A Rational Function \( H \) Is Continuous And Has A Horizontal Asymptote At \( Y=1 \). Which Of The Following Functions Could Be \( H \)? A. \( H(x)=\frac{x^2-16}{x^2+16} \) B. \( H(x)=\frac{x^2+16}{x^2-16} \)

In the realm of mathematics, rational functions hold a significant position, particularly when exploring concepts like continuity and asymptotes. In this comprehensive guide, we will delve into the intricacies of rational functions, with a specific focus on identifying the correct function given certain conditions, such as the presence of a horizontal asymptote. We will address the question: A rational function $h$ h } is continuous, with a horizontal asymptote at ${ y=1$. Which function could be function $h$ h }? A. ${ h(x)=\frac{x^2-16$x^2+16} } B. ${ h(x)=\frac{x^2+16{x^2-16} }$

Defining Rational Functions

At its core, a rational function is a function that can be expressed as the quotient of two polynomials. Mathematically, it can be represented as ${ h(x) = \frac{P(x)}{Q(x)} }$, where ${ P(x) }$ and ${ Q(x) }$ are polynomial functions, and ${ Q(x) }$ is not equal to zero. The behavior of rational functions is governed by the degrees and leading coefficients of the polynomials in the numerator and denominator, especially when analyzing their asymptotes and continuity.

Horizontal Asymptotes: A Key Characteristic

A horizontal asymptote is a horizontal line that the graph of a function approaches as ${ x }$ tends to positive or negative infinity. It provides valuable information about the end behavior of the function. The existence and location of horizontal asymptotes in rational functions are determined by comparing the degrees of the polynomials in the numerator and the denominator.

Rules for Horizontal Asymptotes

To determine the horizontal asymptote of a rational function, we consider three cases:

1. Degree of numerator < Degree of denominator: In this case, the horizontal asymptote is always at ${ y = 0 }$.
2. Degree of numerator = Degree of denominator: The horizontal asymptote is at ${ y = \frac{a}{b} }$, where ${ a }$ is the leading coefficient of the numerator and ${ b }$ is the leading coefficient of the denominator.
3. Degree of numerator > Degree of denominator: There is no horizontal asymptote. Instead, there may be a slant (oblique) asymptote.

Understanding these rules is crucial for identifying the correct rational function that satisfies the given conditions.

Continuity in Rational Functions

Continuity is another essential aspect of rational functions. A function is continuous at a point if there are no breaks, jumps, or holes at that point. For rational functions, continuity is generally maintained across its domain, except at points where the denominator is equal to zero. These points of discontinuity often lead to vertical asymptotes or holes in the graph.

Points of Discontinuity

To find the points of discontinuity, we set the denominator ${ Q(x) }$ equal to zero and solve for ${ x }$. The solutions represent the x-values where the function is not continuous. These points can either be vertical asymptotes or removable discontinuities (holes), depending on whether the factor causing the zero in the denominator can be canceled out by a factor in the numerator.

Analyzing the Given Conditions

The problem states that the rational function ${ h }$ is continuous and has a horizontal asymptote at ${ y = 1 }$. These two conditions are crucial in narrowing down the possible functions. The horizontal asymptote at ${ y = 1 }$ tells us that the degrees of the numerator and denominator must be the same, and the ratio of their leading coefficients must be 1. The continuity condition implies that there are no vertical asymptotes or holes, meaning the denominator should not have any real roots.

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

For option A, ${ h(x) = \frac{x^2 - 16}{x^2 + 16} }$, the degrees of the numerator and the denominator are both 2. The leading coefficients are both 1, so the horizontal asymptote is at ${ y = \frac{1}{1} = 1 }$. To check for continuity, we set the denominator equal to zero:

${ x^2 + 16 = 0 }$

${ x^2 = -16 }$

Since there are no real solutions for ${ x }$, the denominator is never zero, and the function is continuous. Therefore, option A satisfies both conditions.

Evaluating Option B: ${ h(x) = \frac{x^2 + 16}{x^2 - 16} }$

For option B, ${ h(x) = \frac{x^2 + 16}{x^2 - 16} }$, the degrees of the numerator and the denominator are also both 2. The leading coefficients are both 1, so the horizontal asymptote is at ${ y = \frac{1}{1} = 1 }$. However, to check for continuity, we set the denominator equal to zero:

${ x^2 - 16 = 0 }$

${ x^2 = 16 }$

${ x = \pm 4 }$

Since there are real solutions ${ x = 4 }$ and ${ x = -4 }$, the function has vertical asymptotes at these points and is not continuous. Therefore, option B does not satisfy the continuity condition.

Conclusion: Selecting the Correct Function

Based on our analysis, option A, ${ h(x) = \frac{x^2 - 16}{x^2 + 16} }$, satisfies both the horizontal asymptote condition (${ y = 1 }$) and the continuity condition. Option B, ${ h(x) = \frac{x^2 + 16}{x^2 - 16} }$, has a horizontal asymptote at ${ y = 1 }$ but is not continuous due to vertical asymptotes at ${ x = 4 }$ and ${ x = -4 }$. Therefore, the correct answer is A.

In summary, understanding the properties of rational functions, including horizontal asymptotes and continuity, is essential for solving problems like this. By comparing the degrees and leading coefficients of the polynomials in the numerator and denominator, and by checking for points of discontinuity, we can effectively identify the correct function that meets the given criteria. The ability to analyze and interpret these mathematical concepts is invaluable in various fields, making it a crucial skill for students and professionals alike.