Use Quadratic Regression To Find The Equation For The Parabola Going Through Three Points. Find The Values For $a$, $b$, And $c$ In The Equation: $y = Ax^2 + Bx + C$
Finding the equation of a parabola that passes through three given points is a common problem in mathematics, particularly in algebra and calculus. Quadratic regression is the method we use, and it involves determining the coefficients a, b, and c in the quadratic equation y = ax² + bx + c. This article will walk you through the process step by step, using the points (-5, 161), (-2, 29), and (6, 205) as an example. Understanding quadratic regression is essential not only for academic purposes but also for various real-world applications such as physics, engineering, and economics, where parabolic relationships frequently occur.
Understanding Quadratic Equations
To begin, let's clarify the basics of quadratic equations. A quadratic equation is a polynomial equation of the second degree, meaning the highest power of the variable is 2. The general form of a quadratic equation is y = ax² + bx + c, where a, b, and c are constants, and x and y are variables. The graph of a quadratic equation is a parabola, a U-shaped curve that opens upwards if a > 0 and downwards if a < 0. The coefficients a, b, and c determine the shape and position of the parabola in the coordinate plane. The coefficient a affects the width and direction of the parabola, b influences the position of the axis of symmetry, and c determines the y-intercept. Understanding these fundamental properties is crucial for finding the specific equation that fits a given set of points. For instance, if we have three points, we can substitute their x and y coordinates into the general form to create a system of three equations. Solving this system will give us the values of a, b, and c, thus defining the unique parabola that passes through those points. This method is widely used in various fields, from physics, where parabolic trajectories describe the motion of projectiles, to economics, where quadratic functions can model cost curves and revenue functions. Hence, a solid grasp of quadratic equations and their graphical representation as parabolas is not just an academic exercise but a practical skill applicable in diverse scenarios.
Setting Up the Equations
In this specific problem, we are given three points: (-5, 161), (-2, 29), and (6, 205). To find the equation of the parabola that passes through these points, we need to determine the values of a, b, and c in the equation y = ax² + bx + c. We can do this by substituting the x and y coordinates of each point into the equation, creating a system of three linear equations with three unknowns. For the point (-5, 161), substituting x = -5 and y = 161 gives us the equation 161 = a(-5)² + b(-5) + c, which simplifies to 161 = 25a - 5b + c. Similarly, for the point (-2, 29), we substitute x = -2 and y = 29 to get 29 = a(-2)² + b(-2) + c, which simplifies to 29 = 4a - 2b + c. Finally, for the point (6, 205), substituting x = 6 and y = 205 gives us 205 = a(6)² + b(6) + c, which simplifies to 205 = 36a + 6b + c. Now we have a system of three equations:
- 161 = 25a - 5b + c
- 29 = 4a - 2b + c
- 205 = 36a + 6b + c
Solving this system will provide us with the values of a, b, and c, which will define the equation of the parabola. The next step involves using algebraic techniques such as substitution, elimination, or matrix methods to solve this system of equations. The accuracy in setting up these equations is paramount because any error here will propagate through the rest of the solution. Therefore, careful substitution and simplification are essential to ensure the correct quadratic equation is derived.
Solving the System of Equations
Now that we have our system of equations, we need to solve for a, b, and c. There are several methods to solve such systems, including substitution, elimination, and matrix methods. For this example, we will use the elimination method, which involves adding or subtracting multiples of the equations to eliminate variables one by one. First, let’s eliminate c from the equations. Subtract equation (2) from equation (1):
(161 = 25a - 5b + c) - (29 = 4a - 2b + c)
This gives us:
132 = 21a - 3b
Next, subtract equation (2) from equation (3):
(205 = 36a + 6b + c) - (29 = 4a - 2b + c)
This results in:
176 = 32a + 8b
Now we have a system of two equations with two variables:
- 132 = 21a - 3b
- 176 = 32a + 8b
To eliminate b, multiply the first equation by 8 and the second equation by 3:
- 1056 = 168a - 24b
- 528 = 96a + 24b
Add these two equations:
1584 = 264a
Divide by 264 to solve for a:
a = 1584 / 264 = 6
Now that we have a, we can substitute it back into one of the two-variable equations to solve for b. Let's use the equation 132 = 21a - 3b:
132 = 21(6) - 3b
132 = 126 - 3b
6 = -3b
b = -2
Finally, substitute a and b into any of the original three equations to solve for c. Let's use equation (2): 29 = 4a - 2b + c:
29 = 4(6) - 2(-2) + c
29 = 24 + 4 + c
29 = 28 + c
c = 1
Therefore, the values are a = 6, b = -2, and c = 1. This systematic approach to solving the equations ensures accuracy and efficiency in finding the coefficients of the quadratic equation.
