Consider the following situation:

Students in your school are currently required to demonstrate their understanding of their classwork at a level of 75% or higher in order to move on to more advanced material.

The amount of homework time **T** required for each student based on understanding level of **p**% is given by: \begin{align*}T = \frac{(18p)}{(100-p)}\end{align*}

If the school administration decides to raise the minimum level of understanding to 82%, how will this affect the students' homework time?

This is an example of a **rational function** in a real-world situation.

### Analyzing Rational Functions

#### Finding Vertical Asymptotes and Breaks

Recall that **rational functions** are defined as \begin{align*}r(x)=\frac{p(x)}{q(x)}\end{align*} where \begin{align*}p(x)\end{align*} and \begin{align*}q(x)\end{align*} are polynomials.

To find vertical asymptotes and breaks in the domain of a rational function, set the denominator equal to zero and solve for \begin{align*}x\end{align*}. Given \begin{align*}r(x)=\frac{p(x)}{q(x)}\end{align*}, set \begin{align*}q(x)=0\end{align*} and solve for \begin{align*}x\end{align*}.

Note that some rational functions have vertical asymptotes and others do not. Some rational functions have a break in the function, but no vertical asymptote. This usually happens when one term in the numerator cancels with one term in the denominator.

#### Evaluating End Behavior

The **end behavior** of a rational function can often be identified by the horizontal asymptote. That is, as the values of \begin{align*}x\end{align*} get very large or very small, the graph of the rational function will approach (but not reach) the horizontal asymptote.

### Examples

#### Example 1

Earlier, you were asked a question about students' homework time.

The number of minutes of homework time T required for each student based on understanding level of *p*'% is given by: \begin{align*}T = (18p)/(100-p)\end{align*} If the school administration decides to raise the minimum level of understanding to 82%, how will this affect the students' homework time?'

The current time required is: \begin{align*}(18(75))/(100-75)\end{align*} ==> \begin{align*}54min\end{align*}

By substituting in the increased level: \begin{align*}(18(82))/(100-82)\end{align*} ==> \begin{align*}82min\end{align*}

The increase in required proficiency would result in an average of 28mins of added study time per student.

#### Example 2

Consider the following rational function. Find all restrictions on the domain and asymptotes.

\begin{align*}f(x)=\frac{x^{2}+2x-35}{x-5}\end{align*}

Factoring the numerator

\begin{align*}f(x)&=\frac{(x-5)(x+7)}{x+7}\\ f(x)& =\frac{(x-5)\cancel{(x+7)}}{\cancel{x+7}}\end{align*}

Canceling

\begin{align*}f(x)=x+7, x\neq5\end{align*}

Notice that there is no asymptote in this function, but rather a break in the graph at \begin{align*}x=5\end{align*}.

#### Example 3

Find the restrictions on the domain of

\begin{align*}h(x)=\frac{3x}{x^{2}-25}\end{align*}

Setting the denominator equal to 0,

\begin{align*}x^{2}-25 & = 0\\ x^2 & = 25\\ x & = \pm 5\end{align*}

Thus, the domain of \begin{align*}h(x)\end{align*} is the set all real numbers \begin{align*}x\end{align*} with the restriction \begin{align*}x\neq\pm 5\end{align*}. \begin{align*}h(x)\end{align*} has two vertical asymptotes, one at \begin{align*}x=5\end{align*} and one at \begin{align*}x=-5\end{align*}.

**Note for graphing using technology problems:** If you do not have a graphing calculator, there are a number of excellent free apps, and an excellent free online calculator here: Desmos Online Graphing Calculator.

#### Example 4

(Graphing calculator exercise)

Graph \begin{align*}f(x)=\frac{1}{x-3}\end{align*} on the window [-10,10] X [-10,10]. (This means {XMIN}=-10, {XMAX}=10, {YMIN}=-10, and {YMAX}=10).

Notice:

- When graphing a rational function by entering the function in the \begin{align*}Y=\end{align*} screen, remember that you need to use parenthesis to group the numerator and denominator of the rational function.
- Vertical asymptotes are sometimes graphed as vertical lines.
- Graphs of rational functions can be difficult to interpret if the window settings are not chosen carefully.

\begin{align*}y=\frac{1}{x-3}\end{align*} showing vertical line at \begin{align*}x=3\end{align*}

\begin{align*}f(x)\end{align*} is undefined and has a vertical asymptote at \begin{align*}x=3\end{align*}, but the way the graphing calculator draws the graph, it shows a vertical line at \begin{align*}x=3\end{align*}. One way to “fix” this problem is to press **MODE** and select the option “Dot” rather than “Connected”. However, dot graphs can be hard to interpret as well.

