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LESSON
2-2
Creating and Solving Equations
Reteach
To write an equation for a real-world problem, look for words that will
help you solve the problem and translate them into parts of the equation.
For example, the sum of two times a number and 8 is 18 as shown
below.
2n is the same as
“a number times 2.”
2n  8  18
 18 is the same
as “is 18.”
 8 is the same as
“the sum of...and 8.”
Example
The Lions and Tigers played a football game. The Lions scored three
more than three times the number of points the Tigers scored. The total
number of points scored by both teams is 31. How many points were
scored by each team?
points the Tigers scored
t
Lions scored three more than three times
3t  3
total number of points scored by both teams is 31 t  (3t  3)  31
So the equation is t  (3t  3)  31. You can simplify and use Properties
of Equality to solve.
Write an equation and solve each problem.
1. Mako runs m miles each day. De’Anthony runs twice as many miles as
Mako. Altogether, they run 18 miles each day. How many miles does Mako
run each day?
________________________________________________________________________________________
2. Bella’s cell phone costs $18 per month plus $0.15 for every minute
she uses the phone. Last month her bill was $29.25. For how many
minutes did she use her phone last month?
________________________________________________________________________________________
3. Eric and Charlotte collected donations for the food bank. Charlotte
collected 3 times the amount of Eric’s donations minus $20. Eric and
Charlotte combined their donations and spent half of the money on
canned food. The rest of the money, $50, was donated to the food
bank directly. How much money did each student collect?
Solving for a Variable
LESSON
2-3
Reteach
Solving for a variable in a formula can make the formula easier to use.
You can solve a formula, or literal equation, for any one of the variables.
To solve a literal equation or formula, underline the variable you are
solving for, and then undo what has been done to that variable. Use
inverse operations in the same way you do when solving an equation
or inequality.
The formula for finding the circumference of a circle when you know
diameter is C   d. If you know the circumference, you could find the
diameter by using a formula for d.
Examples
Solve C   d for d.
C d



C
C
 d or d 


9
Solve F    C  32 for C.
5
Since d is multiplied by  , use division to undo this.
Divide both sides by  .
What has been done to C? First undo adding 32.
9
F  32    C  32  32
5
Subtract 32 from both sides.
5
 5  9 
 9   F  32   9   5  C
 
  
5
Multiply both sides by   , the reciprocal of
9
5
 9   F  32  C
 
Simplify.
9
 5 .
 
Solve each formula for the indicated variable.
1. A =
1
bh, for b
2
_______________________
4. P  a  b  c, for c
_______________________
2. A  lw, for l
3. R 
________________________
5. I =
1
prt, for t
2
2s  6t  5
, for s
2
________________________
6. G 
________________________
HJ
, for H
K
________________________
Solve each equation for the indicated variable.
7. m  n  p  q  360, for n
_______________________
8. t  rs  s, for s
________________________
9.
a c
 , for a
b d
________________________
LESSON
2-4
Creating and Solving Inequalities
Reteach
An inequality, such as 2x  8, has infinitely many numbers in its solution.
1
For example, 2, 0,  , and 350 are all values of x that make this
2
inequality true.
Solve an inequality by UNDOING what has been done to x, using the
same steps as you would use for an equality, with the exception given
below.
Solving inequalities has one special rule different from solving
equations.
Multiplying or dividing an inequality by a NEGATIVE number
REVERSES the inequality sign.
Try some positive and negative numbers in 3x  15.
Example

