Can the answers to a rational function be found by graphing? How would you go about graphing them?

### Graphing Rational Functions

Any function that has the form

\begin{align*}f(x)=\frac{P(x)}{Q(x)}\end{align*}

where \begin{align*}P(x)\end{align*} and \begin{align*}Q(x)\end{align*} are polynomials and \begin{align*}Q(x)\ne0\end{align*}, is called a **rational function**. The **domain** of any rational function includes all real numbers \begin{align*}x\end{align*} that do not make the denominator zero.

Just like polynomials, rational functions can be graphed using transformations. The main point to remember for graphing rational functions by transformations is that some transformations change the asymptotes while others do not.

- \begin{align*}r(x)+c\end{align*} is a vertical shift which moves each horizontal asymptote up \begin{align*}c\end{align*} units (or down if \begin{align*}c<0\end{align*}).
- \begin{align*}r(x-c)\end{align*} is a horizontal shift which moves each vertical asymptote right \begin{align*}c\end{align*} units (or left if \begin{align*}c<0\end{align*}).
- \begin{align*}a\cdot r(x)\end{align*} is a vertical stretch which moves horizontal asymptotes by a multiple of \begin{align*}a\end{align*} (so this moves the horizontal asymptote closer to the \begin{align*}x-\end{align*}axis if \begin{align*}a<0\end{align*}.
- \begin{align*}r(a\cdot x)\end{align*} is a horizontal compression which moves the vertical asymptotes closer to \begin{align*}y-\end{align*}axis by a factor of \begin{align*}\frac{1}{a}\end{align*}.
- \begin{align*}r(-x)\end{align*} is a reflection about the \begin{align*}y-\end{align*}axis. All vertical asymptotes are also reflected.
- \begin{align*}-r(x)\end{align*} is a reflection about the \begin{align*}x-\end{align*}axis. All horizontal asymptotes are also reflected.

### Examples

#### Example 1

What is the domain of \begin{align*}f(x)=\frac{1}{x}\end{align*}?

Notice that the only input that can make the denominator equal to zero is \begin{align*}x=0\end{align*}. Thus we say that the domain of \begin{align*}f(x)\end{align*} is all real numbers except \begin{align*}x=0\end{align*}. When looking at the graph of \begin{align*}f(x)=\frac{1}{x}\end{align*} (see Example 2 below), you will notice that as \begin{align*}x\end{align*} approaches 0 from the left, \begin{align*}f(x)\end{align*} decreases. But when \begin{align*}x\end{align*} approaches 0 from the right, \begin{align*}f(x)\end{align*} increases. Because of this behavior, the \begin{align*}x-\end{align*}axis and \begin{align*}y-\end{align*}axis play the role of *horizontal* and *vertical asymptotes*, respectively.

#### Example 2

Graph the function \begin{align*}f(x)=\frac{1}{x}\end{align*}.

We know that the domain of \begin{align*}f(x)\end{align*} is all real numbers excluding \begin{align*}x=0\end{align*}. The vertical line \begin{align*}x=0\end{align*} is called a vertical asymptote. For \begin{align*}x<0, f(x)<0,\end{align*} and for \begin{align*}x>0, f(x)>0\end{align*}. Plotting a few sample points should indicate the shape of \begin{align*}f(x)\end{align*}.

\begin{align*}& x && 1 && 2 && 10 && \frac{1}{5} && \frac{1}{10} && -1 && -\frac{1}{2} && -\frac{1}{10} && -2 && -10\\ & f(x)=\frac{1}{x} && 1 && \frac{1}{2} && \frac{1}{10} && 5 && 10 && -1 && -2 && -10 && -\frac{1}{2} && -\frac{1}{10}\end{align*}

#### Example 3

A rational function \begin{align*}r(x)\end{align*} is shown in the figure below. Use the graph of \begin{align*}r(x)\end{align*} to sketch graphs of:

a. \begin{align*}r(x)-3\end{align*}

The horizontal asymptote moves down by three units.

b. \begin{align*}-r(x)\end{align*}

The function is reflected about the \begin{align*}x-\end{align*}axis so the horizontal asymptote is also reflected.

c. \begin{align*}r(3-x)\end{align*}

\begin{align*}r(3-x) = r(-(x-3))\end{align*}. First, graph \begin{align*}r(-x)\end{align*}, and then shift that graph three units to the right to get \begin{align*}r(-(x-3))\end{align*}. The new vertical asymptote is \begin{align*}x = 1\end{align*}.

