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

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

- If the degree of the denominator is greater than the degree of the numerator, then the line
y=0 is a horizontal asymptote.

- If the degree of the numerator and the denominator are equal, then the line
y=ab is a horizontal asymptote, wherea is the leading coefficient off(x) , the numerator, andb is the leading coefficient ofD(x) , the denominator.

- If the degree of the numerator is larger than the degree of the denominator, then the quotient function,
Q(x) , found by dividing the numerator and denominator of the rational function, is an**oblique asymptote**. Recall that for any rational functionf(x)D(X) , you can use polynomial division to re-write that function in the formf(x)D(x)=Q(x)+R(x)D(x) whereQ(x) is the quotient andR(x) 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

First observe that the vertical asymptote is at

Doing the long division here,

So in this case, the function

The above equation tells us that as *oblique asymptote* and it is indicated by the dashed line in the image below.

#### Example 3

Identify the oblique asymptote of

By polynomial division we have,

So

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

#### Example 4

Find the oblique asymptote of

Using polynomial long division,

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

So the two vertical asymptotes are

Finally we use all of this information to make a sketch of the graph of

#### Example 5

Graph

The vertical asymptote here is *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*}

To identify the oblique asymptote, we divide \begin{align*}\frac{x^{2}-x-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*}

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

Looking at \begin{align*}2x+\frac{x}{x^2 - 1}\end{align*}

Looking at the original form: \begin{align*}\frac{1+x^2}{x}+\frac{x^2-1}{x}\end{align*}

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*}
y=3x3x2−1 - \begin{align*} y = \frac{2x^2}{x + 1}\end{align*}
y=2x2x+1 - \begin{align*} y = \frac{2x^3 - 7x^2 - 4}{(x+3)(x - 1)}\end{align*}
y=2x3−7x2−4(x+3)(x−1) - \begin{align*} y = \frac{(2x)(x + 11)}{x - 4}\end{align*}
y=(2x)(x+11)x−4 - \begin{align*} y = \frac{x^3 - x + 3}{x^2 + x - 2}\end{align*}
y=x3−x+3x2+x−2 - \begin{align*}f(x) = \frac{x^2 - 4}{x}\end{align*}
f(x)=x2−4x - \begin{align*}f(x) = \frac{x^3 - 3}{x^2}\end{align*}
f(x)=x3−3x2 - \begin{align*}y = \frac{3x^3 - 3}{2x^2}\end{align*}
y=3x3−32x2

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*}
f(x)=2x21−x - \begin{align*} f(x)= \frac{x^3 - 3x^2}{x^2 - 1}\end{align*}
f(x)=x3−3x2x2−1 - \begin{align*} f(x)= \frac{x^3 - 1}{x^2 - x - 2}\end{align*}
f(x)=x3−1x2−x−2 - \begin{align*} f(x)= \frac{x^3 - 1}{2(x^2 - 1)}\end{align*}
f(x)=x3−12(x2−1) - \begin{align*} y = \frac{2x^2}{x - 3}\end{align*}
y=2x2x−3 - \begin{align*} y = \frac{3x^2}{x+2}\end{align*}
y=3x2x+2

### Review (Answers)

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