Finding Parabola Equation With Quadratic Regression

Hey guys! Ever wondered how to find the equation of a parabola when you're given three points it passes through? It might sound tricky, but with quadratic regression, it's totally doable. In this article, we're going to break down the process step-by-step, so you can confidently tackle these problems. We'll use the points (-7, 134), (-3, 10), and (5, 50) as our example. So, buckle up, and let's dive in!

Understanding Quadratic Regression

Quadratic regression is your best friend when you need to find the equation of a parabola. Remember, a parabola is a U-shaped curve, and its equation is a quadratic equation in the form of y = ax² + bx + c. Our main goal here is to find the values of a, b, and c. Each of these coefficients plays a crucial role in defining the shape and position of the parabola. The coefficient a determines whether the parabola opens upwards (if a > 0) or downwards (if a < 0) and how wide or narrow it is. The coefficient b affects the parabola's axis of symmetry and its horizontal position. The constant c represents the y-intercept, which is the point where the parabola intersects the y-axis.

To find these coefficients, we need at least three points on the parabola. Each point (x, y) gives us an equation when plugged into the quadratic form. So, with three points, we get a system of three equations, which we can solve to find a, b, and c. This system of equations essentially captures the unique relationship between the x and y coordinates of the points and the parabolic shape they define. The process involves setting up these equations, which are linear in terms of a, b, and c, and then using algebraic methods to solve for these unknowns. This method provides a precise way to determine the quadratic equation that fits the given data points, ensuring that the resulting parabola passes exactly through these points.

Setting Up the Equations

Alright, let's get our hands dirty! We have three points: (-7, 134), (-3, 10), and (5, 50). We'll plug each of these into the quadratic equation y = ax² + bx + c to create our system of equations. This is where the magic begins, guys! For each point, we replace x and y with their respective values and form a new equation. This process converts our geometric problem into an algebraic one, where we can use our skills in solving systems of equations to find the coefficients of the parabola.

1. Using (-7, 134):

   • 134 = a(-7)² + b(-7) + c
   • 134 = 49a - 7b + c

2. Using (-3, 10):

   • 10 = a(-3)² + b(-3) + c
   • 10 = 9a - 3b + c

3. Using (5, 50):

   • 50 = a(5)² + b(5) + c
   • 50 = 25a + 5b + c

Now we have a system of three equations:

• 49a - 7b + c = 134
• 9a - 3b + c = 10
• 25a + 5b + c = 50

This system might look intimidating, but don't worry! We've got this. The next step is to solve this system for a, b, and c. There are several methods we can use, such as substitution, elimination, or even matrices. Each method has its advantages, and the choice often depends on personal preference and the specific structure of the equations. The goal is to systematically eliminate variables until we can solve for one, and then back-substitute to find the others. This algebraic process is at the heart of finding the unique parabola that passes through our given points.

Solving the System of Equations

Okay, let's tackle this system of equations. There are several ways to solve it, but we'll use the elimination method. This method involves strategically adding or subtracting multiples of equations to eliminate variables. It’s like a mathematical puzzle where we manipulate the equations to reveal the values of our unknowns. First, we'll eliminate c from the equations. This will reduce our system from three equations in three variables to two equations in two variables, which is much easier to handle. Once we have these two equations, we can apply the elimination method again to solve for one of the remaining variables. Finally, we can back-substitute the values we’ve found to determine the other variables.

Step 1: Eliminate c

Subtract the second equation from the first:

(49a - 7b + c) - (9a - 3b + c) = 134 - 10

40a - 4b = 124

Divide by 4 to simplify:

10a - b = 31 (Equation 4)

Subtract the second equation from the third:

(25a + 5b + c) - (9a - 3b + c) = 50 - 10

16a + 8b = 40

Divide by 8 to simplify:

2a + b = 5 (Equation 5)

Step 2: Solve for a and b

Now we have two equations:

• 10a - b = 31
• 2a + b = 5

Add the equations together to eliminate b:

(10a - b) + (2a + b) = 31 + 5

12a = 36

Divide by 12:

a = 3

Now, plug a = 3 into Equation 5:

2(3) + b = 5

6 + b = 5

b = -1

Step 3: Solve for c

Plug a = 3 and b = -1 into the second original equation:

9(3) - 3(-1) + c = 10

27 + 3 + c = 10

30 + c = 10

c = -20

So, we've found that a = 3, b = -1, and c = -20. Awesome job, guys!

Writing the Equation

Alright, we've got the values for a, b, and c. Now, let's write the equation of the parabola. Remember the general form: y = ax² + bx + c. We just need to substitute our values into this equation to get the specific equation for our parabola. This is the final step where we bring all our hard work together to reveal the equation that perfectly describes the parabola passing through our three points.

Substituting a = 3, b = -1, and c = -20, we get:

y = 3x² - x - 20

There you have it! This is the equation of the parabola that passes through the points (-7, 134), (-3, 10), and (5, 50). You can even graph this equation to visualize the parabola and confirm that it indeed passes through these points. This equation not only represents the parabola but also encapsulates the unique relationship between the x and y coordinates of the points we started with.

Conclusion

So, guys, we've successfully found the equation of a parabola using quadratic regression! We took three points, set up a system of equations, solved for the coefficients, and wrote the equation. How cool is that? This method is super useful for all sorts of applications, from physics to engineering. Understanding quadratic regression empowers you to model and predict parabolic relationships, which appear in various real-world phenomena, such as the trajectory of projectiles or the shape of suspension cables. Keep practicing, and you'll become a parabola pro in no time!