Analyzing Rational Function Asymptotes: A Deep Dive

Nov 17, 2025 by Admin 52 views

Hey math whizzes! Today, we're diving deep into the fascinating world of rational functions, and specifically, we're going to break down how to find those pesky horizontal and vertical asymptotes. Our main focus will be on the function $f(x)= rac{25-x^2}{x^2-4 x-5}$ . Understanding these asymptotes is super crucial for graphing and comprehending the behavior of these functions, guys. We'll be looking for three key things: whether the horizontal asymptote at $y=n$ equals the horizontal asymptote at $y=m$ , if $y=-1$ is indeed the horizontal asymptote, and if there's just one solitary vertical asymptote present. So, grab your calculators, your notebooks, and let's get this mathematical party started!

Deconstructing the Function: First Steps

Before we can even think about asymptotes, we gotta get our function $f(x)= rac{25-x^2}{x^2-4 x-5}$ into its simplest form. This means factoring both the numerator and the denominator. For the numerator, $25-x^2$ , this is a classic difference of squares, so it factors into $(5-x)(5+x)$ . Now, for the denominator, $x^2-4x-5$ , we need two numbers that multiply to -5 and add up to -4. Those numbers are -5 and 1. So, the denominator factors into $(x-5)(x+1)$ .

Now, let's rewrite our function with these factored forms: $f(x) = rac{(5-x)(5+x)}{(x-5)(x+1)}$ .

Here's a super important trick, guys: notice that $(5-x)$ is the same as $-(x-5)$ . So, we can rewrite the function as: $f(x) = rac{-(x-5)(5+x)}{(x-5)(x+1)}$ .

See that $(x-5)$ term in both the numerator and the denominator? That means we have a hole in our graph at $x=5$ , not a vertical asymptote. When we cancel these out, we get the simplified function: $f(x) = rac{-(5+x)}{x+1}$ , or $f(x) = rac{-5-x}{x+1}$ . This simplification is key to finding our asymptotes correctly. Always simplify first, it makes life so much easier!

Navigating Horizontal Asymptotes: The Long-Term Behavior

Now, let's talk about horizontal asymptotes. These guys tell us where the function is heading as $x$ gets super, super large (both positive and negative). For a rational function like ours, $f(x) = rac{a_n x^n + ...}{b_m x^m + ...}$ , we compare the degrees of the numerator and the denominator. Let's look at our simplified function, $f(x) = rac{-x-5}{x+1}$ .

In this simplified form, the degree of the numerator (which is 1, because of the $-x$ term) is the same as the degree of the denominator (which is also 1, because of the $x$ term). When the degrees are the same, the horizontal asymptote is found by taking the ratio of the leading coefficients. The leading coefficient in the numerator is -1, and the leading coefficient in the denominator is 1. So, the horizontal asymptote is $y = rac{-1}{1}$ , which simplifies to $y = -1$ .

Now, let's consider the original function, $f(x)= rac{25-x^2}{x^2-4 x-5}$ . Here, the degree of the numerator ( $x^2$ ) is 2, and the degree of the denominator ( $x^2$ ) is also 2. Since the degrees are equal, the horizontal asymptote is the ratio of the leading coefficients, which are -1 (from $-x^2$ ) and 1 (from $x^2$ ). This gives us $y = rac{-1}{1}$ , so $y = -1$ .

This confirms that our horizontal asymptote is indeed $y=-1$ . This means that as $x$ approaches positive or negative infinity, the graph of the function will get closer and closer to the line $y=-1$ , but never actually touch it. It's like the function is giving up on going up or down and is just settling into this horizontal line.

Some people get confused about $m$ and $n$ . In the context of horizontal asymptotes for rational functions, if we have $f(x) = rac{P(x)}{Q(x)}$ , and the degree of $P(x)$ is equal to the degree of $Q(x)$ , then the horizontal asymptote is $y = rac{ ext{leading coefficient of } P(x)}{ ext{leading coefficient of } Q(x)}$ . If we denote this value as $L$ , then we can say $m=L$ and $n=L$ , which means $m=n$ . So, the statement $m=n$ is correct in this scenario where the degrees are equal.

