Tangent Gradients And Curve Sketching For Y=x(x-6)

In this comprehensive guide, we will delve into the process of determining the gradient of tangents for the curve y = x(x - 6) at specific points. We'll focus on the points where x = 2 1/2, x = 3, and x = 3 1/2. Furthermore, we'll illustrate our findings with a detailed sketch graph, providing a visual representation of the curve and its tangents. This exploration will not only enhance your understanding of calculus concepts but also equip you with the skills to analyze and visualize curves effectively.

Understanding the Curve y = x(x - 6)

To begin, let's thoroughly understand the curve y = x(x - 6). This equation represents a parabola, a fundamental shape in mathematics and physics. Expanding the equation, we get y = x^2 - 6x. This form reveals that the parabola opens upwards because the coefficient of the x^2 term is positive. Identifying the key features of this parabola is crucial for sketching its graph and understanding its behavior. These key features include the roots (where the curve intersects the x-axis), the vertex (the minimum point of the curve), and the axis of symmetry. By carefully analyzing these elements, we can accurately depict the curve's trajectory on a graph. The roots can be found by setting y = 0, which gives us x(x - 6) = 0. This yields roots at x = 0 and x = 6. The x-coordinate of the vertex can be found using the formula x = -b / 2a, where a and b are the coefficients of the x^2 and x terms, respectively. In this case, a = 1 and b = -6, so the x-coordinate of the vertex is x = -(-6) / (2 * 1) = 3. To find the y-coordinate of the vertex, substitute x = 3 into the equation: y = 3(3 - 6) = -9. Therefore, the vertex is at the point (3, -9). The axis of symmetry is a vertical line that passes through the vertex, which in this case is the line x = 3. Understanding these properties allows us to sketch the parabola with precision, capturing its essential characteristics and behavior. This foundational understanding is essential for the subsequent steps, where we will calculate the gradients of tangents at specific points and visualize them on the graph.

Finding the Gradient of Tangents

Finding the gradient of tangents at specific points is a core concept in differential calculus. The gradient of a tangent at a point on a curve represents the instantaneous rate of change of the curve at that point. In mathematical terms, this gradient is given by the derivative of the function at that point. To find the gradient of the tangent for the curve y = x(x - 6), we first need to find the derivative of the function. The derivative, denoted as dy/dx, represents the slope of the tangent line at any point on the curve. For the given function y = x^2 - 6x, we can use the power rule of differentiation, which states that if y = ax^n, then dy/dx = nax^(n-1). Applying this rule to our function, we get dy/dx = 2x - 6. This derivative function gives us the gradient of the tangent at any x-value. Now, we can find the gradient at the specified points: x = 2 1/2, x = 3, and x = 3 1/2. For x = 2 1/2 (or 2.5), we substitute this value into the derivative: dy/dx = 2(2.5) - 6 = 5 - 6 = -1. This means the gradient of the tangent at x = 2.5 is -1. For x = 3, we substitute this value into the derivative: dy/dx = 2(3) - 6 = 6 - 6 = 0. This indicates that the tangent at x = 3 is horizontal, as the gradient is 0. For x = 3 1/2 (or 3.5), we substitute this value into the derivative: dy/dx = 2(3.5) - 6 = 7 - 6 = 1. This means the gradient of the tangent at x = 3.5 is 1. These calculated gradients provide us with crucial information about the slope of the curve at these specific points, which will be essential when we sketch the tangents on the graph.

