Solving Systems Of Equations Quadratic And Linear Functions
In the realm of mathematics, solving systems of equations is a fundamental skill. This article delves into the fascinating intersection of quadratic and linear functions, exploring how to find solutions where these two distinct mathematical entities meet. We will dissect a specific problem involving a quadratic function, f(x) = 5x² + x + 3, and a linear function, g(x), presented in a tabular format. Our goal is to determine the point(s) where these functions intersect, providing a comprehensive understanding of the solution process.
Defining the Functions: Quadratic and Linear
Before we dive into the solution, let's establish a clear understanding of the functions involved. The quadratic function, f(x) = 5x² + x + 3, is characterized by its parabolic shape when graphed. The x² term is the defining feature, causing the function's rate of change to vary, resulting in a curve. On the other hand, the linear function, g(x), represents a straight line when graphed. Its defining characteristic is a constant rate of change, meaning that for every unit increase in x, the value of g(x) changes by a fixed amount. Understanding these fundamental differences is crucial for visualizing and solving the system of equations.
The Quadratic Function: f(x) = 5x² + x + 3
This quadratic function is expressed in the standard form ax² + bx + c, where a = 5, b = 1, and c = 3. The coefficient a determines the parabola's direction (upward if positive, downward if negative) and its width. In this case, a = 5 indicates an upward-opening parabola. The b and c coefficients influence the parabola's position on the coordinate plane. To fully grasp the behavior of f(x), we could further analyze its vertex (the minimum point of the parabola) and its axis of symmetry. However, for the purpose of solving the system of equations, we'll primarily focus on the function's values at specific x-values.
The Linear Function: g(x) from the Table
The linear function g(x) is presented in a tabular format, providing us with specific input-output pairs. This table is a goldmine of information, allowing us to determine the equation of the line. We observe that for each increase of 1 in x, the value of g(x) increases by 2. This constant rate of change is the slope of the line. To find the equation of g(x), we can use the slope-intercept form, y = mx + b, where m is the slope and b is the y-intercept. From the table, we can identify two points, for instance, (-2, 3) and (-1, 5). Using these points, we can calculate the slope as (5 - 3) / (-1 - (-2)) = 2. Thus, m = 2. To find the y-intercept, we can substitute one of the points and the slope into the slope-intercept form and solve for b. Using the point (0, 7), we get 7 = 2(0) + b, so b = 7. Therefore, the equation of the linear function is g(x) = 2x + 7.
Solving the System of Equations: Finding the Intersection Points
The solution to a system of equations represents the point(s) where the graphs of the equations intersect. In our case, we are looking for the x-values where f(x) = g(x). This means we need to solve the equation 5x² + x + 3 = 2x + 7. To solve this equation, we will use algebraic methods, specifically, setting the equation to zero and then applying the quadratic formula.
Setting the Equation to Zero
The first step is to rearrange the equation to have zero on one side. Subtracting 2x and 7 from both sides of the equation 5x² + x + 3 = 2x + 7 gives us: 5x² + x + 3 - 2x - 7 = 0. Simplifying this equation, we get 5x² - x - 4 = 0. This is a quadratic equation in the standard form ax² + bx + c = 0, where a = 5, b = -1, and c = -4. Now we are ready to apply the quadratic formula.
Applying the Quadratic Formula
The quadratic formula is a powerful tool for solving quadratic equations of the form ax² + bx + c = 0. The formula is given by: x = (-b ± √(b² - 4ac)) / (2a). In our case, a = 5, b = -1, and c = -4. Substituting these values into the formula, we get: x = (1 ± √((-1)² - 4 * 5 * -4)) / (2 * 5). Simplifying the expression under the square root, we have: x = (1 ± √(1 + 80)) / 10, which further simplifies to x = (1 ± √81) / 10. Since √81 = 9, we have x = (1 ± 9) / 10. This gives us two possible solutions for x.
Calculating the Solutions for x
The two possible solutions for x are: x₁ = (1 + 9) / 10 = 10 / 10 = 1 and x₂ = (1 - 9) / 10 = -8 / 10 = -0.8. These are the x-coordinates of the intersection points of the quadratic and linear functions. To find the corresponding y-coordinates, we can substitute these x-values into either the equation for f(x) or the equation for g(x). Since g(x) = 2x + 7 is simpler, we'll use that.
