# Finding Slant Asymptotes of Rational Functions

## Slant Asymptote

A **slant** (also called **oblique**) **asymptote** for a function {eq}f(x) {/eq} is a linear function {eq}g(x) {/eq} with the property that the limit as {eq}x {/eq} approaches {eq}\pm\infty {/eq} of {eq}f(x) {/eq} is equal to {eq}g(x) {/eq}. In this lesson, we will see how to find slant asymptotes using either **long division** or **synthetic division**.

### Horizontal, Vertical & Slant Asymptote Graph

There are two other types of asymptotes that usually are taught before slant asymptotes: **vertical** and **horizontal**. A **vertical asymptote** appears whenever a function approaches either positive or negative infinity at a point, and **horizontal asymptotes** occur when a function has a constant limit at either positive or negative infinity. **Slant asymptotes** occur when the limit of a function at positive or negative infinity is a linear function. Let's see some examples of what each of these look like graphically:

## Slant Asymptote Rules

Vertical and horizontal asymptotes can occur in a variety of different types of functions. However, slant asymptotes most commonly appear for two types of functions. The first type of function is a **rational function**, of the form {eq}\frac{f(x)}{g(x)} {/eq}, where {eq}f(x) {/eq} and {eq}g(x) {/eq} are both polynomials, and the degree of {eq}f(x) {/eq} must be exactly one higher than the degree of {eq}g(x) {/eq}. The second type of function is an **irrational function** (a function containing roots) where each root contains a polynomial of degree equal to the power of that root. That might sound a bit confusing, but it just means that if there is a square root it needs to contain a quadratic, a cube root should contain a cubic function, and so on.

Let's look at an example. Suppose we have the function {eq}\sqrt{x^2+x+1} {/eq}. Since this is a quadratic polynomial inside a square root, it will have a slant asymptote. To find the equation of this slant asymptote, we will use the definition of slant asymptote and properties of limits.

First, since a slant asymptote is a linear function, we know it must be equal to {eq}ax+b {/eq}. The formal definition of a slant asymptote is that {eq}\lim_{x\to\pm\infty} f(x) - (ax+b) = 0 {/eq}. Dividing both sides by {eq}x {/eq} gives us that {eq}\lim_{x\to\pm\infty} \frac{f(x)}{x} - a - \frac{b}{x} = 0 {/eq}. Since we're taking the limit at {eq}\pm\infty {/eq}, the {eq}-\frac{b}{x} {/eq} term disappears, and we get {eq}\lim_{x\to\pm\infty} \frac{f(x)}{x} - a = 0 {/eq}. So then {eq}\lim_{x\to\pm\infty} \frac{f(x)}{x} = a {/eq}, and now we have a method for finding {eq}a {/eq}.

Let's apply this method to our function to find {eq}a {/eq}, and then we will explain the method for finding {eq}b {/eq}.

We proceed as follows: {eq}\lim_{x\to\pm\infty} \frac{\sqrt{x^2+x+1}}{x} = \lim_{x\to\pm\infty} \frac{\sqrt{x^2(1+\frac{1}{x}+\frac{1}{x^2})}}{x} = \lim_{x\to\pm\infty} \frac{x\sqrt{(1+\frac{1}{x}+\frac{1}{x^2})}}{x} = \lim_{x\to\pm\infty} \sqrt{(1+\frac{1}{x}+\frac{1}{x^2})} = \lim_{x\to\pm\infty} \sqrt{(1)} = 1 {/eq}

Therefore {eq}a = 1 {/eq}, and our slant asymptote has the form {eq}x + b {/eq}. To solve for {eq}b {/eq} we can simply go back to the definition of slant asymptote, which is that {eq}\lim_{x\to\pm\infty} f(x) - (ax+b) = 0 {/eq}. If we add {eq}b {/eq} to both sides of this equation, we have a method for finding {eq}b {/eq} once we know {eq}a {/eq}: {eq}\lim_{x\to\pm\infty} f(x) -ax = b {/eq}.

Now let's apply this method to our current problem.

{eq}\lim_{x\to\pm\infty} \sqrt{x^2+x+1} -x = \lim_{x\to\pm\infty} (\sqrt{x^2+x+1} -x) (\frac{\sqrt{x^2+x+1} +x}{\sqrt{x^2+x+1} +x}) = \lim_{x\to\pm\infty} \frac{x+1}{\sqrt{x^2+x+1} +x} = \lim_{x\to\pm\infty} \frac{1+\frac{1}{x}}{\sqrt{1+\frac{1}{x}+\frac{1}{x^2}} +1} = \lim_{x\to\pm\infty} \frac{1}{\sqrt{1}+1} = \frac{1}{2} = b {/eq}

Therefore the slant asymptote for {eq}\sqrt{x^2+x+1} {/eq} is {eq}x+\frac{1}{2} {/eq}.

## How to Find the Slant Asymptotes?

The above method works for most types of functions with slant asymptotes, as long as the limits can be evaluated. However, for **rational functions** with slant asymptotes, like {eq}\frac{x^3 + 2x^2 + 3}{x^2 + 4} {/eq}, we can also use polynomial division to find their asymptotes. There are two main methods for doing this: **long division** and** synthetic division**. Polynomial long division is essentially the same as the standard long division for numbers, and synthetic division is a special form of long division that only works for linear factors. However, long division also works for linear factors, so if the method of synthetic division seems confusing, don't worry about it!

