Problem 30 - Algebra and Trigonometry

Question

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

Step-by-Step Solution

Answer

The rational function has a vertical asymptote at \(x = 5\). It also has a hole at the same point \(x = 5\) though it will not be visible in its graph as the hole is at the same position as the vertical asymptote.

Step 1: Simplify Function

Simplify the function as much as possible. The given function is \(f(x) = \frac{x^2 - 25}{x - 5}\). We can simplify this as \(f(x) = \frac{(x - 5)(x + 5)}{x - 5}\) which simplifies to \(f(x) = x + 5\).

Step 2: Find Vertical Asymptotes

The vertical asymptotes of a function are the values of x where the denominator of the function is zero (and not simultaneous zeros of numerator). This function has originally been defined as \(x - 5\) in the denominator, so, setting the denominator equal to zero gives us \(x - 5 = 0\) or \(x = 5\) is a vertical asymptote.

Step 3: Identify Holes

The holes of the function are the values of x where both the numerator and the denominator are zero at the same time. In this case, the numerator and denominator share a denominator \(x - 5\). Thus, \(x = 5\) is not just a vertical asymptote, it is also a hole in the function. However, because the hole precisely coincides with the vertical asymptote, it won't appear in the graph as a hole.

Key Concepts

Vertical Asymptotes

Vertical asymptotes of a rational function occur where the denominator equals zero and the numerator does not simultaneously equal zero. These values are effectively the 'off-limits' zones for the function, as they are points where the function would race off towards infinity (either positive or negative).

For the function \( f(x) = \frac{x^2 - 25}{x - 5} \), you can spot the potential for an asymptote where the denominator \(x - 5 \) is zero. Mathematically, this gives us the equation \( x - 5 = 0 \) or \( x = 5 \). Normally, we would declare \( x = 5 \) as a vertical asymptote, but as you'll see in the following sections, there's more to this story when simplifying rational expressions and identifying holes.

Holes in a Graph

Holes in a graph of a rational function represent points where the function is not defined due to a common factor in both the numerator and the denominator that can be canceled out. These appear as points on the graph where there is a disruption in the curve, often visually represented as a small circle.

In our exercise, the presence of \( x - 5 \) in both the numerator and denominator of \( f(x) = \frac{x^2 - 25}{x - 5} \) indicates that \( x = 5 \) might be such a hole. However, since \( x = 5 \) is also where the vertical asymptote is located, it would typically overshadow the hole. In other circumstances, had there been no vertical asymptote at \( x = 5 \), simplifying the expression would have indeed revealed a hole at this point.

Simplifying Rational Expressions

Simplifying rational expressions involves factoring both the numerator and denominator and canceling out common factors. This step helps to identify any removable discontinuities or 'holes' indicated by factors that appear in both the numerator and denominator.

In the step by step solution, the expression \( f(x) = \frac{x^2 - 25}{x - 5} \) was factored to \( f(x) = \frac{(x - 5)(x + 5)}{x - 5} \). The simplification process allows us to cancel the \( x - 5 \) term on both sides, assuming \( x eq 5 \) to avoid division by zero. Therefore, after simplification, the function becomes \( f(x) = x + 5 \) everywhere except at \( x = 5 \), suggesting that normally there would be a hole at \( x = 5 \).

Analyzing Rational Function Graphs

When analyzing graphs of rational functions, it is crucial to look for key features such as vertical asymptotes, horizontal asymptotes, holes, x-intercepts, and y-intercepts. Graphing these functions helps illustrate behavior, such as how the function approaches asymptotes or where holes might occur.

For our example function \( f(x) = \frac{x^2 - 25}{x - 5} \), if graphed without prior simplification, one would expect a vertical line at \( x = 5 \) indicating the vertical asymptote and a hole at the same point. However, simplifying reveals that the function essentially behaves like the line \( y = x + 5 \), with no asymptotic behavior, but due to the initial form, we still acknowledge \( x = 5 \) as a vertical asymptote when considering the domain restriction of the original rational expression.