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Chapter 3: Problem 123

Determine whether each statement makes sense or does not make sense, and explain your reasoning. My graph of \(y=\frac{x-1}{(x-1)(x-2)}\) has vertical asymptotes at \(x=1\) and \(x=2\)

Short Answer

Expert verified
The statement does not make sense because at \(x=1\), there is a hole and not a vertical asymptote. However, at \(x=2\), the statement is correct in saying there is a vertical asymptote.

Step by step solution

01

- Setup

We have the function \(y = \frac{{x-1}}{{(x-1)(x-2)}} \). To find the vertical asymptotes, we look at where the function is undefined due to division by zero. This happens when the denominator of our function is zero.
02

鈥� Evaluate the denominator

The denominator is zero when either \(x - 1 = 0\) or \(x - 2 = 0\). Solving these gives \(x = 1\) and \(x = 2\).
03

鈥� Consider the Rule

The rule states that there is a hole when a factor is in both the numerator and the denominator. We see that factor \(x-1\) is present in both - this means that for \(x = 1\) there will be a hole and not a vertical asymptote as initially suggested.
04

鈥� Answer the question

The given statement was partially correct, there is a vertical asymptote at \(x = 2\), but not at \(x = 1\). At \(x = 1\) there is a hole instead because the \(x-1\) factor in the numerator and the denominator can cancel each other.

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