Problem 29 - 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. $$f(x)=\frac{x^{2}-9}{x-3}$$

Step-by-Step Solution

Answer

The graph of the function \(f(x)=\frac{x^{2}-9}{x-3}\) has a hole at \(x=3\) and no vertical asymptotes.

Step 1: Identify the denominator and find where it equals zero

The denominator of the function is \(x-3\). We set this equal to zero to find out where the function is undefined. So, \(x-3=0\), which implies \(x=3\).

Step 2: Consider where the numerator equals zero

The numerator of the function is \(x^{2}-9\). We set this equal to zero to find out where it crosses the x-axis. So, \(x^{2}-9=0\). This is equivalent to \((x-3)(x+3)=0\), which tells us \(x=3\) or \(x=-3\). Note that the value \(x=3\) was also found in step 1.

Step 3: Making conclusions about holes and asymptotes

Since \(x=3\) makes both the numerator and the denominator equal to zero, this value corresponds to a hole in the graph of the function. For \(x=-3\), it only makes the numerator zero, not the denominator, so this does not introduce a vertical asymptote nor a hole.