Darnell says that the function

### Graphing Rational Functions

We have already graphed functions in the form

Let's graph

To find the vertical asymptote, it is the same as before, the value that makes the denominator zero. In this case,

When solving for the

The last thing to find is the horizontal asymptote. We know that the function is positive, so the branches will be in the first and third quadrants. Let’s make a table.

3 | |

5 | |

11 | |

0 | |

2 | 0.5 |

5 | 1 |

14 | 1.5 |

It looks like the horizontal asymptote is

Looking back at the original equation, *When the degree of the numerator is equal to the degree of the denominator, the horizontal asymptote is the ratio of the leading coefficients.*

Finally, the domain is all real numbers;

Now, let's graph

From the previous problem above, we can conclude that the horizontal asymptote is at

At this point, put the equation in your calculator to see the general shape. To graph this function using a TI-83 or 84, enter the function into \begin{align*}Y=\end{align*}**GRAPH**. You will need to expand the window to include the bottom portion of the graph. The final graph is below.

The domain is still all real numbers except the vertical asymptotes. For this function, that would be all real numbers; \begin{align*}x \ne -1,x \ne 1\end{align*}

The range is a bit harder to find. Notice the gap in the range from the horizontal asymptote and the \begin{align*}y\end{align*}

The notation above is one way to write a range of numbers called **interval notation** and was already introduced*.* The \begin{align*}\cup\end{align*}

In general, rational functions with quadratics in the denominator are split into six regions and have branches in three of them, like the problem above. However, there are cases when there are no zeros or vertical asymptotes and those look very different. You should always graph the function in a graphing calculator after you find the critical values and make as accurate a sketch as you can.

Finally, let's graph \begin{align*}f(x)=\frac{x^2-8x+12}{x^2-x-6}\end{align*}

Let’s factor the numerator and denominator to find the intercepts and vertical asymptotes.

\begin{align*}f(x)=\frac{x^2-8x+12}{x^2+x-6}=\frac{\left(x-6\right) \left(x-2\right)}{\left(x+3\right) \left(x-2\right)}\end{align*}

Notice that the numerator and denominator both have a factor of \begin{align*}(x - 2)\end{align*}**hole** is created because \begin{align*}x = 2\end{align*}

There is a vertical asymptote at \begin{align*}x = -3\end{align*}

The domain is \begin{align*}x \in \mathbb{R};x \ne 2,-3\end{align*} and the range is \begin{align*}y \in \mathbb{R};y \ne 1,- \frac{2}{3}\end{align*}.

### Examples

#### Example 1

Earlier, you were asked to determine which student is correct.

The vertical asymptote(s) occur(s) when the denominator of the function equals zero. For the function \begin{align*}y=\frac{2x^4+5}{x^4-16}\end{align*}, the denominator equals zero when \begin{align*}x^4-16 = 0\end{align*}.

\begin{align*} x^4-16 = 0 \\ x^4 = 16\end{align*}

\begin{align*}x = 2\end{align*} or \begin{align*}x = -2\end{align*}

Therefore, there are two vertical asymptotes and Darnell is correct.

**Graph the following functions. Find all intercepts, asymptotes, the domain and range. Double-check your answers with a graphing calculator.**

#### Example 2

\begin{align*}y=\frac{4x-5}{2x+7}\end{align*}

\begin{align*}y\end{align*}-intercept: \begin{align*}y=\frac{-5}{7}=- \frac{5}{7}\end{align*}, \begin{align*}x\end{align*}-intercept: \begin{align*}0=4x-5 \rightarrow x=\frac{5}{4}\end{align*}, horizontal asymptote: \begin{align*}y=\frac{4}{2}=2\end{align*}, vertical asymptote: \begin{align*}2x+7=0 \rightarrow x=- \frac{7}{2}\end{align*}, domain: \begin{align*}\mathbb{R};x \ne - \frac{7}{2}\end{align*}, range: \begin{align*}\mathbb{R};y \ne 2\end{align*}

