The end behavior of a rational function is determined by comparing the degrees of these polynomials. The polynomial with the higher degree has a greater effect on the end behavior of the function.

When examining the end behavior of a rational function, we focus on the leading terms of the polynomials in the numerator and denominator. By comparing the degrees and leading coefficients of these polynomials, we can see whether the numerator or denominator will dominate (or take over) the expression. This happens because as x get extremely large, the term with the larger exponent will eventually get so large (or so negative) that the other terms in the polynomial will not make much of a difference. This means that a function like \dfrac{ax^5+bx-c}{dx^3+5x^2} acts like \dfrac{ax^5}{dx^3} for very large values of x.

When analyzing the quotient of the leading terms of a rational function there are several cases that arise:

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r(x)

1. Degree of numerator is greater than degree of denominator.

Find the quotient of the leading terms (the result will be a polynomial)
The end behavior of this polynomial is the same as the end behavior of the rational function
If the polynomial is linear, the rational function has a slant asymptote parallel to the linear polynomial

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r(x)

2. Degree of numerator is equal to degree of denominator

Find the quotient of the leading coefficients (the result will be a constant, b)
There is a horizontal asymptote at y=b
The ends of the rational function will approach this asymptote

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r(x)

3. Degree of numerator is less than degree of denominator

There is a horizontal asymptote at y=0
The ends of the rational function will approach this asymptote

Examples

Example 1

Determine the nature of the horizontal and/or slant asymptotes for each of the following functions:

f\left(x\right) = \dfrac{3x^2 + 2x + 1 }{ x^2 - 4x + 3}

Worked Solution

Create a strategy

Consider the quotient of the leading terms, \dfrac{3x^2}{x^2} and compare the exponents.

Apply the idea

The leading terms of the numerator and denominator have the same degree. This means we need to find the quotient of the leading coefficients which is: \dfrac{3}{1} or 3.

So there is a horizontal asymptote at f\left(x\right)=3.

g\left(x\right) = \dfrac{2x + 5 }{ x^3 + 1}

Worked Solution

Create a strategy

Consider the quotient of the leading terms, \dfrac{2x}{x^3} and compare the exponents.

Apply the idea

The degree of the numerator is less than the degree of the denominator which means there is a horizontal asymptote at g\left(x\right)=0.

h\left(x\right) = \dfrac{-8x^3 - 3x^2 + 2}{ 3x^2 + x - 1}

Worked Solution

Create a strategy

Consider the quotient of the leading terms, \dfrac{-8x^3}{3x^2} and compare the exponents.

Apply the idea

The degree of the numerator is greater than the degree of the denominator, so we should find the quotient of the leading terms.

\dfrac{-8x^3}{3x^2}=\dfrac{-8}{3}x

h\left(x\right) has a slant asymptote that is parallel to h\left(x\right)=-\dfrac{8}{3}x.

k\left(x\right) = \dfrac{5x^4 - 2x^2 + 1}{x^2 + 4}

Worked Solution

Create a strategy

Consider the quotient of the leading terms, \dfrac{5x^4}{x^2} and compare the exponents.

Apply the idea

The degree of the numerator is greater than the degree of the denominator, so we should find the quotient of the leading terms.

\dfrac{5x^4}{x^2}=5x^2

5x^2 is not linear so h\left(x\right) has no horizonal or slant asymptotes.

Example 2

Use the end behavior of each of the following functions to describe the asymptotes of the function.

\lim_{x \to \infty} r\left(x\right) = 3 \\ \lim_{x \to -\infty} r\left(x\right) = 3

Worked Solution

Create a strategy

Consider that, as the values of x get very small and very large the function approaches the same value 3.

Apply the idea

Because both ends of the graph approach 3 this means there must be a horizontal asymptote at y=3 for this function.

\lim_{x \to \infty} r\left(x\right) = 0 \\ \lim_{x \to -\infty} r\left(x\right) = 0

Worked Solution

Create a strategy

Consider that, as the values of x get very small and very large the function approaches the same value 0.

Apply the idea

Because both ends of the graph approach 0 this means there must be a horizontal asymptote at y=0 for this function.

\lim_{x \to \infty} r\left(x\right) = -2 \\ \lim_{x \to -\infty} r\left(x\right) = 2

Worked Solution

Create a strategy

Consider that, as the values of x get very small and very large the function approaches different constant values.

Apply the idea

Because both ends of the graph approach different constant values, this function has two horizontal asymptotes, at least.

\lim_{x \to \infty} r\left(x\right) = \infty \\ \lim_{x \to -\infty} r\left(x\right) = -\infty

Worked Solution

Create a strategy

Consider that the function has no bounds.

Apply the idea

Because the ends of the graph approach \infty and -\infty this means the function has no horizontal asymptotes. It may have a slant asymptote, but we do not have enough information to determine this for sure.

Example 3

Use limit notation to describe the end behavior of the function: r\left(x\right)=\dfrac{2x^3 - 5x^2 + 4x + 7}{x^3 - 3x + 2}.

Worked Solution

Create a strategy

Compare the degree of the numerator to the degree of the denominator.

Apply the idea

Both the numerator and the denominator have a degree of 3.

Next we need to find the quotient of the leading terms which is \dfrac{2x^3}{x^3}=2.

For very large and very small values of x the function approaches 2 which is the location of a horizontal asymptote.

The end behavior is:

\lim_{x \to \infty} r\left(x\right) = 2 \\ \lim_{x \to -\infty} r\left(x\right) = 2

Idea summary

The end behavior of a rational function depends on the locations of its asymptotes (or the absence of them).

There are several cases:

degree	asymptote
\text{num } > \text{ denom}	\text{slant}
\text{num } = \text{ denom}	\text{horizontal at } y=b
\text{num } < \text{ denom}	\text{horizontal at } y=0

1.7 Rational functions and end behavior

Introduction

Ideas

End behavior of rational functions

Examples

Example 1

Example 2

Example 3

Outcomes

1.7.A

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