Working with rational functions is often a matter of finding what the graph **isn't** as much as finding what it **is**. By identifying the values that the function cannot have, either as an input or as an output, we restrict the possibilities of what the graph may look like.

You have previously graphed rational functions using transformations and horizontal and vertical asymptotes. You may have noted that horizontal asymptotes appear in rational equations when the degree of the numerator is less than or equal to the degree of the denominator.

What happens when the degree of the numerator is greater than the degree of the denominator? How does this situation appear in the graph of the function?

### Oblique Asymptotes

Not all asymptotes of rational functions are vertical or horizontal. Let's consider what happens when the degree of the numerator is one greater than the denominator, resulting in a diagonal line known as an **oblique** or **slant** **asymptote**.

If we look at the graph of the rational function \begin{align*}g(x)=\frac{x^{3}}{x^{2}+1}\end{align*}, we can see that there is no horizontal asymptote of this function.

There is no horizontal asymptote in this function because the degree of the numerator is greater than the degree of the denominator.

As a reminder, the following guidelines can help identify the asymptotes of a rational function \begin{align*}r(x)=\frac{f(x)}{D(x)}\end{align*}:

- If the degree of the denominator is greater than the degree of the numerator, then the line \begin{align*}y=0\end{align*} is a horizontal asymptote.

- If the degree of the numerator and the denominator are equal, then the line \begin{align*}y=\frac{a}{b}\end{align*} is a horizontal asymptote, where \begin{align*}a\end{align*} is the leading coefficient of \begin{align*}f(x)\end{align*}, the numerator, and \begin{align*}b\end{align*} is the leading coefficient of \begin{align*}D(x)\end{align*}, the denominator.

- If the degree of the numerator is larger than the degree of the denominator, then the quotient function, \begin{align*}Q(x)\end{align*}, found by dividing the numerator and denominator of the rational function, is an
**oblique asymptote**. Recall that for any rational function \begin{align*}\frac{f(x)}{D(X)}\end{align*}, you can use polynomial division to re-write that function in the form \begin{align*}\frac{f(x)}{D(x)}=Q(x)+\frac{R(x)}{D(x)}\end{align*} where \begin{align*}Q(x)\end{align*} is the quotient and \begin{align*}R(x)\end{align*} is the remainder.

### Examples

#### Example 1

Earlier, you were given a question about an asymptotic situation.

What happens when the degree of the numerator is greater than the degree of the denominator? How does this situation appear in the graph of the function?

You should have no difficulty with this question now: When the degree of the numerator is greater, there is no horizontal asymptote, but rather a **slant** or **oblique** asymptote. It appears as a diagonal line across the graph of the function.

#### Example 2

Graph \begin{align*}g(x)=\frac{x^{2}-1}{x-2}\end{align*}.

First observe that the vertical asymptote is at \begin{align*}x=2\end{align*}. Notice that the degree of the numerator is greater than the degree of the denominator. We can change the form of the rational expression by long division. You may recall from algebra that polynomials can be divided (just like real numbers), and any rational functions can be written as

\begin{align*}\frac{f(x)}{D(x)}=Q(x)+\frac{R(x)}{D(x)}\end{align*}

Doing the long division here,

\begin{align*}& \qquad \qquad x + 2 \\ & x-2 \ \big ) \overline{x^{2} + 0x -1 }\\ & \qquad \quad \underline{x^{2} -2x \ \ \downarrow}\\ & \qquad \qquad \quad 2x -1\\ & \qquad \qquad \quad \underline{2x -4}\\ & \qquad \qquad \qquad \quad \ 3 \end{align*}

So in this case, the function \begin{align*}g(x)\end{align*} can be rewritten as

\begin{align*}g(x)=\frac{x^{2}-1}{x-2}=x+2+\frac{3}{x-2}\end{align*}

The above equation tells us that as \begin{align*}x\to\pm\infty\end{align*}, the graph of \begin{align*}g(x)=x+2+\frac{3}{x-2}\end{align*} gets closer and closer to the line \begin{align*}y=x+2\end{align*}. Why? Suppose we let \begin{align*}x\end{align*} be a big number, i.e. \begin{align*}x=1,000,000\end{align*}. Then the remainder of this rational function becomes \begin{align*}\frac{3}{999,998}\approx 0\end{align*} and we are left with \begin{align*}x+2\end{align*}. We call this line an *oblique asymptote* and it is indicated by the dashed line in the image below.

#### Example 3

Identify the oblique asymptote of \begin{align*}g(x)=\frac{x^{3}}{x^{2}+1}\end{align*}.

By polynomial division we have,

\begin{align*}& \qquad \quad \ \ x \qquad \qquad \qquad \qquad \ \leftarrow\text{Quotient}\\ & x^2+1 \ \big ) \overline{x^3 + 0x^2 + 0x + 0} \quad \ \leftarrow \text{Dividend}\\ & \qquad \quad \ \underline{x^3 \qquad \quad + x}\\ & \qquad \qquad \qquad \quad \ -x \qquad \quad \ \leftarrow \text{Remainder}\end{align*}

So \begin{align*}g(x)=x-\frac{x}{x^{2}+1}\end{align*}. This tells us that the line \begin{align*}y=x\end{align*} is an oblique asymptote of \begin{align*}g(x)\end{align*}.

Notice that the oblique asymptotes of a rational function also describe the *end behavior* of the function. That is, as you “zoom out” from the graph of a rational function it looks like a line or the function defined by \begin{align*}Q(x)\end{align*} in \begin{align*}\frac{f(x)}{D(x)}=Q(x)+\frac{R(x)}{D(x)}\end{align*}.

