Math Problem Statement

Can you create a quadratic function that best estimates the curve passing through the points (0.2 , 0.063), (0.4, 0.167), (0.6, 0.313), (0.8, 0.52), (1, 1)

Solution

To create a quadratic function that best fits the curve passing through the given points, we can use the method of least squares or directly set up the quadratic function as:

f(x)=ax2+bx+cf(x) = ax^2 + bx + c

We have the following points:

(0.2,0.063),(0.4,0.167),(0.6,0.313),(0.8,0.52),(1,1)(0.2, 0.063), \, (0.4, 0.167), \, (0.6, 0.313), \, (0.8, 0.52), \, (1, 1)

These points will give us a system of equations when plugged into the quadratic form f(x)=ax2+bx+cf(x) = ax^2 + bx + c. Let's substitute each point into the general form and solve the resulting system:

Step 1: Form equations using the points

For x=0.2,y=0.063x = 0.2, y = 0.063: a(0.2)2+b(0.2)+c=0.063a(0.2)^2 + b(0.2) + c = 0.063 0.04a+0.2b+c=0.063(Equation 1)0.04a + 0.2b + c = 0.063 \quad \text{(Equation 1)}

For x=0.4,y=0.167x = 0.4, y = 0.167: a(0.4)2+b(0.4)+c=0.167a(0.4)^2 + b(0.4) + c = 0.167 0.16a+0.4b+c=0.167(Equation 2)0.16a + 0.4b + c = 0.167 \quad \text{(Equation 2)}

For x=0.6,y=0.313x = 0.6, y = 0.313: a(0.6)2+b(0.6)+c=0.313a(0.6)^2 + b(0.6) + c = 0.313 0.36a+0.6b+c=0.313(Equation 3)0.36a + 0.6b + c = 0.313 \quad \text{(Equation 3)}

For x=0.8,y=0.52x = 0.8, y = 0.52: a(0.8)2+b(0.8)+c=0.52a(0.8)^2 + b(0.8) + c = 0.52 0.64a+0.8b+c=0.52(Equation 4)0.64a + 0.8b + c = 0.52 \quad \text{(Equation 4)}

For x=1,y=1x = 1, y = 1: a(1)2+b(1)+c=1a(1)^2 + b(1) + c = 1 a+b+c=1(Equation 5)a + b + c = 1 \quad \text{(Equation 5)}

Step 2: Solve the system of equations

Now, we need to solve the system of 5 equations with 3 unknowns (a, b, and c). However, since we have more points than necessary for a quadratic (3 points should ideally be enough), we can use methods like least squares to find the best-fitting solution. Let me calculate that for you. numpy as np

Given points

points = np.array([[0.2, 0.063], [0.4, 0.167], [0.6, 0.313], [0.8, 0.52], [1, 1]])

Separate x and y

x = points[:, 0] y = points[:, 1]

Set up the design matrix for quadratic fitting (Ax = b, where A is the design matrix and x = [a, b, c])

A = np.vstack([x**2, x, np.ones_like(x)]).T

Solve for the coefficients using least squares

coeffs = np.linalg.lstsq(A, y, rcond=None)[0]

Coefficients a, b, c

coeffsThe quadratic function that best fits the given points, using the least squares method, is:

f(x)=1.4518x20.6286x+0.151f(x) = 1.4518x^2 - 0.6286x + 0.151

This function should provide a good estimate of the curve passing through the points (0.2,0.063)(0.2, 0.063), (0.4,0.167)(0.4, 0.167), (0.6,0.313)(0.6, 0.313), (0.8,0.52)(0.8, 0.52), and (1,1)(1, 1).

Would you like a graph of this quadratic curve along with the points to visually inspect the fit? Or do you have any further questions about quadratic regression?

Additional questions to explore:

  1. How do we interpret the coefficients of a quadratic function in real-world applications?
  2. What are the common methods for curve fitting besides least squares?
  3. How can we find the roots (x-intercepts) of the quadratic equation?
  4. How does changing the number of data points affect the fit of the quadratic function?
  5. Can a quadratic function always perfectly fit any set of data points?

Tip: When solving a system of equations from multiple points, the least squares method provides an optimal solution that minimizes the total error between the observed values and the predicted ones.

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Math Problem Analysis

Mathematical Concepts

Quadratic Functions
Least Squares Method
Curve Fitting

Formulas

f(x) = ax^2 + bx + c

Theorems

Least Squares Approximation

Suitable Grade Level

Grades 11-12