Problem 98 - A Graphical Approach to College Algebra

Question

Suppose a friend tells you that the graph of $$f(x)=\frac{x^{2}-25}{x+5}$$ has a vertical asymptote with equation \(x=-5 .\) Is this correct? If not, describe the behavior of the graph at \(x=-5\).

Step-by-Step Solution

Answer

No, the graph has a hole at \(x = -5\), not a vertical asymptote.

Step 1: Finding the Undefined Points

First, determine the values of \(x\) for which the function \(f(x)=\frac{x^{2}-25}{x+5}\) is undefined. This occurs when the denominator equals zero. So, we set \(x+5 = 0\) and solve for \(x\). Thus, \(x = -5\) is the value where the function could be undefined or have an asymptote.

Step 2: Simplifying the Function

Factor the numerator \(x^2 - 25\) using the difference of squares: \((x - 5)(x + 5)\). The function \(f(x)\) simplifies to \(f(x) = \frac{(x-5)(x+5)}{x+5}\). Cancel out the common term \((x+5)\) from the numerator and the denominator. Thus, \(f(x) = x - 5\), except where \(x = -5\).

Step 3: Determining the Graph's Behavior at \(x = -5\)

The initial function was \(f(x)=\frac{x^2-25}{x+5}\), which simplifies, excluding \(x=-5\), to \(f(x) = x - 5\). There is no vertical asymptote because the \((x + 5)\) terms cancel out. Instead, at \(x = -5\), there is a hole in the graph since the function is undefined over that point.

Step 4: Conclusion

Finally, considering the above simplification, the claim that there is a vertical asymptote at \(x = -5\) is incorrect. Evaluating the behavior of the graph, there is a hole at \(x = -5\) because the simplified function \(f(x) = x - 5\) is undefined at \(x = -5\).

Key Concepts

Undefined Points

Undefined points occur in rational functions when the denominator equals zero. For the function \(f(x)=\frac{x^{2}-25}{x+5}\), it's crucial to find when this happens as it will indicate where the function is undefined.

To achieve this, set the denominator \(x+5\) to zero and solve for \(x\). In this case, \(x = -5\) makes the denominator zero, thus \(x = -5\) is an undefined point for this function.

Undefined points are significant as they don't just signify where a graph doesn't exist, but can indicate different behaviors such as vertical asymptotes or holes in the graph. It is the first step in analyzing the function before graphing it.

Simplifying Rational Functions

Simplifying rational functions is crucial in understanding their true behavior. This involves canceling out terms in the numerator and denominator.

For the function \(f(x)=\frac{x^{2}-25}{x+5}\), we begin by factoring the numerator, which is a difference of squares: \(x^2 - 25 = (x-5)(x+5)\). The function then becomes \(\frac{(x-5)(x+5)}{x+5}\).

After canceling the common term \((x+5)\) in both the numerator and the denominator, the function simplifies to \(f(x) = x - 5\). However, this simplification cannot be applied where \(x = -5\) since it was the point of cancellation. Thus, though the expression becomes \(x - 5\), it is not valid at \(x = -5\). Simplifying helps to see the function's form and predict behavior sans undefined points.

Graph Behavior

Graph behavior is deeply tied to the nature of the function, especially around undefined points and singularities like holes or asymptotes.

With \(f(x)=\frac{x^{2}-25}{x+5}\), after simplification, the function turns into \(f(x) = x - 5\) except where \(x = -5\). This means the function behaves linearly overall, with a constant slope and y-intercept below x-axis, except at the undefined point \(x = -5\).

Understanding graph behavior involves recognizing that though \(f(x) = x - 5\) suggests a straight line, the undefined point \(x = -5\) disrupts continuity, causing a distinct disruption reflected as a hole in the graph.

The character of the function must be observed by evaluating these potential disturbances to accurately sketch or interpret its graph.

Holes in Graphs

Holes in graphs are unseen discontinuities that occur when a function is undefined at certain points despite simplification. For \(f(x)=\frac{x^{2}-25}{x+5}\), the numerator and denominator share the factor \((x+5)\), which when canceled out, leads to \(f(x) = x - 5\).

Though this appears to reduce the function to a linear form, the original equation's undefined point \(x = -5\) still exists. This point is a hole because when \((x+5)\) was canceled, \(x = -5\) became an open point of discontinuity.

Holes mean at these points, the function doesn't approach infinity (as with vertical asymptotes) but rather simply doesn’t have a defined value. When graphing or interpreting the function \(x-5\), inclusion of such holes maintains graph accuracy and integrity.