Asymptotes of a Rational Function

Explanation

DEFINITION OF A RATIONAL FUNCTION

A rational function is a mathematical function that can be expressed as the quotient of two polynomials. This means a rational function has the form f(x) = P(x)/Q(x), where P(x) and Q(x) are polynomials, and Q(x) must not be the zero polynomial (Q(x) ≠ 0).

THE CONCEPT OF AN ASYMPTOTE

An asymptote of a rational function is a straight line that describes the behavior of the function's graph as x or f(x) approaches certain values but never actually reaches them. There are three types of asymptotes: vertical, horizontal, and slant (or oblique) asymptotes.

• VERTICAL ASYMPTOTES occur when the function approaches infinity or minus infinity as x approaches a certain value. This happens when the denominator Q(x) is equal to 0, but the numerator P(x) is not equal to 0 at the same value of x (after any common factors have been canceled).
• HORIZONTAL ASYMPTOTES describe the behavior of the function's graph as x approaches infinity or minus infinity. A horizontal asymptote occurs when the value of the function approaches a certain constant.
• SLANT (OR OBLIQUE) ASYMPTOTES are asymptotes that are neither vertical nor horizontal. They occur when the degree of the numerator P(x) is exactly one greater than the degree of the denominator Q(x).

HOW TO FIND THE ASYMPTOTES OF A RATIONAL FUNCTION?

• VERTICAL ASYMPTOTES: Solve the equation Q(x) = 0. Each solution 'c' for which P(c) ≠ 0 (after simplifying the fraction P(x)/Q(x) by canceling common factors) will be the location of a vertical asymptote, x = c.
• HORIZONTAL ASYMPTOTES:
• If the degree of the numerator P(x) is less than the degree of the denominator Q(x), then y = 0 is the horizontal asymptote.If the degrees are equal, the horizontal asymptote is y = (leading coefficient of P(x)) / (leading coefficient of Q(x)).If the degree of P(x) is greater than the degree of Q(x), there is no horizontal asymptote.
• SLANT ASYMPTOTES: If the degree of the numerator P(x) is exactly one greater than the degree of the denominator Q(x), perform polynomial long division to divide P(x) by Q(x). The quotient (which will be a linear equation of the form y = mx + b) represents the slant asymptote. The remainder term should approach zero as x approaches ±∞.

EXAMPLE:

For the function f(x) = (2x² + 3x – 5) / (x – 1), let's find the asymptotes.

VERTICAL ASYMPTOTE:

Set the denominator x – 1 = 0, which gives x = 1. Now, check the value of the numerator P(x) = 2x² + 3x – 5 at x = 1: P(1) = 2(1)² + 3(1) – 5 = 2 + 3 – 5 = 0. Since both the numerator and the denominator are 0 at x = 1, we should factor the numerator: 2x² + 3x – 5 = (x – 1)(2x + 5). So, f(x) = [(x – 1)(2x + 5)] / (x – 1). For x ≠ 1, f(x) = 2x + 5. Because the (x-1) factor cancels, there is no vertical asymptote at x = 1. Instead, there is a hole in the graph at x = 1. The y-coordinate of the hole is 2(1) + 5 = 7.

HORIZONTAL ASYMPTOTE:

The degree of the numerator (2) is greater than the degree of the denominator (1). Therefore, there is no horizontal asymptote.

SLANT ASYMPTOTE:

Since the degree of the numerator is exactly one greater than the degree of the denominator, we perform polynomial long division of (2x² + 3x – 5) by (x – 1). The division yields a quotient of 2x + 5 and a remainder of 0. So, f(x) = 2x + 5, for x ≠ 1. In this particular case, because the remainder is zero, the function itself simplifies to the line y = 2x + 5 (with a hole at x=1). This line is the graph of the function, rather than a line the function approaches asymptotically. (For a typical slant asymptote, the polynomial division would result in a linear quotient mx+b and a non-zero remainder R(x), so f(x) = mx + b + R(x)/Q(x), where R(x)/Q(x) approaches 0 as x approaches ±∞. Then, y = mx + b would be the slant asymptote.)

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