Problem 58 - Calculus: Early Transcendentals

Question

Find a formula for a function that has vertical asymptotes \( x = 1 \) and \( x = 3 \) and horizontal asymptote \( y = 1 \).

Step-by-Step Solution

Answer

The function is \( f(x) = \frac{x^2}{(x-1)(x-3)} \).

Step 1: Understanding the Problem

To find a formula for a function with vertical asymptotes at \( x = 1 \) and \( x = 3 \) and a horizontal asymptote at \( y = 1 \), we need to consider a rational function. Rational functions of the form \( f(x) = \frac{a(x)}{b(x)} \) can have vertical asymptotes where the denominator \( b(x) = 0 \) and horizontal asymptotes based on the degrees of the polynomials in the numerator and denominator.

Step 2: Identify the Structure for Vertical Asymptotes

The vertical asymptotes at \( x = 1 \) and \( x = 3 \) mean the denominator of our rational function should be zero at these points. Thus, the denominator can be \((x-1)(x-3)\).

Step 3: Determine Horizontal Asymptote Conditions

For a horizontal asymptote at \( y = 1 \), the degrees of the numerator and denominator should be the same if the leading coefficients ratio is \( 1 \). If \( b(x) = (x-1)(x-3) \) is a quadratic, then \( a(x) \) should also be a quadratic with a leading coefficient 1, like \( x^2 \).

Step 4: Construct the Function

Combine the numerator and the denominator from the previous steps to form the function. Since the numerator \( a(x) = x^2 \) ensures the horizontal asymptote, the function is \( f(x) = \frac{x^2}{(x-1)(x-3)} \). This satisfies the horizontal and vertical asymptote criteria.

Key Concepts

Vertical Asymptotes

A vertical asymptote occurs in a rational function at values of x where the function tends to infinity or negative infinity. These are the x-values that make the denominator zero.

• For this specific exercise, the vertical asymptotes are at x = 1 and x = 3.
• This means that for these values, the denominator of our rational function is zero.

To put it simply, if a rational function is written as \( f(x) = \frac{a(x)}{b(x)} \), the function has vertical asymptotes at the x-values which make \( b(x) = 0 \).
In this problem, we used a denominator of \((x-1)(x-3)\) ensuring vertical asymptotes at \( x = 1 \) and \( x = 3 \) because \( (x-1)(x-3) = 0 \) at these points.

Horizontal Asymptotes

Horizontal asymptotes describe the behavior of a function as x tends towards infinity or negative infinity.
In the given problem, the horizontal asymptote is at y = 1.

• If the degrees of the numerator and the denominator of a rational function are the same, the horizontal asymptote is the ratio of their leading coefficients. In our function, both are quadratic polynomials with degrees of 2,
• Since the leading coefficients of the numerator and denominator are both 1, the horizontal asymptote is \( y = \frac{1}{1} = 1 \).

Thus, setting the degrees of the numerator and the denominator equal and ensuring they have the same leading coefficient gives a clear path to determining the horizontal asymptote.

Polynomial Degree

The degree of a polynomial tells us the highest power of x present in the polynomial. Understanding this is crucial for analyzing asymptotes in rational functions.

• In the numerator \( a(x) = x^2 \), the degree is 2.
• Similarly, in the denominator \( b(x) = (x-1)(x-3) \) expands to \( x^2 - 4x + 3 \), which also has a degree of 2.

Having the same degree in the numerator and the denominator is a key condition for determining horizontal asymptotes through the leading coefficients.
This balance is crucial for constructing functions that satisfy given asymptotic behaviors, like in our problem where \( f(x) = \frac{x^2}{(x-1)(x-3)} \) neatly fits all given criteria.