Suppose Susan and Victor both graph the function \begin{align*}y= \frac{8}{5-x}\end{align*} on their graphing calculators. Susan says that the graph looks incorrect because there should be a hole in the graph at \begin{align*}x=5\end{align*}. Victor says that the graph looks fine and that the graphing calculator would never make a mistake. Who do you think is right? Why do you think this?

### Graphing Rational Functions

Previously, you learned the basics of graphing an inverse variation function. The **hyperbola** forms two **branches** in opposite quadrants. The axes are **asymptotes** to the graph. This section will compare graphs of inverse variation functions. You will also learn how to graph other rational equations.

#### Graphing Functions for Given Values

Let's graph the function \begin{align*}f(x)=\frac{k}{x}\end{align*} for the following values of \begin{align*}k\end{align*}:

\begin{align*}k=-2, -1, -\frac{1}{2}, 1, 2, 4\end{align*}

The graphs for the values \begin{align*}k = 1, k = 2,\end{align*} and \begin{align*}k = \left(\frac{1}{2} \right )\end{align*} in Quadrant I are shown. Try your hand at filling in Quadrant III for these graphs as well as graphing the other values.

Remember, as mentioned in the previous section, if \begin{align*}k\end{align*} is positive, then the branches of the hyperbola are located in quadrants I and III. If \begin{align*}k\end{align*} is negative, the branches are located in quadrants II and IV. Also notice how the hyperbola changes as \begin{align*}k\end{align*} gets larger.

#### Rational Functions and Asymptotes

A **rational function** is a ratio of two polynomials (a polynomial divided by another polynomial). The formal definition is:

\begin{align*}f(x)=\frac{g(x)}{h(x)}, \text{where} \ h(x) \neq 0\end{align*}

An **asymptote** is a straight line to which a curve gets closer and closer but never intersects. Asymptotes can be vertical, horizontal, or oblique. This text will focus on vertical asymptotes; other math courses will also show you how to find horizontal and oblique asymptotes.

A vertical asymptote occurs when a function is undefined. A function is undefined when the denominator of a fraction is zero. To find the vertical asymptotes, find where the denominator of the rational function is zero. These are called **points of discontinuity** of the function.

Rational functions can also have horizontal asymptotes. The equation of a horizontal asymptote is \begin{align*}y=c\end{align*}, where \begin{align*}c\end{align*} represents the vertical shift of the rational function.

#### Let's find the points of discontinuity and the vertical asymptote for the following function:

\begin{align*}y=\frac{6}{x-5}\end{align*}

First, find the value of \begin{align*}x\end{align*} for which the denominator of the rational function is zero.

\begin{align*}0=x-5 \rightarrow x=5\end{align*}

The point at which \begin{align*}x=5\end{align*} is a point of discontinuity. Therefore, the asymptote has the equation \begin{align*}x=5\end{align*}.

Look at the graph of the function. There is a clear separation of the branches at the vertical line five units to the right of the origin.

The domain is “all real numbers except five” or symbolically written, \begin{align*}x \neq 5\end{align*}.

#### Now, let's identify the vertical and horizontal asymptotes of the following function:

\begin{align*}f(x)=\frac{3}{(x-4)(x+8)}-5\end{align*}

The vertical asymptotes occur where the denominator is equal to zero.

\begin{align*}x-4 &= 0 \rightarrow x=4\\ x+8 &= 0 \rightarrow x=-8\end{align*}

The vertical asymptotes are \begin{align*}x=4\end{align*} and \begin{align*}x=-8\end{align*}.

The rational function has been shifted down five units: \begin{align*}f(x)=\frac{3}{(x-4)(x+8)}-5\end{align*}.

Therefore, the horizontal asymptote is \begin{align*}y=-5\end{align*}.

**Finally, let's solve the following real-world problem:**

Electrical circuits are commonplace is everyday life. For instance, they are present in all electrical appliances in your home. The figure below shows an example of a simple electrical circuit. It consists of a battery that provides a voltage (\begin{align*}V\end{align*}, measured in Volts), a resistor (\begin{align*}R\end{align*}, measured in ohms, \begin{align*}\Omega\end{align*}) that resists the flow of electricity, and an ammeter that measures the current (\begin{align*}I\end{align*}, measured in amperes, \begin{align*}A\end{align*}) in the circuit. Your light bulb, toaster, and hairdryer are all basically simple resistors. In addition, resistors are used in an electrical circuit to control the amount of current flowing through a circuit and to regulate voltage levels. One important reason to do this is to prevent sensitive electrical components from burning out due to too much current or too high a voltage level. Resistors can be arranged in series or in parallel.

