Exercise 15 Page 526

We want to find any holes in the graph of the given rational function as well as any vertical asymptotes. Let's start by factoring the denominator.

y = \frac{x - 1}{x ^{2} - 2 x + 1}

$a^{2} - 2 a b + b^{2} = (a - b)^{2}$

y = \frac{x - 1}{( x - 1 ) ^{2}}

Recall that division by zero is not defined. Therefore, the rational function is undefined where

(x - 1)^{2} = 0 .

(x - 1)^{2} = 0 \Leftrightarrow x = 1

The function is not defined when

x = 1 .

This means we have either a hole or a vertical asymptote. Let's now simplify the function.

y = \frac{x - 1}{( x - 1 ) ^{2}}

Split into factors

y = \frac{x - 1}{( x - 1 ) ( x - 1 )}

Cancel out common factors

y = \frac{1}{x - 1}

After simplifying, the factor

x - 1

is still in the denominator. This indicates that the line

x = 1

is a vertical asymptote.

Horizontal Asymptotes

We are also able to find horizontal asymptotes in rational functions.

$y = \frac{p ( x )}{q ( x )}$
The degree of $p (x)$ is greater than the degree of $q (x)$	There are no horizontal asymptotes
The degree of $q (x)$ is greater than the degree of $p (x)$	$y = 0$ is a horizontal asymptote
The degrees of $p (x)$ and $q (x)$ are equal	$y = \frac{a}{b}$ is a horizontal asymptote, where $a$ and $b$ are the leading coefficients of $p (x)$ and $q (x),$ respectively

Let's now consider the given function.

y = \frac{x ^{1} - 1}{x ^{2} - 2 x + 1}

Since the degree of the denominator is greater than the degree of the numerator, the line

y = 0

is a horizontal asymptote.