3x15
Note that 6 and 8 make this true, but it is not true for 6 or 8.
3 x 15

3 3
Dividing by  3 REVERSES the inequality sign.
x  5
This is still true for 6 and 8, and not true for 6 or 8.
Solve each inequality. Show your work.
1. 3e  10 4
2.
c
 8  11
2
_______________________________________
________________________________________
_______________________________________
________________________________________
_______________________________________
________________________________________
Solve each inequality.
3. 15  3  4s
_______________________________________
5. 8c  4  4(c  3)
_______________________________________
7. 8  4a 12  2a  10
4.
3
j14
4
________________________________________
6. 5(x 1)  3x  10  8x
________________________________________
8. 0.6t  3  15
Graphing Functions
LESSON
3-4
Reteach
To check whether an equation is a function, isolate the y variable on one side of the equation
and simplify.
4x  2y  8
2y  4x  8
Subtract 4x from both sides.
y  2x  4
Divide each term by 2.
Write the equation as a function: f(x)  2x  4.
If the domain  {2, 3, 4}, substitute each value into the function and simplify to find the values
of the range.
f(x)  2(2)  4
f(x)  2(3)  4
f(x)  2(4)  4
f(x)  4  4
f(x)  6  4
f(x)  8  4
f(x)  8
f(x)  10
f(x)  12
Use a table to find the ordered pairs. Graph the ordered pairs.
x
y
2
8
3
10
4
12
This is a function because
each value of x has only
one value of y. Any
vertical line would pass
through only one point.
Is this equation a function? Use the domain values to find the values
for the range. Graph the points and tell whether it is a function or not.
1. f(x)  3x  2 for the domain {1, 2, 3, 4}
x
y
function? _________________
Use the vertical line test to tell if each graph shows a function.
2.
3.
function? ____________
4.
function? ____________
function? _____________
LESSON
5-2
Using Intercepts
Reteach
Doug has $12 to spend on popcorn and peanuts. The peanuts are $4
and popcorn is $2. If he spends all his money, the equation 4x  2y 12
shows the amount of peanuts, x, and popcorn, y, he can buy. Here is
the graph of 4x  2y 12.
To find the y-intercept, substitute x  0. Solve for y.
He can buy 6 boxes of popcorn (y) if he buys 0 peanuts.
To find the x-intercept, substitute y  0. Solve for x.
He can buy 3 bags of peanuts (x) if he buys 0 popcorn.
4(0)  2y  12
y6
4x  2(0)  12
x3
Find each x- and y-intercept.
1.
2.
_______________________
3.
________________________
________________________
Find each intercept. Use these two points to graph each line.
4. 3x  9y  9
5. 4x  6y  12
6. 2x  y  4
LESSON
5-3
Interpreting Rate of Change and Slope
Reteach
Find the rate of change, or slope, for the graph of a straight line by finding
change in y
.
change in x
Step 1: First choose any two points on the line.
Step 2: Begin at one of the points.
Step 3: Count vertically until you are even with the
second point.
This is the rise. If you go down the rise will be
negative. If you go up the rise will be positive.
Step 4: Count over until you are at the second point.
This is the run. If you go left the run will be
negative. If you go right the run will be positive.
Step 5: Divide to find the slope.
rise
6
slope 
   3
run
2
The slope of a horizontal line is zero. A horizontal line has no steepness at all.
The slope of a vertical line is undefined. A vertical line is infinitely steep.
Find the slope of each line.
1.
2.
_______________________
4.
3.
________________________
5.
_______________________
________________________
6.
________________________
________________________
LESSON
5-4
Direct Variation
Reteach
When a situation can be described by an equation of the form y  kx, where k is a constant,
we say that y varies directly with x. The equation y  kx is an equation of direct variation.
When k is greater than zero (k is a positive number) the equation y  kx implies that as x
increases, y also increases. k is called the variation constant and can be found if one pair of
(x, y) values are known.
Example
In the graph, y  4 when x  1. Substituting these values
in the equation y  kx gives 4  k x 1 so k  4. The
equation of direct variation for this graph is y  4x. This
equation can be used to find values for y for known values
of x. Use the equation to complete the table.
y  4x
8  42
12  4  3
x
2
3
5
10
y
8
12
?
?

?  4  5  20
?  4  10  40
Determine k, the constant of variation for each (x, y) pair, and write
the equation of direct variation.
1. (3, 6)
2. (6, 24)
_______________________
3. (4, 28)
________________________
________________________
Determine k, the constant of variation that relates x and y for each
table, and write the equation of direct variation.
4.
x
1
3
5
y
6
18
30
5.
_______________________
x
2
4
8
y
6
12
24
6.
________________________
x
1
4
9
y
5
20
45
________________________
For each table, y varies directly with x. Determine the missing value.
7.
x
2
4
6
y
8
16
?
_______________________
8.
x
3
7
11
y
12
28
?
________________________
9.
x
5
7
9
y
10
14
?
________________________

