How To Find A Quadratic Equation With Three Points

How to Find a Quadratic Equation with Three Points

Finding the equation of a quadratic function given three points is a common problem in algebra and has applications in various fields, including physics, engineering, and data analysis. A quadratic equation, represented generally as y = ax² + bx + c, describes a parabola. With three points, we have enough information to solve for the three unknowns: a, b, and c. This article will guide you through various methods to achieve this, catering to different levels of mathematical understanding.

Understanding the Fundamentals

Before we delve into the methods, let's solidify our understanding of the core concepts:

• Quadratic Equation: A quadratic equation is an equation of the form y = ax² + bx + c, where 'a', 'b', and 'c' are constants, and 'a' is not equal to zero. The graph of a quadratic equation is a parabola.
• Points: A point on the graph of a quadratic equation is represented by an ordered pair (x, y) that satisfies the equation. Given three points, we have three equations with three unknowns.
• Solving Systems of Equations: The core of finding the quadratic equation lies in solving a system of three simultaneous linear equations. These equations arise by substituting the x and y coordinates of each point into the general quadratic equation.

Method 1: Using a System of Equations

This method involves creating a system of three linear equations and then solving them using techniques like substitution, elimination, or matrix methods.

1. Set up the Equations:

Let's assume our three points are (x₁, y₁), (x₂, y₂), and (x₃, y₃). Substituting these points into the general quadratic equation y = ax² + bx + c, we get the following system of equations:

• a(x₁)² + b(x₁) + c = y₁
• a(x₂)² + b(x₂) + c = y₂
• a(x₃)² + b(x₃) + c = y₃

2. Solve the System of Equations:

This system of equations can be solved using various methods:

• Substitution: Solve one equation for one variable and substitute it into the other two equations. This reduces the system to two equations with two variables. Repeat the process until you have a solution for a, b, and c.
• Elimination: Manipulate the equations to eliminate one variable at a time. This involves adding or subtracting multiples of the equations to cancel out terms.
• Matrix Methods: Represent the system of equations in matrix form (Ax = B), where A is the coefficient matrix, x is the variable matrix (containing a, b, and c), and B is the constant matrix. Solve for x using matrix inversion or Gaussian elimination (row reduction). This method is particularly efficient for larger systems of equations.

Example:

Let's find the quadratic equation for the points (1, 2), (2, 3), and (3, 6).

1. Equations:

   • a(1)² + b(1) + c = 2 => a + b + c = 2
   • a(2)² + b(2) + c = 3 => 4a + 2b + c = 3
   • a(3)² + b(3) + c = 6 => 9a + 3b + c = 6

2. Solving (using elimination): Subtract the first equation from the second and third equations:

   • 3a + b = 1
   • 8a + 2b = 4

   Simplify the second equation by dividing by 2:

   • 4a + b = 2

   Subtract the equation 3a + b = 1 from 4a + b = 2:

   • a = 1

   Substitute a = 1 into 3a + b = 1:

   • 3(1) + b = 1 => b = -2

   Substitute a = 1 and b = -2 into a + b + c = 2:

   • 1 + (-2) + c = 2 => c = 3

Therefore, the quadratic equation is y = x² - 2x + 3.

Method 2: Using Lagrange Interpolation

Lagrange interpolation is a powerful method for finding a polynomial that passes through a given set of points. While it can be applied to polynomials of any degree, it's particularly useful for finding quadratic equations.

The Lagrange interpolating polynomial for three points (x₁, y₁), (x₂, y₂), and (x₃, y₃) is given by:

y = y₁ * L₁(x) + y₂ * L₂(x) + y₃ * L₃(x)

where:

• L₁(x) = [(x - x₂)(x - x₃)] / [(x₁ - x₂)(x₁ - x₃)]
• L₂(x) = [(x - x₁)(x - x₃)] / [(x₂ - x₁)(x₂ - x₃)]
• L₃(x) = [(x - x₁)(x - x₂)] / [(x₃ - x₁)(x₃ - x₂)]

Example:

Let's use the same points as before: (1, 2), (2, 3), and (3, 6).

1. Calculate L₁(x), L₂(x), and L₃(x):

   • L₁(x) = [(x - 2)(x - 3)] / [(1 - 2)(1 - 3)] = (x² - 5x + 6) / 2
   • L₂(x) = [(x - 1)(x - 3)] / [(2 - 1)(2 - 3)] = (x² - 4x + 3) / (-1) = -x² + 4x - 3
   • L₃(x) = [(x - 1)(x - 2)] / [(3 - 1)(3 - 2)] = (x² - 3x + 2) / 2

2. Substitute into the Lagrange interpolation formula:

   y = 2 * [(x² - 5x + 6) / 2] + 3 * [-x² + 4x - 3] + 6 * [(x² - 3x + 2) / 2]

3. Simplify:

   y = x² - 5x + 6 - 3x² + 12x - 9 + 3x² - 9x + 6 y = x² - 2x + 3

This confirms the result obtained using the system of equations method.

Method 3: Using a Calculator or Software

Many graphing calculators and mathematical software packages (like MATLAB, Python with NumPy/SciPy, etc.) have built-in functions or tools to fit a quadratic curve to a set of points. These tools often use more sophisticated numerical methods than the ones described above, providing a quick and efficient solution, especially for a large number of points or complex datasets. These tools typically employ techniques like least squares regression, which finds the best-fitting quadratic curve even if the points don't perfectly lie on a parabola.

Choosing the Right Method

The best method depends on several factors:

• Mathematical Background: If you're comfortable with solving systems of equations, that's a straightforward and readily understandable approach.
• Number of Points: For three points, all methods are relatively easy. For a larger number of points, Lagrange interpolation becomes less efficient, while least-squares regression methods employed by calculators or software are preferable.
• Accuracy: If the points are obtained experimentally and may contain some error, least squares regression provides a more robust solution by minimizing the overall error.
• Available Tools: If you have access to a calculator or software with curve-fitting capabilities, that is the most convenient option.

Applications of Finding Quadratic Equations

The ability to find a quadratic equation from three points has numerous practical applications:

• Projectile Motion: In physics, the trajectory of a projectile under the influence of gravity follows a parabolic path, which can be modeled using a quadratic equation.
• Engineering Design: Quadratic equations are used in designing curves for roads, bridges, and other structures.
• Data Analysis: Fitting a quadratic curve to data points helps identify trends and make predictions.
• Computer Graphics: Quadratic curves are used extensively in computer graphics to create smooth, curved shapes.
• Economics: Quadratic functions can model cost, revenue, or profit functions in various economic scenarios.

Conclusion

Finding a quadratic equation given three points is a fundamental skill with significant practical applications. This article has presented three methods: solving systems of equations, using Lagrange interpolation, and utilizing calculators/software. Choosing the appropriate method depends on your mathematical background, the number of points involved, the desired accuracy, and the available resources. Regardless of the method chosen, the ability to determine a quadratic equation from given data points is a valuable tool across diverse fields. Remember to always check your solution by substituting the original points back into the equation to verify accuracy.