### Horizontal and Vertical Asymptotes

We said that a horizontal asymptote is the value of \begin{align*}y\end{align*}

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

If we plug in a large value of \begin{align*}x\end{align*}

\begin{align*}y = \frac{2(100)^2 + (100) - 1}{3(100)^2 - 4(100) + 3} = \frac{20000 + 100 - 1}{30000 - 400 + 2}\end{align*}

We can see that the beginning terms in the numerator and denominator are much bigger than the other terms in each expression. One way to find the horizontal asymptote of a rational function is to ignore all terms in the numerator and denominator except for the highest powers.

In this example the horizontal asymptote is \begin{align*}y = \frac{2x^2}{3x^2}\end{align*}

In the function above, the highest power of \begin{align*}x\end{align*} was the same in the numerator as in the denominator. Now consider a function where the power in the numerator is less than the power in the denominator:

\begin{align*}y = \frac{x}{x^2 + 3}\end{align*}

As before, we ignore all the terms except the highest power of \begin{align*}x\end{align*} in the numerator and the denominator. That gives us \begin{align*}y = \frac{x}{x^2}\end{align*}, which simplifies to \begin{align*}y = \frac{1}{x}\end{align*}.

For large values of \begin{align*}x\end{align*}, the value of \begin{align*}y\end{align*} gets closer and closer to zero. Therefore the horizontal asymptote is \begin{align*}y = 0\end{align*}.

To summarize:

- Find vertical asymptotes by setting the denominator equal to zero and solving for \begin{align*}x\end{align*}.
- For horizontal asymptotes, we must consider several cases:
- If the highest power of \begin{align*}x\end{align*} in the numerator is less than the highest power of \begin{align*}x\end{align*} in the denominator, then the horizontal asymptote is at \begin{align*}y = 0\end{align*}.
- If the highest power of \begin{align*}x\end{align*} in the numerator is the same as the highest power of \begin{align*}x\end{align*} in the denominator, then the horizontal asymptote is at \begin{align*}y = \frac{\text{coefficient of highest power of } x}{\text{coefficient of highest power of } x}\end{align*}.
- If the highest power of \begin{align*}x\end{align*} in the numerator is greater than the highest power of \begin{align*}x\end{align*} in the denominator, then we don’t have a horizontal asymptote; we could have what is called an oblique (slant) asymptote, or no asymptote at all.

#### Finding Asymptotes

1. Find the vertical and horizontal asymptotes for \begin{align*}y = \frac {1}{x-1}\end{align*}.

**Vertical asymptotes:**

Set the denominator equal to zero. \begin{align*}x-1=0\Rightarrow x=1\end{align*} is the vertical asymptote.

**Horizontal asymptote:**

Keep only the highest powers of \begin{align*}x\end{align*}. \begin{align*}y=\frac {1}{x}\Rightarrow y=0\end{align*} is the horizontal asymptote.

2. Find the vertical and horizontal asymptotes for \begin{align*}y= \frac {3x}{4x+2}\end{align*}.

**Vertical asymptotes:**

Set the denominator equal to zero. \begin{align*}4x+2=0\Rightarrow x=-\frac{1}{2}\end{align*} is the vertical asymptote.

**Horizontal asymptote:**

Keep only the highest powers of \begin{align*}x\end{align*}. \begin{align*}y=\frac {3x}{4x}\Rightarrow y=\frac{3}{4}\end{align*} is the horizontal asymptote.

3. Find the vertical and horizontal asymptotes for \begin{align*}y=\frac {x^3}{x^2-3x+2}\end{align*}.

**Vertical asymptotes:**

Set the denominator equal to zero: \begin{align*}x^2 - 3x + 2 =0\end{align*}

Factor: \begin{align*}(x - 2)(x - 1) = 0\end{align*}

Solve: \begin{align*}x = 2\end{align*} and \begin{align*}x = 1\end{align*} are the vertical asymptotes.

**Horizontal asymptote.** There is no horizontal asymptote because the power of the numerator is larger than the power of the denominator.

Notice the function in part *d* had more than one vertical asymptote. Here’s another function with two vertical asymptotes.

#### Graphing Functions

Graph the function \begin{align*}y = \frac{-x^2}{x^2 - 4}\end{align*}.

\begin{align*}\text{Set the denominator equal to zero:} \qquad x^2 - 4 = 0\!\\ \\ \text{Factor:} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad (x - 2)(x + 2) = 0\!\\ \\ \text{Solve:} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ x = 2, x = -2\end{align*}

We find that the function is undefined for \begin{align*}x = 2\end{align*} **and** \begin{align*}x = -2\end{align*}, so we know that there are vertical asymptotes at these values of \begin{align*}x\end{align*}.