Uncovering Vertical Asymptotes: Where the Function Explodes

Next up, vertical asymptotes! These are vertical lines that the function approaches but never crosses. They happen where the denominator of the simplified rational function is equal to zero. Remember our simplified function? It's $f(x) = rac{-5-x}{x+1}$ .

To find the vertical asymptotes, we set the denominator equal to zero: $x+1 = 0$ . Solving for $x$ , we get $x = -1$ . So, we have a vertical asymptote at $x=-1$ . This means that as $x$ gets closer and closer to -1 (from either the left or the right), the value of $f(x)$ will shoot off towards positive or negative infinity.

It's super important to remember that we look for vertical asymptotes after we've canceled out any common factors. In our case, the factor $(x-5)$ was common to both the numerator and the denominator. This cancellation removed a potential vertical asymptote at $x=5$ and instead created a hole at that point. Holes are different from asymptotes, guys; they are just single points missing from the graph, whereas asymptotes are lines the graph approaches.

So, after simplifying, we only have one factor left in the denominator, $(x+1)$ . Setting this to zero gives us $x=-1$ . This means there is indeed only one vertical asymptote. If we had forgotten to simplify and just set the original denominator $x^2-4x-5$ to zero, we would have gotten $(x-5)(x+1)=0$ , which gives $x=5$ and $x=-1$ . But, as we saw, $x=5$ corresponds to a hole because the factor $(x-5)$ cancels out. So, we must always simplify first to correctly identify vertical asymptotes.

Putting It All Together: The Correct Statements

Alright, let's recap what we've found for our function $f(x)= rac{25-x^2}{x^2-4 x-5}$ :

Simplified function: $f(x) = rac{-x-5}{x+1}$
Horizontal Asymptote: We found that the degrees of the numerator and denominator in the original and simplified functions are equal (both are 2 in the original, and both are 1 in the simplified). The ratio of the leading coefficients is $rac{-1}{1} = -1$ . So, the horizontal asymptote is $y=-1$ .
Vertical Asymptote: After simplifying, the only factor remaining in the denominator is $(x+1)$ . Setting $x+1=0$ gives us $x=-1$ . Therefore, there is one vertical asymptote at $x=-1$ .
Hole: The factor $(x-5)$ canceled out, indicating a hole at $x=5$ .

Now, let's evaluate the given statements:

A. $m=n$ : When the degrees of the numerator and denominator of a rational function are equal, the horizontal asymptote is given by the ratio of the leading coefficients. If we call this value $L$ , then we can think of $m=L$ and $n=L$ , implying $m=n$ . Since the degrees are equal in our function, this statement is correct.
B. $y=-1$ is the horizontal asymptote: As we calculated, when the degrees of the numerator and denominator are equal, the horizontal asymptote is the ratio of the leading coefficients. In $f(x)= rac{25-x^2}{x^2-4 x-5}$ , the leading coefficients are -1 and 1. Thus, the horizontal asymptote is $y = rac{-1}{1} = -1$ . This statement is correct.
C. There is only one vertical asymptote: After simplifying the function to $f(x) = rac{-x-5}{x+1}$ , we found that the only factor in the denominator that causes the function to be undefined (and doesn't cancel out) is $(x+1)$ . Setting $x+1=0$ gives us $x=-1$ . The factor $(x-5)$ from the original denominator resulted in a hole, not a vertical asymptote. Therefore, there is indeed only one vertical asymptote at $x=-1$ . This statement is correct.

So, the three correct answers are A, B, and C. It's all about carefully factoring, simplifying, and then applying the rules for finding asymptotes, guys! Keep practicing, and you'll be asymptote masters in no time!