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? In this Concept, you'll learn about the graphs of rational functions and their attributes so that you can settle arguments like this one.

### Watch This

**Multimedia Link:** For further explanation about asymptotes, read through this PowerPoint presentation presented by North Virginia Community College or watch this: CK-12 Basic Algebra: Finding Vertical Asymptotes of Rational Functions

- YouTube video.

### Guidance

In the previous Concept, 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 Concept will compare graphs of inverse variation functions. You will also learn how to graph other rational equations.

#### Example A

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*}

**Solution:**

*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 Concept, 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**

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 value for which the equation or function is undefined. 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 function is undefined when the denominator of a fraction is zero. To find the asymptotes, find where the denominator of the rational function is zero. These are called **points of discontinuity** of the function.

The formal definition for asymptote is as follows.

An **asymptote** is a straight line to which, as the distance from the origin gets larger, a curve gets closer and closer but never intersects.

#### Example B

*Find the points of discontinuity and the asymptote for the function \begin{align*}y=\frac{6}{x-5}\end{align*}.*

**Solution:**

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*}.

**Horizontal Asymptotes**

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.

#### Example C

*Identify the vertical and horizontal asymptotes of \begin{align*}f(x)=\frac{3}{(x-4)(x+8)}-5\end{align*}.*

**Solution:** 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*}.

**Real-World Rational Functions**

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*}

#### Example D

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

**Solution:**

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*}

### Video Review

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### Guided Practice

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

**Solution:**

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*}.

### Explore More

- 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.

### Answers for Explore More Problems

To view the Explore More answers, open this PDF file and look for section 12.2.