Problem 116 - A Graphical Approach to Precalculus with Limits

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

There is a removable discontinuity (a hole) at \(x=-5\), not a vertical asymptote.

Step 1: Identify the Function

We are given the function \( f(x) = \frac{x^2 - 25}{x + 5} \). Our goal is to determine if there is a vertical asymptote and specifically at \(x=-5\).

Step 2: Factor the Numerator

The numerator can be factored as a difference of squares: \( x^2 - 25 = (x-5)(x+5) \). The function can therefore be rewritten as \( f(x) = \frac{(x-5)(x+5)}{x+5} \).

Step 3: Simplify the Function

Since \((x+5)\) is both in the numerator and the denominator, we can cancel it from the expression for \(x eq -5\). Thus, \( f(x) = x-5 \) for \(x eq -5\).

Step 4: Analyze the Behavior at \(x=-5\)

At \(x = -5\), the function is undefined due to division by zero. Thus, \((x+5)\) actually cancels out, resulting in a hole at \(x = -5\) instead of a vertical asymptote.

Step 5: Conclusion

There is no vertical asymptote at \(x = -5\). Instead, the function has a hole (removable discontinuity) at that point.

Key Concepts

Hole in Graph

When examining the graph of a function, a "hole" represents a point where the function is not defined even though it could intuitively continue smoothly through that location. Holes occur when a particular factor in the numerator and the denominator cancels out, leaving the graph undefined at that specific point. For the function given, \( f(x) = \frac{x^2 - 25}{x+5} \), this happens at \( x = -5 \). After factoring the numerator as \((x-5)(x+5)\), the common factor \((x+5)\) cancels with the denominator, indicating a hole at \( x = -5 \).

• This point, \( x = -5 \), is not visible on the graph but affects the continuity.
• It represents an instance where the function has a potential value, but is rendered undefined due to division by zero.
• Feel free to consider the graph as complete, except at this single point.

Understanding this concept helps explain why some graphs may seem continuous except for missing points or disruptions.

Removable Discontinuity

A removable discontinuity is a point on the graph where the function is not defined, but the limit exists, and the function can be made continuous by redefining it at this point. For example, for the function \( f(x) = \frac{x^2 - 25}{x+5} \), after cancelling \((x+5)\), it simplifies to \( f(x) = x - 5 \), except at \( x = -5 \).

• This represents a removable discontinuity because the function can be redefined at \( x = -5 \) to make the function continuous across all real numbers.
• Graphically, it simply appears as a hole.
• By evaluating the limit from both sides as \( x \) approaches \(-5\), you'll find that the limit of \( f(x) \) approaches \(-10\), even though the function itself does not exist at this point.

This demonstrates the large gaps removable discontinuities can fill just by redefining the function value explicitly at the point.

Difference of Squares

The difference of squares is a fundamental algebraic identity that allows for the simplification of expressions such as \( a^2 - b^2 \) into \((a-b)(a+b)\). In the given function, \( f(x) = \frac{x^2 - 25}{x+5} \), the numerator is written in the form \( x^2 - 25 \), which translates to \((x-5)(x+5)\).

• This simplification is crucial for identifying the behavior of the function.
• It allows you to cancel terms with the denominator, which highlights potential discontinuities.
• By understanding this concept, you can quickly identify holes in graphs or simplify complicated algebraic expressions.

Mastering the difference of squares not only simplifies factoring but also aids in solving equations and understanding the underlying structure of polynomials.

Undefined Functions

A function is considered undefined at points where it results in operations that are not mathematically permissible, such as division by zero. In the function \( f(x) = \frac{x^2 - 25}{x+5} \), it becomes undefined at \( x = -5 \), because it would require dividing by zero, which is impossible.

• Undefined points often lead to discontinuities in graphs.
• Functions often become undefined at critical points that involve a denominator equaling zero.
• While some undefined functions signify vertical asymptotes, in this case, a cancelling term creates a hole instead.

Understanding where and why functions are undefined is key in math, as it helps in sketching accurate graphs and solving equations.