Transcript Chapter Six 6.1

Chapter 6
Rational Expressions,
Functions, and Equations
§ 6.1
Rational Expressions and Functions:
Multiplying and Dividing
Rational Expressions
A rational expression consists of a polynomial divided by a
nonzero polynomial (denominator cannot be equal to 0).
A rational function is a function defined by a formula that is a
rational expression. For example, the following is a rational
function:
130 x
f x  
100  x
Blitzer, Intermediate Algebra, 5e – Slide #3 Section 6.1
Rational Expressions
EXAMPLE
The rational function
130 x
f x  
100  x
models the cost, f (x) in millions of dollars, to inoculate x% of
the population against a particular strain of flu. The graph of
the rational function is shown. Use the function’s equation to
solve the following problem.
Find and interpret f (60). Identify your solution as a point on
the graph.
p 393
Blitzer, Intermediate Algebra, 5e – Slide #4 Section 6.1
Rational Expressions
CONTINUED
1000
900
800
700
600
500
400
300
200
100
0
0
20
40
60
Blitzer, Intermediate Algebra, 5e – Slide #5 Section 6.1
80
100
Rational Expressions
CONTINUED
SOLUTION
We use substitution to evaluate a rational function, just as we
did to evaluate other functions in Chapter 2.
f x  
130 x
100  x
130 60 
f 60  
100  60
7800

 195
40
This is the given rational
function.
Replace each occurrence of
x with 60.
Perform the indicated
operations.
Blitzer, Intermediate Algebra, 5e – Slide #6 Section 6.1
Rational Expressions
CONTINUED
Thus, f (60) = 195. This means that the cost to inoculate 60%
of the population against a particular strain of the flu is $195
million. The figure below illustrates the solution by the point
(60,195) on the graph of the rational function.
1000
900
800
700
600
(60,195)
500
400
300
200
100
0
0
20
40
60
80
Blitzer, Intermediate Algebra, 5e – Slide #7 Section 6.1
100
Rational Expressions - Domain
EXAMPLE
Find the domain of f if
f x  
3x
.
2
x  13 x  36
SOLUTION
The domain of f is the set of all real numbers except those for
which the denominator is zero. We can identify such numbers
by setting the denominator equal to zero and solving for x.
x 2  13x  36  0
Set the denominator equal to 0.
x  4x  9  0
Factor.
p 393
Blitzer, Intermediate Algebra, 5e – Slide #8 Section 6.1
Rational Expressions - Domain
CONTINUED
x40
x4
or
x9  0
x9
Set each factor equal to 0.
Solve the resulting equations.
Because 4 and 9 make the denominator zero, these are the
values to exclude. Thus,
Domainof f  x | x is a real number and x  4 and x  9
or
Domainof f  - ,4  4,9  9,  .
Blitzer, Intermediate Algebra, 5e – Slide #9 Section 6.1
Rational Expressions - Domain
CONTINUED
Domainof f  - ,4  4,9  9,  .
3x
f x  
.
( x  4)(x  9)
In this example, we excluded 4 and 9 from the domain. Unlike the
graph of a polynomial which is continuous, this graph has two breaks
in it – one at each of the excluded values. Since x cannot be 4 or 9,
there is not a function value corresponding to either of those x values.
At 4 and at 9, there will be dashed vertical lines called vertical
asymptotes.
The graph of the function will approach these vertical lines on each
side as the x values draw closer and closer to each of them, but will
not touch (cross) the vertical lines. The lines x = 4 and x = 9 each
represent vertical asymptotes for this particular function.
p 395
Blitzer, Intermediate Algebra, 5e – Slide #10 Section 6.1
Rational Expressions
Asymptotes
Vertical
Asymptotes
A vertical line that the graph of a function
approaches, but does not touch.
Horizontal
Asymptotes
A horizontal line that the graph of a function
approaches as x gets very large or very small. The
graph of a function may touch/cross its horizontal
asymptote.
Simplifying Rational Expressions
1) Factor the numerator and the denominator completely.
2) Divide both the numerator and the denominator by any common
factors.
p 395
Blitzer, Intermediate Algebra, 5e – Slide #11 Section 6.1
Rational Expressions - Domain
Check Point 2
Find the domain of f if
x 5
f x   2
.
2 x  5x  3
SOLUTION
2 x 2  5x  3  0
Set the denominator equal to 0.
2x 1x  3  0
Factor.
2x 1  0
1
x
2
or
or
x3 0
x  3
Set each factor equal to 0.
Solve the resulting equations.
1

Domainof f   x | x is a real number and x  3 and x  
2

1 1 

Domainof f  - ,-3    3,    ,   .
2 2 
p 394

Blitzer, Intermediate Algebra, 5e – Slide #12 Section 6.1
Simplifying Rational Expressions
EXAMPLE
Simplify:
x 1
.
2
x  2x  3
SOLUTION
x 1
1 x  1

x 2  2 x  3 x  3x  1

1 x  1
x  3x  1
1

x3
Factor the numerator and
denominator.
Divide out the common factor,
x + 1.
Simplify.
Blitzer, Intermediate Algebra, 5e – Slide #13 Section 6.1
Simplifying Rational Expressions
Check Point 3
2
x
 7 x  10
Simplify:
.
x2
SOLUTION
x 2  7 x  10 x  5x  2

x2
1x  2

x  5x  2
1x  2
 x5
Factor the numerator and
denominator.
Divide out the common factor,
x + 1.
Simplify.
p 397
Blitzer, Intermediate Algebra, 5e – Slide #14 Section 6.1
Simplifying Rational Expressions
EXAMPLE
2
x
 7 x  12
Simplify:
.
2
9 x
SOLUTION
x 2  7 x  12 x  3x  4

