Learn on PengiYoshiwara Intermediate AlgebraChapter 4: Applications of Quadratic Models

Lesson 3: Curve Fitting

In this Grade 7 lesson from Yoshiwara Intermediate Algebra, students learn how to fit a quadratic equation to a set of data points by solving a system of three linear equations using Gaussian reduction to find the parameters a, b, and c in y = ax² + bx + c. The lesson builds on prior work with linear regression and extends it to quadratic models, walking students through both the algebraic process and real-world applications such as modeling driving costs at different speeds. By the end, students can select three data points, set up a 3×3 system, and determine the unique parabola passing through those points.

Section 1

📘 Curve Fitting

New Concept

Curve fitting finds a quadratic model, y=ax2+bx+cy = ax^2 + bx + c, for a set of data. Just as two points define a line, three non-collinear points define a unique parabola by creating a system of equations we can solve.

What’s next

Next, you’ll master this concept through practice cards on solving these systems, then use interactive examples to perform quadratic regression on real-world data.

Section 2

Quadratic equation through three points

Property

A quadratic equation can be written in the form

y=ax2+bx+cy = ax^2 + bx + c
To find the three parameters aa, bb, and cc, you need three data points. Substituting the coordinates of each of the three points into the equation of the parabola creates a system of three linear equations in the three unknowns aa, bb, and cc, which can then be solved.

Examples

  • To find the parabola through (1,3)(1, 3), (3,5)(3, 5), and (4,9)(4, 9), we solve the system: a+b+c=3a + b + c = 3, 9a+3b+c=59a + 3b + c = 5, and 16a+4b+c=916a + 4b + c = 9. The solution is a=1,b=3,c=5a=1, b=-3, c=5, so the equation is y=x23x+5y = x^2 - 3x + 5.
  • A parabola passes through (0,8)(0, 8), (1,5)(1, 5), and (2,6)(2, 6). The system is c=8c = 8, a+b+c=5a+b+c=5, and 4a+2b+c=64a+2b+c=6. Substituting c=8c=8 gives a+b=3a+b=-3 and 4a+2b=24a+2b=-2. The solution is a=2,b=5,c=8a=2, b=-5, c=8, so y=2x25x+8y = 2x^2 - 5x + 8.

Section 3

Fitting a parabola to data

Property

The simplest way to fit a parabola to a set of data points is to pick three of the points and find the equation of the parabola that passes through those three points. This creates a quadratic model, y=ax2+bx+cy = ax^2 + bx + c, for the relationship shown in the data.

Examples

  • A ball's height is measured at three times: (1,21)(1, 21), (2,24)(2, 24), and (3,21)(3, 21). Fitting a parabola gives the model h=3t2+12t+12h = -3t^2 + 12t + 12, which describes the ball's trajectory.
  • Data for driving cost at different speeds are (50,6.20)(50, 6.20), (60,7.80)(60, 7.80), and (70,10.60)(70, 10.60). We can fit a parabola C=av2+bv+cC = av^2 + bv + c to model how cost changes with speed, finding C=0.006v20.5v+16.2C = 0.006v^2 - 0.5v + 16.2.

Section 4

Quadratic regression with a calculator

Property

A graphing calculator can find the quadratic regression equation, y=ax2+bx+cy = ax^2 + bx + c, that provides the best fit for a set of data. This is done by entering the data points into lists and using the quadratic regression command (often QuadReg). The calculator computes the parameters aa, bb, and cc that minimize the overall error for all data points.

Examples

  • Given data for speed versus driving cost, enter speeds in list L1 and costs in L2. Using the QuadReg function yields the regression equation C=0.0057v20.47v+15.56C = 0.0057v^2 - 0.47v + 15.56, which models the overall trend.
  • For data on radiation dosage versus egg survival, a calculator's QuadReg provides the best-fit model, such as y=(3.65×107)x20.001x+1.02y = (3.65 \times 10^{-7})x^2 - 0.001x + 1.02, without needing to solve equations manually.

Section 5

Choosing an appropriate model

Property

Using the wrong type of function to fit the data is a common error in making predictions. Even though a calculator can always compute a regression equation, that equation is not necessarily appropriate for your data. The chosen model must be reasonable for the situation it describes.

Examples

  • Modeling the height of a clock's minute hand with a quadratic function works for a short time but fails over an hour. The motion is periodic, so a different type of function would be a more appropriate model.
  • A quadratic regression might fit a company's stock price for one week, but it cannot be trusted for long-term prediction because market behavior is far too complex and not inherently parabolic.

Book overview

Jump across lessons in the current chapter without opening the full course modal.

