Rational Functions: Definition, Identification, Domain, Asymptotes and Graphs, Lecture notes of Algebra

The concepts of rational functions, including their definition, identification, finding the domain, the Big-Little Principle, and the graphs with emphasis on horizontal and vertical asymptotes. It includes examples and exercises.

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12 Rational Functions
Concepts:
โ€ขThe Definition of a Rational Function
โ€ขIdentifying Rational Functions
โ€ขFinding the Domain of a Rational Function
โ€ขThe Big-Little Principle
โ€ขVertical and Horizontal Asymptotes
โ€ขThe Graphs of Rational Functions
(Section 4.5)
Definition 12.1
Arational function is a function that is equivalent to a function of the following form.
r(x) = polynomial
polynomial
Example 12.2
Which of the following are rational functions? If the function is rational, find its domain.
โ€ขf(x) = x2+ 2x+ 1
xโˆ’3Rational
f(x) is defined if x6= 3, so the domain of f(x) is (โˆ’โˆž,3) โˆช(3,โˆž)
โ€ขg(x) = 2โˆšx+ 1
x2+ 2xNot rational since the numerator is not a polynomial.
โ€ขh(x) = โˆš6 + 3x
x5Rational
h(x) is defined for all xexcept 0, so the domain of h(x) is (โˆ’โˆž,0) โˆช(0,โˆž).
โ€ขj(x) = x+ 1
xโˆ’2+x=x+ 1
xโˆ’2+x(xโˆ’2)
xโˆ’2=x+ 1 + x2โˆ’2x
xโˆ’2=x2โˆ’x+ 1
xโˆ’2Rational
j(x) is defined if x6= 2, so the domain of j(x) is (โˆ’โˆž,2) โˆช(2,โˆž).
โ€ขk(x) = x5+ 2x+ 1 =x5+ 2x+ 1
1Rational
k(x) is defined for all xso its domain is (โˆ’โˆž,โˆž).
Note 12.3
Polynomials are a special case of rational functions.
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12 Rational Functions

Concepts:

  • The Definition of a Rational Function
  • Identifying Rational Functions
  • Finding the Domain of a Rational Function
  • The Big-Little Principle
  • Vertical and Horizontal Asymptotes
  • The Graphs of Rational Functions

(Section 4.5)

Definition 12. A rational function is a function that is equivalent to a function of the following form.

r(x) =

polynomial polynomial

Example 12. Which of the following are rational functions? If the function is rational, find its domain.

  • f (x) =

x^2 + 2x + 1 x โˆ’ 3

Rational f (x) is defined if x 6 = 3, so the domain of f (x) is (โˆ’โˆž, 3) โˆช (3, โˆž)

  • g(x) =

x + 1 x^2 + 2x

Not rational since the numerator is not a polynomial.

  • h(x) =

6 + 3x x^5

Rational h(x) is defined for all x except 0, so the domain of h(x) is (โˆ’โˆž, 0) โˆช (0, โˆž).

  • j(x) =

x + 1 x โˆ’ 2

  • x =

x + 1 x โˆ’ 2

x(x โˆ’ 2) x โˆ’ 2

x + 1 + x^2 โˆ’ 2 x x โˆ’ 2

x^2 โˆ’ x + 1 x โˆ’ 2

Rational j(x) is defined if x 6 = 2, so the domain of j(x) is (โˆ’โˆž, 2) โˆช (2, โˆž).

  • k(x) = x^5 + 2x + 1 =

x^5 + 2x + 1 1

Rational k(x) is defined for all x so its domain is (โˆ’โˆž, โˆž).

Note 12. Polynomials are a special case of rational functions.

12.1 Graphs of Some Simple Rational Functions

Let f (x) =

xn^

. Below is a chart of the basic shapes that the graph of f (x) can take on.

n odd n even

Examples: f (x) =

x^5

, g(x) =

x^7

Example: f (x) =

x^6

, h(x) =

x^10

Property 12.4 (The Big-Little Principle) If c is a number that is close to 0 on the number line, then (^1) c is a number that is far from 0 on the number line. If c is a number that is far from 0 on the number line, then (^1) c is a number that is close to 0 on the number line.

How should you remember the Big-Little Principle?

If c is big then

c

is little and if

c

is big then c is little.

