Quadratic Equation: Find Y = Ax² + Bx + C

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Understanding how to derive the equation of a quadratic function is a fundamental concept in mathematics. This article will guide you through the process of finding the specific quadratic equation in the form y = ax² + bx + c, given three points that the parabola passes through. We'll break down each step, making it easy to follow and understand. Let’s dive in!

Understanding Quadratic Functions

Before we jump into solving the problem, let's first understand what a quadratic function is. At its core, quadratic functions are polynomial functions of degree two. The general form of a quadratic function is y = ax² + bx + c, where a, b, and c are constants, and x is the variable. The graph of a quadratic function is a parabola, a U-shaped curve that opens either upwards or downwards, depending on the sign of the coefficient a. When a is positive, the parabola opens upwards, and when a is negative, it opens downwards. The vertex of the parabola is the point where the curve changes direction, representing either the minimum or maximum value of the function.

Key characteristics of quadratic functions include their symmetry. Parabolas are symmetrical about a vertical line passing through their vertex, known as the axis of symmetry. This symmetry is a crucial property that simplifies many calculations and graphical representations. The coefficients a, b, and c play significant roles in determining the shape and position of the parabola. The coefficient a not only dictates the direction of the parabola's opening but also affects its width; a larger absolute value of a results in a narrower parabola. The coefficients b and c, on the other hand, influence the horizontal and vertical positioning of the parabola, respectively. The constant c represents the y-intercept, which is the point where the parabola intersects the y-axis. Understanding these properties is essential for analyzing and solving problems involving quadratic functions. In this article, we will use these properties to determine the equation of a parabola given specific points it passes through, effectively reverse-engineering the function from its graphical behavior. By grasping these foundational concepts, you'll be well-equipped to tackle more complex problems and applications of quadratic functions in various fields of mathematics and beyond.

Problem Statement: Finding the Equation

Our main objective is to find the equation of a quadratic function in the form y = ax² + bx + c. We are given three points through which the parabola passes: (-1, 5), (0, 1), and (1, 3). These points provide us with the necessary information to determine the constants a, b, and c. Each point (x, y) on the parabola satisfies the equation y = ax² + bx + c. By substituting the coordinates of each point into the equation, we will obtain a system of three linear equations with three unknowns (a, b, and c). Solving this system will give us the values of a, b, and c, thereby defining the specific quadratic function that represents the parabola. This method is a classic application of algebra in solving geometric problems, demonstrating the powerful connection between algebraic equations and graphical curves. The points provided are carefully chosen to ensure a unique solution exists, meaning there is only one parabola that passes through these three points. This uniqueness is a fundamental concept in quadratic functions and their graphical representations. Understanding how to use given points to define a quadratic equation is crucial for various applications, including curve fitting, optimization problems, and modeling real-world phenomena that exhibit parabolic behavior. In the following sections, we will systematically substitute these points into the general equation and then solve the resulting system of equations to find the specific quadratic function that fits these points.

Step 1: Substituting the Points

Let's begin by substituting the given points into the general quadratic equation y = ax² + bx + c. This will generate three equations that we can use to solve for a, b, and c. This step is crucial as it translates the geometric information (points on the parabola) into algebraic equations that we can manipulate. The accuracy of this substitution is paramount, as any errors here will propagate through the rest of the solution. Each substitution provides a unique piece of information that contributes to the final solution. The points are chosen to ensure that the resulting system of equations is solvable, meaning that there is a unique solution for a, b, and c. By performing these substitutions carefully, we set the stage for solving the system of equations and finding the equation of the quadratic function. This process demonstrates the power of algebra in describing geometric shapes and relationships.

  1. Using the point (-1, 5): Substituting x = -1 and y = 5 into the equation, we get: 5 = a(-1)² + b(-1) + c Which simplifies to: 5 = a - b + c

  2. Using the point (0, 1): Substituting x = 0 and y = 1 into the equation, we get: 1 = a(0)² + b(0) + c Which simplifies to: 1 = c

    This immediately gives us the value of c, which is a significant simplification.

  3. Using the point (1, 3): Substituting x = 1 and y = 3 into the equation, we get: 3 = a(1)² + b(1) + c Which simplifies to: 3 = a + b + c

Now we have three equations:

  1. 5 = a - b + c
  2. 1 = c
  3. 3 = a + b + c

Step 2: Solving the System of Equations

Now that we have our system of equations, we can solve for the constants a, b, and c. We already know that c = 1 from the second equation, which simplifies our task considerably. The strategy here is to use the known value of c to reduce the number of unknowns in the other equations. This approach is a common technique in solving systems of equations, making the process more manageable and efficient. By substituting the value of c into the remaining equations, we will obtain a system of two equations with two unknowns, which can be solved using various methods such as substitution or elimination. This step is crucial for determining the specific coefficients of the quadratic function, allowing us to fully define the parabola.

  1. Substitute c = 1 into the first equation: 5 = a - b + 1 Subtract 1 from both sides: 4 = a - b

  2. Substitute c = 1 into the third equation: 3 = a + b + 1 Subtract 1 from both sides: 2 = a + b

Now we have a system of two equations:

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

We can solve this system using either substitution or elimination. Let's use the elimination method.

Add the two equations together:

(4 = a - b) + (2 = a + b)

This gives us:

6 = 2a

Divide both sides by 2:

a = 3

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 equation 2 = a + b:

2 = 3 + b

Subtract 3 from both sides:

b = -1

So, we have found that a = 3, b = -1, and c = 1.

Step 3: Writing the Quadratic Equation

With the values of a, b, and c determined, we can now write the specific quadratic equation that represents the parabola. This step is the culmination of our efforts, where we combine the constants we found into the general form y = ax² + bx + c. This final equation is the algebraic representation of the parabola that passes through the given points, and it allows us to predict the y-value for any x-value on the curve. This process highlights the elegance of mathematics in capturing geometric shapes with algebraic expressions. The resulting equation can be used for various applications, such as graphing the parabola, finding its vertex, and analyzing its behavior. The accuracy of the equation depends on the precision of the calculations made in the previous steps, emphasizing the importance of careful substitution and solving of the system of equations. The final equation serves as a concise summary of the parabola's properties and its relationship to the given points.

Substitute a = 3, b = -1, and c = 1 into the equation y = ax² + bx + c:

y = 3x² + (-1)x + 1

Simplify the equation:

y = 3x² - x + 1

Therefore, the quadratic equation that defines the parabola passing through the points (-1, 5), (0, 1), and (1, 3) is y = 3x² - x + 1. This is the final solution to our problem.

Conclusion

In this article, we successfully found the quadratic equation y = 3x² - x + 1 that defines the parabola passing through the points (-1, 5), (0, 1), and (1, 3). We achieved this by substituting the given points into the general quadratic equation y = ax² + bx + c, setting up a system of equations, and then solving for the constants a, b, and c. This process underscores the connection between algebraic equations and geometric curves, demonstrating how algebra can be used to describe and analyze geometric shapes. Understanding how to derive quadratic equations from given points is a valuable skill in mathematics, with applications ranging from graphing functions to solving real-world problems involving parabolic paths. The systematic approach we followed—substitution, solving the system of equations, and writing the final equation—provides a clear and reliable method for tackling similar problems. This example highlights the power and versatility of quadratic functions in mathematics and their relevance in various practical contexts. Remember, practice is key to mastering these concepts, so try applying this method to different sets of points to solidify your understanding.

For further exploration of quadratic functions and their properties, visit Khan Academy's Quadratic Functions Section.