Vertical asymptotes describe the behavior of a function as the values of \begin{align*}x\end{align*} approach a specific number. Horizontal asymptotes describe the behavior of a function as the values of \begin{align*}x\end{align*} become infinitely large and infinitely small. Since functions cannot touch vertical asymptotes, are they not allowed to touch horizontal asymptotes either?

#### Watch This

Watch the following video, focusing on the parts about horizontal asymptotes.

http://www.youtube.com/watch?v=wBZxVxiJS9I James Sousa: Determining Horizontal and Vertical Asymptotes of Rational Functions

#### Guidance

**Horizontal asymptotes** are a means of describing end behavior of a function. End behavior essentially is a description of what happens on either side of the graph as the function continues to the right and left infinitely. When you are determining the horizontal asymptotes, it is important to consider both the right and the left hand sides, because the horizontal asymptotes will not necessarily be the same in both places. Consider the reciprocal function and note how as \begin{align*}x\end{align*} goes to the right and left it flattens to the line \begin{align*}y=0\end{align*}.

Sometimes functions flatten out and other times functions increase or decrease without bound. There are basically three cases.

**Case 1: ** The first case is the function flattens out to 0 as \begin{align*}x\end{align*} gets infinitely large or infinitely small. This happens when the degree of the numerator is less than the degree of the denominator. The degree is determined by the greatest exponent of \begin{align*}x\end{align*}.

\begin{align*}f(x)=\frac{2x^8+3x^2+100}{x^9-12}\end{align*}

One way to reason through why this makes sense is because when \begin{align*}x\end{align*} is a ridiculously large number then most parts of the function hardly make any impact. The 100 for example is nothing in comparison and neither is the \begin{align*}3x^2\end{align*}. The two important terms to compare are \begin{align*}x^8\end{align*} and \begin{align*}x^9\end{align*}. The 2 isn’t even important now because if \begin{align*}x\end{align*} is even just a million than the \begin{align*}x^9\end{align*} will be a million times bigger than the \begin{align*}x^8\end{align*} and the 2 hardly matters again. Essentially, when \begin{align*}x\end{align*} gets big enough, this function acts like \begin{align*}\frac{1}{x}\end{align*} which has a horizontal asymptote of 0.

**Case 2: ** The degree of the numerator is equal to the degree of the denominator. In this case, the horizontal asymptote is equal to the ratio of the leading coefficients.

\begin{align*}f(x)=\frac{6x^4-3x^3+12x^2-9}{3x^4+144x-0.001}\end{align*}

Notice how the degree of both the numerator and the denominator is 4. This means that the horizontal asymptote is \begin{align*}y=\frac{6}{3}=2\end{align*}. One way to reason through why this makes sense is because when \begin{align*}x\end{align*} gets to be a very large number all the smaller powers will not really make much of an impact. The biggest contributors are only the biggest powers. Then the value of the numerator will be about twice that of the denominator. As \begin{align*}x\end{align*} gets even bigger, then the function will get even closer to 2.

**Case 3: ** The degree of the numerator is greater than the degree of the denominator. In this case there does not exist a horizontal asymptote and you must determine if the function increases or decreases without bound in both the left and right directions.

**Example A**

Identify the horizontal asymptotes of the following 3 functions:

- \begin{align*}f(x)=\dfrac{4x^3+99}{3x^4-99}\end{align*}
- \begin{align*}h(x)=\dfrac{234x^{45}-45x^{234}+100}{x^{235}}\end{align*}
- \begin{align*}g(x)=\dfrac{x^3+3x^6}{x^3-6x^6}\end{align*}

**Solution: **

- \begin{align*}y=0\end{align*} because the degree of the numerator is 3 and the degree of the denominator is 4, so the denominator gets bigger eventually and the fraction approaches 0.
- \begin{align*}y=0\end{align*} because the degree of the numerator is 234 which is smaller than the degree of the denominator (235).
- \begin{align*}y=-\frac{1}{2}\end{align*} because the degree of both the numerator and the denominator is 6 so the horizontal asymptote is the ratio of the leading coefficients. Note that leading refers to the coefficients of the greatest powers of \begin{align*}x\end{align*} not the coefficients that happen to be written out front. Convention usually tells you to write the powers of \begin{align*}x\end{align*} in descending order.

