Algebra 2 Interactive Chalkboard
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Lesson 9-1 Multiplying and Dividing
Rational Expressions
Lesson 9-2 Adding and Subtracting Rational Expressions
Lesson 9-3 Graphing Rational Functions
Lesson 9-4 Direct, Joint, and Inverse Variation
Lesson 9-5 Classes of Functions
Lesson 9-6 Solving Rational Equations and Inequalities
Example 1 Simplify a Rational Expression
Example 2 Use the Process of Elimination
Example 3 Simplify by Factoring Out –1
Example 4 Multiply Rational Expressions
Example 5 Divide Rational Expressions
Example 6 Polynomials in the Numerator
and Denominator
Example 7 Simplify a Complex Fraction
Simplify
Look for common factors.
1
Factor.
1
Answer:
Simplify.
Under what conditions is this expression undefined?
A rational expression is undefined if the denominator
equals zero. To find out when this expression is
undefined, completely factor the denominator.
Answer: The values that would make the denominator
equal to 0 are –7, 3, and –3. So the expression is
undefined at y = –7, y = 3, and y = –3. These values are
called excluded values.
a. Simplify
Answer:
b. Under what conditions is this expression
undefined?
Answer: undefined for x = –5, x = 4, x = –4
Multiple-Choice Test Item
For what values of p is
A 5
B –3, 5
undefined?
C 3, –5
D 5, 1, –3
Read the Test Item
You want to determine which values of p make the
denominator equal to 0.
Solve the Test Item
Look at the possible answers. Notice that the p term
and the constant term are both negative, so there will
be one positive solution and one negative solution.
Therefore, you can eliminate choices A and D. Factor
the denominator.
Factor the denominator.
or
Zero Product Property
Solve each equation.
Answer: B
Multiple-Choice Test Item
For what values of p is
A –5, –3, –2
Answer: D
B –5
undefined?
C 5
D –5, –3
Simplify
Factor the numerator
and the denominator.
1
a
or
1
Answer:
1
or –a
Simplify.
Simplify
Answer: –x
Simplify
1
1
1
1
1
1
1
1
1
1
1
1
1
1
Factor.
Simplify.
Answer:
Simplify.
Simplify
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
Factor.
Answer:
Simplify.
Simplify each expression.
a.
Answer:
b.
Answer:
Simplify
Multiply by the
reciprocal of divisor.
1
1
1
1
1
1
1
1
1
1
1
1
Factor.
Simplify.
Answer:
Simplify.
1
1
Simplify
Answer:
Simplify
Multiply by
the reciprocal
of the divisor.
Answer:
1
–1
1
1
1
1
Simplify.
Simplify
Multiply
by the
reciprocal
of the
divisor.
1
1
Factor.
1
Answer:
1
Simplify.
Simplify each expression.
a.
Answer: 1
b.
Answer:
Simplify
Express as a
division expression.
Multiply by the
reciprocal of divisor.
1
1
1
–1
1
1
Factor.
Answer:
Simplify.
Simplify
Answer:
Example 1 LCM of Monomials
Example 2 LCM of Polynomials
Example 3 Monomial Denominators
Example 4 Polynomial Denominators
Example 5 Simplify Complex Fractions
Example 6 Use a Complex Fraction to Solve a Problem
Find the LCM of 15a2bc3, 16b5c2, and 20a3c6.
Factor the first
monomial.
Factor the second
monomial.
Factor the third
monomial.
Use each factor the
greatest number of
times it appears as a
factor and simplify.
Answer:
Find the LCM of 6x2zy3, 9x3y2z2, and 4x2z.
Answer: 36x3y3z2
Find the LCM of x3 – x2 – 2x and x2 – 4x + 4.
Factor the first
polynomial.
Factor the second
polynomial.
Answer:
Use each factor the
greatest number of
times it appears as
a factor.
Find the LCM of x3 + 2x2 – 3x and x2 + 6x + 9.