Can you use the Big-Little Principle to explain the shape of the graph of y = f (x) =

xn^

How do you describe the end behavior of the graph of y = f (x) =

xn^

If x is big (in either the positive or negative sense) then

x

is little so

xn^

is also little. As x

gets bigger,

xn^

gets littler - that is, it gets closer to zero. So, the end behavior of f (x) is

y โ†’ 0 as x โ†’ โˆž

y โ†’ 0 as x โ†’ โˆ’โˆž

12.2 The Graphs of Rational Functions - Some Examples

We have used a graphing calculator (TI-83 Plus) to approximate the graphs of a few rational functions. Graphing calculators do not do a very good job of sketching asymptotes. Nevertheless, we can use them to better understand the graphs of rational functions. For each graph, look at the algebraic description of the function and the approximate graph to better understand its asymptotes. Show algebraically how you would find the asymptotes of each graph. Draw a better sketch of the graph that includes all asymptotes. Make sure that asymptotes are drawn with dotted lines.

Note: Each graph is in a [โˆ’ 10 , 10] ร— [โˆ’ 10 , 10] viewing window.

  • The graph of y = f (x) =^6 x^ โˆ’^5 2 x โˆ’ 1 .

2

4

6

8

โˆ’ 2 โˆ’ 4 โˆ’ 6 โˆ’ 8

โˆ’ 10 โˆ’ 8 โˆ’ 6 โˆ’ 4 โˆ’ 2 2 4 6 8

x

y

The leading term of the numerator is 6x and the leading term of the denominator is 2x. The ratio of the leading terms is 6 x 2 x = 3. Thus, f (x) has the same end behavior as y = 3 which is y โ†’ 1 as x โ†’ โˆž andy โ†’ 1 as x โ†’ โˆ’โˆž So, y = 3 is a horizontal asymptote of f (x). Notice that x = 1 2 is a zero of the denominator but not the numerator. So x = 1 2 is a vertical asymptote of f (x).

  • The graph of y = f (x) = x^ + 4 x^2 + 5x + 6 .

2

4

6

8

โˆ’ 2 โˆ’ 4 โˆ’ 6 โˆ’ 8 โˆ’ 10

โˆ’ 10 โˆ’ 8 โˆ’ 6 โˆ’ 4 โˆ’ 2 2 4 6 8

x

y

The leading term of the numerator is x and the leading term of the denominator is x^2. The ratio of the leading terms is x x^2 = 1 x

. Thus, f (x) has the same end behavior as y = 1 x which is y โ†’ 0 as x โ†’ โˆž and y โ†’ 0 as x โ†’ โˆ’โˆž So, y = 0 is a horizontal asymptote of g(x). The denominator x^2 + 5x + 6 = (x + 3)(x + 2) is zero when x = โˆ’3 and when x = โˆ’2. Since neither of these make the numerator zero then x = โˆ’3 and x = โˆ’2 are vertical asymptotes of f (x).

Example 12.8 (Graphs of Rational Functions)

Let h(x) =

x^2 + x โˆ’ 2 2 x^2 โˆ’ 8 x โˆ’ 10

. Sketch the graph of h(x) without using your calculator. Be sure

to label all asymptotes and intercepts of the graph.

The leading term of the numerator is x^2 and the leading term of the denominator is 2x^2.

The ratio of the leading terms is

x^2 2 x^2

. Thus, h(x) has the same end behavior as y =

which is

y โ†’

as x โ†’ โˆž and y โ†’

as x โ†’ โˆ’โˆž

So, y =

is a horizontal asymptote. Notice that the numerator x^2 + x โˆ’ 2 = (x + 2)(x โˆ’ 1)

and the denominator 2x^2 โˆ’ 8 x โˆ’ 10 = 2(x + 1)(x โˆ’ 5). So, x = โˆ’1 and x = 5 are zeros of the denominator but not the numerator. So x = โˆ’1 and x = 5 are vertical asymptotes of h(x).

The y-intercept of the graph is y =

2(0)^2 โˆ’ 8(0) โˆ’ 10

. The x-intercepts are

found by setting h(x) = 0. The x-intercepts are where the numerator of h(x) are 0. Since weโ€™ve already factored the numerator above, its easy to see that the x-intercepts are x = โˆ’ 2 and x = 1.

-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

0

1

2

3

4

5

6

7

8

9

10

y = โˆ’ 0. 5

x = โˆ’ (^1) x = 5

b b b x

y