X- Intercepts And Gradients Of The Curve Y = X(x² - 1)

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Introduction

In the realm of calculus and coordinate geometry, understanding the behavior of curves is paramount. This exploration delves into the specifics of the curve defined by the equation y = x(x² - 1), focusing on two key aspects: the points where it intersects the x-axis and the gradient of the curve at these intersection points. This analysis not only provides insights into the curve's geometrical properties but also reinforces fundamental concepts in calculus, such as finding roots of a function and determining the slope of a tangent line. To fully grasp the nuances of this curve, we will systematically approach each question, employing algebraic techniques and calculus principles. The curve y = x(x² - 1) presents an interesting case study due to its polynomial nature and the implications of its factored form. By identifying the points of intersection with the x-axis, we gain a clear understanding of the curve's zeros, which are crucial in sketching its graph and analyzing its behavior. Moreover, calculating the gradient at these points provides valuable information about the curve's steepness and direction at these specific locations. This exploration serves as a practical application of calculus concepts, demonstrating how mathematical tools can be used to analyze and interpret the characteristics of curves. Understanding these concepts is essential for students and professionals alike, as they form the foundation for more advanced topics in mathematics, physics, and engineering. This detailed analysis will not only answer the specific questions posed but also provide a broader understanding of how to approach similar problems in the future. So, let's embark on this journey of mathematical discovery and unravel the mysteries of the curve y = x(x² - 1). Let's begin by answering the first question: At what points does the curve y = x(x² - 1) cut the x-axis?

(a) Determining the Points of Intersection with the x-axis

To determine the points where the curve y = x(x² - 1) intersects the x-axis, we need to find the values of x for which y equals zero. This is because the x-axis is defined as the line where y = 0. The equation we need to solve is therefore: 0 = x(x² - 1). This equation can be readily solved by recognizing that the product of two factors is zero if and only if at least one of the factors is zero. In this case, the factors are x and (x² - 1). Setting each factor equal to zero gives us two equations: x = 0 and x² - 1 = 0. The first equation, x = 0, immediately gives us one solution. The second equation, x² - 1 = 0, can be further simplified by factoring the left-hand side as a difference of squares: (x - 1)(x + 1) = 0. This equation yields two more solutions: x = 1 and x = -1. Therefore, the curve intersects the x-axis at three points: x = -1, x = 0, and x = 1. To express these intersections as coordinate points, we pair each x-value with its corresponding y-value, which is 0 since these points lie on the x-axis. Thus, the points of intersection are (-1, 0), (0, 0), and (1, 0). These points represent the locations where the curve crosses or touches the x-axis, providing crucial information about its behavior. The fact that there are three intersection points suggests that the curve is a cubic function with three real roots, which is consistent with the form of the equation y = x(x² - 1). These roots are essential for sketching the curve and understanding its overall shape. Moreover, the symmetry of the roots around x = 0 hints at the symmetry of the curve itself, which is a characteristic feature of odd functions. By finding these points of intersection, we have laid the groundwork for a more comprehensive analysis of the curve's properties, including its gradient at these points, which we will explore in the next section. This methodical approach to finding the roots of the equation is a fundamental technique in algebra and calculus, applicable to a wide range of problems involving curve sketching and function analysis. So, we've successfully pinpointed where the curve crosses the horizontal axis. Now, let's examine the steepness of the curve at these very points. What is the gradient at x = -1, 0, and 1?

(b) Finding the Gradient of the Curve at the Points of Intersection

Now that we have identified the points where the curve y = x(x² - 1) intersects the x-axis, the next step is to determine the gradient of the curve at these points. The gradient, also known as the slope, represents the rate of change of y with respect to x and is given by the derivative of the function. To find the gradient, we first need to find the derivative of the function y = x(x² - 1). We can rewrite the function as y = x³ - x. Now, we differentiate y with respect to x using the power rule, which states that the derivative of xⁿ is nxⁿ⁻¹. Applying this rule, we get: dy/dx = 3x² - 1. This expression gives us the gradient of the curve at any point x. To find the gradient at the points of intersection, we substitute the x-values we found earlier (-1, 0, and 1) into the derivative. At x = -1, the gradient is: dy/dx = 3(-1)² - 1 = 3(1) - 1 = 2. This means that at the point (-1, 0), the curve has a positive slope of 2, indicating that the curve is increasing as we move from left to right. At x = 0, the gradient is: dy/dx = 3(0)² - 1 = 0 - 1 = -1. At the point (0, 0), the curve has a negative slope of -1, indicating that the curve is decreasing as we move from left to right. This point is also a stationary point, as the gradient changes sign around it. At x = 1, the gradient is: dy/dx = 3(1)² - 1 = 3(1) - 1 = 2. Similar to the point x = -1, the curve has a positive slope of 2 at the point (1, 0), indicating that the curve is increasing as we move from left to right. These gradient values provide valuable information about the behavior of the curve at the points of intersection. The positive gradients at x = -1 and x = 1 suggest that the curve is moving upwards as it crosses the x-axis at these points, while the negative gradient at x = 0 indicates a downward movement. The change in gradient from negative to positive around x = 0 suggests that this point is a local minimum. By finding the gradient at these specific points, we gain a deeper understanding of the curve's shape and direction. This analysis not only answers the specific question but also demonstrates the power of calculus in revealing the properties of curves and functions. Understanding how to find the derivative and evaluate it at specific points is a fundamental skill in calculus, with applications in various fields such as physics, engineering, and economics. So, we've successfully calculated the steepness of the curve at its intersection points. This completes our exploration of the curve y = x(x² - 1).

Conclusion

In this exploration, we have meticulously analyzed the curve defined by the equation y = x(x² - 1). We began by identifying the points where the curve intersects the x-axis, which are (-1, 0), (0, 0), and (1, 0). This involved setting the equation equal to zero and solving for x, a fundamental technique in algebra. Next, we delved into the realm of calculus to determine the gradient of the curve at these intersection points. By finding the derivative of the function, dy/dx = 3x² - 1, and substituting the x-values, we calculated the gradients to be 2 at x = -1, -1 at x = 0, and 2 at x = 1. These gradient values provided insights into the curve's behavior at these specific locations, revealing its steepness and direction. The positive gradients at x = -1 and x = 1 indicated an upward movement as the curve crosses the x-axis, while the negative gradient at x = 0 suggested a downward movement and identified this point as a potential local minimum. This comprehensive analysis demonstrates the power of combining algebraic techniques with calculus principles to understand the properties of curves. The ability to find intersection points and gradients is crucial in various applications, including curve sketching, optimization problems, and modeling real-world phenomena. This exploration has not only answered the specific questions posed but also reinforced fundamental concepts in mathematics, such as finding roots of a function and determining the slope of a tangent line. By systematically approaching each aspect of the problem, we have gained a deeper understanding of the curve's characteristics and its behavior. This knowledge serves as a solid foundation for further studies in calculus and related fields. The process of analyzing a curve, such as y = x(x² - 1), involves a combination of algebraic manipulation and calculus techniques. Understanding these techniques and their applications is essential for anyone pursuing a career in mathematics, science, or engineering. From finding roots to calculating gradients, each step in the analysis provides valuable information about the curve's shape and behavior. This detailed exploration serves as a practical example of how mathematical tools can be used to solve real-world problems and gain insights into complex systems. So, we conclude our journey with a clear understanding of the curve y = x(x² - 1), its intersections, and its gradients, equipped with the knowledge and skills to tackle similar challenges in the future. This exercise not only reinforces our understanding of the specific problem but also enhances our overall mathematical proficiency and problem-solving abilities.