College Algebra - St. Cloud Technical and Community College Algebra 4.6.pdf · College Algebra...

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College Algebra Chapter 4, section 6 Created by Lauren Atkinson Mary Stangler Center for Academic Success This review is meant to highlight basic concepts from Chapter 4. It does not cover all concepts presented by your instructor. Refer back to your notes, handouts, the book, MyMathLab, etc. for further prepare for your exam.

Transcript of College Algebra - St. Cloud Technical and Community College Algebra 4.6.pdf · College Algebra...

Page 1: College Algebra - St. Cloud Technical and Community College Algebra 4.6.pdf · College Algebra Chapter 4, section 6 Created by Lauren Atkinson ... Rational Functions Rational functions

College Algebra

Chapter 4, section 6 Created by Lauren Atkinson

Mary Stangler Center for Academic Success

This review is meant to highlight basic concepts from Chapter 4. It does not cover all concepts presented by your instructor. Refer back to your notes, handouts, the book, MyMathLab, etc. for further prepare for your exam.

Page 2: College Algebra - St. Cloud Technical and Community College Algebra 4.6.pdf · College Algebra Chapter 4, section 6 Created by Lauren Atkinson ... Rational Functions Rational functions

4.6 Rational Functions and Models

• This section covers:

– Rational functions

– Vertical asymptotes

– Horizontal asymptotes

– Slant asymptotes

– Graphing

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Rational Functions

Rational functions are defined:

A function that can be written 𝑃(𝑥)

𝑄(𝑥) where 𝑄(𝑥) ≠ 0 and both

𝑃 𝑥 , 𝑄(𝑥) are polynomials

To determine if a function is a rational function, look at 𝑃(𝑥) and 𝑄(𝑥) separately and determine if they are polynomials. If both

are polynomials then 𝑃(𝑥)

𝑄(𝑥) is a rational function.

Domain of a rational function: The domain is all real numbers except where the denominator (𝑄(𝑥)) equals zero.

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Examples:

Are these functions rational functions?

𝑓 𝑥 =6

𝑥2

𝑓 𝑥 =𝑥2 + 1

𝑥 − 8

𝑓 𝑥 =4

𝑥+ 1

𝑓 𝑥 =𝑥 + 1

𝑥 + 1

𝑓 𝑥 =𝑥3 − 3𝑥 + 1

𝑥2 − 5

𝑓 𝑥 = 5𝑥3 − 4𝑥

YES NO YES NO YES YES

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Vertical Asymptotes:

These occur when the graph is undefined (when 𝑄 𝑥 = 0). To find, we have to set the denominator = 0 and solve for x. The values you find for x are your vertical asymptotes.

The graph will never touch or cross a vertical asymptote!

Notice that all vertical asymptotes are written in terms of 𝑥.

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Horizontal Asymptotes:

There are three different scenarios for horizontal asymptotes:

• Degree of 𝑃(𝑥) > 𝑄(𝑥), there is no horizontal asymptote

• Degree of 𝑃 𝑥 < 𝑄 𝑥 , the horizontal asymptote is 𝑦 = 0

• Degree of 𝑃 𝑥 = 𝑄(𝑥), we take the leading coefficients of both the numerator and the denominator and divide the two.

Notice the all horizontal asymptotes are written in terms of 𝑦.

The graph can cross or intersect the horizontal asymptotes (this is different from vertical asymptotes).

The function is 𝑓 𝑥 =𝑃(𝑥)

𝑄(𝑥)

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Vertical Asymptotes Vertical and Horizontal Asymptotes

Vertical and Slant Asymptotes

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Examples: Find any horizontal or vertical asymptotes:

𝑓 𝑥 = 3𝑥2

𝑥2 − 9

𝑓 𝑥 = 𝑥

𝑥3 − 𝑥

𝑓 𝑥 =𝑥4 + 1

𝑥2 + 3𝑥 − 10

Write a formula for 𝑓(𝑥) for a rational function so that its graph has the specified asymptotes: Vertical: 𝑥 = ±3 Horizontal: 𝑦 = 5