Answer:
Simplify
The LCD is 42a2b2.
Find equivalent
fractions that have
this denominator.
Simplify each
numerator and
denominator.
Answer:
Add the
numerators.
Simplify
Answer:
Simplify
Factor the
denominators.
The LCD is
6(x – 5).
Subtract the
numerators.
Distributive
Property
Combine like
terms.
1
Simplify.
1
Answer:
Simplify.
Simplify
Answer:
Simplify
The LCD of the
numerator is ab.
The LCD of the
denominator is b.
Simplify the numerator
and denominator.
Write as a division
expression.
1
Multiply by the
reciprocal of the divisor.
1
Answer:
Simplify.
Simplify
Answer:
Coordinate Geometry Find the slope of the line that
passes through
and
Definition of slope
The LCD of the numerator
is 3k. The LCD of the
denominator is 2k.
Write as a division
expression.
Simplify.
Answer: The slope is
Coordinate Geometry Find the slope of the line that
passes through
Answer:
and
Example 1 Vertical Asymptotes and Point Discontinuity
Example 2 Graph with a Vertical Asymptote
Example 3 Graph with Point Discontinuity
Example 4 Use Graphs of Rational Functions
Determine the equations of any vertical asymptotes
and the values of x for any holes in the graph of
First factor the numerator and denominator of the
rational expression.
Answer: The function is undefined for x = –2 and –3.
1
x = –3 is a vertical
Since
1
asymptote and x = –2 is a hole in the graph.
Determine the equations of any vertical asymptotes
and the values of x for any holes in the graph of
Answer: vertical asymptote: x = –5; hole: x = –3
Graph
Answer:
The function is undefined for
x = –1. Since
is in its
simplest form, x = –1 is a
vertical asymptote. Draw the
vertical asymptote.
Make a table of values.
x
f (x)
–4
1.33
–3
–2
0
1
1.5
2
0
0.5
2
3
0.67
0.75
Plot the points and
draw the graph.
Answer:
As |x| increases, it appears
that the y values of the
function get closer and
closer to 1. The line with
the equation f(x) = 1 is a
horizontal asymptote of
the function.
Answer:
Graph
Answer:
Graph
Notice that
the graph of
with a hole at
or
is the graph of
Therefore,
Answer:
Graph
Answer:
Transportation A train travels at one velocity V1 for
a given amount of time t1 and then another velocity
V2 for a different amount of time t2. The average
velocity is given by
Let t1 be the independent variable and let V be the
dependent variable. Draw the graph if V1 = 50 miles
per hour, V2 = 30 miles per hour, and t2 = 1 hour.
The function is
The vertical asymptote is
Graph the vertical
asymptote and the function.
Notice that the horizontal
asymptote is
Answer:
What is the V-intercept of the graph?
Answer: The V-intercept
is 30.
What values of t1 and V are meaningful in the
context of the problem?
Answer: In the problem context, time and velocity are
positive values. Therefore, positive values
of t1 and V values between 30 and 60 are
meaningful.
Transportation A train travels at one velocity V1 for
a given amount of time t1 and then another velocity
V2 for a different amount of time t2. The average
velocity is given by
a. Let t1 be the independent variable and let V be the
dependent variable. Draw the graph if V1 = 60 miles
per hour, V2 = 30 miles per hour, and t2 = 1 hour.
Answer:
b. What is the V-intercept of the graph?
Answer: The V-intercept is 30.
c. What values of t1 and V are
meaningful in the context of
the problem?
Answer: t1 is positive and V is
between 30 and 60.
Example 1 Direct Variation
Example 2 Joint Variation
Example 3 Inverse Variation
Example 4 Use Inverse Variation
If y varies directly as x and y = –15 when x = 5,
find y when x = 3.
Use a proportion that relates the values.
Direct proportion
y1 = –15; x1 = 5, and x2 = 3
Cross multiply.
Simplify.
Divide each side by 5.