Calculating Gradients at Specific Points: x = 2 1/2, x = 3, and x = 3 1/2

Now, let's meticulously calculate the gradients at the specific points mentioned: x = 2 1/2, x = 3, and x = 3 1/2. This process involves substituting these x-values into the derivative function we previously calculated, dy/dx = 2x - 6. This derivative represents the instantaneous rate of change of the curve at any given point, and by plugging in the specific x-values, we can determine the exact gradient of the tangent at those points. At x = 2 1/2 (which is equivalent to 2.5), we substitute 2.5 into the derivative: dy/dx = 2(2.5) - 6 = 5 - 6 = -1. This result tells us that the tangent to the curve at the point where x = 2.5 has a negative slope of -1. A negative slope indicates that the tangent line is decreasing as we move from left to right. At x = 3, we substitute 3 into the derivative: dy/dx = 2(3) - 6 = 6 - 6 = 0. A gradient of 0 signifies that the tangent line is horizontal. This is a critical point on the curve, as it represents either a local maximum or a local minimum. In this case, it is the vertex of the parabola, where the curve changes direction. At x = 3 1/2 (which is equivalent to 3.5), we substitute 3.5 into the derivative: dy/dx = 2(3.5) - 6 = 7 - 6 = 1. This result shows that the tangent to the curve at the point where x = 3.5 has a positive slope of 1. A positive slope indicates that the tangent line is increasing as we move from left to right. These calculations provide us with a clear understanding of the behavior of the curve at these specific points. The negative gradient at x = 2.5 suggests that the curve is decreasing, the zero gradient at x = 3 indicates a turning point, and the positive gradient at x = 3.5 shows that the curve is increasing. This information will be invaluable when we proceed to sketch the graph and illustrate these tangents.

Sketching the Graph and Tangents

Sketching the graph is a crucial step in visualizing the curve and its tangents. A sketch of the graph provides a visual representation of the function's behavior, making it easier to understand the relationship between the curve and its tangents. To accurately sketch the graph of y = x(x - 6), we need to consider the key features we identified earlier: the roots, the vertex, and the axis of symmetry. The roots are at x = 0 and x = 6, which means the curve intersects the x-axis at these points. The vertex is at (3, -9), which is the minimum point of the parabola. The axis of symmetry is the vertical line x = 3, which divides the parabola into two symmetrical halves. Now, let's plot these key points on a coordinate plane. We have the roots at (0, 0) and (6, 0), and the vertex at (3, -9). Using these points as a guide, we can sketch a smooth, U-shaped curve that represents the parabola. The curve opens upwards because the coefficient of the x^2 term is positive. Next, we need to illustrate the tangents at the points x = 2 1/2, x = 3, and x = 3 1/2. At x = 2 1/2, the gradient is -1. To draw the tangent, we first find the y-coordinate at this point: y = 2.5(2.5 - 6) = 2.5(-3.5) = -8.75. So, the point is (2.5, -8.75). We then draw a line that passes through this point with a slope of -1. This means for every one unit we move to the right, we move one unit down. At x = 3, the gradient is 0, which means the tangent is a horizontal line. We already know the vertex is at (3, -9), so we draw a horizontal line through this point. At x = 3 1/2, the gradient is 1. We first find the y-coordinate at this point: y = 3.5(3.5 - 6) = 3.5(-2.5) = -8.75. So, the point is (3.5, -8.75). We then draw a line that passes through this point with a slope of 1. This means for every one unit we move to the right, we move one unit up. By sketching these tangents, we can visually confirm our calculations and gain a deeper understanding of how the gradient changes along the curve. The tangent at x = 2.5 slopes downwards, the tangent at x = 3 is horizontal, and the tangent at x = 3.5 slopes upwards, which aligns with our calculated gradients of -1, 0, and 1, respectively. This graphical representation is a powerful tool for understanding the concepts of derivatives and tangents.

Conclusion

In conclusion, we've successfully navigated the process of finding the gradients of tangents for the curve y = x(x - 6) at the points x = 2 1/2, x = 3, and x = 3 1/2. We began by understanding the nature of the curve, recognizing it as a parabola and identifying its key features such as roots and vertex. Next, we delved into the concept of derivatives, using the power rule to find the derivative function dy/dx = 2x - 6, which represents the gradient of the tangent at any point on the curve. By substituting the specific x-values into the derivative, we calculated the gradients at the points of interest: -1 at x = 2 1/2, 0 at x = 3, and 1 at x = 3 1/2. Finally, we illustrated these findings with a detailed sketch graph, plotting the curve and its tangents. The sketch visually confirmed our calculations, showing the tangent at x = 2 1/2 sloping downwards, the tangent at x = 3 being horizontal, and the tangent at x = 3 1/2 sloping upwards. This comprehensive exercise underscores the importance of understanding calculus concepts and their practical applications in analyzing and visualizing curves. By mastering these techniques, you can effectively solve a wide range of problems involving tangents, gradients, and curve sketching. This ability is not only valuable in mathematics but also in various fields such as physics, engineering, and computer graphics, where understanding the behavior of functions and their rates of change is crucial.