𝑓 𝑥 =5𝑥2

𝑥 − 3 (𝑥 + 3)

𝐻𝐴: 𝑦 = 3 𝑉𝐴: 𝑥 = ±3 𝐻𝐴: 𝑦 = 0 𝑉𝐴: 𝑥 = 0, ±1 HA: none 𝑉𝐴: 𝑥 = −5,2

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Slant Asymptotes

A special case of asymptotes that occur when the degree of the numerator (𝑃(𝑥)) is one degree larger than the degree of the denominator (𝑄(𝑥)).

To find these, we divide our rational function using either long division or synthetic division.

When dividing, this is the one and only time that you can ignore the remainder.

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Example: Find the vertical and slant asymptotes:

𝑓 𝑥 =4𝑥2

2𝑥 − 1

VA: set the denominator equal to zero and solve:

2𝑥 − 1 = 0

𝑥 =1

2

Slant A: use long division and the quotient (ignoring the remainder) is the slant asymptote:

𝑦 = 2𝑥 + 1

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Graphing By Hand

To graph by hand it is best to go step by step: 1. Identify any Vertical Asymptotes (draw them in

as a dashed line) 2. Identify any Horizontal or Slant Asymptotes

(draw them in as a dashed line) 3. Find any y-intercepts 4. Find any x-intercepts 5. Make a t-chart containing values to the left and

to the right of your vertical asymptotes 6. Plot the points and graph

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When we graph, we can also use the transformations rules from previous chapters to visualize rational

functions. For instance, just like before, constants that are added or subtracted INSIDE the parenthesis’ are

horizontal shifts (shifts to the left or to the right).

Constants that are just “hanging out” in front or on the end are vertical shifts (shifts up or down). Reflections occur just as before, we reflect the function over the

x-axis when we have a negative sign out in front.

𝑓 𝑥 = 1 −2

(𝑥 + 3)

reflection (reflect over the x-axis)

horizontal shift (shift three units to the left)

vertical shift (shift one unit up)

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Special Cases:

We can have a hole in the graph. This can occur if and only if 𝑃 𝑥 = 0 and 𝑄 𝑥 = 0 at exactly the same point. This point is then referred to as a whole, not a vertical asymptote.

An example of this is as follows:

𝑓 𝑥 =𝑥+5

𝑥2−25

We can rewrite this function: 𝑓 𝑥 =(𝑥+5)

(𝑥+5)(𝑥−5) and then

can cross out the factors that are the same:

𝑓 𝑥 =(𝑥 + 5)

(𝑥 + 5)(𝑥 − 5)

Therefore our function reduced to: 𝑓 𝑥 =1

(𝑥−5) and

we conclude that there is a hole, not a vertical asymptote when x − 5 = 0 which occurs at 𝑥 = 5.

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Another “special case” occurs in terms of horizontal asymptotes. There are some instances where the graph might intersect or cross over their horizontal asymptote and then re-approach it from below. An example of this is as follows:

𝑓 𝑥 =3𝑥2−3𝑥−6

𝑥2+8𝑥+16

We still graph this by the same process as before.

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Examples: Graph by hand: You should be able to complete these BY HAND. Your hand drawings should look like the following:

𝑓 𝑥 =2𝑥2 − 3𝑥 − 2

𝑥2 − 4𝑥 + 4

𝑓 𝑥 =2𝑥 + 3

𝑥 + 1

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Solve

• Lastly in this section, we solve rational equations. These are best shown through examples:

3

2𝑥 + 1= −1

2𝑥 + 13

2𝑥 + 1= −1 2𝑥 + 1

3 = −1 2𝑥 + 1 3 = −2𝑥 − 1 4 = −2𝑥 −2 = 𝑥

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Solve: 𝑥 + 1

𝑥= 2

𝑥𝑥 + 1

𝑥= 2 𝑥

𝑥 + 1 = 2𝑥 1 = 𝑥