Finding Parabola Equation Using Quadratic Regression

by ADMIN 53 views

How to find the equation of the parabola going through the points (-1, -23), (-3, -67), and (2, 13) using quadratic regression?

To find the equation of a parabola that passes through three given points, we can use the method of quadratic regression. This method involves setting up a system of equations based on the standard form of a quadratic equation and then solving for the coefficients. Let's delve into the process using the points (-1, -23), (-3, -67), and (2, 13).

Understanding Quadratic Regression

Quadratic regression is a statistical method used to model the relationship between a dependent variable and one or more independent variables by fitting a quadratic equation to the observed data. In simpler terms, it helps us find the best-fit parabola for a given set of points. The general form of a quadratic equation is:

y=ax2+bx+cy = ax^2 + bx + c

where:

  • y is the dependent variable.
  • x is the independent variable.
  • a, b, and c are the coefficients we need to determine.

To find the specific quadratic equation that passes through the points (-1, -23), (-3, -67), and (2, 13), we need to substitute the x and y values of each point into the general form and create a system of three equations with three unknowns (a, b, and c). Solving this system will give us the values of the coefficients, thus defining the parabola.

Setting Up the Equations

Let's start by plugging in the coordinates of the first point, (-1, -23), into the quadratic equation:

23=a(1)2+b(1)+c-23 = a(-1)^2 + b(-1) + c

Simplifying, we get:

23=ab+c(Equation1)-23 = a - b + c (Equation 1)

Next, we substitute the coordinates of the second point, (-3, -67):

67=a(3)2+b(3)+c-67 = a(-3)^2 + b(-3) + c

Simplifying, we have:

67=9a3b+c(Equation2)-67 = 9a - 3b + c (Equation 2)

Finally, we plug in the coordinates of the third point, (2, 13):

13=a(2)2+b(2)+c13 = a(2)^2 + b(2) + c

Which simplifies to:

13=4a+2b+c(Equation3)13 = 4a + 2b + c (Equation 3)

Now we have a system of three linear equations:

  1. ab+c=23a - b + c = -23

  2. 9a3b+c=679a - 3b + c = -67

  3. 4a+2b+c=134a + 2b + c = 13

Solving the System of Equations

There are several methods to solve this system of equations, including substitution, elimination, and matrix methods. Let's use the elimination method for this example. The goal is to eliminate one variable at a time until we can solve for the remaining variables.

Step 1: Eliminate 'c'

We can eliminate c by subtracting Equation 1 from Equation 2 and Equation 1 from Equation 3.

Subtracting Equation 1 from Equation 2:

(9a3b+c)(ab+c)=67(23)(9a - 3b + c) - (a - b + c) = -67 - (-23)

8a2b=44(Equation4)8a - 2b = -44 (Equation 4)

Subtracting Equation 1 from Equation 3:

(4a+2b+c)(ab+c)=13(23)(4a + 2b + c) - (a - b + c) = 13 - (-23)

3a+3b=36(Equation5)3a + 3b = 36 (Equation 5)

Step 2: Simplify the Equations

We can simplify Equation 4 by dividing both sides by 2:

4ab=22(Equation6)4a - b = -22 (Equation 6)

And we can simplify Equation 5 by dividing both sides by 3:

a+b=12(Equation7)a + b = 12 (Equation 7)

Step 3: Eliminate 'b'

Now we can eliminate b by adding Equation 6 and Equation 7:

(4ab)+(a+b)=22+12(4a - b) + (a + b) = -22 + 12

5a=105a = -10

Dividing both sides by 5, we get:

a=2a = -2

Step 4: Solve for 'b'

Substitute the value of a into Equation 7:

2+b=12-2 + b = 12

Adding 2 to both sides, we get:

b=14b = 14

Step 5: Solve for 'c'

Substitute the values of a and b into Equation 1:

214+c=23-2 - 14 + c = -23

16+c=23-16 + c = -23

Adding 16 to both sides, we get:

c=7c = -7

The Quadratic Equation

Now that we have found the values of a, b, and c, we can write the quadratic equation:

y=2x2+14x7y = -2x^2 + 14x - 7

This is the equation of the parabola that passes through the points (-1, -23), (-3, -67), and (2, 13).

Verification

To verify our solution, we can plug the coordinates of the original points into the equation and see if they hold true. Let’s test each point:

  1. Point (-1, -23):

    y=2(1)2+14(1)7y = -2(-1)^2 + 14(-1) - 7

    y=2147y = -2 - 14 - 7

    y=23y = -23

    The equation holds true for the first point.

  2. Point (-3, -67):

    y=2(3)2+14(3)7y = -2(-3)^2 + 14(-3) - 7

    y=2(9)427y = -2(9) - 42 - 7

    y=18427y = -18 - 42 - 7

    y=67y = -67

    The equation holds true for the second point.

  3. Point (2, 13):

    y=2(2)2+14(2)7y = -2(2)^2 + 14(2) - 7

    y=2(4)+287y = -2(4) + 28 - 7

    y=8+287y = -8 + 28 - 7

    y=13y = 13

    The equation holds true for the third point.

Since the equation holds true for all three points, we have successfully found the quadratic equation that passes through these points using quadratic regression.

Applications of Quadratic Regression

Quadratic regression isn't just a mathematical exercise; it has numerous real-world applications. Here are a few examples:

  • Physics: Projectile motion often follows a parabolic path. Quadratic regression can be used to model the trajectory of a projectile, such as a ball thrown in the air.
  • Economics: Cost curves and revenue curves in economics can sometimes be modeled using quadratic functions. Quadratic regression can help businesses analyze their costs and revenues.
  • Engineering: In civil engineering, the shape of arches and bridges can be approximated using parabolas. Quadratic regression can be used in the design and analysis of these structures.
  • Data Analysis: In general, quadratic regression is a powerful tool for modeling data that exhibits a curved relationship between variables. It's used in various fields to find the best-fit parabolic curve for a given dataset.

Conclusion

In this article, we have demonstrated how to use quadratic regression to find the equation of a parabola that passes through three given points. The process involves setting up a system of equations and solving for the coefficients of the quadratic equation. This method is not only mathematically sound but also has practical applications in various fields. Understanding quadratic regression provides a valuable tool for modeling and analyzing curved relationships in data.

By following the steps outlined, you can confidently find the equation of a parabola given three points. Remember, practice makes perfect, so try applying this method to different sets of points to solidify your understanding. The resulting equation, in the form $y = ax^2 + bx + c$, provides a powerful representation of the parabolic relationship between x and y.