LESSON
6-1
Slope-Intercept Form
Reteach
An equation is in slope-intercept form if it is written as:
m is the slope.
b is the y-intercept.
y  mx  b.
You can use the slope and y-intercept to graph a line.
Write 2x  6y  12 in slope-intercept form.
Then graph the line.
Step 3: Graph the line.
Step 1: Solve for y.
 Then count 1 down (because the rise is
negative) and 3 right (because the run is
positive) and plot another point.
2x  6y  12
2x
Subtract 2x from
both sides.
2x
 Plot (0, 2).
 Draw a line connecting the points.
6y  2 x  12
6y 2 x  12

6
6
1
y  x2
3
Plot (0, 2).
Divide both sides by 6.
Count 1 down.
Simplify.
Step 2: Find the slope and y-intercept.
slope: m  
1 1

3 3
Count 3 right.
y-intercept: b  2
Find each slope and y-intercept. Then graph each equation.
1. y 
1
x 3
2
2. 3x  y 2
3. 2x  y  3
slope:______________
slope: ____________
slope: __________________
y-intercept: _________
y-intercept: ________
y-intercept: _____________
LESSON
6-2
Point-Slope Form
Reteach
An equation is in slope-intercept form if it is written as:
m is the slope.
b is the y-intercept.
y  mx  b.
A line has a slope of 4 and a y-intercept of 3. Write the
equation in slope-intercept form.
y  mx  b
Substitute the given values for m and b.
y  4x  3
Find values for m and b from a table or graph.
change in y -values
between any two points on the line.
change in x -values
At the y-intercept, x  0 and y  b.
Slope  m 
Example: The linear function in this graph is f(x)  2x  5.
Example: The linear function in this table is f(x)  3x  6.
x
0
1
2
3
f(x)
6
3
0
3
Write an equation for each linear function f(x) using the given
information.
1
1. slope  ,y-intercept  3
4
__________________________________
2. slope  5, y-intercept  0
__________________________________
3. slope  7, y-intercept  2
__________________________________
4.
__________________________________
5.
x
1
0
1
2
f(x)
4
2
0
2
__________________________________
LESSON
6-4
Transforming Linear Functions
Reteach
For a linear function f(x)  mx  b, changing the value of b moves the
graph up or down.
Description
Equation
Parent function
Translate up 2
Translate down 4
f(x)  x
g(x)  x  2
h(x)  x  4
y-intercept or b
0
2
4
Changing the absolute value of the slope m
makes the line more or less steep.
If m is positive, the line goes up
from left to right.
If m is negative, the line goes down
from left to right.
Predict the change in the graph from f(x) to g(x).
Then graph both lines to check your prediction.
1. f(x)  x; g(x)  x  5
_______________________________________
2. f(x)  3x 1; g(x)  3x  1
________________________________________
LESSON
7-1
Parallel and Perpendicular Lines
Reteach
The chart shows how parallel and perpendicular lines differ.
Lines
Slopes
y-intercept
parallel
same
different
perpendicular
negative reciprocals of
each other
can be the same or
different
The examples show equations for parallel and perpendicular lines in
slope-intercept and point-slope form.
Examples – parallel
slope-intercept
point-slope
y  3x  6
y  5  3(x  2)
y  3x  4
y  6  3(x  1)
Examples – perpendicular
slope-intercept
point-slope
y  4x  2
y  3  2(x  4)
y 
1
x3
4
1
y  2   ( x  3)
2
Find each slope and y-intercept.
1. Are the lines whose equations are y  2x  4 and y  
1
x  3 parallel or perpendicular?
2
Explain.
________________________________________________________________________________________
________________________________________________________________________________________
2. Are the lines whose equations are y  4  7(x  3) and y  2  7(x  5) parallel or
perpendicular? Explain.
________________________________________________________________________________________
________________________________________________________________________________________
LESSON
7-2
Using Functions to Solve One-Variable Equations
Reteach
Having several methods to solve one-variable equations can help.
Depending on the problem, often one method is easier than the others.
Algebraic Solution is often fastest and can handle decimals and fractions most easily.
f(x)  19  8x and g(x)  17  10x