#### Example 4

Find the oblique asymptote of \begin{align*}f(x)=\frac{3x^{3}-5x^{2}+2}{x^{2}-3x}\end{align*}. Sketch a graph of \begin{align*}f(x)\end{align*}.

Using polynomial long division, \begin{align*}f(x)=3x+4+\frac{4x^{2}}{x^{2}-3x}\end{align*}. Thus, the line \begin{align*}y=3x+4\end{align*} is an oblique asymptote of \begin{align*}f(x)\end{align*}.

To sketch the graph we can find the vertical asymptotes by setting the denominator equal to zero,

\begin{align*}x^{2}-3x & = 0\\ x(x-3) & = 0\end{align*}

So the two vertical asymptotes are \begin{align*}x=0\end{align*} and \begin{align*}x=3\end{align*}.

\begin{align*}f(0)\end{align*} is undefined, so there is no \begin{align*}y-\end{align*}intercept. Also, there is no simple way to solve for the roots (setting the numerator equal to zero), but we can see by inspection that \begin{align*}f(1)=0\end{align*}. To get an idea of the shape of the graph we will make a table of a few test points. We used a calculator to evaluate decimal values of \begin{align*}x\end{align*} in \begin{align*}f(x)\end{align*}.

\begin{align*}& x && -2 && -1 && -0.1 && 1 && \ \ \ 2 && \ 4 && 3.5 && \ 5 && \ 6\\ & f(x) && -4.2 && -1.5 && 66.48 && 0 && -3 && 28.5 && 39.6 && 25.2 && 26.1\end{align*}

Finally we use all of this information to make a sketch of the graph of \begin{align*}f(x)\end{align*}:

#### Example 5

Graph \begin{align*}f(x)=\frac{x^{2}-x-2}{x-1}\end{align*}.

The vertical asymptote here is \begin{align*}x=1\end{align*} since a *1* for *x* in the denominator makes the fraction undefined.

To find the *x*-intercepts, factor the numerator:\begin{align*}f(x)=\frac{x^{2}-x-2}{x-1}=\frac{(x-2)(x+1)}{x-1}\end{align*}

Notice that the \begin{align*}x-\end{align*}intercepts are at \begin{align*}x=2\end{align*} and \begin{align*}x=-1\end{align*} since those are the values which make \begin{align*}\frac{(x-2)(x+1)}{x-1} = 0\end{align*} true.

To identify the oblique asymptote, we divide \begin{align*}\frac{x^{2}-x-2}{x-1}\end{align*} using polynomial long division, yielding:\begin{align*}f(x)=x-\frac{2}{x-1}\end{align*}

Recall that the whole (non-fractional) part of the quotient indicates the oblique asymptote, so we have \begin{align*}y=x\end{align*}.

Make a table of points, then sketch the graph using those points and the asymptote.

x |
f(x) |
---|---|

2 | 0 |

(-1) | 0 |

0 | 2 |

3 | 0 |

#### Example 6

Identify the asymptote(s) of \begin{align*}\frac{1+x^2}{x}+\frac{x^2-1}{x}+\frac{x}{x^2-1}\end{align*}.

First, we need to simplify the expressions \begin{align*}\frac{1+x^2}{x}+\frac{x^2-1}{x}\end{align*}

Since the denominators are the same, we can just add the numerators, yielding: \begin{align*}\frac{1 + x^2 + x^2 -1}{x} ==> 2x\end{align*}

Now we have: \begin{align*}2x+\frac{x}{x^2 - 1}\end{align*} convenient! No polynomial division necessary!

The slant asymptote is \begin{align*}y = 2x\end{align*}

Looking at \begin{align*}2x+\frac{x}{x^2 - 1}\end{align*} we can see that the vertical asymptotes are at \begin{align*}x = 1\end{align*} and \begin{align*}x = -1\end{align*}

Looking at the original form: \begin{align*}\frac{1+x^2}{x}+\frac{x^2-1}{x}\end{align*} we can see a hole in the graph at (0, 0)

The sketch of the graph would look like this:

### Review

- What has to be true of the degree of the numerator and the denominator for an asymptote to be called oblique or slant?

Find the slant asymptotes:

- \begin{align*} y = \frac{3x^3}{x^2 - 1}\end{align*}
- \begin{align*} y = \frac{2x^2}{x + 1}\end{align*}
- \begin{align*} y = \frac{2x^3 - 7x^2 - 4}{(x+3)(x - 1)}\end{align*}
- \begin{align*} y = \frac{(2x)(x + 11)}{x - 4}\end{align*}
- \begin{align*} y = \frac{x^3 - x + 3}{x^2 + x - 2}\end{align*}
- \begin{align*}f(x) = \frac{x^2 - 4}{x}\end{align*}
- \begin{align*}f(x) = \frac{x^3 - 3}{x^2}\end{align*}
- \begin{align*}y = \frac{3x^3 - 3}{2x^2}\end{align*}

Find all intercepts and asymptotes for the graphs of the following rational functions and use that information to help you sketch the graphs of the functions.

- \begin{align*} f(x)= \frac{2x^2}{1 - x}\end{align*}
- \begin{align*} f(x)= \frac{x^3 - 3x^2}{x^2 - 1}\end{align*}
- \begin{align*} f(x)= \frac{x^3 - 1}{x^2 - x - 2}\end{align*}
- \begin{align*} f(x)= \frac{x^3 - 1}{2(x^2 - 1)}\end{align*}
- \begin{align*} y = \frac{2x^2}{x - 3}\end{align*}
- \begin{align*} y = \frac{3x^2}{x+2}\end{align*}

### Review (Answers)

To see the Review answers, open this PDF file and look for section 2.6.