Answer: When x = 3, the value of y is –9.
If y varies directly as x and y = 12 when x = –3,
find y when x = 7.
Answer: –28
Suppose y varies jointly as x and z. Find y when
x = 10 and z = 5 if y = 12 when x = 3 and z = 8.
Use a proportion that relates the values.
Joint variation
y1 = 12, x1 = 3, z1 = 8,
x2 = 10, and z2 = 5
Cross multiply.
Simplify.
Divide each side by 24.
Answer: When x = 10 and z = 5, y = 25.
Suppose y varies jointly as x and z. Find y when
x = 3 and z = 2 if y = 11 when x = 5 and z = 22.
Answer:
If a varies inversely as b and a = –6 when b = 2,
find a when b = –7.
Use a proportion that relates the values.
Inverse variation
a1 = –6; b1 = 2, and b2 = –7
Cross multiply.
Simplify.
Divide each side by –7.
Answer: When b = –7, a is
If a varies inversely as b and a = 3 when b = 8,
find a when b = 6.
Answer: 4
Space The apparent length of an object is inversely
proportional to one’s distance from the object. Earth
is about 93 million miles from the Sun and Venus is
about 67 million miles away. How much larger would
the diameter of the Sun appear on Venus than on
Earth?
Explore
You know that the apparent diameter of the
Sun varies inversely with the distance from
the Sun. You also know the distance from
the Sun to Venus and from the Sun to Earth.
You want to determine how much larger the
diameter of the Sun appears on Venus than
on Earth.
Plan
Let the apparent diameter of the Sun from
Earth equal 1 unit and the apparent diameter
of the Sun from Venus equal v. Then use a
proportion that relates the values.
Solve
Substitution
Cross multiply.
Divide each
side by 67
million miles.
Simplify.
Answer:
From Venus, the diameter of the Sun will
appear about 1.39 times as large as it
appears from Earth.
Examine
Since the distance between the Sun and
Earth is between 1 and 2 times the distance
between the Sun and Venus, the answer
seems reasonable.
Space Mars is
about 142.5 million
miles from the Sun
and Earth is about
93 million miles away. How much smaller would the
diameter of the Sun appear on Mars than on Earth?
Answer: about 0.65 of the size
Example 1 Identify a Function Given the Graph
Example 2 Match Equation with Graph
Example 3 Identify a Function Given its Equation
Identify the type of function
represented by the graph.
Answer: The graph is a V
shape. So, it is
an absolute
value function.
Identify the type of function
represented by the graph.
Answer: The graph is a
parabola, so it is a
quadratic function.
Identify the type of function represented by
each graph.
a.
b.
Answer: greatest integer
function
Answer: inverse variation
Shipping Charges A chart gives the shipping rates for
an Internet company. They charge $3.50 to ship less
than 1 pound, $3.95 for 1 pound and over up to 2
pounds, and $5.20 for 2 pounds and over up to 3
pounds. Which graph depicts these rates?
a.
b.
c.
a.
b.
c.
The shipping rate is constant for x values from 0 to 1.
Then it jumps at x = 1 and remains constant until x = 2.
The graph jumps again at x = 2 and remains constant
until x = 3.
Answer: The graph of this function looks like steps, so
this is c, the step or greatest integer function.
A ball is thrown into the air. The path of the ball is
represented by the equation
Which graph best represents this situation?
a.
b.
Answer: b; quadratic function
c.
Identify the type of function represented by
Then graph the equation.
Since the equation has no x-intercept, it is the constant
function. Determine some points on the graph and graph it.
Answer:
Identify the type of function represented by
Then graph the equation.
Since the equation includes an expression with a square
root, it is a square root function. Plot some points and use
what you know about square root graphs to graph it.
Answer:
Identify the type of function represented by each
equation. Then graph the equation.
a.
b.
