Identifying Rational Functions With Horizontal Asymptotes At Y=1

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In mathematics, rational functions play a crucial role in various fields, from calculus to engineering. Understanding their behavior, especially concerning asymptotes, is fundamental. This article delves into identifying rational functions based on their properties, particularly the presence of horizontal asymptotes. We will explore how to determine the correct function from a given set of options, focusing on the characteristics that define a rational function with a specific horizontal asymptote. Let's consider a scenario: a rational function $h$ is continuous and has a horizontal asymptote at $y=1$. Which function could be function $h$? We will analyze this problem, providing a comprehensive explanation and solution.

What are Rational Functions?

To effectively address the question, it's essential to first define what rational functions are. A rational function is any function that can be written as the ratio of two polynomials. Mathematically, it is expressed as: $f(x) = \frac{P(x)}{Q(x)}$, where $P(x)$ and $Q(x)$ are polynomial functions, and $Q(x) \neq 0$. The domain of a rational function includes all real numbers except for the values of $x$ that make the denominator $Q(x)$ equal to zero, as division by zero is undefined.

Rational functions exhibit a variety of behaviors, including the presence of vertical, horizontal, and oblique asymptotes. Asymptotes are lines that the graph of the function approaches but does not touch or cross. They provide valuable information about the function's behavior as $x$ approaches infinity or specific values.

Understanding Horizontal Asymptotes

A horizontal asymptote is a horizontal line that the graph of a function approaches as $x$ tends to positive or negative infinity. The presence and location of a horizontal asymptote are determined by the degrees of the polynomials in the numerator and the denominator of the rational function. There are three primary scenarios to consider:

  1. Degree of Numerator < Degree of Denominator: If the degree of the polynomial in the numerator is less than the degree of the polynomial in the denominator, the horizontal asymptote is always $y = 0$. This is because, as $x$ becomes very large, the denominator grows much faster than the numerator, causing the overall value of the function to approach zero.
  2. Degree of Numerator = Degree of Denominator: If the degree of the polynomial in the numerator is equal to the degree of the polynomial in the denominator, the horizontal asymptote is the line $y = \frac{a}{b}$, where $a$ is the leading coefficient of the numerator and $b$ is the leading coefficient of the denominator. In this case, the function approaches a constant value as $x$ goes to infinity, determined by the ratio of the leading coefficients.
  3. Degree of Numerator > Degree of Denominator: If the degree of the polynomial in the numerator is greater than the degree of the polynomial in the denominator, there is no horizontal asymptote. Instead, there may be an oblique (or slant) asymptote, which is a diagonal line that the function approaches. The behavior of the function in this scenario is more complex and involves polynomial division to determine the equation of the oblique asymptote.

Identifying the Correct Function with a Horizontal Asymptote at $y=1$

Now, let's focus on the specific problem: identifying a rational function $h$ that is continuous and has a horizontal asymptote at $y = 1$. Given the understanding of horizontal asymptotes, we know that for a rational function to have a horizontal asymptote at $y = 1$, the degree of the numerator must be equal to the degree of the denominator, and the ratio of their leading coefficients must be 1. This is the key criterion for selecting the correct function.

Consider the given option: A. $h(x) = \frac{x^2 - 16}{x^2 + 16}$. To determine if this function meets the criteria, we need to analyze the degrees and leading coefficients of the numerator and denominator.

The numerator, $x^2 - 16$, is a polynomial of degree 2, and its leading coefficient is 1. The denominator, $x^2 + 16$, is also a polynomial of degree 2, and its leading coefficient is 1. Since the degrees are equal, and the ratio of the leading coefficients is $\frac{1}{1} = 1$, this function has a horizontal asymptote at $y = 1$. Therefore, option A, $h(x) = \frac{x^2 - 16}{x^2 + 16}$, could be the correct function.

To further solidify our understanding, let's consider why other types of functions might not satisfy the given conditions. For instance, a function with the degree of the numerator less than the denominator would have a horizontal asymptote at $y = 0$, not $y = 1$. Similarly, a function with the degree of the numerator greater than the denominator would not have a horizontal asymptote at all.

