Rational Function (Domain, x & y - Intercepts, Zeros, Vertical and Horizontal Asymptotes and Hole)

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Summary

This tutorial video explains how to find the domain, x-intercepts, y-intercepts, zeros, vertical asymptotes, horizontal asymptotes, and holes of rational functions through detailed examples.

Highlights

Finding the Y-intercept
00:05:28

The y-intercept is found by setting x to zero in the function. Substituting x=0 into the original function results in y = (-6) / (-3) = 2. Thus, the y-intercept is (0, 2).

Finding the Zeros
00:06:42

The zeros are the same as the x-intercepts. Considering the domain restriction, the only zero for this function is x = -2.

Introduction to Rational Functions and First Example
00:00:00

The video introduces rational functions as a ratio of two polynomial functions. The first example problem, f(x) = (x^2 - x - 6) / (x - 3), is presented to demonstrate how to find various properties.

Finding the Domain
00:00:55

The domain is defined as all real numbers except for the zeros of the denominator. For the first example, setting the denominator (x - 3) to zero yields x = 3, so the domain is all real numbers except 3.

Finding the X-intercept
00:02:12

To find the x-intercept, the function f(x) (or y) is set to zero. After cross-multiplication, the numerator (x^2 - x - 6) is factored into (x - 3)(x + 2). Setting these factors to zero gives x = 3 and x = -2. However, since x=3 is restricted by the domain, the only x-intercept is (-2, 0).

Finding the Horizontal Asymptote
00:07:31

Rules for horizontal asymptotes based on the degrees of the numerator and denominator are explained. In the first example, the degree of the numerator (2) is greater than the degree of the denominator (1), meaning there is no horizontal asymptote.

Finding the Hole (Removable Discontinuities)
00:09:06

A hole occurs when there is a common factor in both the numerator and denominator. In the first example, (x - 3) is a common factor. Setting x - 3 = 0 gives x = 3. Substituting x = 3 into the simplified function (x + 2) gives y = 5. So, the hole is at (3, 5).

Second Example: f(x) = 1 / (x - 1)
00:11:17

The domain is x ≠ 1. Setting y = 0 for x-intercept leads to 0 = 1, which is impossible, so there's no x-intercept and no zeros. Setting x = 0 for y-intercept gives y = 1 / (-1) = -1, so the y-intercept is (0, -1). The vertical asymptote is x = 1 (from the domain restriction). The degree of the numerator (0) is less than the denominator (1), so the horizontal asymptote is y = 0. There are no common factors, so there is no hole.

Third Example: f(x) = (x^2 - 4) / (x^2 - 9)
00:15:37

The factored form is f(x) = (x + 2)(x - 2) / (x + 3)(x - 3). The domain is x ≠ 3 and x ≠ -3. The x-intercepts are (-2, 0) and (2, 0) because 0 = (x + 2)(x - 2). The y-intercept is found by setting x = 0, resulting in y = (-4) / (-9) = 4/9, or (0, 4/9). The zeros are x = ±2. The vertical asymptotes are x = 3 and x = -3. Since the degrees of the numerator and denominator are equal (both 2), the horizontal asymptote is y = (leading coefficient of numerator) / (leading coefficient of denominator) = 1/1 = 1. There are no common factors, so there is no hole.

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