Answer:
Answer:
rational
function
direct
variation
Example 1 Solve a Rational Equation
Example 2 Elimination of a Possible Solution
Example 3 Work Problem
Example 4 Rate Problem
Example 5 Solve a Rational Inequality
Solve
Check your solution.
The LCD for the three denominators is
Original equation
Multiply each side
by 24(3 – x).
6
1
1
1
1
1
Simplify.
Simplify.
Add.
Check
Original equation
Simplify.
Simplify.
The solution
is correct.
Answer: The solution is –45.
Solve
Answer:
Solve
Check your solution.
The LCD is
Original
equation
p–1
1
1
1
Multiply by the
LCD, (p2 – 1).
Distributive
Property
Simplify.
Simplify.
Add
(2p2 – 2p + 1)
to each side.
Divide each
side by 3.
Factor.
or
Zero Product
Property
Solve each
equation.
Check
Original equation
Simplify.
Simplify.
Original equation
Simplify.
Since p = –1 results in a zero in the denominator,
eliminate –1.
Answer: The solution is p = 2.
Solve
Answer:
Mowing Lawns Tim and Ashley mow
lawns together. Tim working alone
could complete the job in 4.5 hours,
and Ashley could complete it alone in
3.7 hours. How long does it take to
complete the job when they work
together?
In 1 hour, Tim could complete
In 1 hour, Ashley could complete
of the job.
of the job.
In t hours, Tim could complete
In t hours, Ashley could complete
Part completed
by Tim
plus
part completed
by Ashley
or
of the job.
or
of the job.
equals
entire job.
1
Solve the equation.
Original equation
Multiply each
side by 16.65.
Distributive
Property
Simplify.
Simplify.
Divide each
side by 8.2.
Answer: It would take them about 2 hours
working together.
Cleaning Libby and Nate clean together. Nate working
alone could complete the job in 3 hours, and Libby
could complete it alone in 5 hours. How long does it
take to complete the job when they work together?
Answer: about 2 hours
Swimming Janine swims for 5 hours in a stream that
has a current of 1 mile per hour. She leaves her dock
and swims upstream for 2 miles and then back to her
dock. What is her swimming speed in still water?
Words
The formula that relates distance, time,
and rate is
Variables Let r be her speed in still water. Then her
speed with the current is r + 1 and her speed
against the current is r – 1.
Time going with
time going against
total
the current
plus
the current
equals time.
5
Equation
Solve the equation.
Original
equation
Multiply each
side by r2 – 1.
r–1
r+1
1
Distributive
Property
1
Simplify.
Simplify.
Subtract 4r
from each side.
Use the Quadratic Formula to solve for r.
Quadratic Formula
x = r, a = 5, b = –4,
and c = –5
Simplify.
Simplify.
Use a calculator.
Answer: Since the speed must be positive, the answer
is about 1.5 miles per hour.
Swimming Lynne swims for 1 hour in a stream that
has a current of 2 miles per hour. She leaves her dock
and swims upstream for 3 miles and then back to her
dock. What is her swimming speed in still water?
Answer: about 6.6 mph
Solve
Step 1
Values that make the denominator equal to 0
are excluded from the denominator. For this
inequality the excluded value is 0.
Step 2
Solve the related equation.
Related equation
Multiply each
side by 9s.
Simplify.
Add.
Divide each
side by 6.
Step 3
Draw vertical lines at the excluded value and
at the solution to separate the number line into
regions.
Now test a sample value in each region to
determine if the values in the region satisfy the
inequality.
Test
is a solution.
Test
is not a solution.
Test
is a solution.
Answer: The solution
Solve
Answer:
Explore online information about the
information introduced in this chapter.
Click on the Connect button to launch your browser
and go to the Algebra 2 Web site. At this site, you
will find extra examples for each lesson in the
Student Edition of your textbook. When you finish
exploring, exit the browser program to return to this
presentation. If you experience difficulty connecting
to the Web site, manually launch your Web browser
and go to www.algebra2.com/extra_examples.
Click the mouse button or press the
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