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Rational Functions and Holes

Alice White

Alice White

7 min read

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Study Guide Overview

This study guide covers rational functions and holes, focusing on identifying and calculating hole coordinates. Key concepts include factoring, canceling common factors, and evaluating limits related to holes. The guide also connects holes to limits, emphasizing how to find the x and y coordinates of a hole. Examples, practice questions, and exam tips are provided to reinforce learning.

1.10 Rational Functions and Holes

🏌🏽 Hole in One!

Key Concept

The presence of a real zero in both the numerator and denominator of a rational function can create a hole. This occurs when a factor (x - a) is present in both the numerator and the denominator. The function is undefined at x = a, but can be 'repaired' by canceling the common factor.

If the multiplicity of the zero 'a' in the numerator is greater than or equal to its multiplicity in the denominator, then the function has a hole at x = a. Remember, multiplicity is the number of times a zero appears in the factorization. For example, if 'a' appears 3 times in the numerator and 2 times in the denominator, there's a hole at x = a.

To find the y-value of the hole, cancel the common factor and then evaluate the simplified function at x = a. Let's look at an example:

r(x) = (x² - 9) / (x - 3)

The numerator has zeros at x = -3 and x = 3, while the denominator has a zero at x = 3. The zero at x=3 has a multiplicity of 2 in the numerator and 1 in the denominator, so there is a hole at x=3. First, factor the numerator:

r(x) = (x + 3)(x - 3) / (x - 3)

Then, cancel the common factor (x - 3):

r(x) = x + 3

Finally, plug in x = 3 to find the y-value of the hole:

r(3) = 3 + 3 = 6

So, there's a hole at (3, 6). 🤓

hole_ex794723468239872505.png

Graph displaying a hole at the intersection (1,1)

Source: Jed Q

🌉 Connecting to Limits

When a rational function, r(x), has a hole at x = c, the function approaches a finite limit as x approaches c. This limit, L, is the y-coordinate of t...

Question 1 of 7

Which of the following rational functions has a hole? 🧐

f(x)=x+2x+3f(x) = \frac{x+2}{x+3}

g(x)=x24x+2g(x) = \frac{x^2 - 4}{x+2}

h(x)=x2x+3h(x) = \frac{x-2}{x+3}

k(x)=x2+4x2k(x) = \frac{x^2 + 4}{x-2}