Eight Steps to Discovering the Equation of a Tangent Line

1

Plot the function and the line that touches it at one point (recommended). A visual representation will facilitate the understanding of the problem and allow for the verification of the solution's plausibility. Sketch the function on a sheet of graph paper, consulting a graphing calculator as needed. Sketch the line tangent to the function at the specified point. (Remember, the tangent line passes through that point and has the same slope as the function at that point.)

• Example 1: Illustrate the graph of the parabola f(x)=0.5x2 3x−1{\displaystyle f(x)=0.5x^ 3x-1}. Draw the tangent line passing through the point (-6, -1). You may not be familiar with the equation of the tangent line just yet, but you can already infer that its slope is negative, and its y-intercept is negative as well (well below the vertex of the parabola with a y-value of -5.5). If your final answer does not align with these observations, it serves as an indicator to review your work for errors.

2

Calculate the first derivative to determine the equation for the slope of the tangent line.[1]

For function f(x), the first derivative f'(x) represents the equation for the slope of the tangent line at any given point on the graph of f(x). Multiple methods exist to calculate derivatives. Here's a straightforward example utilizing the power rule:[2]

• 

  Example 1 (continued): The graph is described by the function f(x)=0.5x2 3x−1{\displaystyle f(x)=0.5x^ 3x-1}. Recall the power rule when calculating derivatives: ddxxn=nxn−1{\displaystyle {\frac {d}{dx}}x^{n}=nx^{n-1}}. The first derivative of the function = f'(x) = (2)(0.5)x 3 - 0. f'(x) = x 3. By substituting any value a for x into this equation, the outcome will represent the slope of the line tangent to f(x) at the point where x = a.

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3

Provide the x-coordinate of the point under scrutiny.[3]

Examine the problem to ascertain the coordinates of the point for which you are seeking the tangent line. Input the x-coordinate of this point into f'(x). The output will correspond to the slope of the tangent line at this specific point.

• Example 1 (continued): The problem states that the point is (-6, -1). Let's use the x-coordinate -6 as the input for f'(x): f'(-6) = -6 * 3 = -3. Therefore, the slope of the tangent line is -3.

4

Let's now derive the equation of the tangent line in point-slope form. The point-slope form of a linear equation is y−y1 = m(x−x1), where m represents the slope and (x1, y1) is a point on the line.[4] With the information we have, we can proceed to write the equation in this form.

• Example 1 (continued): Using the formula y−y1 = m(x−x1), as the slope of the line is -3, we get y−y1 = -3(x−x1). Since the tangent line passes through the point (-6, -1), the final equation becomes y−(-1) = -3(x−(-6)). Simplifying further, we have y+1 = -3(x+6). Finally, simplifying the equation gives us y = -3x - 18. Hence, the equation of the tangent line is y = -3x - 19.

5

To validate our equation, let's check it on our graph. If you have access to a graphing calculator, plot the original function and the tangent line to ensure that our answer is correct. If you are working on paper, refer to your previous graph to verify that there are no obvious errors in your solution.

6

Try a more challenging problem. Here's a step-by-step walkthrough of the entire process once again. This time, the objective is to determine the tangent line for f(x)=x3 2x2 5x 1{\displaystyle f(x)=x^ 2x^ 5x 1} at x = 2:

• Utilizing the power rule, we calculate the first derivative f′(x)=3x2 4x 5{\displaystyle f'(x)=3x^ 4x 5}. This function provides us with the slope of the tangent.
• Given x = 2, we evaluate f′(2)=3(2)2 4(2) 5=25{\displaystyle f'(2)=3(2)^ 4(2) 5=25}. This value corresponds to the slope at x = 2.
• Observe that this time we lack a specific point, only an x-coordinate. To determine the y-coordinate, we substitute x = 2 into the initial function: f(2)=23 2(2)2 5(2) 1=27{\displaystyle f(2)=2^ 2(2)^ 5(2) 1=27}. Thus, the coordinates of the point are (2,27).
• Express the equation of the tangent line in point-slope form: y−y1=m(x−x1){\displaystyle y-y_=m(x-x_)}y−27=25(x−2){\displaystyle y-27=25(x-2)}. If necessary, simplify it to y = 25x - 23.

Discovering the Extreme Points on a Graph"

Looking for the utmost points on a graph? These are locations where the graph reaches either a local maximum (a point higher than the surrounding points) or a local minimum (a point lower than the surrounding points). However, simply having a slope of 0 on the tangent line does not guarantee the presence of an extreme point. To identify these points, follow these steps:

1. Derive the first derivative of the function to obtain f'(x), the equation for the slope of the tangent.

2. Solve for f'(x) = 0 to identify potential extreme points.

3. Determine the second derivative, f''(x), which reveals the rate at which the slope of the tangent is changing.

4. For each potential extreme point, substitute the x-coordinate (a) into f''(x). If f''(a) is positive, there is a local minimum at that point. Conversely, if f''(a) is negative, a local maximum exists. However, if f''(a) is 0, an inflection point is present rather than an extreme point.

5. If a maximum or minimum is identified at point a, find f(a) to ascertain the y-coordinate.

"Finding the Equation for the Normal"

The "normal" to a curve refers to a line passing through a specific point on the curve while possessing a slope perpendicular to that of the tangent. To determine the equation for the normal, utilize the relationship that the product of the slopes of the tangent and the normal at the same point equals -1. Here's how:

1. Calculate f'(x), which corresponds to the slope of the tangent line.

2. If the point lies at x = a, determine f'(a) to establish the slope of the tangent at that particular point.

3. Compute −1f′(a) to ascertain the slope of the normal.

4. Express the normal equation in slope-point form.

"Add New Question"

Have a query? Feel free to ask:

- How can you determine the slope of a tangent line?

Finding the equations of two tangent lines to a given graph can be achieved by following a particular process. In general, a line can be represented by the equation y=mx+c, where m represents the slope. To determine the equations of the tangent lines, it is necessary to find the values of m and c. The slope (m) can be determined by examining the relationship between the change in y and the change in x. For instance, if the line moves up two units in the y-direction for every three units it moves across in the x-direction, then the slope (m) would be equal to 2/3.

Once the slope (m) is obtained, the next step is to find the y-intercept (c) of the line. The y-intercept can be determined by utilizing a known point (x, y) on the line. In this case, because the line is tangent to a graph, the point of tangency (x, y) can be used. By using this point along with the slope (m), it is possible to utilize algebra to calculate the value of c. Therefore, the equation of the line can be expressed as y=mx+c, where c equals y-mx.

To address the question of finding the equation for a line tangent to the graph of f(x) and running parallel to the line y=2x-3, several steps need to be followed. Parallel lines always share the same slope. Therefore, since the line y=2x-3 has a slope of 2, the tangent line will also have a slope of 2. Additionally, it is known that the derivative of f(x), denoted as f'(x), will equal 2 at the point where the tangent line intersects. By differentiating f(x) to obtain f'(x) and equating it to 2, it then becomes possible to solve for x. Once the x-coordinate is determined, it can be substituted into f(x) to discover the y-coordinate. With this information, it is then feasible to write the equation of the tangent line in point-slope form, considering all the data that has been collected.

To find further answers and information on the topic, it is recommended to visit the following source: [link: https://www.wikihow.com/Questions/Find-the-Equation-of-a-Tangent-Line#offset=3]. If you have any additional questions, feel free to inquire.

Last updated: March 11, 2023

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Categories: Calculus

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