Problem 123 - Algebra and Trigonometry

Question

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 \(x=2\)

Step-by-Step Solution

Answer

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 1 - 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.

Step 2 – Evaluate the denominator

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

Step 3 – 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.

Step 4 – 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.