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Chapter 3: Problem 30

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

Short Answer

Expert verified
The rational function \(f(x) = \frac{x^2 - 25}{x - 5}\) has a hole at the point \(5, 10\) and has no vertical asymptotes.

Step by step solution

01

Simplify the Function

Firstly, factorize the numerator and the denominator. The numerator \(x^{2}-25\) is a difference of squares and can be factored as \((x - 5)(x + 5)\). The denominator is \(x - 5\). This gives us the factored function \(\frac{(x - 5)(x + 5)}{x - 5}\).
02

Identify Holes

You can now see that the factors \(x - 5\) in the numerator and denominator cancel each other. However, because \(x - 5\) was in both the numerator and denominator in the original form of the function, the point \(x = 5\) is a hole in the graph. The y-coordinate of the hole is found by substituting \(x = 5\) into the reduced function \(y = x + 5\). This gives us the hole at the point: \(5, 10\).
03

Identify Vertical Asymptotes

After cancelling the common factors, the denominator has no \(x\) terms, and therefore, there are no vertical asymptotes for this rational function.

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