Detailed Analysis of Option A: $h(x) = \frac{x^2 - 16}{x^2 + 16}$

Let’s delve deeper into why option A, $h(x) = \frac{x^2 - 16}{x^2 + 16}$, fits the criteria. As we established, the degrees of the numerator and the denominator are both 2, and their leading coefficients are both 1. This immediately suggests a horizontal asymptote at $y = 1$. However, a comprehensive analysis requires us to examine the function's behavior as $x$ approaches infinity.

To confirm the horizontal asymptote, we can divide both the numerator and the denominator by the highest power of $x$, which in this case is $x^2$:

h(x)=x2βˆ’16x2+16=x2x2βˆ’16x2x2x2+16x2=1βˆ’16x21+16x2h(x) = \frac{x^2 - 16}{x^2 + 16} = \frac{\frac{x^2}{x^2} - \frac{16}{x^2}}{\frac{x^2}{x^2} + \frac{16}{x^2}} = \frac{1 - \frac{16}{x^2}}{1 + \frac{16}{x^2}}

As $x$ approaches infinity, the terms $\frac{16}{x^2}$ in both the numerator and the denominator approach 0. Therefore, the function $h(x)$ approaches:

lim⁑xβ†’βˆžh(x)=1βˆ’01+0=1\lim_{x \to \infty} h(x) = \frac{1 - 0}{1 + 0} = 1

This confirms that the horizontal asymptote is indeed at $y = 1$. Furthermore, the function is continuous because the denominator, $x^2 + 16$, is never equal to zero for any real value of $x$. The term $x^2$ is always non-negative, and adding 16 ensures that the denominator is always positive.

The function $h(x) = \frac{x^2 - 16}{x^2 + 16}$ also has interesting characteristics regarding its intercepts and symmetry. The x-intercepts occur when the numerator is zero:

x2βˆ’16=0β€…β€ŠβŸΉβ€…β€Šx2=16β€…β€ŠβŸΉβ€…β€Šx=Β±4x^2 - 16 = 0 \implies x^2 = 16 \implies x = \pm 4

Thus, the x-intercepts are at $x = 4$ and $x = -4$. The y-intercept occurs when $x = 0$:

h(0)=02βˆ’1602+16=βˆ’1616=βˆ’1h(0) = \frac{0^2 - 16}{0^2 + 16} = \frac{-16}{16} = -1

So, the y-intercept is at $y = -1$. The function is also an even function because $h(-x) = h(x)$, indicating symmetry about the y-axis.

Why Other Functions May Not Fit

To further illustrate the concept, let's consider scenarios where other functions might not satisfy the condition of having a horizontal asymptote at $y = 1$. For example, a function like $f(x) = \frac{x}{x^2 + 1}$ has a horizontal asymptote at $y = 0$ because the degree of the numerator (1) is less than the degree of the denominator (2). As $x$ approaches infinity, the denominator grows much faster than the numerator, causing the function to approach zero.

Another example is a function like $g(x) = \frac{x^3 + 1}{x^2 + 1}$. Here, the degree of the numerator (3) is greater than the degree of the denominator (2). This type of function does not have a horizontal asymptote; instead, it has an oblique or slant asymptote. The function's behavior is characterized by its approach to a diagonal line as $x$ goes to infinity.

In summary, for a rational function to have a horizontal asymptote at $y = 1$, the degrees of the numerator and the denominator must be equal, and their leading coefficients must have a ratio of 1. This ensures that as $x$ approaches infinity, the function approaches the value of 1.

Conclusion

Identifying rational functions with specific horizontal asymptotes requires a clear understanding of the relationship between the degrees and leading coefficients of the polynomials in the numerator and the denominator. In the case of a horizontal asymptote at $y = 1$, the degrees must be equal, and the leading coefficients must have a ratio of 1. Option A, $h(x) = \frac{x^2 - 16}{x^2 + 16}$, perfectly fits this criterion, making it a valid solution. This analysis underscores the importance of understanding the fundamental properties of rational functions and their asymptotic behavior in various mathematical contexts. By carefully examining the degrees and coefficients of the polynomials, we can accurately determine the horizontal asymptotes and gain valuable insights into the behavior of rational functions. This knowledge is crucial for solving a wide range of problems in calculus, algebra, and beyond.

In conclusion, mastering the identification of rational functions and their asymptotes is a critical skill in mathematics. By understanding the principles discussed in this article, you can confidently tackle similar problems and deepen your understanding of rational functions.