Chapter 6

Rational Expressions and Functions

6.1 Rational Functions and Equations

Rational Function  Let p(x) and q(x) be polynomials. Then a

rational function is given by   f(x) =

The domain of f includes all x-values such that q(x) = 0

Examples - 4 , x , 3x2 –6x + 1 x x – 5 3x - 7 

)(

)(

xQ

xP

Identify the domain of rational function(Ex 2 pg 405)

b) g(x) = 2x

x2 - 3x + 2

Denominator = x2 - 3x + 2 = 0

(x – 1)(x –2) = 0 Factor x = 1 or x = 2 Zero product property

Thus D = { x / x is any real number except 1 and 2 }

Using technology( ex 73, pg 414 )

[ -4.7 , 4.7 , 1 ] by [ -3.1, 3.1, 1 ]

(ex 80, pg 414)

Highway curve ( ex 100, page 415 )

R(m) =

0 0.2 0.4 0.6 0.8 slope

500

400

300

200

100

a) R(0.1) = = 457About 457 : a safe curve with a slope of 0.1 will have a minimum radius of 457 ftb) As the slope of banking increases , the radius of the curve decreasesc) 320 = , 320( 15m + 2) = 1600 , 4800m +640 = 1600

4800m = 960, m = = 0.2

Rad

ius

2)1.0(15

1600

215

1600

m

215

1600

m

4800

960

Evaluating a rational function Evaluate f(-1), f(1), f(2)

Numerical value x -3 -2 -1 0 1 2 3

y 3/2 4/3 1 0 __ 4 3

f(x) = 2x

x - 1

-4 -3 -2 -1 1 2 3

f(-1) = 1 f(1) = undefined and f(2) = 4

Vertical asymptote

4

3

2

1

6.2 Multiplications and Divisions of Rational Expression 

To multiply two rational expressions, multiplynumerators and multiply denominators.  , B and C not zero = . , B and D are nonzero.

Example  To divide two rational expressions, multiply bythe reciprocal of the divisor. 

÷ = , B, C, and D are nonzeroExample

B

A

D

CBD

AC

CB

CA

.

.

B

A

B

AD

C

21

10

7

5.

3

2

5

3

4

5

4

3

BC

AD

6.3 Addition and Subtraction of RationalExpressions

 To add (or subtract) two rational expressions withlike denominators, add (or subtract) theirnumerators. The denominator does not change.

Example , C is not zero

, C is not zero

Example

5

3

5

2

5

1

)(

C

BA

C

B

C

A

C

BA

C

B

C

A )(

5

1

5

)23(

5

2

5

3

Finding the Least Common Multiple

Step 1: Factor each polynomial completely

Step 2: List each factor the greatest number of

times that it occurs in either factorization.

Step 3: Find the product of this list of factors. The

result is the LCM

6.4 Solving rational equations graphically and numerically ( Ex- 3, pg 442 )

Solution- The LCD for 2,3, and 5 is their product, 30.

(Multiply by the LCD )

(Distributive property)

15 + 10x = 6x (Reduce)

4x = -15 (Subtract 6x and 15)

x = (Solve)

Graphically Y1 = Y2 =

[ -9, 9, 1] by [ -6, 6, 1]

532

1 xx

30.532

130

xx

5

30

3

30

2

30 xx

4

15

32

1 x 5

x

Determining the time required to empty a pool (Ex 6.4, pg 450, no.88)

A pump can empty a pool in 40 hours. It can empty of the pool in1 hour.In 2 hour, can empty a pool in th of the poolGenerally in t hours it can empty a pool in of the pool.Second pump can empty the pool in 70 hours. So it can empty a poolin of the pool in t hours.Together the pumps can empty of the pool in t hours.

The job will complete when the fraction of the pool is empty equals 1. The equation is = 1

(40)(70) = 1 (40)(70) Multiply (40)(70)70t + 40t = 2800 110t = 2800t = = 25.45 hr. Two pumps can empty a pool in 25.45 hr

7040

tt

7040

tt

7040

tt

40

1

40

2

40

t

70

t

110

2800

Ex 93 Pg 416 A tugboat can travel 15 miles per hour in still water36 miles upstream ( 15 – x) Total time 5 hours downstream (15 + x)t =

So the equation is = 5The LCD is (15-x)(15 + x)Multiply both sides by LCD, we get

(15 – x)(15 + x)[ ] = 5 (15 – x)(15 + x)

540 + 36x + 540 - 36x = 1125 – 5x 2

5x2 – 45 = 0 5x2 = 45x = + 9, x = 3 mph

x 15

36

15

36

x

x 15

36

15

36

x

r

d

Modeling electrical resistance

R1 = 120 ohms

R2 = 160 ohms

R

) reciprocal ( ohms 69 7

480 R

480

7

480

3

480

4

480 LCD , 3

3 .

160

1

4

4 .

120

1

160

1

120

1

1

2

1

1

1

RRR

6.6 Modelling with Proportion

a c is equivalent to ad = bc

b d

Example 6 8

5 x

6x = 40

or x =

6 feet h feet

4 feet 44 feet

3

20

6

40

feet

h

664

44.64

6

64

Modeling AIDS cases[ 1980, 1997, 2] by [-10000, 800000, 100000] Y = 1000 (x – 1981)2