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# 12.6: Determining Asymptotes by Division

Difficulty Level: At Grade Created by: CK-12
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Practice Determining Asymptotes by Division

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What if you had a function like y=3x22x+1x+2\begin{align*}y=\frac{3x^2 - 2x + 1}{x + 2}\end{align*}? How could you rewrite it to find its asymptotes? After completing this Concept, you'll be able to rewrite rational functions like this one using division.

### Guidance

In the last section we saw how to find vertical and horizontal asymptotes. Remember, the horizontal asymptote shows the value of y\begin{align*}y\end{align*} that the function approaches for large values of x\begin{align*}x\end{align*}. Let’s review the method for finding horizontal asymptotes and see how it’s related to polynomial division.

When it comes to finding asymptotes, there are basically four different types of rational functions.

Case 1: The polynomial in the numerator has a lower degree than the polynomial in the denominator.

#### Example A

Find the horizontal asymptote of y=2x1\begin{align*}y=\frac{2}{x-1}\end{align*}.

Solution:

We can’t reduce this fraction, and as x\begin{align*}x\end{align*} gets larger the denominator of the fraction gets much bigger than the numerator, so the whole fraction approaches zero.

The horizontal asymptote is y=0\begin{align*}y = 0\end{align*}.

Case 2: The polynomial in the numerator has the same degree as the polynomial in the denominator.

#### Example B

Find the horizontal asymptote of y=3x+2x1\begin{align*}y=\frac{3x+2}{x-1}\end{align*}.

Solution:

In this case we can divide the two polynomials:

x1)3x+2¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ 33x+35\begin{align*}& \overset{\qquad \qquad \ 3}{x-1 \overline{ ) 3x+2 \;}}\\ & \qquad \underline{-3x+3}\\ & \qquad \qquad \quad 5\end{align*}

So the expression can be written as y=3+5x1\begin{align*}y=3+\frac{5}{x-1}\end{align*}.

Because the denominator of the remainder is bigger than the numerator of the remainder, the remainder will approach zero for large values of x\begin{align*}x\end{align*}. Adding the 3 to that 0 means the whole expression will approach 3.

The horizontal asymptote is y=3\begin{align*}y = 3\end{align*}.

Case 3: The polynomial in the numerator has a degree that is one more than the polynomial in the denominator.

#### Example C

Find any asymptotes of y=4x2+3x+2x1\begin{align*}y=\frac{4x^2+3x+2}{x-1}\end{align*}.

Solution:

We can do long division once again and rewrite the expression as y=4x+7+9x1\begin{align*}y=4x+7+\frac{9}{x-1}\end{align*}. The fraction here approaches zero for large values of x\begin{align*}x\end{align*}, so the whole expression approaches 4x+7\begin{align*}4x + 7\end{align*}.

When the rational function approaches a straight line for large values of x\begin{align*}x\end{align*}, we say that the rational function has an oblique asymptote. In this case, then, the oblique asymptote is y=4x+7\begin{align*}y = 4x + 7\end{align*}.

Case 4: The polynomial in the numerator has a degree that is two or more than the degree in the denominator.

#### Example D

Find any asymptotes of y=x3x1\begin{align*}y=\frac{x^3}{x-1}\end{align*}.

This is actually the simplest case of all: the polynomial has no horizontal or oblique asymptotes.

Notice that a rational function will either have a horizontal asymptote, an oblique asymptote or neither kind. In other words, a function can’t have both; in fact, it can’t have more than one of either kind. On the other hand, a rational function can have any number of vertical asymptotes at the same time that it has horizontal or oblique asymptotes.

Watch this video for help with the Examples above.

### Vocabulary

• When the rational function approaches a straight line for large values of x\begin{align*}x\end{align*}, we say that the rational function has an oblique asymptote.

### Guided Practice

Find the horizontal or oblique asymptotes of the following rational functions.

a) y=3x2x2+4\begin{align*}y=\frac{3x^2}{x^2+4}\end{align*}

b) y=x13x26\begin{align*}y=\frac{x-1}{3x^2-6}\end{align*}

c) y=x4+1x5\begin{align*}y=\frac{x^4+1}{x-5}\end{align*}

d) y=x33x2+4x1x22\begin{align*}y=\frac{x^3-3x^2+4x-1}{x^2-2}\end{align*}

Solution

a) When we simplify the function, we get y=312x2+4\begin{align*}y=3-\frac{12}{x^2+4}\end{align*}. There is a horizontal asymptote at y=3\begin{align*}y = 3\end{align*}.

b) We cannot divide the two polynomials. There is a horizontal asymptote at y=0\begin{align*}y = 0\end{align*}.

c) The power of the numerator is 3 more than the power of the denominator. There are no horizontal or oblique asymptotes.

d) When we simplify the function, we get y=x3+6x7x22\begin{align*}y=x-3+\frac{6x-7}{x^2-2}\end{align*}. There is an oblique asymptote at y=x3\begin{align*}y = x - 3\end{align*}.

### Practice

Find all asymptotes of the following rational functions:

1. x2x2\begin{align*}\frac{x^2}{x-2}\end{align*}
2. 1x+4\begin{align*}\frac{1}{x+4}\end{align*}
3. x21x2+1\begin{align*}\frac{x^2-1}{x^2+1}\end{align*}
4. x4x29\begin{align*}\frac{x-4}{x^2-9}\end{align*}
5. x2+2x+14x1\begin{align*}\frac{x^2+2x+1}{4x-1}\end{align*}
6. x3+14x1\begin{align*}\frac{x^3+1}{4x-1}\end{align*}
7. xx3x26x7\begin{align*}\frac{x-x^3}{x^2-6x-7}\end{align*}
8. x42x8x+24\begin{align*}\frac{x^4-2x}{8x+24}\end{align*}

Graph the following rational functions. Indicate all asymptotes on the graph:

1. x2x+2\begin{align*}\frac{x^2}{x+2}\end{align*}
2. x31x24\begin{align*}\frac{x^3-1}{x^2-4}\end{align*}
3. x2+12x4\begin{align*}\frac{x^2+1}{2x-4}\end{align*}
4. xx23x+2\begin{align*}\frac{x-x^2}{3x+2}\end{align*}

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Color Highlighted Text Notes

### Vocabulary Language: English

Function

A function is a relation where there is only one output for every input. In other words, for every value of $x$, there is only one value for $y$.

Horizontal Asymptote

A horizontal asymptote is a horizontal line that indicates where a function flattens out as the independent variable gets very large or very small. A function may touch or pass through a horizontal asymptote.

Ohm's Law

Ohm's Law states that a current through a conductor that connects two points is directly proportional to the potential difference between its ends. Consider $V=IR$, where $V$ is the voltage or the potential difference, $I$ is the current, and $R$ is the resistance of the conductor.

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