**Example B**

Identify the vertical and horizontal asymptotes of the following rational function.

\begin{align*}f(x)=\dfrac{(x-2)(4x+3)(x-4)}{(x-1)(4x+3)(x-6)}\end{align*}

**Solution: ** There is factor that cancels that is neither a horizontal or vertical asymptote. The vertical asymptotes occur at \begin{align*}x=1\end{align*} and \begin{align*}x=6\end{align*}. To obtain the horizontal asymptote you could methodically multiply out each binomial, however since most of those terms do not matter, it is more efficient to first determine the relative powers of the numerator and the denominator. In this case they both happen to be 3. Next determine the coefficient of the cubic terms only. The numerator will have \begin{align*}4x^3\end{align*} and the denominator will have \begin{align*}4x^3\end{align*} and so the horizontal asymptote will occur at \begin{align*}y=\frac{4}{4}=1\end{align*}.

**Example C**

Describe the right hand end behavior of the following function.

**Solution: ** Notice how quickly this dampening wave function settles down. There seems to be an obvious horizontal axis on the right at \begin{align*}y=1\end{align*}

**Concept Problem Revisited**

As in Example C, functions may touch and pass through horizontal asymptotes without limit. This is a difference between vertical and horizontal asymptotes. In calculus, there are rigorous proofs to show that functions like the one in Example C do become arbitrarily close to the asymptote.

#### Vocabulary

A ** horizontal asymptote** is a flat dotted line that indicates where a function goes as \begin{align*}x\end{align*} get infinitely large or infinitely small.

** End behavior** is a term that asks you to describe the horizontal asymptotes.

A ** vertical asymptote** is a dashed vertical line that indicates that as a function approaches, it shoots off to positive or negative infinity without ever actually touching the line.

A ** rational function** is a function with at least one rational expression.

A ** rational expression** is a ratio of two polynomial expressions.

#### Guided Practice

1. Identify the horizontal asymptotes of the following function.

\begin{align*}f(x)=\frac{(x-3)(x+2)}{|(x-5)| \cdot (x-1)}\end{align*}

2. Identify the vertical and horizontal asymptotes and the holes of the following function.

\begin{align*}f(x)=\dfrac{(x^4-9)(x-1)}{(x^2+3)(x-3)}\end{align*}

3. Identify the horizontal asymptotes if they exist for the following 3 functions.

- \begin{align*}f(x)=\dfrac{3x^6-72x}{x^6+999}\end{align*}
- \begin{align*}h(x)=\dfrac{ax^4+bx^3+cx^2+dx+e}{fx^4+gx^3+hx^2}\end{align*}
- \begin{align*}g(x)=\dfrac{f(x)}{h(x)}\end{align*}

**Answers:**

1. First notice the absolute value surrounding one of the terms in the denominator. The degrees of both the numerator and the denominator will be 2 which means that the horizontal asymptote will occur at a number. As \begin{align*}x\end{align*} gets infinitely large, the function is approximately:

\begin{align*}f(x)=\frac{x^2}{x^2}\end{align*}

So the horizontal asymptote is \begin{align*}y=-1\end{align*} as \begin{align*}x\end{align*} gets infinitely large.

On the other hand, as \begin{align*}x\end{align*} gets infinitely small the function is approximately:

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

So the horizontal asymptote is \begin{align*}y=-1\end{align*} as \begin{align*}x\end{align*} gets infinitely small.

In this case, you cannot blindly use the leading coefficient rule because the absolute value changes the sign.

