Lesson 8-1: Multiplying and Dividing Rational Expressions

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Transcript Lesson 8-1: Multiplying and Dividing Rational Expressions

Lesson 8-1: Multiplying and Dividing
Rational Expressions
Rational Expression

Definition: a ratio of two polynomial expressions
8 x
13  x
To Simplify A Rational Expression
1. Make sure both the numerator and
denominator are factored completely!!!
2. Look for common factors and cancel
–
Remember factors are things that are being
multiplied you can NEVER cancel things that are
being added or subtracted!!!
3. Find out what conditions make the
expression undefined and state them.
Examples: Simplify and state the values for x that result
in the expression being undefined
1.
3 y  y  6
 y  6( y 2  8 y  12)

3
2
4
x
x
 7x  8
2.
12 x x 2  64



Examples Cont… Simplify
3.
z 2w  z 2
z 3  z 3w
4.
a 4b  2a 4
2a 3  a 3b
5.
xy  3x
3x 2  x 2 y
Operations with Rational Expressions
To Multiply Rational Expressions:
Factor and cancel where possible. Then
multiply numerators and denominators
To Divide Rational Expressions:
Rewrite the problem as a multiplication
problem with the first expression times the
reciprocal of the second expression. Then factor
and cancel where possible. Multiply numerators
and denominators
Define xvalues for
which the
expression
is undefined
Examples:
6.
8.
8t 2 s 15sr

2
5r 12t 3 s 2
4a 15b 2

5a 16a 3
Simplify
7.
9.
18ab 2
9b

2 3
25 x y 10 xy
14 pq 4 21 p 3q

7 3
15w z 35w3 z 8
Polynomials in Numerator and Denominator

Rules are the same as before…
1. Make sure everything is factored completely
2. Cancel common factors
3. Simplify and define x values for which the
expression is undefined.
Examples:
Simplify and define x values for which it
is undefined
10.
y 1
y2  5y  6
 2
5 y  15 y  4 y  5
11.
a  2 a 2  a  12

a3
a2  9
Examples:
12.
2d  6
d 3

d 2  d  2 d 2  3d  2
13.
k 3
1 k 2
 2
k  1 k  4k  3
Simplifying complex fractions

A complex fraction is a rational expression whose
numerator and/or denominator contains a rational
expression
a
5
3b
3
t
t 3
3t
t
6
t
2
To simplify complex fractions

Same rules as before
–
–
–
–
Rewrite as multiplication with the reciprocal
Factor and cancel what you can
Simplify everything
Multiply to finish
Examples:
14.
x  3
2
x 2  16
x3
x4
15.
x2
9x2  4 y2
x3
2 y  3x
Lesson 8-2: Adding and
Subtracting Rational Expressions
Adding and Subtracting Rational
Expressions

Finding Least Common Multiples and Least
Common Denominators!
–
Use prime factorization
–
Example: Find the LCM of 6 and 4



6 = 2·3
4 = 22
LCM= 22·3 = 12
Find the LCM
1. 18r2s5, 24r3st2, and 15s3t
2. 15a2bc3, 16b5c2, 20a3c6
3. a2 – 6a + 9 and a2 + a -12
4. 2k3 – 5k2 – 12k and k3 – 8k2 +16k
Add and Subtract Rational
Expressions

Same as fractions…
–
To add two fractions we find the LCD, the same
things is going to happen with rational
expressions
Examples:
5.
Simplify
8a
1

9b 7 ab 2
6.
7.
1
1

2
8m n mn2
7x
y

2
15 y 18 xy
8. w  12  w  4
4 w  16
2w  8
9.
x6
x6

6 x  18 2 x  6
10.
x 1
x 1

3 x 2  8 x  5 12 x  20
11.
1 1

y x
1 1

y x
12.
1 1

x y
1
1
x
4
x
4
x
x
2
13.
x
3
x
x
6
x
14.
Lesson 8-3: Graphing Rational
Functions
Definitions



Continuity: graph may not be able to be
traced without picking up pencil
Asymptote: a like that the graph of the
function approaches, but never touches (this
line is graphed as a dotted line)
Point discontinuity: a hole in the graph
Vertical Asymptote
How to find a Vertical Asymptote:
x = the value that makes the rational expression undefined
*Set the denominator of the rational expression equal to zero and solve.
Point Discontinuity
How to find point discontinuity:
* Factor completely
* Set any factor that cancels equal to zero and solve. Those are the
x values that are points of discontinuity
Graphing Rational Functions
f(x) =
3
x 1
Graphing Rational Functions
f(x) =
2
x
Graphing Rational Functions
f(x) =
2x 1
x3
Graphing Rational Functions
f(x)=
2x
( x  4) 2
Graphing Rational Functions
f(x) = x 2  9
x3
Graphing Rational Functions
f(x) = x 2  6 x  5
x 1
Lesson 8-4: Direct, Joint, and
Inverse Variation
Direct Variation
y varies directly as x if there is a nonzero
constant, k, such that y = kx
*k is called the constant of variation
1.
2.
Plug in the two values you have and solve for
the missing variable
Plug in that variable and the other given value to
solve for the requested answer
Example

If y varies directly as x and y = 12 when
x = -3, find y when x = 16.
Joint Variation

y varies jointly as x and z if there is a
nonzero constant, k, such that y = kxz
* Follow the same directions as before
Example

Suppose y varies jointly as x and z. Find y when x =
8 and z = 3, if y = 16 when z = 2 and x = 5.
Inverse Variation

y varies inversely as x if there is a nonzero
constant, k, such that xy = k or y= k
x
Example

If y varies inversely as x and y = 18 when x = -3,
find y when x = -11
Lesson 8-6: Solving Rational
Equations and Inequalities
Let’s review some old skills

How do you find the LCM of two monomials
–
8x2y3 and 18x5
* Why do we find LCM’s with rational expressions?
Old Skills Cont…

What is it called when two fractions are equal
to each other?

What process do we use to solve a problem
like this?
To solve a rational equation:
1. Make sure the problem is written as a
proportion
2. Cross Multiply
3. Solve for x
4. Check our answer
Examples

Solve
1 x 1
 
2 5 9
2
3
  10
x3 2
Let’s put those old skills to new
use…
Solve
.
Solve
.
Check your solution.
2x
x 2  x  10
3
 2

x  5 x  8 x  15 x  3
Examples

Solve
7n
5
3n


3n  3 4n  4 2n  2
Lesson 8-6: Day #2
Rational Inequalities
Solving Rational Inequalities
Step 1: State any excluded values (where the
denominator of any fraction could equal zero)
Step 2: Solve the related equation
Step 3: Divide a numberline into intervals using
answers from steps 1 and 2 to create intervals
Step 4: Test a value in each interval to determine which
values satisfy the inequality
Examples

Solve
x
1
x 1


3 x2
4
Examples

Solve
5
6
2


x 5x 3
Examples

Solve
x2
1
x4


x2 x2 x2
Examples

Solve
x
1
3


x  2 x 1 2