# Oblique (or skew) asymptotes

Chpt: Rational functions
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## 1 The reason why

The idea behind finding all non-vertical asymptotes is the following theorem.

Key idea
If $f(x)=\frac{p(x)}{d(x)}$ is a rational function, and the degree of the numerator is less than

that of the denominator, then $f(x)\;$ has the asymptote $y=0\;$ on both sides.

Now take any rational function $f(x)=\frac{p(x)}{d(x)}$ and divide the denominator $d(x)\;$ into the numerator $q(x)\;$ to get a quotient $q(x)\;$ and a remainder $r(x)\;$. This means

$\frac{p(x)}{d(x)} = q(x) + \frac{r(x)}{d(x)}.$

Since the remainder has degree less than the divisor, when $x\;$ is large (positive or negative)

$\frac{p(x)}{d(x)} \approx q(x).$

So when the quotient is a constant (0 or otherwise), we have a horizontal asymptote. If the quotient is a linear function, then we have an oblique (or skew) asymptote, the main topic of this page. Finally if the quotient is a higher degree polynomial, then we have a non-linear asymptote--ignored in most College Algebra courses, but useful enough that we have a section about them below.

## 2 Oblique (or skew) asymptotes

If we divide $d(x)\;$ into $p(x)\;$ and get a linear quotient, then $y=\frac{p(x)}{d(x)}$ has a oblique (or skew) asymptotes. Note this means we have a oblique asymptote if and only if the degree of the numerator is one more than the degree of the denominator.

For a simple example consider $y=\frac{x^3-x+1}{x^2}$, dividing $x^2\;$ into $x^3-x+1\;$ we get a quotient of $x\;$ (and the remainder $-x+1\;$). So this function has the oblique asymptote $y = x\;$. This is clear in the graph on the right.

Thats all! Divide, and if linear, you have a oblique asymptote!

### 2.1 Test yourself

Now why don't you try some? For the rational functions below, determine the oblique asymptotes, if any, of the following examples. Slide you mouse over the black rectangles to check your answers. If you have trouble, listen to the pencast on the right.

function oblique
asymptote
1 $y=\frac{x^3}{x^2+2}$ y = x
2 $y=\frac{x+2}{3x^2+2}$ no oblique, but
horizontal y = 0
3 $y=\frac{4x^3}{2x^2+3}$ y = 2x
4 $y=\frac{2x(3x-2)}{2-x}$ y = – 6x – 8
5 $y=\frac{5x^2+2x^6}{(x^2+2)^3}$ no oblique, but
horizontal y = 2
6 $y=\frac{x^3}{x^5+3x^2-10}$ no oblique, but
horizontal y = 0
7 $y=\frac{2x^3(3x-2x^3)}{5x^5-x^2}$ y = – 4x5

### 2.2 Example graph

Lets end our discussion of oblique asymptotes by graphing the function $y=\frac{x^3+1}{x^2-1}$. To see if it has vertical asymptotes we look to see where the denominator is zero (x2–1=0, so x=±1), but the numerator is not. This means x=1 is a vertical asymptote, but x=–1 is not, in fact it is a common factor of the numerator and denominator. This means the graph has a hole (undefined point) at (−1,−3/2).

Next we divide x2–1 into x3+1 to get x, so we have the oblique asymptote y = x.

Finally we need a few points. This graph has the y-intercept (0,–1), no x-intercepts (numerator is only zero at the excluded value x=–1). Another point is (2,3). Plotting these points along with the two asymptotes we get the graph on the left.

## 3 Non-linear "asymptotes"

### 3.1 Example One

An asymptote of a curve is a line such that the distance between the curve and the line approaches zero as they tend to infinity, but why restrict ourselves to lines? For example, consider the rational function

$y=\frac{x^3-1}{x+1}.$

If we divide the denominator into the numerator we get a quotient of $x^2-x+1\;$ with a remainder of -2, that is

$y=x^2-x+1 +\frac{-2}{x+1}.$
On the right we graph this quadratic (the quotient from the division above) as the blue dashed curve. The rational function also has a vertical asymptote at $x=-1\;$ (plotted as the vertical red dashed line). Notice how close to these curves the actual graph is (black).

### 3.2 Example Two

Lets look at another example,

$y=\frac{x^4+2x^3+x^2}{x+2}.$

This clearly has a vertical asymptote at $x=-2\;$. If we divide we find

$y= x^3+x -\frac{2}{x+2}.$

Plotting the vertical asymptote as a red dashed line, and the cubic quotient as a blue-dashed curve, we get the graph on the right. Below e have a closer look at the origin. Click on any of these graphs to see larger versions.

### 3.3 Moral

The process we learned for finding oblique asymptotes: divide the denominator into the numerator of the rational function, gives us very useful information anytime the degree of the numerator is great then that of the denominator.

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