We have investigated rational functions in prior lessons, and described them loosely as functions that involve the division of polynomials.

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

### Watch This

James Sousa: Graph Rational Functions

### Guidance

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.

#### Example A

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

*Solution:*

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*} (Figure 13), 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 B

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

*Solution:*

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 C

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

*Solution:*

- a) The horizontal asymptote moves down by three units

- b) 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)=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*}.

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### Guided Practice

For each of the rational functions below, determine the domain, the asymptotes, the \begin{align*}x-\end{align*} and \begin{align*}y-\end{align*}intercepts and then sketch the graph.

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

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

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

*Answers*

1) To find the domain limitations, identify the value(s) which make the denominator = 0: \begin{align*}x\ne 1\end{align*}

- That gives us a vertical asymptote of \begin{align*}x=1\end{align*}
- The horizontal asymptote becomes apparent as
*x*becomes truly huge and the "+5" and "-1" no longer matter. At that point, we have \begin{align*}f(x) = \frac{2x}{x} \to f(x) = 2\end{align*} So the horizontal asymptote is \begin{align*}y=2\end{align*}

2) 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*

3) 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*}.

- 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

- 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-existant 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!

### Explore More

- 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 Q's 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 function #s 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*}

### Answers for Explore More Problems

To view the Explore More answers, open this PDF file and look for section 2.7.