline the process, especially for larger systems, though our scenario involves only three equations.

• The system derived from the quadratic polynomial equations can be represented in matrix form. Consider a matrix \(A\) formed by the coefficients of \(A, B, C\), and a column matrix \(X\) representing these unknown coefficients.
• The third element is a matrix \(Y\) that contains the constants from the right-hand side of our equations, i.e., \(y_1, y_2, y_3\).

This leads to the matrix equation \(AX = Y\), where \(A\) is a matrix of known values, \(X\) represents the coefficients \(A, B, C\), and \(Y\) is the matrix of outputs.
Using methods such as Gaussian elimination or matrix inversions, we can solve \(AX = Y\) and find our coefficients. Matrix operations not only simplify calculations but also provide a framework for visualizing and solving complex algebraic systems efficiently, even in dimensions beyond simple polynomials.