Finding Quadratic Function Equation Passing Through Points

Hey guys! Today, we're diving into the fascinating world of quadratic functions. Ever wondered how to find the equation of a parabola when you're given a few points it passes through? Well, you're in the right place! We're going to break it down step by step, making it super easy to understand. Let's get started!

Understanding Quadratic Functions

Before we jump into solving the problem, let's quickly recap what a quadratic function actually is. A quadratic function is a polynomial function of degree two, meaning the highest power of the variable x is 2. The general form of a quadratic function is:

$$y = ax^2 + bx + c$$

Where a, b, and c are constants, and a is not equal to zero (otherwise, it would be a linear function). The graph of a quadratic function is a parabola, a U-shaped curve that can open upwards or downwards depending on the sign of a. If a is positive, the parabola opens upwards, and if a is negative, it opens downwards.

The coefficients a, b, and c play crucial roles in determining the shape and position of the parabola. The coefficient a dictates the parabola's width and direction. A larger absolute value of a makes the parabola narrower, while a smaller absolute value makes it wider. As we mentioned, the sign of a determines whether the parabola opens upwards or downwards. The coefficient b influences the parabola's position along the x-axis. It affects the axis of symmetry, which is the vertical line that divides the parabola into two symmetrical halves. The axis of symmetry is given by the equation x = -b/(2a). Finally, the coefficient c represents the y-intercept of the parabola. This is the point where the parabola intersects the y-axis, and it occurs when x = 0. Understanding these roles is fundamental to graphing and manipulating quadratic functions effectively.

Now, imagine you're given a few points on a graph, and you need to find the specific quadratic function that passes through those points. This is where the magic happens! We'll use the given points to create a system of equations, and then solve for the unknowns a, b, and c. This process allows us to pinpoint the exact quadratic function that fits the given data. It’s like being a detective, using clues to uncover the equation that represents the parabolic path. So, with a solid grasp of these fundamentals, let’s move on to the exciting part: tackling the problem and finding the quadratic function that fits our specific set of points.

The Problem: Finding the Quadratic Function

Alright, let's get to the problem at hand. We're given three points: (-3, 2), (1, -2), and (2, 12). Our mission, should we choose to accept it (and we do!), is to find the quadratic function in the form $y = ax^2 + bx + c$ that passes through all these points.

So, what does it mean for a graph to pass through a point? It simply means that the coordinates of the point satisfy the equation of the function. In other words, if we plug in the x-coordinate into the equation, we should get the y-coordinate as the output. This is the key to solving our problem. Each point gives us a piece of the puzzle, and by combining these pieces, we can reveal the complete picture.

Think of it like this: each point is a constraint that the quadratic function must satisfy. The more points we have, the more constrained the function becomes. In our case, with three points, we have enough information to uniquely determine the three coefficients a, b, and c. It's like having three locks, and each point provides a key that unlocks one of the locks. Once we have all three keys, we can open the treasure chest and reveal the quadratic function. The process of using points to define a function is fundamental in many areas of mathematics and engineering. It allows us to model real-world phenomena, such as the trajectory of a projectile or the shape of a bridge. By understanding how to find the equation of a function that passes through given points, we gain a powerful tool for analyzing and predicting behavior in various systems. So, let's dive into the step-by-step solution and see how we can use these points to find our quadratic function. Get ready to transform these points into equations and unlock the values of a, b, and c!

Setting Up the System of Equations

The first step in solving this problem is to create a system of equations. Remember, each point (x, y) gives us an equation by plugging the x and y values into the quadratic function $y = ax^2 + bx + c$. Let's do it for each of our points:

• Point (-3, 2): Plugging in x = -3 and y = 2, we get:

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

  Simplifying, we have:

  $$2 = 9a - 3b + c$$

• Point (1, -2): Plugging in x = 1 and y = -2, we get:

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

  Simplifying, we have:

  $$-2 = a + b + c$$

• Point (2, 12): Plugging in x = 2 and y = 12, we get:

  $$12 = a(2)^2 + b(2) + c$$

  Simplifying, we have:

  $$12 = 4a + 2b + c$$

Now we have a system of three linear equations with three unknowns (a, b, and c):

1. $$9a - 3b + c = 2$$

2. $$a + b + c = -2$$

3. $$4a + 2b + c = 12$$

This is our system of equations! This is the foundation upon which we will build our solution. Each equation represents a constraint imposed by one of the points, and the solution to this system will give us the values of a, b, and c that define our quadratic function. It's like having three pieces of a puzzle, and now we need to figure out how they fit together to reveal the complete picture. The beauty of this method is that it transforms a geometric problem (finding a curve that passes through points) into an algebraic one (solving a system of equations). This is a common theme in mathematics, where problems are often approached from different perspectives to find the most effective solution. So, now that we have our system of equations, let's move on to the next step: solving this system to find the values of a, b, and c.