2. The numerator of the function factors to be:

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

Note that a factor does cancel and also notice that this factor is never equal to zero. Not all factors that cancel indicate a hole. A horizontal asymptote does not exist because the degree of the numerator is greater than the degree of the denominator. The vertical asymptote is at \begin{align*}x=3\end{align*}.

3.

- The degrees of the numerator and the denominator are equal so the horizontal asymptote is \begin{align*}y=3\end{align*}.
- The degrees of the numerator and the denominators are equal again so the horizontal asymptote is \begin{align*}y=\frac{a}{f}\end{align*}.
- As \begin{align*}x\end{align*} gets infinitely large,

\begin{align*}g(x)=\frac{f(x)}{h(x)}=\frac{\frac{3x^6-72x}{x^6+999}}{\frac{ax^4+bx^3+cx^2+dx+e}{fx^4+gx^3+hx^2}} \approx \frac{3}{\frac{a}{f}}=\frac{3f}{a}\end{align*}

When you study calculus, you will learn the rigorous techniques that enable you to feel more confident about results like this.

#### Practice

Identify the horizontal asymptotes, if they exist, for the following functions.

- \begin{align*}f(x)=\dfrac{5x^4-2x}{x^4+32}\end{align*}
- \begin{align*}g(x)=\dfrac{3x^4-2x^6}{-x^4+2}\end{align*}
- \begin{align*}h(x)=\dfrac{3x^4-5x}{8x^3+3x^4}\end{align*}
- \begin{align*}j(x)=\dfrac{2x^3-15x}{-x^4+3}\end{align*}
- \begin{align*}k(x)=\dfrac{2x^5-3x}{5x^2+3x^4+2x-7x^5}\end{align*}
- \begin{align*}f(x)=\dfrac{ax^{14}+bx^{23}+cx^{12}+dx+e}{fx^{24}+gx^{23}+hx^{21}}\end{align*}
- \begin{align*}g(x)=\dfrac{(x-1)(x+4)}{|(x-2)| \cdot (x-1)}\end{align*}
- Write a function that fits the following criteria:
- Vertical asymptotes at \begin{align*}x=1\end{align*} and \begin{align*}x=4\end{align*}
- Zeroes at 3 and 5
- Hole when \begin{align*}x=6\end{align*}
- Horizontal asymptote at \begin{align*}y=\frac{2}{3}\end{align*}

- Write a function that fits the following criteria:
- Vertical asymptotes at \begin{align*}x=-2\end{align*} and \begin{align*}x=2\end{align*}
- Zeroes at 1 and 5
- Hole when \begin{align*}x=3\end{align*}
- Horizontal asymptote at \begin{align*}y=1\end{align*}

- Write a function that fits the following criteria:
- Vertical asymptotes at \begin{align*}x=0\end{align*} and \begin{align*}x=3\end{align*}
- Zeroes at 1 and 2
- Hole when \begin{align*}x=8\end{align*}
- Horizontal asymptote at \begin{align*}y=2\end{align*}

- Write a function that fits the following criteria:
- Vertical asymptotes at 2 and 6
- Zero at 5
- Hole when \begin{align*}x=4\end{align*}
- Horizontal asymptote at \begin{align*}y=0\end{align*}

- Write a function that fits the following criteria:
- Vertical asymptote at 4
- Zeroes at 0 and 3
- Hole at when \begin{align*}x=5\end{align*}
- No horizontal asymptotes

Identify the vertical and horizontal asymptotes of the following rational functions.

- \begin{align*}f(x)=\dfrac{(x-5)(2x+1)(x-3)}{(x-3)(4x+5)(x-6)}\end{align*}
- \begin{align*}g(x)=\dfrac{x(x-1)(x+3)(x-5)}{3x(x-1)(4x+3)}\end{align*}
- \begin{align*}h(x)=\dfrac{(x-2)(x+3)(x-6)}{(x-4)(x+3)^2(x+2)}\end{align*}