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Chapter 3: Problem 31

Find the vertical asymptotes, if any, and the values of \(x\) corresponding to holes, if any, of the graph of each rational function. $$g(x)=\frac{x-3}{x^{2}-9}$$

Short Answer

Expert verified
The graph of the function \(g(x)= \frac{x-3}{x^{2}-9}\) has a vertical asymptote at \(x=-3\) and no holes since the factor that could have caused a hole, \(x - 3\), is cancelled out during simplification.

Step by step solution

01

Factorization

The first thing to do is to factor the denominator, as it is a difference of squares. That gives \(g(x) = \frac{x - 3} { (x - 3) ( x + 3 ) }\)
02

Simplifying the expression

The next step would be to simplify the expression. Notice that \(x - 3\) appears in both the numerator and the denominator, so these terms cancel out, leaving the simplified function \(g(x) = \frac{1}{ x + 3 }\)
03

Identify vertical asymptotes

The vertical asymptotes are given by values of x where the denominator equals 0, causing the function to approach infinity. Solving \(x + 3 = 0\) gives the vertical asymptote at \(x = -3\)
04

Identify holes

A hole occurs at a given x-value if the same x-value also makes the numerator 0, so before simplifying, the fraction \(\frac{x-3}{(x-3)(x+3)}\) would give a hole at \x = 3. But after simplifying the expression, we end up with no x-values which results 0 in the numerator, so there are no holes for the given function. The hole is actually filled since the numerator and denominator shared a factor, which got cancelled out and simplified the function.

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