We can also find the horizontal asymptote by the method we outlined above. It’s at \begin{align*}y = \frac{-x^2}{x^2}\end{align*}, or \begin{align*}y = -1\end{align*}.

So, we start plotting the function by drawing the vertical and horizontal asymptotes on the graph.

Now, let’s make a table of values. Because our function has a lot of detail we must make sure that we pick enough values for our table to determine the behavior of the function accurately. We must make sure especially that we pick values close to the vertical asymptotes.

\begin{align*}x\end{align*} | \begin{align*}y = \frac {-x^2}{x^2-4}\end{align*} |
---|---|

\begin{align*}-5\end{align*} | \begin{align*}y = \frac {-(-5)^2}{(-5)^2-4} = \frac {-25}{21} = -1.19\end{align*} |

-4 | \begin{align*}y = \frac {-(-4)^2}{(-4)^2-4} = \frac {-16}{12} = -1.33\end{align*} |

-3 | \begin{align*}y = \frac {-(-3)^2}{(-3)^2-4} = \frac {-9}{5} = -1.8\end{align*} |

-2.5 | \begin{align*}y = \frac {-(-2.5)^2}{(-2.5)^2-4} = \frac {-6.25}{2.25} = -2.8\end{align*} |

-1.5 | \begin{align*}y = \frac {-(-1.5)^2}{(-1.5)^2-4} = \frac {-2.25}{-1.75} = 1.3\end{align*} |

-1 | \begin{align*}y = \frac {-(-1)^2}{(-1)^2-4} = \frac {-1}{-3} = 0.33\end{align*} |

0 | \begin{align*}y = \frac {-0^2}{(0)^2-4} = \frac {0}{-4} = 0\end{align*} |

1 | \begin{align*}y= \frac {-1^2}{(1)^2-4} = \frac {-1}{-3} = 0.33\end{align*} |

1.5 | \begin{align*}y = \frac {-1.5^2}{(1.5)^2-4} = \frac {-2.25}{-1.75} = 1.3\end{align*} |

2.5 | \begin{align*}y = \frac {-2.5^2}{(2.5)^2-4} = \frac {-6.25}{2.25} = -2.8\end{align*} |

3 | \begin{align*}y = \frac {-3^2}{(3)^2-4} = \frac {-9}{5} = -1.8\end{align*} |

4 | \begin{align*}y = \frac {-4^2}{(4)^2-4} = \frac {-16}{12} = -1.33\end{align*} |

5 | \begin{align*}y = \frac {-5^2}{(5)^2-4} = \frac {-25}{21} = -1.19\end{align*} |

Here is the resulting graph.

### Example

#### Example 1

Find the vertical and horizontal asymptotes for \begin{align*}y=\frac {x^2-2}{2x^2+3}\end{align*}.

**Vertical asymptotes:**

Set the denominator equal to zero: \begin{align*} 2x^2+3 = 0 \Rightarrow 2x^2 = -3 \Rightarrow x^2 = -\frac{3}{2}\end{align*}. Since there are no solutions to this equation, there is no vertical asymptote.

**Horizontal asymptote:**

Keep only the highest powers of \begin{align*}x\end{align*}. \begin{align*}y=\frac {x^2}{2x^2} \Rightarrow y= \frac {1}{2}\end{align*} is the horizontal asymptote.

### Review

Find all the vertical and horizontal asymptotes of the following rational functions.

- \begin{align*}y=\frac {4}{x+2}\end{align*}
- \begin{align*}y=\frac {5x-1}{2x-6}\end{align*}
- \begin{align*}y=\frac {10}{x}\end{align*}
- \begin{align*}y=\frac {2}{x}-5\end{align*}
- \begin{align*}y=\frac {x + 1}{x^2}\end{align*}
- \begin{align*}y=\frac {4x^2}{4x^2+1}\end{align*}
- \begin{align*}y=\frac {2x}{x^2-9}\end{align*}
- \begin{align*}y=\frac {3x^2}{x^2-4}\end{align*}
- \begin{align*}y=\frac {1}{x^2+4x+3}\end{align*}
- \begin{align*}y=\frac {2x+5}{x^2-2x-8}\end{align*}

### Review (Answers)

To view the Review answers, open this PDF file and look for section 12.3.