Finding Tangent Line Equation To Y=(x+2)/(x-3) At (2,-4)

Finding the equation of a tangent line to a curve at a specific point is a fundamental problem in calculus. This article will delve into the step-by-step process of determining the tangent line equation for the curve Y = (x+2)/(x-3) at the point (2, -4). We will explore the necessary calculus concepts, including derivatives and point-slope form, to arrive at the solution. This exploration not only provides a practical application of calculus but also enhances understanding of the relationship between a function and its derivative.

Understanding Tangent Lines and Derivatives

At its core, finding the tangent line involves determining the slope of the curve at the given point. This is where the concept of the derivative comes into play. The derivative of a function, denoted as dy/dx or f'(x), gives the instantaneous rate of change of the function with respect to its variable. Geometrically, the derivative at a point represents the slope of the tangent line to the curve at that point. Therefore, to find the tangent line equation, our primary goal is to calculate the derivative of the given function and evaluate it at the specified x-coordinate.

In this particular problem, our main keyword lies in calculating the tangent line. To do this effectively, we need to grasp the essence of the derivative, which signifies the curve's instantaneous rate of change. Imagine zooming in on the curve at the point (2, -4). As we zoom in closer and closer, the curve starts to resemble a straight line. This straight line is the tangent line, and its slope mirrors the derivative's value at that specific point. Therefore, the derivative serves as a crucial tool for pinpointing the tangent line's slope, a vital component in formulating its equation. The derivative helps us understand the behavior of the function at a specific point and enables us to draw the tangent line, providing a visual and mathematical representation of the function's slope at that instant.

Step 1: Finding the Derivative

The given function is Y = (x+2)/(x-3). To find its derivative, we will use the quotient rule, a fundamental rule in calculus for differentiating rational functions (functions that are a ratio of two other functions). The quotient rule states that if we have a function Y = u(x)/v(x), then its derivative dy/dx is given by:

dy/dx = [v(x) * u'(x) - u(x) * v'(x)] / [v(x)]^2

In our case, u(x) = x+2 and v(x) = x-3. Let's find the derivatives of u(x) and v(x):

u'(x) = d/dx (x+2) = 1 v'(x) = d/dx (x-3) = 1

Now, we can apply the quotient rule:

dy/dx = [(x-3) * 1 - (x+2) * 1] / (x-3)^2

Simplify the expression:

dy/dx = (x - 3 - x - 2) / (x-3)^2 dy/dx = -5 / (x-3)^2

Therefore, the derivative of the function Y = (x+2)/(x-3) is dy/dx = -5 / (x-3)^2. This derivative represents the slope of the tangent line at any point x on the curve. By finding the derivative of the function, we've unlocked the key to understanding how the function's slope changes across its domain. The quotient rule, a cornerstone of calculus, enables us to differentiate rational functions, where one function is divided by another. Applying this rule meticulously ensures we accurately capture the slope's behavior. This calculated derivative serves as the foundation for our next step, where we'll evaluate it at the specific point (2, -4) to determine the tangent line's slope at that location. Grasping the concept of the derivative is paramount for tackling tangent line problems and gaining deeper insights into function behavior.

Step 2: Evaluating the Derivative at x=2

To find the slope of the tangent line at the point (2, -4), we need to evaluate the derivative dy/dx at x = 2:

dy/dx |_(x=2) = -5 / (2-3)^2 dy/dx |_(x=2) = -5 / (-1)^2 dy/dx |_(x=2) = -5 / 1 dy/dx |_(x=2) = -5

So, the slope of the tangent line at the point (2, -4) is -5. The slope of the tangent line, derived from the function's derivative, is a crucial piece of information for constructing the tangent line's equation. Evaluating the derivative at the specified x-coordinate, in this case x = 2, provides us with the precise slope at that point. This slope represents the steepness and direction of the tangent line as it grazes the curve at the point (2, -4). A negative slope, as we found here, signifies that the tangent line is decreasing or sloping downwards as we move from left to right. This numerical value, the slope of -5, will be used in conjunction with the point-slope form to define the tangent line's equation completely. Understanding how to evaluate the derivative at a given point is essential for solving tangent line problems and for gaining a deeper understanding of the function's behavior at that specific location.

Step 3: Using the Point-Slope Form

Now that we have the slope of the tangent line (-5) and the point (2, -4), we can use the point-slope form of a line to find the equation of the tangent line. The point-slope form is given by:

y - y_1 = m(x - x_1)

where m is the slope and (x_1, y_1) is the point. Plugging in our values, we get:

y - (-4) = -5(x - 2) y + 4 = -5x + 10

Now, we can rewrite the equation in slope-intercept form (y = mx + b):

y = -5x + 10 - 4 y = -5x + 6

Thus, the equation of the tangent line to the curve Y = (x+2)/(x-3) at the point (2, -4) is y = -5x + 6. The point-slope form serves as a bridge, connecting the calculated slope and the given point to the equation of the line. This form, y - y_1 = m(x - x_1), elegantly captures the relationship between a line's slope, a point on the line, and the line's overall equation. By substituting the slope of -5 and the point (2, -4) into the point-slope form, we initiated the process of defining our tangent line. The subsequent simplification and transformation into the slope-intercept form, y = -5x + 6, provide a clear and concise representation of the tangent line's equation. This final equation allows us to visualize the line's position on the coordinate plane, with a slope of -5 and a y-intercept of 6. Mastering the application of the point-slope form is crucial for solving various linear equation problems, including finding tangent lines.

Conclusion

In conclusion, we successfully found the equation of the tangent line to the curve Y = (x+2)/(x-3) at the point (2, -4). We achieved this by first finding the derivative of the function using the quotient rule, then evaluating the derivative at x = 2 to find the slope of the tangent line, and finally using the point-slope form to construct the equation of the tangent line. The equation of the tangent line is y = -5x + 6. This problem highlights the power of calculus in analyzing the behavior of functions and finding geometric properties such as tangent lines. Understanding these concepts is crucial for further studies in mathematics and related fields. The journey of finding the tangent line demonstrates the interconnectedness of calculus concepts. We began by applying the quotient rule to find the derivative, representing the instantaneous rate of change. Next, we evaluated the derivative at the specific point to determine the slope of the tangent line. Finally, we utilized the point-slope form to translate the slope and point into the tangent line's equation. This step-by-step process showcases the elegance and efficiency of calculus in solving geometric problems. The ability to find tangent lines has wide-ranging applications, from optimization problems in engineering to modeling physical phenomena. This exercise not only reinforced our understanding of derivatives and tangent lines but also highlighted the broader applicability of calculus in various fields.