Problem 34 - College Algebra

Question

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

Step-by-Step Solution

Answer

For \(h(x) = \frac{x+6}{x^{2}+2x-24}\), \(x=4\) is a vertical asymptote and \(x=-6\) corresponds to a hole in the graph.

Step 1: Identify potential vertical asymptotes and holes

Vertical asymptotes and holes occur where the denominator of the rational function is equal to zero. We begin by setting the denominator of \(h(x)\) equal to zero and solving the resulting quadratic equation.\(x^{2}+2x-24=0\)

Step 2: Finding the zeroes of the denominator

The quadratic equation from step 1 can be factored and then solved to find the zeroes.\(x^{2}+2x-24=(x-4)(x+6)\)Setting these factors to 0:\(x-4=0\) gives \(x=4\)\(x+6=0\) gives \(x=-6\)

Step 3: Identifying if these are vertical asymptotes or holes

To be a vertical asymptote, \(x=4\) or \(x=-6\) has to make the numerator (\(x+6\)) equal to a non-zero number. For \(x=4\), \(x+6 = 4+6 = 10\), which is not zero. To be a hole, \(x=4\) or \(x=-6\) has to make both the numerator and the denominator equal to zero. We have seen that for \(x=-6\), both the numerator (\(x+6\)) and the denominator (\(x^{2}+2x-24\)) equal zero.