Solve using algebra.
Set f(x)  g(x) and solve for x.
19  8x  17  10x
36  18x
2x
Graphic Solution is often easier with graphing calculators and can give a visual
understanding of a problem. This method also works
well with decimals.
Solve using graphs.
1
f ( x )   x  2 and g( x )  3x  3
2
For f(x)
slope  
1
and y-intercept  2
2
For g(x)
slope  3 and y-intercept  3
Plot f(x) and g(x) and find the
intersection.
Tables work well with graphing calculators and
are usually easiest for integers.
Solve with a table.
f(x)  3  2x and g(x)  12  3x
x
f (x)  3  2x
g(x)  12  3x
0
3
12
1
1
9
2
1
6
3
3
3
4
5
0
1. In the algebraic example above, if x  2, what does f(x) equal? _____ What does g(x)
equal? ______
2. In the graphic example above, at what point do the lines for the equations intersect? ______
3. What does 3, 3 in the table indicate? ____________________________________________________
LESSON
7-3
Linear Inequalities in Two Variables
Reteach
To graph a linear inequality:
Step 1: Solve the inequality for y.
Step 2: Graph the boundary line. If  or  use a solid line. If  or  use a dashed line.
Step 3: Determine which side to shade.
Graph the solutions of 2x  y  4.
Step 1: Solve for y.
2x  y  4
2 x
2 x
y  2 x  4
Step 2: Graph the boundary line.
Use a solid line for .
Step 3: Determine which side to shade.
Substitute (0, 0) into 2x  y  4.
2x  y  4
?
2(0)  0  4
?
0  4. The statement is true. Shade the side that contains the point (0, 0).
Summary for Graphing Linear Inequalities in Two Variables
• The boundary line is solid for  and  and dashed for  and .
When the inequality is written with y alone on the left, then:
• shade below and to the left of the line for  and ,
• shade above and to the right of the line for  and .
Graph the solutions of each linear inequality.
In the first one, the boundary line is already drawn.
1. y  2x  9
2. y  x  3
3. x  y  2  0
LESSON
8-1
Scatter Plots and Trend Lines
Reteach
Correlation is one way to describe the relationship between two sets of data.
Positive Correlation
Data: As one set increases, the other set increases.
Graph: The graph goes up from left to right.
Negative Correlation
Data: As one set increases, the other set decreases.
Graph: The graph goes down from left to right.
No Correlation
Data: There is no relationship between the sets.
Graph: The graph has no pattern.
Example
Correlation
Correlation Coefficient
(estimated )
1st graph above
strong positive
1
2nd graph above
strong negative
1
3rd graph above
no correlation
0
4th graph beside
weak positive
0.5
5th graph beside
weak negative
0.5
Estimate the correlation coefficient for each scatter plot as 1, 0.5, 0, 0.5, or 1.
1.
2.
_______________________
3.
________________________
________________________
LESSON
9-1
Solving Linear Systems by Graphing
Reteach
The solution to a system of linear equations can be found by graphing. Write both equations
so that they are in slope-intercept form and draw their lines on a coordinate graph. The point
of intersection is the solution. If the lines have the same slope but different y-intercepts they
won’t intersect and there is no solution. If the graphs are the same line then there are an
infinite number of solutions.
Example
Solve the system.
y  2x  2
y  2x  6
4 x  2y  4
6 x  3y  18

Rewrite each equation in slope-intercept form.
Graph the lines and look for the point of intersection.
The lines intersect at (2, 2).
The solution to the system is (2, 2).
Solve each linear system of equations by graphing.
 x  y  1
1. 
4 x  2y  16
3 x  y  10
2. 
2 x  4 y  0
____________
2 x  y  4
3. 
x  y  2
____________
6 x  3 y  12
4. 
2 x  2y  10
____________
____________

LESSON
9-2
Solving Linear Systems by Substitution
Reteach
You can use substitution to solve a system of equations if one of the
equations is already solved for a variable.
y  x  2
Solve 
3 x  y  10
Step 1: Choose the equation to use as the
substitute.
Use the first equation y  x  2
because it is already solved for a
variable.
Step 2: Solve by substitution.
x2
3x y 10
3x (x 2) 10
4x  2  10
2 2