#### Example 4

For the following rational function, determine the domain, the asymptotes, the \begin{align*}x-\end{align*} and \begin{align*}y-\end{align*}intercepts and then sketch the graph.

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

To identify domain limitations, find value(s) which make the denominator = 0: In this case, where \begin{align*}x^2\end{align*}, the only variable in the denominator, is added to *0*, any value for *x* will be positive. So the domain is all real numbers.

With no limitations on the domain, there are no vertical asymptotes.

The horizontal asymptote: \begin{align*}y=0\end{align*} becomes apparent as *x* becomes huge and the "+2" and "+1" no longer have an effect, giving: \begin{align*}f(x) = \frac{x}{x^2} \to f(x) = \frac{1}{x} \to f(x) = 0\end{align*} So the horizontal asymptote is *0*

#### Example 5

For the following rational function, determine the domain, the asymptotes, the \begin{align*}x-\end{align*} and \begin{align*}y-\end{align*}intercepts and then sketch the graph.

\begin{align*}f(x)=\frac{9x^{2}-4}{3x+2}\end{align*}

For this problem, the first step is to recognize the **difference of squares** in the numerator: \begin{align*}9x^2 - 4\end{align*} which factors easily into \begin{align*}(3x - 2)(3x + 2)\end{align*}

This gives: \begin{align*}f(x) = \frac{(3x - 2)(3x + 2)}{(3x + 2)}\end{align*}

NOTE: Although we can cancel the (3x + 2) to evaluate the general behavior of the function, giving: \begin{align*}f(x) = 3x + 2\end{align*} (which is just a line), that does not change the fact that **a zero in the denominator of the equation is undefined**, it just means that the rest of the function behaves like \begin{align*}y = 3x - 2\end{align*}.

To identify domain limitations, find value(s) which make the denominator = 0: In this case: \begin{align*}3x + 2 = 0 \to 3x = -2 \to x = -\frac{2}{3}\end{align*} which means the domain cannot include a value of -2/3 for *x*: \begin{align*}x\ne-\frac{2}{3}\end{align*}

Since the equation behaves as a line with a single non-existent value, there is no vertical asymptote.

As *x* gets huge, we end up with \begin{align*}f(x) = \frac{9x^2}{3x} \to f(x) = 3x\end{align*} which is a **slant asymptote** that parallels the line of the function!

### Review

- What is the hole in the graph of a function called?
- How are polynomial graphs different from rational function graphs?
- What does the graph of the simplified function that is continuous everywhere not have?
- Why can't rational function be simplified to determine the domain?
- Put the following steps into the correct order for graphing rational functions. a) Find all asymptotes. b) Sketch a smooth graph based on the information. c) Factor numerator and denominator completely and put in lowest terms. Identify any holes. d) Determine the behavior around the vertical asymptotes using a table of signs. e) Find all intercepts. f) Find the places where the function crosses the horizontal asymptote/oblique asymptote.
- Simplify \begin{align*}\frac{6x^2 + 21x + 9}{4x^2 - 1}\end{align*}
- Given: \begin{align*}y = \frac{x^2 + 2x - 8}{x}\end{align*}. Find the intercepts, any asymptotes, and identify end behavior.
- Graph the equation and state the domain.

Simplify each function in questions 9 and 10 and state any value(s) of *x* that make the function undefined.

- \begin{align*}f(x)=\frac{3 - x}{x^2 - 3x}\end{align*}
- \begin{align*}f(x) = \frac{x^2 - 2x - 15}{x - 5}\end{align*}

For each function in questions 11-13, identify points of discontinuity and the form of the graph of the simplified function (linear or quadratic).

- \begin{align*}f(x) = \frac{ 2x^2 + x - 1}{2x - 1}\end{align*}
- \begin{align*}f(x) = \frac{-2x^3 + 9x^2 - 10x + 3}{x - 3}\end{align*}
- \begin{align*}f(x) = \frac{x^3 - 13x -12}{x^2 - 3x - 4}\end{align*}
- Write a quadratic function where the domain cannot include 5 or -5, and the graph has two asymptotes
- Simplify, identify the asymptotes and intercepts, and sketch the graph of \begin{align*}f(x) = \frac{x^3 - 64x}{x^2 - 16}\end{align*}

### Review (Answers)

To see the Review answers, open this PDF file and look for section 2.7.