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# 2.8: Analysis of Rational Functions

Created by: CK-12

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: $T = (18p)/(100-p)$

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. Can you figure out the answer before the review at the end of the lesson?

Embedded Video:

### Guidance

Finding Vertical Asymptotes and Breaks

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

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

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 $x$ get very large or very small, the graph of the rational function will approach (but not reach) the horizontal asymptote.

#### Example A

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

$f(x)=\frac{x^{2}+2x-35}{x-5}$

Solution

Factoring the numerator

$f(x)&=\frac{(x-5)(x+7)}{x+7}\\f(x)& =\frac{(x-5)\cancel{(x+7)}}{\cancel{x+7}}$

Canceling

$f(x)=x+7, x\neq5$

Notice that there is no asymptote in this function, but rather a break in the graph at $x=5$ .

#### Example B

Find the restrictions on the domain of

$h(x)=\frac{3x}{x^{2}-25}$

Solution

Setting the denominator equal to 0,

$x^{2}-25 & = 0\\x^2 & = 25\\x & = \pm 5$

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

#### Example C

(Graphing calculator exercise)

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

Solution

Be Aware:

1. When graphing a rational function by entering the function in the $Y=$ screen, remember that you need to use parenthesis to group the numerator and denominator of the rational function.
2. Vertical asymptotes are sometimes graphed as vertical lines.
3. Graphs of rational functions can be difficult to interpret if the window settings are not chosen carefully.

$y=\frac{1}{x-3}$ showing vertical line at $x=3$

$f(x)$ is undefined and has a vertical asymptote at $x=3$ , but the way the graphing calculator draws the graph, it shows a vertical line at $x=3$ . 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.

Were you able to answer the question at the beginning of the lesson?

'The number of minutes of homework time T required for each student based on understanding level of p % is given by: $T = (18p)/(100-p)$ 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: $(18(75))/(100-75)$ ==> $54min$

By substituting in the increased level: $(18(82))/(100-82)$ ==> $82min$

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

### Vocabulary

End Behavior: A description of what happens at the 'ends' of a rational function graph. Generally a description of the graph as $x\rightarrow\pm\infty$

Break or Hole: A single point that is not possible on the graph of an otherwise continuous function is often represented by a gap or a hole in the line or curve of the graph.

### Guided Practice

Questions

1) What are the vertical asymptotes of the function

$k(x)=\frac{x^{2}}{x^{2}+5}$

2) Find the vertical asymptotes of

$g(x)=\frac{x^{3}}{x^{2}+1}$

3) Solve the equation and find any points of discontinuity using technology: $f(x) = \frac{6}{x-2} + \frac{4}{x+7}$

4) Using technology, find all intercepts, asymptotes, and discontinuities, and graph: $\frac{2x^3 + 14x^2 - 9x + 6}{x + 3}$

5) Using technology, simplify, find discontinuities and limitations, then graph: $\frac{6x^2 + 21x +9}{4x^2 -1}$

Solutions

1) Set the denominator equal to zero,

$x^{2}+5 & = 0\\x^2 & = -5$
There are no real solutions. The horizontal asymptote is $y=1$ .

2) Setting the denominator equal to zero,

$x^{2}+1 & = 0\\x^2 & = -1\\x & = \sqrt{-1}$
There are no real solutions, so there are no vertical asymptotes and no restrictions on the domain of this function.
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

3) 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 $x = 2$ and $x = 7$ and $y = 0$

4) Input the function with care, it should look something like: $(2x^3 + 14x^2 - 9x + 6) / (x + 3)$ before you press the calc or view button.

The graph should look like:
The intercepts are at (apx) $(-7.64, 0)$ and (exactly) $(0, 2)$
There is a vertical asymptote at $x = 3$ and a curved asymptote described by: $y = 2x^2 +8x -33$

5) Input the function accurately, The graph should look like:

The function simplifies to: $\frac{3(2x^2 + 7x +3)}{4(x^2-\frac{1}{4})} \to \frac{3(2x + 1)(x + 3)}{4(x - \frac{1}{2})(x + \frac{1}{2})}$
There are asymptotes at $x = \frac{1}{2}$ and at $y = \frac{1}{2}$
There is a hole at $y = -3\frac{3}{4}$

### Practice

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

1. $y = \frac{x^2 + 3x - 10}{x - 2}$
2. $f(x) = \frac{x^2 + 2x -24}{x - 4}$
3. $f(x) = \frac{x^2 - 12x +32}{x - 4}$
4. $y = \frac{x^2 +\frac{21}{20}}{x + \frac{4}{5}}$
5. $y = \frac{x^2 +13x + 42}{x +7}$

For each problem, 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.

1. $y = \frac{x^3+5x^2+3x+7}{x - 1}$
2. $y = \frac{9x^2 + 6}{x}$
3. $f(x) = \frac{-8x^3 - 8x^2 + 2x + 8}{x + 2}$
4. $f(x) = \frac{5x^3 - 9x^2 - 7x + 1}{x^2 - 4}$
5. $y = \frac{(-2x^3 + 2x^2 + 5x + 2}{(x-2)(x+7)}$
6. $y = \frac{4x^3 + 2x^2 +7}{(x + 2)^2}$
7. $f(x) = \frac{7x^3+2x^2-7x-3}{x^3}$
8. $f(x) = \frac{-6x^3+8x^2+7}{x^2}$
9. $y = \frac{-5x^2-2x-5}{x^2+2}$
10. $y = \frac{x - 1}{x^3 - 2}$

Nov 01, 2012

Dec 08, 2014

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