2.10 Functions with oblique asymptotes

Functions with oblique asymptotes

Rational functions are of the form $\frac{P\left(x\right)}{Q\left(x\right)}$P(x)Q(x)​ where $P\left(x\right)$P(x) and $Q\left(x\right)$Q(x) are polynomials and $Q\left(x\right)\ne0$Q(x)≠0. Some rational functions include:

Functions of these forms are known to have asymptotes. There are three types of asymptotes: horizontal, vertical and oblique.

Horizontal asymptotes occur when the function approaches a constant value as $x$x tends towards positive or negative infinity.

Vertical asymptotes occur when the function values tend towards positive or negative infinity as $x$x approaches a constant.

It is also possible for a function to have an oblique asymptote or slanted asymptote. Graphically this means that the function approaches a straight line with some non-zero slope as $x$x tends towards positive or negative infinity.

The graph of $f\left(x\right)=\frac{x^2+1}{x-1}$f(x)=x2+1x−1​ has the oblique asymptote $y=x+1$y=x+1.

Worked example

Question 1

Let's consider the function $f\left(x\right)=\frac{x^2+1}{x-1}$f(x)=x2+1x−1​.

Think:  To determine the equation of the oblique asymptote, we can use polynomial long division to rewrite the function into a different form.

In this form, we can determine which terms go to zero as $x$x tends to infinity and then identify the equation of the oblique asymptote.

Do:  We can use long division as follows.

The quotient is $x+1$x+1 and the remainder is $2$2. To rewrite the expression $\frac{x^2+1}{x-1}$x2+1x−1​, we express the remainder as a fraction of the divisor and add this to the quotient to give $\frac{x^2+1}{x-1}=x+1+\frac{2}{x-1}$x2+1x−1​=x+1+2x−1​.

Reflect:  To find the equation of the oblique asymptote, we take $x$x to infinity and observe which terms vanish.

Notice that the last term $\frac{2}{x-1}$2x−1​ goes to zero, so the equation of the oblique asymptote is $y=x+1$y=x+1.

Notice as $x$x gets large, the term $\frac{2}{x-1}$2x−1​ decreases and the graph of the function approaches the oblique asymptote $y=x+1$y=x+1. So the sign of $\frac{2}{x-1}$2x−1​ for large values of $x$x determines whether the graph approaches the asymptote from above or below.

This process is quite involved if all we want to know is whether or not a function has an oblique asymptote.

To determine whether a function has an oblique asymptote, without finding the actual equation of the asymptote, we can subtract the degree of the polynomial in the denominator from the degree of the polynomial in the numerator. If the result is $1$1, then the function has an oblique asymptote.

This is because when we use polynomial division, the resulting quotient has a degree that is equal to the difference between the degrees of the numerator and denominator.

Worked example

Question 2

Consider the rational function $f\left(x\right)=\frac{x^4-5x+1}{\left(x-1\right)\left(x^2+3\right)}$f(x)=x4−5x+1(x−1)(x2+3)​.

Does the graph of $f\left(x\right)$f(x) have an oblique asymptote?

Think: A rational function has an oblique asymptote if the degree of the numerator minus the degree of the denominator is $1$1.

Do: The degree of the polynomial in the numerator is $4$4 and the degree of the polynomial in the denominator is $3$3.

The difference is $4-3=1$4−3=1, so the function does have an oblique asymptote.

Practice questions

question 3

Question 4

question 5