Deriving Equations of Tangent Lines for Specific Functions at Given Points

In this article, we delve into the process of finding the equations of tangent lines for given functions at specific points. This is a fundamental concept in calculus, particularly useful in understanding the behavior of functions near specific points. We will go through three distinct examples and explain the steps involved.

Step 1: Understanding the Process

To find the equation of a tangent line at a specific point, follow these steps:

Find the derivative of the function. The derivative at a point gives the slope of the tangent line at that point. Evaluate the derivative at the given point. This will give you the slope of the tangent line. Use the point-slope form of the equation of a line. The point-slope form is: y - y1 m(x - x1), where (x1, y1) is the point and m is the slope. Simplify the equation to get the final form.

Example 1: fx 3x2 - 12x - 1 at the Point (0, -1)

Step 1: Find the derivative

fx' 6x - 12

Step 2: Evaluate the derivative at the point (0, -1)

fx'(0) 6(0) - 12 -12

The slope of the tangent line at (0, -1) is -12.

Step 3: Use the point-slope form of the equation

The point-slope form for the line is: y - y1 m(x - x1)

Substituting the point (0, -1) and the slope -12:

y - (-1) -12(x - 0)

y 1 -12x

y -12x - 1

Final Equation: y -12x - 1

Example 2: fx x2 1 at the Point (0, 1)

Step 1: Find the derivative

fx' 2x

Step 2: Evaluate the derivative at the point (0, 1)

fx'(0) 2(0) 0

The slope of the tangent line at (0, 1) is 0.

Step 3: Use the point-slope form of the equation

The point-slope form for the line is: y - y1 m(x - x1)

Substituting the point (0, 1) and the slope 0:

y - 1 0(x - 0)

y - 1 0

y 1

Final Equation: y 1

Example 3: fx x2 - 2x 1 at the Point (2, 1)

Step 1: Find the derivative

fx' 2x - 2

Step 2: Evaluate the derivative at the point (2, 1)

fx'(2) 2(2) - 2 2

The slope of the tangent line at (2, 1) is 2.

Step 3: Use the point-slope form of the equation

The point-slope form for the line is: y - y1 m(x - x1)

Substituting the point (2, 1) and the slope 2:

y - 1 2(x - 2)

y - 1 2x - 4

2x - y - 1 0

Final Equation: 2x - y - 1 0

Understanding these steps and practicing them is crucial for mastering calculus. The ability to find the equation of a tangent line is not only useful in calculus but also in fields such as physics, engineering, and economics where these concepts are frequently applied.