How To Find A Quadratic Equation From 3 Points

How to Find a Quadratic Equation from 3 Points

Finding the equation of a quadratic function given three points is a common problem in algebra and has various applications in fields like physics, engineering, and data analysis. A quadratic equation, in its standard form, is represented as ax² + bx + c = y, where 'a', 'b', and 'c' are constants, and 'a' is not equal to zero. Determining these constants is the key to finding the specific quadratic equation that passes through your given points. This article will guide you through several methods to achieve this, explaining the underlying principles and offering practical examples.

Understanding the Problem

Before diving into the solution methods, let's clarify the problem. We are given three distinct points, (x₁, y₁), (x₂, y₂), and (x₃, y₃). Our goal is to find the values of 'a', 'b', and 'c' such that these three points satisfy the quadratic equation ax² + bx + c = y. This means substituting each point's x and y coordinates into the equation should result in a true statement. This leads to a system of three simultaneous linear equations.

Method 1: Solving a System of Equations

This is the most straightforward approach. By substituting each point into the quadratic equation, we obtain three equations:

• Equation 1: a(x₁)² + b(x₁) + c = y₁
• Equation 2: a(x₂)² + b(x₂) + c = y₂
• Equation 3: a(x₃)² + b(x₃) + c = y₃

This system of three equations with three unknowns (a, b, and c) can be solved using various techniques, including substitution, elimination, or matrices (using Cramer's Rule or Gaussian elimination). Let's illustrate with an example using elimination:

Example:

Let's say we have the points (1, 2), (2, 3), and (3, 6). Substituting these points into the general quadratic equation, we get:

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

Now, we can solve this system using elimination. Subtracting Equation 1 from Equation 2 and Equation 2 from Equation 3 gives us:

• Equation 4: 3a + b = 1 (Subtracting Equation 1 from Equation 2)
• Equation 5: 5a + b = 3 (Subtracting Equation 2 from Equation 3)

Subtracting Equation 4 from Equation 5 gives:

• Equation 6: 2a = 2 => a = 1

Substituting a = 1 into Equation 4:

• 3(1) + b = 1 => b = -2

Finally, substituting a = 1 and b = -2 into Equation 1:

• 1 + (-2) + c = 2 => c = 3

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

Method 2: Using Matrices and Determinants (Cramer's Rule)

This method provides a more systematic and efficient way to solve the system of equations, especially when dealing with more complex scenarios. We can represent the system of equations in matrix form:

| x₁²  x₁  1 |   | a |   | y₁ |
| x₂²  x₂  1 | * | b | = | y₂ |
| x₃²  x₃  1 |   | c |   | y₃ |

Cramer's Rule allows us to solve for a, b, and c using determinants. The solution is given by:

• a = Det(A₁)/Det(A)
• b = Det(A₂)/Det(A)
• c = Det(A₃)/Det(A)

Where:

• Det(A) is the determinant of the coefficient matrix.
• Det(A₁), Det(A₂), and Det(A₃) are determinants of matrices formed by replacing the respective column of the coefficient matrix with the constant vector (y₁, y₂, y₃).

Calculating determinants can be tedious for larger systems, but for a 3x3 matrix, it's manageable. This method is particularly useful when utilizing computational tools or programming languages like Python (with NumPy) or MATLAB.

Method 3: Lagrange Interpolation (Generalized Approach)

While the previous methods directly solve for the coefficients, Lagrange interpolation offers a more generalized approach for finding polynomials that pass through a given set of points. It's especially useful when dealing with polynomials of higher degrees. For a quadratic equation (degree 2), the Lagrange interpolation formula is:

P₂(x) = y₁L₁(x) + y₂L₂(x) + y₃L₃(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)]

Substituting the given points (x₁, y₁), (x₂, y₂), and (x₃, y₃) into these formulas, we can obtain the quadratic equation P₂(x). This approach avoids solving simultaneous equations, offering an alternative route to finding the quadratic.

Example:

Using the same points as before (1, 2), (2, 3), and (3, 6):

1. Calculate L₁(x), L₂(x), and L₃(x) using the formulas above.
2. Substitute the values of L₁(x), L₂(x), L₃(x), and the y-coordinates into the Lagrange interpolation formula.
3. Simplify the resulting expression to obtain the quadratic equation.

This will again yield the same quadratic equation: y = x² - 2x + 3.

Choosing the Best Method

The best method depends on the context and your preference.

• System of Equations: Suitable for simple problems and easy to understand conceptually. However, it can become cumbersome for complex systems.
• Matrices and Determinants: Efficient for larger systems and easily implemented using computational tools. Requires familiarity with matrix operations.
• Lagrange Interpolation: Provides a generalized approach for polynomial interpolation, but the calculations can be more complex than directly solving the equations, especially for higher-degree polynomials.

Verifying your Solution

After finding the quadratic equation, it's crucial to verify your solution by substituting the original points back into the equation. If all three points satisfy the equation, then your solution is correct. This step ensures accuracy and helps catch any potential calculation errors.

Applications

Finding a quadratic equation from three points has wide-ranging applications:

• Physics: Modeling projectile motion, where the trajectory can often be approximated by a parabola.
• Engineering: Designing curves for roads, bridges, and other structures.
• Data Analysis: Fitting a quadratic model to data points to understand trends and make predictions.
• Computer Graphics: Creating smooth curves and surfaces in computer-generated images.
• Economics: Modelling certain economic relationships that show quadratic behavior.

Conclusion

Determining the quadratic equation from three points is a fundamental concept with practical implications across various disciplines. This article presented three distinct methods – solving a system of equations, using matrices and determinants, and employing Lagrange interpolation. Choosing the most suitable method depends on individual preference and the complexity of the problem. Remember to always verify your solution to ensure accuracy. Mastering these techniques will enhance your problem-solving skills and expand your understanding of quadratic functions and their applications. By understanding these methods, you can confidently tackle problems involving quadratic equations and their real-world applications. Remember that practice is key; working through various examples will solidify your understanding and build your confidence in applying these techniques.