Solving the System of Equations

Okay, guys, we've got our system of equations. Now comes the fun part: solving for a, b, and c. There are several methods we can use to solve a system of linear equations, such as substitution, elimination, or matrix methods. For this problem, let's use the elimination method, as it's quite straightforward and efficient.

Step 1: Eliminate c from two pairs of equations.

Let's start by eliminating c from equations (1) and (2). To do this, we can subtract equation (2) from equation (1):

$$(9a - 3b + c) - (a + b + c) = 2 - (-2)$$

Simplifying, we get:

$$8a - 4b = 4$$

We can further simplify this equation by dividing both sides by 4:

$$2a - b = 1$$

Let's call this equation (4).

Now, let's eliminate c from equations (2) and (3). To do this, we can subtract equation (2) from equation (3):

$$(4a + 2b + c) - (a + b + c) = 12 - (-2)$$

Simplifying, we get:

$$3a + b = 14$$

Let's call this equation (5).

Step 2: Solve the new system of equations (4) and (5).

We now have a system of two equations with two unknowns:

4. $$2a - b = 1$$

5. $$3a + b = 14$$

Notice that the coefficients of b in these equations are opposites. This makes it easy to eliminate b by adding the two equations together:

$$(2a - b) + (3a + b) = 1 + 14$$

Simplifying, we get:

$$5a = 15$$

Dividing both sides by 5, we find:

$$a = 3$$

Step 3: Substitute the value of a back into either equation (4) or (5) to solve for b.

Let's use equation (5):

$$3(3) + b = 14$$

$$9 + b = 14$$

Subtracting 9 from both sides, we find:

$$b = 5$$

Step 4: Substitute the values of a and b back into any of the original equations (1), (2), or (3) to solve for c.

Let's use equation (2):

$$3 + 5 + c = -2$$

$$8 + c = -2$$

Subtracting 8 from both sides, we find:

$$c = -10$$

We've done it! We've found the values of a, b, and c: a = 3, b = 5, and c = -10. It's like cracking a code, where each step brings us closer to the solution. The elimination method is a powerful tool for solving systems of equations, and it's widely used in various fields of mathematics and science. By systematically eliminating variables, we can reduce a complex problem into simpler steps, making it easier to find the solution. Now that we have the values of a, b, and c, we're just one step away from finding our quadratic function. Let's put these values together and see the final result!

The Solution: The Quadratic Function

Alright, we've done the hard work! We've found that a = 3, b = 5, and c = -10. Now, all that's left is to plug these values back into the general form of the quadratic function:

$$y = ax^2 + bx + c$$

Substituting our values, we get:

$$y = 3x^2 + 5x - 10$$

And there you have it! This is the quadratic function whose graph passes through the points (-3, 2), (1, -2), and (2, 12). This function uniquely describes the parabola that connects these three points. It's like finding the missing piece of a puzzle, and seeing how everything fits together perfectly. We started with a geometric problem – finding a curve that passes through given points – and we transformed it into an algebraic problem – solving a system of equations. By systematically applying the elimination method, we were able to find the coefficients of the quadratic function. This is a powerful demonstration of how different areas of mathematics can be used to solve real-world problems.

To double-check our answer, we can plug the coordinates of the given points into our function and make sure they satisfy the equation. For example, let's check the point (-3, 2):

$$y = 3(-3)^2 + 5(-3) - 10 = 3(9) - 15 - 10 = 27 - 15 - 10 = 2$$

It works! We can do the same for the other two points and confirm that they also lie on the parabola defined by our function. So, congratulations! You've successfully found the quadratic function that passes through the given points. You've mastered a valuable skill that can be applied in various mathematical and scientific contexts. Now, go forth and conquer more quadratic function challenges!

Conclusion

So, guys, we've reached the end of our journey! We've successfully navigated the world of quadratic functions and learned how to find the equation of a parabola given three points. We started by understanding the general form of a quadratic function, then set up a system of equations based on the given points, and finally solved the system using the elimination method. It's been quite the adventure!

Finding the quadratic function that passes through given points is a fundamental problem in mathematics with applications in various fields, such as physics, engineering, and computer graphics. This method allows us to model and analyze parabolic trajectories, design curved structures, and create smooth curves in computer-generated images. The ability to connect discrete points with a continuous function is a powerful tool for understanding and manipulating the world around us.

Remember, the key to solving problems like this is to break them down into smaller, manageable steps. Don't be intimidated by the complexity of the problem; instead, focus on understanding each step and applying the appropriate techniques. With practice, you'll become a pro at solving quadratic function problems! We hope you found this guide helpful and easy to follow. Now you're equipped to tackle similar problems with confidence. Keep practicing, keep exploring, and keep learning! You've got this! And until next time, happy problem-solving!