For resistors placed in a series, the total resistance is just the sum of the resistances of the individual resistors.

\begin{align*}R_{tot} =R_1+R_2\end{align*}

For resistors placed in parallel, the reciprocal of the total resistance is the sum of the reciprocals of the resistances of the individual resistors.

\begin{align*}\frac{1}{R_c}=\frac{1}{R_1}+\frac{1}{R_2}\end{align*}

**Ohm’s Law** gives a relationship between current, voltage, and resistance. It states that:

\begin{align*}I=\frac{V}{R}\end{align*}

Find the value of \begin{align*}x\end{align*} marked in the diagram.

Using Ohm’s Law, \begin{align*}I=\frac{V}{R}\end{align*}, and substituting the appropriate information yields:

\begin{align*}2= \frac{12}{R}\end{align*}

Using the cross multiplication of a proportion yields:

\begin{align*}2R=12 \rightarrow R=6 \ \Omega\end{align*}

### Examples

#### Example 1

Earlier, you were asked about the graph of the function \begin{align*}y= \frac{8}{5-x}\end{align*}. Susan and Victor both graphed it on a calculator, but Susan thought it looked wrong because it was missing at hole at \begin{align*}x=5\end{align*}. Victor disagreed and said that there is no way the calculator is wrong. Who is right?

When you graph the function \begin{align*}y= \frac{8}{5-x}\end{align*}, you get the following graph:

As you can see, Susan was correct. There is a gap in the graph at \begin{align*}x=5\end{align*}, so the calculator was wrong.

#### Example 2

Determine the asymptotes of \begin{align*}t(x)=\frac{2}{(x-2)(x+3)}\end{align*}.

Using the Zero Product Property, there are two cases for asymptotes, where each set of parentheses equals zero.

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

The two asymptotes for this function are \begin{align*}x=2\end{align*} and \begin{align*}x=-3\end{align*}.

Check your solution by graphing the function.

The domain of the rational function above has two points of discontinuity. Therefore, its domain cannot include the numbers 2 or –3. The following is the *domain:* \begin{align*}x \neq 2, x \neq -3\end{align*}.

### Review

- What is a rational function?
- Define
*asymptote.*How does an asymptote relate algebraically to a rational equation? - Which asymptotes are described in this Concept? What is the general equation for these asymptotes?

Identify the vertical and horizontal asymptotes of each rational function.

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

Graph each rational function. Show the vertical asymptote and horizontal asymptote as a dotted line.

- \begin{align*}y=-\frac{6}{x}\end{align*}
- \begin{align*}y=\frac{x}{2-x^2}-3\end{align*}
- \begin{align*}f(x)=\frac{3}{x^2}\end{align*}
- \begin{align*}g(x)=\frac{1}{x-1}+5\end{align*}
- \begin{align*}y=\frac{2}{x+2}-6\end{align*}
- \begin{align*}f(x)=\frac{-1}{x^2+2}\end{align*}
- \begin{align*}h(x)=\frac{4}{x^2+9}\end{align*}
- \begin{align*}y=\frac{-2}{x^2+1}\end{align*}
- \begin{align*}j(x)=\frac{1}{x^2-1}+1\end{align*}
- \begin{align*}y=\frac{2}{x^2-9}\end{align*}
- \begin{align*}f(x)=\frac{8}{x^2-16}\end{align*}
- \begin{align*}g(x)=\frac{3}{x^2-4x+4}\end{align*}
- \begin{align*}h(x)=\frac{1}{x^2-x-6}-2\end{align*}

Find the quantity labeled \begin{align*}x\end{align*} in the following circuit.

**Mixed Review**

- A building 350 feet tall casts a shadow \begin{align*}\frac{1}{2}\end{align*}
*mile*long. How long is the shadow of a person five feet tall? - State the Cross Product Property.
- Find the slope between (1, 1) and (–4, 5).
- The amount of refund from soda cans in Michigan is directly proportional to the number of returned cans. If you earn a $12.00 refund for 120 cans, how much do you get per can?
- You put the letters from VACATION into a hat. If you reach in randomly, what is the probability you will pick the letter \begin{align*}A\end{align*}?
- Give an example of a sixth-degree binomial.

### Review (Answers)

To see the Review answers, open this PDF file and look for section 12.2.