Equation of a Tangent Line in Standard Form: A Comprehensive Guide

The equation of a tangent line in standard form is a crucial concept in calculus, particularly in analytical geometry. A tangent line touches a curve at a single point and shares the same slope as the curve at that point. In this article, we will explore different methods to determine the equation of a tangent line, apply these methods to specific problems, and discuss related concepts such as extreme points and the normal line.

Method 1: Finding the Equation of a Tangent Line

To find the equation of a tangent line to a given function, you can follow these steps:

Sketch the function and tangent line:
Select a point on the curve and draw the tangent line through that point. A graph can help visualize the problem and verify your solution. Take the first derivative:
The first derivative of the function gives the slope of the tangent line at any point. For the given function, the process involves applying the power rule to determine the derivative. Substitute the x-coordinate:
Use the x-coordinate of the given point to find the slope of the tangent line. Write in point-slope form:
The point-slope form of a linear equation is y - y1 m(x - x1). Substitute the slope and the given point into this form. Confirm the equation:
Graph the original function and the tangent line to ensure accuracy. Check the slope and y-intercept to match the expected attributes.

Example 1: Finding the Tangent Line to a Parabola

Consider the parabola y 0.5x^2 - 3x - 1. We will find the tangent line passing through the point (-6, -1).

Find the slope:
Take the first derivative: y' x - 3. Substitute x -6: y'(-6) -6 - 3 -9 (This was incorrectly stated as -3 in the original text). Write in point-slope form:
The equation is y 1 -9(x 6). Simplify to: y -9x - 55.

Example 2: Finding the Tangent Line to a Cubic Function

Now, we will find the line tangent to the function y x^3 - 2x^2 - 5x 1 at x 2.

Find the first derivative:
y' 3x^2 - 4x - 5. Substitute x 2:
y'(2) 3(2)^2 - 4(2) - 5 12 - 8 - 5 -1. Find the y-coordinate:
y(2) 2^3 - 2(2)^2 - 5(2) 1 8 - 8 - 10 1 -9. The point is (2, -9). Write in point-slope form:
y 9 -1(x - 2). Simplify to: y -x - 7.

Method 2: Solving Related Problems

This method involves finding extreme points and the normal line:

Find extreme points:
The first derivative gives the slope of the tangent at any point. To find extreme points, set the first derivative to zero and solve. The second derivative can help determine if these points are maxima, minima, or inflection points. Find the equation of the normal:
Once the slope of the tangent line is known, the slope of the normal (perpendicular to the tangent) can be found by taking the negative reciprocal. The normal line can be written in slope-point form.

Example: Finding Extreme Points on a Graph

Consider the function y x^3 - 2x^2 - 5x 1. To find the extreme points:

Take the first derivative:
y' 3x^2 - 4x - 5. Solve for y' 0:
Set the derivative to zero and solve: 3x^2 - 4x - 5 0. This can be solved using the quadratic formula. Take the second derivative:
y'' 6x - 4. Use this to determine the nature of the points.

By following these methods and applying the steps carefully, you can effectively solve a variety of problems related to tangent lines, extreme points, and normal lines.