The Quadratic Equation
Having found the values of a, b, and c, we can now write the quadratic equation for the parabola that passes through the points (-5, 161), (-2, 29), and (6, 205). We found that a = 6, b = -2, and c = 1. Substituting these values into the general form of a quadratic equation, y = ax² + bx + c, we get:
y = 6x² - 2x + 1
This equation represents the unique parabola that goes through the three given points. To verify this, we can substitute the x-coordinates of the given points into the equation and check if the resulting y-coordinates match the original points. For example, let's check the point (-5, 161):
y = 6(-5)² - 2(-5) + 1
y = 6(25) + 10 + 1
y = 150 + 10 + 1
y = 161
The y-coordinate matches, so the equation holds true for the point (-5, 161). Similarly, we can check the other two points:
For (-2, 29):
y = 6(-2)² - 2(-2) + 1
y = 6(4) + 4 + 1
y = 24 + 4 + 1
y = 29
For (6, 205):
y = 6(6)² - 2(6) + 1
y = 6(36) - 12 + 1
y = 216 - 12 + 1
y = 205
In all three cases, the calculated y-coordinates match the given y-coordinates, confirming that the equation y = 6x² - 2x + 1 is indeed the correct quadratic equation for the parabola passing through the specified points. This verification step is crucial to ensure the accuracy of the solution and demonstrates the practical application of the equation.
Applications of Quadratic Regression
Quadratic regression is a powerful tool with numerous applications in various fields. Beyond the mathematical exercises, understanding how to derive a quadratic equation from given points is highly valuable in real-world scenarios. In physics, for instance, the trajectory of a projectile under the influence of gravity follows a parabolic path. If we know the projectile's position at three different times, we can use quadratic regression to determine the equation of its path. This can help in predicting where the projectile will land or at what time it will reach a certain point. Engineering also benefits significantly from quadratic regression. Engineers often need to model curves and shapes, such as the curve of a bridge arch or the shape of a reflective surface in a telescope. If they have a few data points, quadratic regression can provide a precise equation for these curves, which is crucial for design and construction. Economics is another field where quadratic regression is useful. Cost and revenue curves, for example, can often be modeled using quadratic functions. By analyzing historical data points, economists can use regression to estimate these curves and make predictions about future costs and revenues. This is essential for business planning and financial analysis. In statistics and data analysis, quadratic regression is used to model relationships between variables when a linear model is not sufficient. For instance, the relationship between advertising expenditure and sales might follow a quadratic curve, with sales increasing rapidly up to a point and then leveling off or even decreasing as expenditure increases further. By fitting a quadratic regression model, analysts can gain a better understanding of these complex relationships. Furthermore, quadratic regression finds applications in computer graphics and animation, where smooth curves are often needed to represent shapes and movements. By defining a few key points, animators can use quadratic equations to generate smooth, natural-looking curves. Therefore, understanding and applying quadratic regression is not only an important mathematical skill but also a versatile tool for problem-solving in many diverse fields.
Conclusion
In conclusion, quadratic regression is a valuable technique for finding the equation of a parabola that passes through a given set of points. By setting up a system of equations and solving for the coefficients a, b, and c in the quadratic equation y = ax² + bx + c, we can accurately model parabolic relationships. In our example, we successfully found the equation y = 6x² - 2x + 1 for the parabola passing through the points (-5, 161), (-2, 29), and (6, 205). This process involves several key steps: first, understanding the general form of a quadratic equation and the properties of a parabola; second, setting up a system of linear equations by substituting the given points into the general form; third, solving the system of equations using methods such as elimination or substitution; and finally, verifying the solution by plugging the original points back into the derived equation. The applications of quadratic regression are widespread, spanning from physics and engineering to economics and computer graphics. Its ability to model curved relationships makes it an essential tool for analyzing and predicting real-world phenomena. Mastering quadratic regression not only enhances mathematical proficiency but also provides a practical skill applicable in numerous professional fields. Whether it's predicting projectile motion, designing curved structures, or modeling economic trends, the principles of quadratic regression offer a powerful approach to problem-solving. Therefore, a thorough understanding of this technique is highly beneficial for students and professionals alike, enabling them to tackle complex challenges with greater confidence and accuracy.