4x
8
4x 8

4
4
x2
Step 3: Now substitute x  2 back into
one of the original equations to find the
value of y.
y x 2
y22
y 4
The solution is (2, 4).
Check:
Substitute (2, 4) into both equations.
y  x2
3xy10
Substitute x 2 for y.
Combine like terms.
4  22
?
4  4
?
3(2)4  10
?
64  10
?
10  10 
?
You may need to solve one of the equations for a variable before solving with substitution.
Solve each system by substitution.
y  x  2
1. 
y  2x  5
_______________________________________
 x  y  3
3. 
2 x  y  12
_______________________________________
 x  y  10
2. 
 x  2y  3
________________________________________
y  x  8
4. 
5 x  2y  9
________________________________________
LESSON
9-3
Solving Linear Systems by Adding or Subtracting
Reteach
To use the elimination method to solve a system of linear equations:
1. Add or subtract the equations to eliminate one variable.
2. Solve the resulting equation for the other variable.
3. Substitute the value for the known variable into one of the original equations.
4. Solve for the other variable.
5. Check the values in both equations.
3 x  2y  7
The y-terms have

Use the elimination
opposite coefficients,
5 x  2y  1
method when the
so add.
coefficients of one of
3 x  2y  7
the variables are the
Add the equations.
 5 x  2y  1
same or opposite.
8x  8
Solve for x.
x1
Substitute x  1 into 3x  2y  7 and
solve for y:
3x  2y  7
3(1)  2y  7
2y  4
y2
The solution to the system is the ordered pair (1, 2).
Check using both equations:
3x  2y  7
5x  2y  1
3(1)  2(2)  7
5(1)  2(2)  1
?
7  7
?
1  1
Solve each system by adding or subtracting.
 2 x  y  5
1. 
 3 x  y  1
__________________________________
 x  y  12
3. 
 2x  y  6
__________________________________
 3 x  2y  10
2. 
 3 x  2y  14
________________________________
 2x  y  1
4. 
 2 x  3 y  5
________________________________
LESSON
9-4
Solving Linear Systems by Multiplying First
Reteach
To solve a system by elimination, you may first need to multiply one of the equations to
make the coefficients match.
2 x  5 y  9

 x  3 y  10
Multiply bottom
equation by 2.
2x  5y  9
2( x  3 y )  2(10)
2x  5y  9
2 x  6 y  20
0  11y  11
Solve for y: 11y  11
11
11

y  1
Substitute 1 for y in x  3y  10.
x  3(1)  10
x  3  10
 3  3
x7
The solution to the system is the ordered pair (7, 1).
You may need to multiply both of the equations to make the coefficients match.
5 x  3 y  2

4 x  2y  10
Multiply the top by 2
and the bottom by 3.
-2(5 x + 3y = 2)
3(4 x + 2y = 10)
10 x  ( 6 y )  4
12 x  6 y  30
2 x  0  26
x  13
The solution to this system is the ordered pair (13, 21).
After you multiply, add or subtract the two equations.
Solve for the variable that is left.
Substitute to find the value of the other variable.
Check in both equations.
Solve each system by multiplying first. Check your answer.
2 x  3 y  5
1. 
 x  2y  1
__________________________________
2 x  5 y  22
3. 
10 x  3 y  22
3 x  y  2
2. 
8 x  2y  4
________________________________
4 x  2y  14
4. 
7 x  3 y  8
LESSON
10-2
Graphing Systems of Linear Inequalities
Reteach
You can graph a system of linear inequalities by combining the graphs of the inequalities.
Graph of y  2x  3
Graph of y  x  6
All solutions are in this
double shaded area.
Graph of the system
y  2 x  3