#### Example 5

Solve the equation and find any points of discontinuity using technology: \begin{align*}f(x) = \frac{6}{x-2} + \frac{4}{x+7}\end{align*}.

Input the function into the grapher, don't forget to use "(" ")" to group polynomials. Your function should look something like this on your input screen (at least until you tell the calc to process the function): ** 6/(x - 2) + 4/(x + 7)**. Note that some calculators require the "

*y*= " or "

*f*(

*x*) = " and some do not.

Once you are sure you have the information entered correctly, press "calculate", or the equivalent. The graph should look like:

There are asymptotes at \begin{align*}x = 2\end{align*} and \begin{align*}x = 7\end{align*} and \begin{align*}y = 0\end{align*}.

#### Example 6

Using technology, find all intercepts, asymptotes, and discontinuities, and graph: \begin{align*}\frac{2x^3 + 14x^2 - 9x + 6}{x + 3}\end{align*}.

Input the function with care, it should look something like: \begin{align*}(2x^3 + 14x^2 - 9x + 6) / (x + 3)\end{align*} before you press the calc or view button.

The graph should look like:

The intercepts are at (apx) \begin{align*}(-7.64, 0)\end{align*} and (exactly) \begin{align*}(0, 2)\end{align*}.

There is a vertical asymptote at \begin{align*}x = 3\end{align*} and a curved asymptote described by: \begin{align*}y = 2x^2 +8x -33\end{align*}.

#### Example 7

Using technology, simplify, find discontinuities and limitations, then graph: \begin{align*}\frac{6x^2 + 21x +9}{4x^2 -1}\end{align*}.

Input the function accurately. The graph should look like:

The function simplifies to: \begin{align*}\frac{3(2x^2 + 7x +3)}{4(x^2-\frac{1}{4})} \to \frac{3(2x + 1)(x + 3)}{4(x - \frac{1}{2})(x + \frac{1}{2})}\end{align*}

There are asymptotes at \begin{align*}x = \frac{1}{2}\end{align*} and at \begin{align*}y = \frac{1}{2}\end{align*}.

There is a hole at \begin{align*}y = -3\frac{3}{4}\end{align*}.

### Review

For questions 1 - 5, factor the numerator and denominator, then set the denominator equal to zero and solve to find restrictions on the domain.

- \begin{align*}y = \frac{x^2 + 3x - 10}{x - 2}\end{align*}
- \begin{align*}f(x) = \frac{x^2 + 2x -24}{x - 4}\end{align*}
- \begin{align*}f(x) = \frac{x^2 - 12x +32}{x - 4}\end{align*}
- \begin{align*}y = \frac{x^2 +\frac{21}{20}}{x + \frac{4}{5}}\end{align*}
- \begin{align*}y = \frac{x^2 +13x + 42}{x +7}\end{align*}

For questions 6 - 15, input the function into your graphing tool carefully and accurately. Record any asymptotes or holes and record *x* and *y* intercepts. Finally either copy and print or sketch the image of the graphed function.

- \begin{align*}y = \frac{x^3+5x^2+3x+7}{x - 1}\end{align*}
- \begin{align*}y = \frac{9x^2 + 6}{x}\end{align*}
- \begin{align*}f(x) = \frac{-8x^3 - 8x^2 + 2x + 8}{x + 2}\end{align*}
- \begin{align*}f(x) = \frac{5x^3 - 9x^2 - 7x + 1}{x^2 - 4}\end{align*}
- \begin{align*}y = \frac{(-2x^3 + 2x^2 + 5x + 2}{(x-2)(x+7)}\end{align*}
- \begin{align*}y = \frac{4x^3 + 2x^2 +7}{(x + 2)^2}\end{align*}
- \begin{align*}f(x) = \frac{7x^3+2x^2-7x-3}{x^3}\end{align*}
- \begin{align*}f(x) = \frac{-6x^3+8x^2+7}{x^2}\end{align*}
- \begin{align*}y = \frac{-5x^2-2x-5}{x^2+2}\end{align*}
- \begin{align*}y = \frac{x - 1}{x^3 - 2}\end{align*}

### Review (Answers)

To see the Review answers, open this PDF file and look for section 2.8.