2
3  x 3  x
9 x
Factor the numerator and
denominator.

x  3x  4

3  x  1 3  x 
Rewrite 3 – x as (-1)(-3 + x).

x  3x  4

3  x  1x  3
Rewrite -3 + x as x – 3.
Blitzer, Intermediate Algebra, 5e – Slide #15 Section 6.1
Simplifying Rational Expressions
CONTINUED

x  3x  4

3  x  1x  3

x  4
3  x  1
Divide out the common factor,
x – 3.
Simplify.
Do Check 4a and 4b on page 397
Blitzer, Intermediate Algebra, 5e – Slide #16 Section 6.1
Multiplying Rational Expressions
Multiplying Rational Expressions
1) Factor all numerators and denominators completely.
2) Divide numerators and denominators by common
factors.
3) Multiply the remaining factors in the numerators and
multiply the remaining factors in the denominators.
p 398
Blitzer, Intermediate Algebra, 5e – Slide #17 Section 6.1
Multiplying Rational Expressions
EXAMPLE
2 x 2  3xy  2 y 2 3x 2  2 xy  y 2
Multiply:
 2
.
2
2
2
3x  4 xy  y
x  xy  6 y
SOLUTION
2 x 2  3xy  2 y 2 3x 2  2 xy  y 2
 2
2
2
3x  4 xy  y
x  xy  6 y 2

2 x  y x  2 y  3x  y x  y 


3x  y x  y  x  2 y x  3 y 

2 x  y x  2 y   3x  y x  y 
3x  y x  y  x  2 y x  3 y 
This is the original expression.
Factor the numerators and
denominators completely.
Divide numerators and
denominators by common
factors.
Blitzer, Intermediate Algebra, 5e – Slide #18 Section 6.1
Multiplying Rational Expressions
CONTINUED

2 x  y 3x  y 

3x  y x  3 y 
Multiply the remaining factors
in the numerators and in the
denominators.
Note that when simplifying rational expressions or multiplying rational
expressions, we just used factoring.
With one additional step that is provided in the following
Definition for Division, division of rational expressions promises to be just as
straightforward.
Blitzer, Intermediate Algebra, 5e – Slide #19 Section 6.1
Multiplying Rational Expressions
EXAMPLE
8y  2 3  y
 2
.
Multiply: 2
y  9 4y  y
SOLUTION
8y  2 3  y
 2
2
y 9 4y  y

24 y  1
3 y

 y  3 y  3 y4 y  1

24 y  1
 1( y  3)


 y  3 y  3 y4 y  1
This is the original expression.
Factor the numerators and
denominators completely.
Divide numerators and
denominators by common
factors. Because 3 – y and y -3
are opposites, their quotient is -1.
Blitzer, Intermediate Algebra, 5e – Slide #20 Section 6.1
Multiplying Rational Expressions
CONTINUED
2

 y  3 y
or
2

 y  3 y
Now you may multiply the
remaining factors in the
numerators and in the
denominators.
Blitzer, Intermediate Algebra, 5e – Slide #21 Section 6.1
Multiplying Rational Expressions
Check Point 5
Multiply:
x  4 x 2  4 x  21

.
2
x7
x  16

x  3

x  4
pages 398-399
Blitzer, Intermediate Algebra, 5e – Slide #22 Section 6.1
Multiplying Rational Expressions
Check Point 6
Multiply:
4 x  8 3x 2  4 x  4

.
2
2
6 x  3x
9x  4
 4 x  2 

3 x3 x  2 
4 x  2 

3 x3 x  2 
pages 398-399
Blitzer, Intermediate Algebra, 5e – Slide #23 Section 6.1
Dividing Rational Expressions
Simplifying Rational Expressions with Opposite
Factors in the Numerator and Denominator
The quotient of two polynomials that have opposite signs and are additive
inverses is -1.
Dividing Rational Expressions
If P, Q, R, and S are polynomials, where Q  0, R  0, and S  0, then
P R P S PS
   
.
Q S Q R QR
R
Change division
to multiplication.
Replace S with its reciprocal by
interchanging its numerator and
denominator.
Blitzer, Intermediate Algebra, 5e – Slide #24 Section 6.1
Dividing Rational Expressions
EXAMPLE
xy  y 2
2 x 2  xy  3 y 2
Divide: 2
 2
.
2
x  2 x  1 2 x  5xy  3 y
SOLUTION
xy  y 2
2 x 2  xy  3 y 2
 2
2
x  2 x  1 2 x  5xy  3 y 2
This is the original expression.
xy  y 2 2 x 2  5 xy  3 y 2
 2

x  2 x  1 2 x 2  xy  3 y 2
Invert the divisor and multiply.
yx  y  2 x  3 y x  y 


2
x  1 2 x  3 y x  y 
Factor.
Blitzer, Intermediate Algebra, 5e – Slide #25 Section 6.1
Dividing Rational Expressions
CONTINUED
yx  y  2 x  3 y x  y 


2
x  1 2 x  3 y x  y 
y x  y 

x  12
Divide numerators and
denominators by common
factors.
Multiply the remaining factors
in the numerators and in the
denominators.
Blitzer, Intermediate Algebra, 5e – Slide #26 Section 6.1
Multiplying Rational Expressions
Check Point 7a
Divide:
9 x
2

 49 
3x  7
.
9
 93x  7 
Blitzer, Intermediate Algebra, 5e – Slide #27 Section 6.1
Multiplying Rational Expressions
Check Point 7b
Divide:
x 2  x  12 x 2  10x  24

.
2
5x
x  6x

x  3

5
Blitzer, Intermediate Algebra, 5e – Slide #28 Section 6.1
DONE