A method to find coefficients of a quadratic given 3 points in 6 multiplies and 2 divisions

find_quadratic_coeffs.py

def calculate_quadratic_coefficients(points):
    # Extract x and y values from the points
    x1, y1 = points[0]
    x2, y2 = points[1]
    x3, y3 = points[2]

    # Set up the determinant of the matrix
    D = (x1**2 * (x2 - x3) + 
         x2**2 * (x3 - x1) + 
         x3**2 * (x1 - x2))

    if D == 0:
        raise ValueError("The points do not form a valid quadratic polynomial (D is zero).")

    # Determinants for a, b, c
    Da = (y1 * (x2 - x3) + 
          y2 * (x3 - x1) + 
          y3 * (x1 - x2))

    Db = (x1**2 * (y2 - y3) + 
          x2**2 * (y3 - y1) + 
          x3**2 * (y1 - y2))

    Dc = (x1**2 * (x2 * y3 - x3 * y2) + 
          x2**2 * (x3 * y1 - x1 * y3) + 
          x3**2 * (x1 * y2 - x2 * y1))

    # Coefficients
    a = Da / D
    b = Db / D
    c = Dc / D

    return a, b, c

def calculate_quadratic_coefficients_2(points):
    # Extract x and y values from the points
    x1, y1 = points[0]
    x2, y2 = points[1]
    x3, y3 = points[2]

    # Fit a linear polynomial (a=0) to the endpoints 
    b = (y3-y1) / (x3 - x1)
    c = y1 - b*x1

    # Reduce y2 by the linear polynomial
    y2 = y2 - b*x2 - c

    # Because reduction caused endpoints to equal 0, they're roots of a polynomial y2 = a(x2-x1)(x2-x3). Calculate a
    a = y2 / ((x2-x1)*(x2-x3))
    # Calculate rest of coefficients and add back linear polynomial
    b = -a *(x1+x3) + b
    c = a*x1*x3 + c

    return a, b, c

# Example usage:
points = [(1, 2), (2, 3), (3, 5)]  # Replace with your points

try:
    a, b, c = calculate_quadratic_coefficients(points)
    print(f"Coefficients with Cramer's rule: a = {a}, b = {b}, c = {c}")
    a, b, c = calculate_quadratic_coefficients_2(points)
    print(f"Coefficients with new method: a = {a}, b = {b}, c = {c}")
except ValueError as e:
    print(e)