Find Parabola Equation With Quadratic Regression Tutorial

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Finding the equation of a parabola that passes through three given points is a common problem in mathematics, particularly in algebra and calculus. This article will explore how to solve this problem using quadratic regression. Quadratic regression is a statistical method used to find the quadratic equation that best fits a set of data points. In this case, the data points are the three coordinates $(-4, 7)$, $(6, -33)$, and $(10, -105)$. The general form of a quadratic equation is $y = ax^2 + bx + c$, where a, b, and c are constants. Our goal is to determine the values of these constants that define the parabola passing through the given points.

Understanding Quadratic Regression

In quadratic regression, we aim to find the quadratic equation that minimizes the sum of the squared differences between the observed values and the values predicted by the equation. This method is also known as the least squares method. For a parabola, this means finding the curve that best fits the given points by minimizing the vertical distances between the points and the curve. Unlike linear regression, which fits a straight line to the data, quadratic regression fits a curve, specifically a parabola, to the data. This is particularly useful when the data points exhibit a curved relationship, which is characteristic of quadratic functions. The process involves setting up a system of equations using the given points and then solving for the coefficients a, b, and c. These coefficients define the specific parabola that passes through the points, allowing us to predict other values along the curve. Understanding quadratic regression is crucial not only for solving mathematical problems but also for various real-world applications, such as modeling physical phenomena, economic trends, and engineering designs where curved relationships are prevalent.

Setting Up the Equations

The first step in finding the equation of the parabola is to set up a system of equations using the given points. We substitute the x and y coordinates of each point into the general form of the quadratic equation, $y = ax^2 + bx + c$. This will give us three equations with three unknowns: a, b, and c. Let's start with the first point, $(-4, 7)$. Substituting these values into the equation, we get $7 = a(-4)^2 + b(-4) + c$, which simplifies to $7 = 16a - 4b + c$. This is our first equation. Next, we use the second point, $(6, -33)$. Substituting these values, we get $-33 = a(6)^2 + b(6) + c$, which simplifies to $-33 = 36a + 6b + c$. This is our second equation. Finally, we use the third point, $(10, -105)$. Substituting these values, we get $-105 = a(10)^2 + b(10) + c$, which simplifies to $-105 = 100a + 10b + c$. This gives us our third equation. Now we have a system of three linear equations:

  1. 16a−4b+c=716a - 4b + c = 7

  2. 36a+6b+c=−3336a + 6b + c = -33

  3. 100a+10b+c=−105100a + 10b + c = -105

This system of equations can be solved using various methods, such as substitution, elimination, or matrix methods. The goal is to find the values of a, b, and c that satisfy all three equations simultaneously. Once we have these values, we can plug them back into the general quadratic equation to obtain the specific equation for the parabola passing through the given points. Setting up these equations correctly is a crucial step in solving the problem, as the accuracy of the solution depends on the correct formulation of the system of equations.

Solving the System of Equations

Now that we have the system of equations, we need to solve for a, b, and c. One common method for solving such systems is the elimination method. This involves eliminating one variable at a time by performing algebraic operations on the equations. Let's start by eliminating c from the first two equations. We can subtract the first equation from the second equation to achieve this: $(36a + 6b + c) - (16a - 4b + c) = -33 - 7$. This simplifies to $20a + 10b = -40$. We can further simplify this equation by dividing all terms by 10, resulting in $2a + b = -4$. Now, let's eliminate c from the second and third equations. We subtract the second equation from the third equation: $(100a + 10b + c) - (36a + 6b + c) = -105 - (-33)$. This simplifies to $64a + 4b = -72$. We can further simplify this equation by dividing all terms by 4, resulting in $16a + b = -18$. Now we have two equations with two variables:

  1. 2a+b=−42a + b = -4

  2. 16a+b=−1816a + b = -18

We can eliminate b by subtracting the first equation from the second equation: $(16a + b) - (2a + b) = -18 - (-4)$. This simplifies to $14a = -14$, so $a = -1$. Now that we have the value of a, we can substitute it back into one of the equations to find b. Let's use the first equation: $2(-1) + b = -4$, which gives $-2 + b = -4$, so $b = -2$. Finally, we can substitute the values of a and b into one of the original three equations to find c. Let's use the first original equation: $16(-1) - 4(-2) + c = 7$, which gives $-16 + 8 + c = 7$, so $-8 + c = 7$, and $c = 15$. Thus, we have found the values a = -1, b = -2, and c = 15. Solving the system of equations using the elimination method is a systematic approach that ensures we find the correct coefficients for the quadratic equation.

Final Equation and Verification

Now that we have found the values of a, b, and c, we can write the final equation for the parabola. Recall that the general form of a quadratic equation is $y = ax^2 + bx + c$. Substituting the values we found, $a = -1$, $b = -2$, and $c = 15$, we get the equation $y = -1x^2 - 2x + 15$, or simply $y = -x^2 - 2x + 15$. This is the equation of the parabola that passes through the points $(-4, 7)$, $(6, -33)$, and $(10, -105)$. To ensure that our solution is correct, it's important to verify the equation by plugging in the original points and checking if they satisfy the equation. Let's start with the point $(-4, 7)$. Substituting $x = -4$ into the equation, we get $y = -(-4)^2 - 2(-4) + 15 = -16 + 8 + 15 = 7$, which matches the given y-coordinate. Next, let's check the point $(6, -33)$. Substituting $x = 6$ into the equation, we get $y = -(6)^2 - 2(6) + 15 = -36 - 12 + 15 = -33$, which also matches the given y-coordinate. Finally, let's check the point $(10, -105)$. Substituting $x = 10$ into the equation, we get $y = -(10)^2 - 2(10) + 15 = -100 - 20 + 15 = -105$, which again matches the given y-coordinate. Since all three points satisfy the equation, we can confidently say that $y = -x^2 - 2x + 15$ is the correct equation for the parabola passing through these points. Verifying the solution is a crucial step in the problem-solving process, as it helps to catch any potential errors and ensures the accuracy of the final answer.

Conclusion

In conclusion, finding the equation of a parabola that passes through three given points involves using quadratic regression to determine the coefficients a, b, and c in the general quadratic equation $y = ax^2 + bx + c$. This process involves setting up a system of three linear equations by substituting the coordinates of the given points into the general equation. The system of equations can then be solved using methods such as substitution or elimination to find the values of a, b, and c. Once these values are found, they are substituted back into the general equation to obtain the specific equation for the parabola. It is crucial to verify the solution by plugging the original points back into the equation to ensure accuracy. This method is not only useful in mathematical contexts but also has practical applications in various fields where quadratic relationships are observed. Understanding and applying quadratic regression allows us to model and predict outcomes based on curved relationships in data, making it a valuable tool in mathematics and beyond. By following these steps, we can accurately determine the equation of a parabola given three points, enhancing our understanding of quadratic functions and their applications.