Quadratic Function Through Points: Find F(x)

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Finding the Quadratic Function Through Given Points

Hey guys! Ever wondered how to find the equation of a quadratic function when you're given a few points it passes through? It's like solving a mini-mystery, and we're going to crack the code together. In this article, we'll walk through the steps to determine the quadratic function that passes through specific points. Let's dive in!

Understanding Quadratic Functions

Before we jump into solving, let's quickly recap what a quadratic function is. A quadratic function is a polynomial function of degree two, generally written in the form:

f(x) = ax^2 + bx + c

Where a, b, and c are constants, and a is not equal to zero. The graph of a quadratic function is a parabola, a U-shaped curve. Our goal here is to find the specific values of a, b, and c that make the parabola pass through the given points.

To really nail this, it's important to understand the role each coefficient plays in shaping the parabola. The coefficient a dictates the parabola's direction (whether it opens upwards or downwards) and its 'width'. A larger absolute value of a means a narrower parabola, while a smaller value makes it wider. If a is positive, the parabola opens upwards, and if it's negative, it opens downwards. This is a crucial first step in visualizing what the function might look like and helps in confirming your solution later. The coefficient b influences the position of the parabola's axis of symmetry, which is a vertical line that runs through the vertex (the highest or lowest point) of the parabola. The vertex's x-coordinate can be found using the formula -b/(2a). This gives you a sense of how the parabola is shifted horizontally. Lastly, the constant c is the y-intercept of the parabola, meaning it's the point where the parabola crosses the y-axis. This is a straightforward clue, as it tells you the value of the function when x is zero. Grasping these individual roles will make the process of finding the quadratic function much more intuitive and less like a mechanical exercise.

Setting up the Equations

Okay, let's get to the fun part – solving! We're given three points: (2,-3), (-2,-7), and (-1,0). Since each point lies on the graph of the quadratic function, we can plug the x and y values into our general form f(x) = ax^2 + bx + c to create three equations.

  1. For the point (2, -3): -3 = a(2)^2 + b(2) + c => -3 = 4a + 2b + c
  2. For the point (-2, -7): -7 = a(-2)^2 + b(-2) + c => -7 = 4a - 2b + c
  3. For the point (-1, 0): 0 = a(-1)^2 + b(-1) + c => 0 = a - b + c

Now we have a system of three linear equations with three unknowns (a, b, and c). This is a classic setup for solving using various methods, like substitution, elimination, or matrices. The key here is to be organized and methodical in your approach to avoid making simple arithmetic errors. Each equation represents a constraint that the quadratic function must satisfy, and by solving this system, we're essentially finding the unique parabola that threads its way through all three points. Before diving into the algebra, it's often helpful to take a moment to look at the equations and see if there's an obvious first step. For example, are there any equations that can be easily subtracted or added to eliminate a variable? This strategic thinking can save time and reduce the chance of mistakes. Understanding this initial setup is crucial because it translates the geometric problem (finding a curve through points) into an algebraic one (solving a system of equations), making it more tractable.

Solving the System of Equations

We've got our system of equations. Let's solve it! There are a few ways to tackle this, but I'm going to use elimination. First, let's label our equations:

  1. 4a + 2b + c = -3
  2. 4a - 2b + c = -7
  3. a - b + c = 0

Step 1: Eliminate 'c'

Subtract equation (3) from equation (1) and equation (2) to eliminate c:

  • (1) - (3): (4a + 2b + c) - (a - b + c) = -3 - 0 => 3a + 3b = -3 => a + b = -1 (Divide by 3) [Equation 4]
  • (2) - (3): (4a - 2b + c) - (a - b + c) = -7 - 0 => 3a - b = -7 [Equation 5]

Now we have two equations with two variables, which is much easier to handle. This step is a great example of how simplifying a problem can make it more manageable. By strategically eliminating one variable, we've reduced the complexity of the system and brought ourselves closer to a solution. It's like peeling away layers of an onion – each step reveals something simpler underneath. The act of subtracting equations might seem like a simple trick, but it's a powerful tool in linear algebra. It allows us to combine the information from multiple equations in a way that cancels out unwanted terms, revealing hidden relationships between the remaining variables. Recognizing these kinds of opportunities is a key skill in solving mathematical problems efficiently.

Step 2: Eliminate 'b'

Add equation (4) and equation (5):

  • (a + b) + (3a - b) = -1 + (-7) => 4a = -8 => a = -2

Step 3: Solve for 'b'

Substitute a = -2 into equation (4):

  • -2 + b = -1 => b = 1

Step 4: Solve for 'c'

Substitute a = -2 and b = 1 into equation (3):

  • -2 - 1 + c = 0 => c = 3

We've found our values! a = -2, b = 1, and c = 3. It's like finding all the pieces of a puzzle and finally seeing the full picture. Each step we took, from setting up the equations to eliminating variables, was a necessary part of the process. Now, it's time to put these values back into the original equation and see what our quadratic function looks like. This moment of truth is where we confirm that our hard work has paid off and we've successfully navigated the algebraic landscape to arrive at the correct solution.

The Quadratic Function

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

F(x) = -2x^2 + x + 3

This is the quadratic function that passes through the points (2,-3), (-2,-7), and (-1,0). Awesome, right? But before we celebrate too much, it's always a good idea to check our work. This is a critical step in the problem-solving process because it's easy to make small errors along the way. Checking our solution not only gives us confidence in our answer but also helps us develop a deeper understanding of the problem.

Verifying the Solution

To verify, let's plug our original points back into the function and see if they hold true:

  1. For (2, -3): F(2) = -2(2)^2 + 2 + 3 = -8 + 2 + 3 = -3 (Correct!)
  2. For (-2, -7): F(-2) = -2(-2)^2 + (-2) + 3 = -8 - 2 + 3 = -7 (Correct!)
  3. For (-1, 0): F(-1) = -2(-1)^2 + (-1) + 3 = -2 - 1 + 3 = 0 (Correct!)

It works! All three points satisfy the equation. We've successfully found the quadratic function. Verifying the solution is like putting the final stamp of approval on your work. It's the moment where you can confidently say, "Yes, I got it right!" This step not only confirms the correctness of the answer but also reinforces the process you used to get there. By plugging the original points back into the function, we're essentially testing whether our solution is consistent with the given information. If even one point doesn't fit, it's a sign that there might be an error somewhere in our calculations, and we need to go back and review our steps. This practice of verification is a cornerstone of mathematical problem-solving and is invaluable in building accuracy and confidence.

Conclusion

So, there you have it! Finding a quadratic function that passes through given points involves setting up a system of equations and solving for the coefficients. It might seem daunting at first, but with a systematic approach, it becomes a fun challenge. Remember, the key is to break down the problem into smaller, manageable steps, and always double-check your work. Keep practicing, and you'll become a quadratic function pro in no time! Guys, I hope this breakdown helped you out. Happy solving!