Finding the Corresponding y-coordinates
For x₁ = 1, we have g(1) = 2(1) + 7 = 9. Thus, one intersection point is (1, 9). For x₂ = -0.8, we have g(-0.8) = 2(-0.8) + 7 = -1.6 + 7 = 5.4. Thus, the other intersection point is (-0.8, 5.4). These two points represent the solutions to the system of equations.
Verifying the Solution: Checking the Intersection Points
To ensure the accuracy of our solution, it's essential to verify that the points we found indeed lie on both the quadratic and linear functions. We can do this by substituting the x-values into both f(x) and g(x) and confirming that the y-values match.
Verification for (1, 9)
For the point (1, 9), we have: f(1) = 5(1)² + 1 + 3 = 5 + 1 + 3 = 9 and g(1) = 2(1) + 7 = 9. Since f(1) = g(1) = 9, the point (1, 9) is indeed a solution.
Verification for (-0.8, 5.4)
For the point (-0.8, 5.4), we have: f(-0.8) = 5(-0.8)² + (-0.8) + 3 = 5(0.64) - 0.8 + 3 = 3.2 - 0.8 + 3 = 5.4 and g(-0.8) = 2(-0.8) + 7 = -1.6 + 7 = 5.4. Since f(-0.8) = g(-0.8) = 5.4, the point (-0.8, 5.4) is also a solution. Therefore, our solution is verified.
The Solution Set: The Intersection Points
In conclusion, the solution to the system of equations consisting of the quadratic function f(x) = 5x² + x + 3 and the linear function g(x) (defined by the table) are the points (1, 9) and (-0.8, 5.4). These points represent the locations where the parabola defined by f(x) intersects the line defined by g(x). Solving systems of equations involving different types of functions is a critical skill in mathematics, with applications in various fields, including physics, engineering, and economics.
Analyzing the Given Options: Identifying the Correct Solution
The question provides a set of options, and we need to identify which one represents a solution to the system of equations. Option A, (-1, 7), is given as a potential solution. Let's evaluate this option by substituting x = -1 into both f(x) and g(x).
Evaluating Option A: (-1, 7)
For the quadratic function, f(-1) = 5(-1)² + (-1) + 3 = 5 - 1 + 3 = 7. For the linear function, we can use the equation we derived earlier, g(x) = 2x + 7. So, g(-1) = 2(-1) + 7 = -2 + 7 = 5. Since f(-1) = 7 and g(-1) = 5, the point (-1, 7) does not lie on both functions. Therefore, option A is not a solution to the system of equations.
The Correct Solution: A Detailed Explanation
As we determined earlier, the actual solutions to the system are (1, 9) and (-0.8, 5.4). The option (-1, 7) is incorrect because, while it lies on the quadratic function f(x), it does not lie on the linear function g(x). To be a solution to the system, a point must satisfy both equations simultaneously. This highlights the importance of verifying solutions by substituting them back into all equations in the system.
Visualizing the Solution: Graphing the Functions
To gain a deeper understanding of the solution, it's helpful to visualize the functions graphically. If we were to plot the quadratic function f(x) = 5x² + x + 3 and the linear function g(x) = 2x + 7 on the same coordinate plane, we would see a parabola and a line intersecting at two points. These intersection points are precisely the solutions we calculated: (1, 9) and (-0.8, 5.4). The graph provides a visual confirmation of our algebraic solution and helps to solidify the concept of solving systems of equations.
Graphing can be done by hand, plotting points for each function, or using graphing software or calculators. The visual representation makes it clear that the solutions are the points where the two graphs overlap, satisfying both equations simultaneously.
Conclusion: Mastering Systems of Equations
Solving systems of equations involving quadratic and linear functions is a crucial skill in mathematics. This article has provided a comprehensive guide to solving such systems, including: understanding the characteristics of quadratic and linear functions, deriving the equation of a linear function from a table, applying the quadratic formula, and verifying solutions. By mastering these techniques, you can confidently tackle a wide range of mathematical problems and gain a deeper appreciation for the interconnectedness of different mathematical concepts. The ability to solve these systems has wide applications in various fields, making it an invaluable tool in your mathematical arsenal. Remember, practice is key to mastering any mathematical skill, so continue to work through examples and explore different types of systems of equations.