#### Example 3

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

\begin{align*}y\end{align*}-intercept: \begin{align*}y=\frac{-9}{1}=-9\end{align*}, \begin{align*}x\end{align*}-intercepts: \begin{align*}0=x^2-9 \rightarrow x= \pm 3\end{align*}, horizontal asymptote: \begin{align*}y=1\end{align*}, vertical asymptote: none, domain: \begin{align*}\mathbb{R}\end{align*}, range: \begin{align*}\mathbb{R};y \ne 1\end{align*}

**Special Note:** When there are no vertical asymptotes and the numerator and denominator are both quadratics, this is the general shape. It could also be reflected over the horizontal asymptote.

#### Example 4

\begin{align*}y=\frac{2x^2+7x+3}{x^2+3x+2}\end{align*}

\begin{align*}y\end{align*}-intercept: \begin{align*}\left(0, \frac{3}{2}\right)\end{align*}, \begin{align*}x\end{align*}-intercepts: \begin{align*}(-3,0)\end{align*} and \begin{align*}\left(- \frac{1}{2},0\right)\end{align*}, horizontal asymptote: \begin{align*}y = 2\end{align*}, vertical asymptotes: \begin{align*}x = -2, x = -1\end{align*}.

domain: \begin{align*}\mathbb{R};x \ne -1,-2\end{align*}

range: \begin{align*}y \in (- \infty,2.1] \cup [12, \infty)\end{align*}

#### Example 5

\begin{align*}y=\frac{x^2-4}{2x^2-5x+2}\end{align*}

horizontal asymptote: \begin{align*}y=\frac{1}{2}\end{align*}, \begin{align*}y\end{align*}-intercept: \begin{align*}(0,-2)\end{align*}

vertical asymptotes: \begin{align*}x=\frac{1}{2}\end{align*}, \begin{align*}x\end{align*}-intercept: \begin{align*}(-2,0)\end{align*}

hole: \begin{align*}x = 2, f(2)=\frac{4}{3}\end{align*}

domain: \begin{align*}\mathbb{R};x \ne \frac{1}{2},2\end{align*}

range: \begin{align*}\mathbb{R};y \ne \frac{1}{2}, \frac{4}{3}\end{align*}

### Review

- What are the vertical and horizontal asymptotes for \begin{align*}y=\frac{x-2}{x+7}\end{align*}?
- What is the domain of this function?
- What is the range of this function?
- Are there any
*x*-intercepts? If so, what are they? - Is there a
*y*-intercept? If so, what is it?

Graph the following rational functions. Write down the equations of the asymptotes, the domain and range, \begin{align*}x\end{align*} and \begin{align*}y\end{align*} intercepts and identify any holes.

- \begin{align*}y=\frac{x+3}{x-5}\end{align*}
- \begin{align*}y=\frac{5x+2}{x-4}\end{align*}
- \begin{align*}y=\frac{3-x}{2x+10}\end{align*}
- \begin{align*}y=\frac{x^2+5x+6}{x^2-8x+12}\end{align*}
- \begin{align*}y=\frac{x^2+4}{2x^2+x-3}\end{align*}
- \begin{align*}y=\frac{2x^2-x-10}{3x^2+10x+8}\end{align*}
- \begin{align*}y=\frac{x^2-4}{x^2+3x-10}\end{align*}
- \begin{align*}y=\frac{6x^2-7x-3}{4x^2-1}\end{align*}
- \begin{align*}y=\frac{x^3-8}{x^3+x^2-4x-4}\end{align*}
- Graph \begin{align*}y=\frac{1}{x-2}+3\end{align*} and \begin{align*}y=\frac{3x-5}{x-2}\end{align*} on the same set of axes. Compare the two. What do you notice? Explain your results.

### Answers for Review Problems

To see the Review answers, open this PDF file and look for section 9.5.