Continue this chapter

Chapter 4: Applications of Quadratic Models

  1. Lesson 1

    Lesson 1: Quadratic Formula

  2. Lesson 2

    Lesson 2: The Vertex

  3. Lesson 3Current

    Lesson 3: Curve Fitting

  4. Lesson 4

    Lesson 4: Quadratic Inequalities

  5. Lesson 5

    Lesson 5: Chapter Summary and Review

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

📘 Curve Fitting

New Concept

Curve fitting finds a quadratic model, y=ax2+bx+cy = ax^2 + bx + c, for a set of data. Just as two points define a line, three non-collinear points define a unique parabola by creating a system of equations we can solve.

What’s next

Next, you’ll master this concept through practice cards on solving these systems, then use interactive examples to perform quadratic regression on real-world data.

Section 2

Quadratic equation through three points

Property

A quadratic equation can be written in the form

y=ax2+bx+cy = ax^2 + bx + c
To find the three parameters aa, bb, and cc, you need three data points. Substituting the coordinates of each of the three points into the equation of the parabola creates a system of three linear equations in the three unknowns aa, bb, and cc, which can then be solved.

Examples

  • To find the parabola through (1,3)(1, 3), (3,5)(3, 5), and (4,9)(4, 9), we solve the system: a+b+c=3a + b + c = 3, 9a+3b+c=59a + 3b + c = 5, and 16a+4b+c=916a + 4b + c = 9. The solution is a=1,b=3,c=5a=1, b=-3, c=5, so the equation is y=x23x+5y = x^2 - 3x + 5.
  • A parabola passes through (0,8)(0, 8), (1,5)(1, 5), and (2,6)(2, 6). The system is c=8c = 8, a+b+c=5a+b+c=5, and 4a+2b+c=64a+2b+c=6. Substituting c=8c=8 gives a+b=3a+b=-3 and 4a+2b=24a+2b=-2. The solution is a=2,b=5,c=8a=2, b=-5, c=8, so y=2x25x+8y = 2x^2 - 5x + 8.

Section 3

Fitting a parabola to data

Property

The simplest way to fit a parabola to a set of data points is to pick three of the points and find the equation of the parabola that passes through those three points. This creates a quadratic model, y=ax2+bx+cy = ax^2 + bx + c, for the relationship shown in the data.

Examples

  • A ball's height is measured at three times: (1,21)(1, 21), (2,24)(2, 24), and (3,21)(3, 21). Fitting a parabola gives the model h=3t2+12t+12h = -3t^2 + 12t + 12, which describes the ball's trajectory.
  • Data for driving cost at different speeds are (50,6.20)(50, 6.20), (60,7.80)(60, 7.80), and (70,10.60)(70, 10.60). We can fit a parabola C=av2+bv+cC = av^2 + bv + c to model how cost changes with speed, finding C=0.006v20.5v+16.2C = 0.006v^2 - 0.5v + 16.2.

Section 4

Quadratic regression with a calculator

Property

A graphing calculator can find the quadratic regression equation, y=ax2+bx+cy = ax^2 + bx + c, that provides the best fit for a set of data. This is done by entering the data points into lists and using the quadratic regression command (often QuadReg). The calculator computes the parameters aa, bb, and cc that minimize the overall error for all data points.

Examples

  • Given data for speed versus driving cost, enter speeds in list L1 and costs in L2. Using the QuadReg function yields the regression equation C=0.0057v20.47v+15.56C = 0.0057v^2 - 0.47v + 15.56, which models the overall trend.
  • For data on radiation dosage versus egg survival, a calculator's QuadReg provides the best-fit model, such as y=(3.65×107)x20.001x+1.02y = (3.65 \times 10^{-7})x^2 - 0.001x + 1.02, without needing to solve equations manually.

Section 5

Choosing an appropriate model

Property

Using the wrong type of function to fit the data is a common error in making predictions. Even though a calculator can always compute a regression equation, that equation is not necessarily appropriate for your data. The chosen model must be reasonable for the situation it describes.

Examples

  • Modeling the height of a clock's minute hand with a quadratic function works for a short time but fails over an hour. The motion is periodic, so a different type of function would be a more appropriate model.
  • A quadratic regression might fit a company's stock price for one week, but it cannot be trusted for long-term prediction because market behavior is far too complex and not inherently parabolic.

Book overview

Jump across lessons in the current chapter without opening the full course modal.

Continue this chapter

Chapter 4: Applications of Quadratic Models

  1. Lesson 1

    Lesson 1: Quadratic Formula

  2. Lesson 2

    Lesson 2: The Vertex

  3. Lesson 3Current

    Lesson 3: Curve Fitting

  4. Lesson 4

    Lesson 4: Quadratic Inequalities

  5. Lesson 5

    Lesson 5: Chapter Summary and Review