### Horizontal and Vertical Asymptotes

We said that a horizontal asymptote is the value of

If we plug in a large value of

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

In the function above, the highest power of

As before, we ignore all the terms except the highest power of

For large values of

To summarize:

- Find vertical asymptotes by setting the denominator equal to zero and solving for
x . - For horizontal asymptotes, we must consider several cases:
- If the highest power of
x in the numerator is less than the highest power ofx in the denominator, then the horizontal asymptote is aty=0 . - If the highest power of
x in the numerator is the same as the highest power ofx in the denominator, then the horizontal asymptote is aty=coefficient of highest power of xcoefficient of highest power of x . - If the highest power of
x in the numerator is greater than the highest power ofx 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.

- If the highest power of

#### Finding Asymptotes

1. Find the vertical and horizontal asymptotes for

**Vertical asymptotes:**

Set the denominator equal to zero.

**Horizontal asymptote:**

Keep only the highest powers of

2. Find the vertical and horizontal asymptotes for

**Vertical asymptotes:**

Set the denominator equal to zero.

**Horizontal asymptote:**

Keep only the highest powers of

3. Find the vertical and horizontal asymptotes for

**Vertical asymptotes:**

Set the denominator equal to zero:

Factor:

Solve:

**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

We find that the function is undefined for **and**

We can also find the horizontal asymptote by the method we outlined above. It’s at

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*}-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*}

**Horizontal asymptote:**

Keep only the highest powers of \begin{align*}x\end{align*}

### Review

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

- \begin{align*}y=\frac {4}{x+2}\end{align*}
y=4x+2 - \begin{align*}y=\frac {5x-1}{2x-6}\end{align*}
y=5x−12x−6 - \begin{align*}y=\frac {10}{x}\end{align*}
y=10x - \begin{align*}y=\frac {2}{x}-5\end{align*}
y=2x−5 - \begin{align*}y=\frac {x + 1}{x^2}\end{align*}
y=x+1x2 - \begin{align*}y=\frac {4x^2}{4x^2+1}\end{align*}
y=4x24x2+1 - \begin{align*}y=\frac {2x}{x^2-9}\end{align*}
y=2xx2−9 - \begin{align*}y=\frac {3x^2}{x^2-4}\end{align*}
y=3x2x2−4 - \begin{align*}y=\frac {1}{x^2+4x+3}\end{align*}
y=1x2+4x+3 - \begin{align*}y=\frac {2x+5}{x^2-2x-8}\end{align*}
y=2x+5x2−2x−8

### Review (Answers)

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