### Slant Asymptote Example 1

Let's briefly review polynomial long division and show how to find the slant asymptote of a function using this method. Using our function {eq}\frac{x^3 + 2x^2 + 3}{x^2 + 4} {/eq} from earlier, we place the numerator and denominator into the following positions:

{eq}\begin{align*} &\text{ }\text{ }\text{ }\\ x^2 + 0x + 4 &\overline{\big)x^3+2x^2+0x+3}\\ \end{align*} {/eq}

Note that we write out a {eq}0x {/eq} term for both polynomials in order to avoid mistakes during the subtraction step. Since {eq}x^2 {/eq} needs to be multiplied by {eq}x {/eq} to equal {eq}x^3 {/eq}, we place an {eq}x {/eq} at the top, and then multiply {eq}x^2 + 0x + 4 {/eq} by {eq}x {/eq} and place it below {eq}x^3 + 2x^2 + 0x + 3 {/eq}, preparing it for subtraction. Then, we subtract {eq}x^3 + 0x^2 + 4x + 0 {/eq} from {eq}x^3 + 2x^2 + 0x + 3 {/eq}, obtaining {eq}2x^2 - 4x + 3 {/eq}.

{eq}\begin{align*} &\text{ }\text{ }\text{ }x\\ x^2+0x+4 &\overline{\big)x^3+2x^2+0x+3}\\ &\underline{\text{ }x^3 + 0x^2 + 4x + 0}\\ &\text{ }\text{ }\text{ }\text{ }2x^2-4x+3\\ \end{align*} {/eq}

Now we add a {eq}2 {/eq} at the top, since {eq}x^2 * 2 = 2x^2 {/eq}. We multiply {eq}x^2 + 0x + 4 {/eq} by {eq}2 {/eq}, and repeat the process of subtraction.

{eq}\begin{align*} &\text{ }\text{ }\text{ }x+2\\ x^2+0x+4 &\overline{\big)x^3+2x^2+0x+3}\\ &\underline{\text{ }x^3 + 0x^2 + 4x + 0}\\ &\text{ }\text{ }\text{ }\text{ }2x^2-4x+3\\ &\text{ }\text{ }\underline{\text{ }\text{ }2x^2+0x+8}\\ &\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }-4x-5 \end{align*} {/eq}

We end up with the term {eq}-4x-5 {/eq}, and now we are done, since there is nothing we can multiply {eq}x^2 {/eq} by to get {eq}-4x {/eq}. This means that {eq}-4x-5 {/eq} is our remainder, and {eq}x+2 {/eq} is the slant asymptote. To see this more clearly, what this division tells us is that the function {eq}f(x) = \frac{x^3 + 2x^2 + 3}{x^2 + 4} {/eq} is equal to {eq}\frac{(x+2)(x^2+4) - 4x -5}{x^2+4} {/eq}. In this form, we can simplify it to {eq}f(x) = x+2 + \frac{-4x-5}{x^2+4} {/eq}. Now we can see that the rational term will go to {eq}0 {/eq} as {eq}x {/eq} goes to {eq}\pm\infty {/eq}, since the degree of the denominator is larger. This means that the limit of {eq}f(x)-(x+2) {/eq} will be equal to {eq}0 {/eq}, so it is indeed the slant asymptote.

### Slant Asymptote Example 2

Let's try another example of finding a slant asymptote, this time with synthetic division. Synthetic division is similar to long division, but it requires that the denominator is a linear function of the form {eq}x+c {/eq} (note that {eq}c {/eq} may be negative). Let's use synthetic division to find the slant asymptote of {eq}\frac{2x^2+4x+1}{x+3} {/eq}. First, we place the coefficients of the numerator as follows, and to the left of this we then place the negative of the constant term in the denominator, which in this case would be {eq}-3 {/eq}.

{eq}\begin{array}{c|rrr}&2&4&1\\-3\\\hline\\\end{array} {/eq}

Now we can proceed with the synthetic division algorithm. First, drop the leading coefficient {eq}2 {/eq} to the bottom.

{eq}\begin{array}{c|rrr}&2&4&1\\-3\\\hline\\&2\\\end{array} {/eq}

Now multiply this {eq}2 {/eq} by the {eq}-3 {/eq} we have on the left, and place it in the second column.

{eq}\begin{array}{c|rrr}&2&4&1\\-3&&-6\\\hline\\&2\\\end{array} {/eq}

Now add the numbers in the second column, and repeat the process by taking {eq}-3 {/eq} times our new number {eq}-2 {/eq} and placing it in the third column.

{eq}\begin{array}{c|rrr}&2&4&1\\-3&&-6&6\\\hline\\&2&-2&7\\\end{array} {/eq}

Now add the numbers in the third column, and we are done. The number in the third column, {eq}7 {/eq}, represents our remainder, and the {eq}2 {/eq} and {eq}-2 {/eq} in our first two columns tell us that our slant asymptote is {eq}2x-2 {/eq}.

This problem can also be solved with long division as in Example 1, or using the limit method.

## Lesson Summary

**Slant asymptotes** are a special type of asymptote which have the equation of a line. They occur in **rational functions** when the degree of the numerator is exactly one higher than the degree of the denominator, as well as in other types of functions. Slant asymptotes of rational functions can be found by using **long division**, or in some cases **synthetic division**.

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#### How do you find the slant asymptote?

The slant asymptote can be found using the limit method shown in the lesson, or for rational functions by using either polynomial long division or synthetic division.

#### What is the rule for oblique asymptotes?

A rational function will have a slant or oblique asymptote if the degree of the numerator is exactly one higher than the degree of the denominator.

#### Are slant and oblique asymptotes the same?

Slant asymptotes and oblique asymptotes are two different terms for the same thing: an asymptote which has a linear equation.

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