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Find the vertical asymptotes, if any, and the values of \(x\) corresponding to holes, if any, of the graph of each rational function. $$r(x)=\frac{x^{2}+2 x-24}{x+6}$$

Short Answer

Expert verified
The given rational function \(r(x)=\frac{x^{2}+2 x-24}{x+6}\) does not have any vertical asymptotes. It has a hole on the graph at \(x = -6\).

Step by step solution

01

Factorize the Function

The first step involves factorizing the rational function. To factorize \(r(x)=\frac{x^{2}+2 x-24}{x+6}\), the numerator \(x^{2}+2 x-24\) factors into \((x-4)(x+6)\). Therefore, \(r(x)=\frac{(x-4)(x+6)}{x+6}\).
02

Find the Vertical Asymptotes

Vertical asymptotes occur where the denominator of a rational function is zero, and the numerator isn't zero. Here, it can be observed that the denominator of the function is \(x+6 = 0\), which implies \(x = -6\). However, at \(x = -6\), the numerator also becomes zero since \(x+6\) is a common factor between the numerator and denominator. Therefore, no vertical asymptotes exist for this function.
03

Identify Possible Holes in the Graph

Holes in the graph of a rational function occur at values of \(x\) where both the numerator and denominator become zero. From our factoring, it's observed that at \(x = -6\), both the numerator and denominator become zero. Therefore, \(x = -6\) corresponds to a hole in the graph of the rational function.

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