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Chapter 3: Problem 32

Find the vertical asymptotes, if any, and the values of \(x\) corresponding to holes, if any, of the graph of each rational function. $$g(x)=\frac{x-5}{x^{2}-25}$$

Short Answer

Expert verified
The function \(g(x)=\frac{x-5}{x^{2}-25}\) has a vertical asymptote at \(x=-5\) and a hole at \(x=5\).

Step by step solution

01

Factor the Function

Firstly, factor the denominator as well as the numerator. The given function \(g(x)=\frac{x-5}{x^{2}-25}\) can be factored as \(g(x)=\frac{x-5}{(x-5)(x+5)}\). By comparing the numerator and the denominator, it can be observed that \(x-5\) is common in both the numerator and the denominator.
02

Find the Vertical Asymptotes

To find the vertical asymptotes, equate the denominator to zero and solve for \(x\). The denominator is \((x-5)(x+5)\). Set it equal to zero, that gives: \((x-5)(x+5)=0\). Solving for \(x\), we get \(x=5\) and \(x=-5\). However, since \(x-5\) is in both the numerator and denominator, \(x=5\) will be a hole instead of a vertical asymptote. Hence, there is one vertical asymptote at \(x=-5\).
03

Find the Holes

A hole exists when a value makes both the numerator and denominator of the function zero. As observed in Step 1, \(x-5\) is common to both the numerator and the denominator. Thus, \(x=5\) is a hole in the function.

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