Problem 35 - 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. $$r(x)=\frac{x^{2}+4 x-21}{x+7}$$

Step-by-Step Solution

Answer

The rational function \(r(x)=\frac{x^{2}+4 x-21}{x+7}\) has a hole at \(x=-7\), and it does not have any vertical asymptotes.

Step 1: Find the value of x that makes the denominator zero

To find the vertical asymptote, which is the value of \(x\) that makes the denominator zero, solve the equation \(x+7=0\). Solving this gives \(x=-7\).

Step 2: Determine if it's an asymptote or hole

Substitute \(x=-7\) into the numerator \(x^{2}+4 x-21\) to see if it becomes zero as well. The result is \((-7)^{2}+4(-7)-21 = 0\). Therefore, both the numerator and denominator become zero when \(x=-7\), indicating this is a hole, not an asymptote.

Step 3: Simplify the function

To confirm this, rewrite the function \(r(x)=\frac{x^{2}+4 x-21}{x+7}\) in a factored form. The numerator can be factored as \((x-3)(x+7)\). So, the function becomes \(r(x)=\frac{(x-3)(x+7)}{x+7}\). When we cancel out \(x+7\) from the numerator and denominator, it leaves a hole at \(x=-7\).

Step 4: Conclude the solution

So, the function has a hole at \(x=-7\), and there are no vertical asymptotes.