The standard form of a rational function is given by the equation

### Rational Function

A **rational function** is in the form **hyperbola**.

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1 | 1 |

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4 |

Notice the following properties of this hyperbola: the

The two parts of the graph are called **branches**. In the case with a hyperbola, the branches are always symmetrical about the point where the asymptotes intersect. In this example, they are symmetrical about the origin.

In this lesson, all the rational functions will have the form

Let's graph

Let’s make a table of values.

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4 |

Notice that these branches are in the second and fourth quadrants. This is because of the negative sign in front of the 2, or

Now, let's graph

For

Therefore, for the general form of a rational function,

The domain is all real numbers;

To find the

Finally, let's find the equation of the hyperbola below.

We know that the numerator will be negative because the branches of this hyperbola are in the second and fourth quadrants. The asymptotes are

\begin{align*}0&=\frac{a}{-3.75+3}-4 \\ 4&=\frac{a}{-0.75} && \text{The equation is} \ y=\frac{-3}{x+3}-4 \\ -3&=a\end{align*}

### Examples

#### Example 1

Earlier, you were asked to find the asymptotes of the equation \begin{align*}f(x)=\frac{a}{x-h}+k\end{align*}.

From the previous problems, we've seen that the vertical asymptote occurs when the denominator of the equation equals zero and the horizontal asymptote occurs when the range is undefined.

When \begin{align*}x = h\end{align*}, the denominator of \begin{align*}f(x)=\frac{a}{x-h}+k\end{align*} is zero, so \begin{align*}x = h\end{align*} is the vertical asymptote.

The value of the function at \begin{align*}x = h\end{align*} would be \begin{align*}y=\frac{a}{0}+k\end{align*}, making the range undefined at \begin{align*}y = k\end{align*}. Therefore, \begin{align*}y=k\end{align*} is the horizontal asymptote.

#### Example 2

What are the asymptotes for \begin{align*}f(x)=\frac{-1}{x+6}+9\end{align*}? Is \begin{align*}(-5, -8)\end{align*} on the graph?

The asymptotes are \begin{align*}x = -6\end{align*} and \begin{align*}y = 9\end{align*}. To see if the point \begin{align*}(-5, -8)\end{align*} is on the graph, substitute it in for \begin{align*}x\end{align*} and \begin{align*}y\end{align*}.

\begin{align*}-8&=\frac{-1}{-5+6}+9 && -8 \ne 8, \ \text{therefore, the point} \ (-5, -8) \ \text{is not on the graph}. \\ -8&=-1+9\end{align*}

**For Examples 3 & 4, graph the rational functions. Find the zero, \begin{align*}y\end{align*}-intercept, asymptotes, domain and range.**

#### Example 3

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

There is no \begin{align*}y\end{align*}-intercept because the \begin{align*}y\end{align*}-axis is an asymptote. The other asymptote is \begin{align*}y = -2\end{align*}. The domain is all real numbers; \begin{align*}x \ne 0\end{align*}. The range is all real numbers; \begin{align*}y \ne -2\end{align*}. The zero is:

\begin{align*}0&=\frac{4}{x}-2 \\ 2&=\frac{4}{x} \\ 2x&=4 \\ x&=2\end{align*}

#### Example 4

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

The asymptotes are \begin{align*}x = 1\end{align*} and \begin{align*}y = 3\end{align*}. Therefore, the domain is all real numbers except 1 and the range is all real numbers except 3. The \begin{align*}y\end{align*}-intercept is \begin{align*}y=\frac{2}{0-1}+3=-2+3=1\end{align*} and the zero is:

\begin{align*}0&=\frac{2}{x-1}+3 \\ -3&=\frac{2}{x-1} \\ -3x+3&=2 \\ -3x&=-1 \rightarrow x=\frac{1}{3}\end{align*}

#### Example 5

Determine the equation of the hyperbola.

The asymptotes are \begin{align*}x = -1, y = 3,\end{align*} making the equation \begin{align*}y=\frac{a}{x+1}+3\end{align*}. Taking the \begin{align*}y\end{align*}-intercept, we can solve for \begin{align*}a\end{align*}.

\begin{align*}4&=\frac{a}{0+1}+3 && \text{The equation is} \ y=\frac{1}{x+1}+3. \\ 1&=a\end{align*}

### Review

- What are the asymptotes for \begin{align*}y=\frac{2}{x+8}-3\end{align*}?
- Is \begin{align*}(-6, -2)\end{align*} a point on the graph from #1?
- What are the asymptotes for \begin{align*}y=6- \frac{1}{x-4}\end{align*}?
- Is \begin{align*}(5, 4)\end{align*} a point on the graph from #3?

For problems 5-13, graph each rational function, state the equations of the asymptotes, the domain and range and the intercepts.

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

Write the equations of the hyperbolas. You may assume that *a* = 1.

### Answers for Review Problems

To see the Review answers, open this PDF file and look for section 9.4.