Plotting The Quadratic Function Y=(x+1)(x-3) A Step-by-Step Guide
In this article, we will delve into the process of plotting the quadratic function y = (x+1)(x-3). Understanding how to plot quadratic functions is a fundamental skill in mathematics, with applications spanning various fields, from physics to economics. Quadratic functions, characterized by their parabolic shapes, can be easily visualized by identifying key features such as intercepts and vertices. This guide provides a step-by-step approach to plotting the given function, making the process clear and accessible. We'll start by finding the x-intercepts, then locate the axis of symmetry, and finally determine the vertex of the parabola. By following these steps, you'll gain a solid understanding of how to plot quadratic functions effectively and accurately. This comprehensive guide is designed for students, educators, and anyone interested in enhancing their mathematical skills. Let's embark on this journey to master the art of plotting quadratic functions.
Step 1: Plot the x-intercepts of the function
To begin plotting the quadratic function y = (x+1)(x-3), the first critical step is to identify and plot the x-intercepts. The x-intercepts are the points where the parabola intersects the x-axis, and they occur when the function's value, y, is equal to zero. In other words, we need to find the values of x that satisfy the equation (x+1)(x-3) = 0. This equation is already factored, making it straightforward to find the solutions. According to the zero-product property, if the product of two factors is zero, then at least one of the factors must be zero. Thus, we set each factor equal to zero and solve for x.
So, we have two equations to solve: x + 1 = 0 and x - 3 = 0. Solving the first equation, x + 1 = 0, involves subtracting 1 from both sides, which gives us x = -1. This means that the function has an x-intercept at the point (-1, 0). Solving the second equation, x - 3 = 0, involves adding 3 to both sides, which gives us x = 3. This indicates that the function has another x-intercept at the point (3, 0). Now that we have found both x-intercepts, we can plot these points on the Cartesian plane. The points (-1, 0) and (3, 0) serve as crucial anchors for sketching the parabola. These intercepts provide a clear indication of where the curve crosses the x-axis, which is essential for understanding the overall shape and position of the quadratic function. By accurately plotting these points, we lay a solid foundation for the subsequent steps in graphing the function. Understanding the significance of x-intercepts is paramount in grasping the behavior of quadratic functions, as they reveal key aspects of the parabola's symmetry and orientation.
Step 2: Plot the point on the x-axis that is halfway between the intercepts
After plotting the x-intercepts, the next step in graphing the quadratic function y = (x+1)(x-3) is to find the point on the x-axis that lies exactly halfway between these intercepts. This point is crucial because it represents the axis of symmetry of the parabola. The axis of symmetry is a vertical line that passes through the vertex of the parabola, dividing it into two symmetrical halves. To find the x-coordinate of this point, we simply calculate the midpoint of the x-intercepts. The x-intercepts we found in the previous step are -1 and 3. The midpoint formula is given by (x1 + x2) / 2, where x1 and x2 are the x-coordinates of the intercepts. In our case, x1 = -1 and x2 = 3.
Applying the midpoint formula, we get ((-1) + 3) / 2 = 2 / 2 = 1. This tells us that the x-coordinate of the point halfway between the intercepts is 1. Therefore, the axis of symmetry is the vertical line x = 1. Now, we plot this point on the x-axis, which is (1, 0). This point is not on the parabola itself, but it is a critical reference point for sketching the parabola because it indicates the line around which the parabola is symmetric. Understanding the axis of symmetry is vital in plotting quadratic functions, as it simplifies the process of finding the vertex and sketching the overall shape of the parabola. The symmetry property of parabolas means that for every point on one side of the axis of symmetry, there is a corresponding point on the other side at the same vertical distance. By identifying the axis of symmetry, we gain a deeper insight into the function's behavior and can more accurately plot its graph. This step serves as a bridge between identifying the intercepts and determining the vertex, which is the next critical point to plot.
Step 3: Plot the vertex
The final step in plotting the quadratic function y = (x+1)(x-3) is to find and plot the vertex. The vertex is the point where the parabola changes direction; it is the minimum point if the parabola opens upwards or the maximum point if it opens downwards. The vertex lies on the axis of symmetry, which we found in the previous step to be x = 1. To find the y-coordinate of the vertex, we substitute x = 1 into the original equation of the function, y = (x+1)(x-3).
Substituting x = 1, we get y = (1+1)(1-3) = (2)(-2) = -4. Therefore, the vertex of the parabola is at the point (1, -4). Now, we plot this point on the Cartesian plane. The vertex is a crucial point because it defines the extreme value of the function and the turning point of the parabola. Once we have plotted the vertex, along with the x-intercepts and the axis of symmetry, we have enough information to sketch the entire parabola. The parabola will pass through the x-intercepts (-1, 0) and (3, 0), with the axis of symmetry x = 1 running vertically through the vertex (1, -4). Since the coefficient of the x^2 term in the expanded form of the equation (y = x^2 - 2x - 3) is positive, the parabola opens upwards. This means that the vertex (1, -4) is the minimum point of the function. By connecting the plotted points in a smooth, parabolic curve, we can accurately represent the function y = (x+1)(x-3) graphically. Plotting the vertex is the culmination of the previous steps and provides a complete picture of the quadratic function's behavior. Understanding how to find and plot the vertex is essential for mastering the graphing of quadratic functions and for analyzing their properties and applications.
In conclusion, plotting the quadratic function y = (x+1)(x-3) involves a series of straightforward steps that, when followed methodically, lead to a clear and accurate graphical representation. We began by identifying and plotting the x-intercepts, which are the points where the function intersects the x-axis. This involved setting the function equal to zero and solving for x, giving us the points (-1, 0) and (3, 0). Next, we determined the axis of symmetry by finding the midpoint between the x-intercepts, which is the line x = 1. Plotting the point (1, 0) on the x-axis helped visualize this crucial line of symmetry. The final step was to find and plot the vertex, which is the turning point of the parabola. By substituting x = 1 into the original equation, we found the y-coordinate of the vertex to be -4, giving us the point (1, -4). With the x-intercepts, axis of symmetry, and vertex plotted, we were able to sketch the complete parabola. The parabola opens upwards, as indicated by the positive coefficient of the x^2 term in the expanded form of the equation, and the vertex (1, -4) represents the minimum point of the function. This step-by-step guide provides a solid foundation for understanding how to plot quadratic functions. The techniques discussed here are applicable to a wide range of quadratic equations, making them invaluable for students, educators, and anyone working with mathematical functions. By mastering these steps, you can confidently graph quadratic functions and analyze their key features, enhancing your mathematical skills and problem-solving abilities. The ability to visualize quadratic functions graphically is a powerful tool in various fields, including physics, engineering, and economics, where quadratic relationships often arise. This comprehensive approach ensures a thorough understanding of the process and empowers you to tackle more complex mathematical challenges.