y   x  6
Two ordered pairs that are
solutions: (3, 4) and (5, 2)
Solve each system of linear inequalities by graphing. Check your
answer by testing an ordered pair from each region of your graph.
y  x  3
1. 
y   x  6
_______________________
y  x
2. 
 y   2x  1
________________________
y  2x  2
3. 
y  2x  3
________________________
LESSON
10-3
Modeling with Linear Systems
Reteach
Mrs. Hathaway bought a total of 12 items made up of some sticky notes
and some pens. The sticky notes cost $4 each and the pens cost $2 each.
She spent a total of $40 on all items. How many pens and how many
sticky notes did she buy?
Organize the
information.
Number of Items
Cost
Sticky Notes
n
4n
Pens
p
2p
Total
12
40
Write two equations. Use the information in each row of the chart.
Number of Items
n
p
12
n  p  12
Cost
4n
2p
40
4n  2p  40
Write each equation in slope-intercept form.
n  p  12
n  p  12
4n  2p  40
4n  2p  40
1
n   p   10
2
Set the equations equal to each other and solve.
1
p  12  - p  10
2
1
12  p  10
2
1
2 p
2
4p
She bought 4 pens.
n  p  12
n  4  12
n8
She bought 8 sticky notes.
Solve.
1. Tia has 25 china figures in her collection. The horse figures cost $2 each,
and the cat figures cost $1 each. She paid $39 for all the figures in the
collection. How many horses and how many cats does she have?
Equations: ___________________________________
Solution: ____________________________________
2. Mr. Wallace has 32 models of antique cars. The Hupmobile models
cost $5 each, and the Duesenberg models cost $18 each. He paid a
total of $264 for all the models. How many Hupmobile models and how
many Duesenberg models does he have?
Equations: ___________________________________
Solution: ____________________________________
LESSON
12-1
Understanding Geometric Sequences
Reteach
It is important to understand the difference between arithmetic and
geometric sequences.
Arithmetic sequences are based on adding a common difference, d.
Geometric sequences are based on multiplying a common ratio, r.
• If the first term of an arithmetic sequence, a1, is 2 and the
common difference is 3, the arithmetic sequence is: 2, 5, 8, 11, …
• If the first term of an arithmetic sequence, a1, is 72 and the common
difference is  3, the arithmetic sequence is: 72, 69, 66, 63, …
• If the first term of a geometric sequence, a1, is 2 and the common
ratio is 3, the geometric sequence is: 2, 6, 18, 54, …
• If the first term of a geometric sequence, a1, is 72 and the
1
8
common ratio is , the geometric sequence is: 72, 24, 8, , …
3
3
Complete each table.
1. An arithmetic sequence has a1  4 and d  3:
an
a1
a2
a3
a4
a5
Value
2. A geometric sequence has a1  4 and r  3:
an
a1
a2
a3
a4
a5
Value
3. An arithmetic sequence has a1  96 and d  4:
an
a1
a2
a3
a4
a5
Value
4. A geometric sequence has a1  96 and r 
an
Value
a1
a2
a3
a4
a5
1
:
4


LESSON
13-2
Modeling Exponential Growth and Decay
Reteach
In the exponential growth and decay formulas, y final amount, a original amount,
r rate of growth or decay, and t time.
Exponential growth: y a (1 r)t
Exponential decay: y  a (1  r)t
The population of a city is increasing at a
rate of 4% each year. In 2000, there were
236,000 people in the city. Write an
exponential growth function to model this
situation. Then find the population in 2009.
The population of a city is decreasing at a
rate of 6% each year. In 2000, there were
35,000 people in the city. Write an
exponential decay function to model this
situation. Then find the population in 2012.
Step 1: Identify the variables.
Step 1: Identify the variables.
a  236,000——r  0.04
Step 2: Substitute for a and r.
a  35,000——r  0.06
Step 2: Substitute for a and r.
y  a (1  r)t
y  a (1  r)t
y  236,000 (1  0.04)t
y  35,000 (1  0.06)t
The exponential growth function is
y  236,000 (1.04)t.
Growth;the growth factor is
greater than 1.
Step 3: Substitute for t.
y  236,000 (1.04)
The exponential decay function is
y  35,000 (0.94)t.
Decay; the growth factor is less than
1 and greater than 0.
Step 3: Substitute for t.
9
 335,902
The 2009 population was about
335,902 people.
y  35,000 (0.94)12
 16,657
The 2009 population was about
16,657 people.
Write an exponential growth function to model each situation. Then
find the value of the function after the given amount of time.
1. Annual sales at a company are $372,000 and increasing
at a rate of 5% per year; 8 years
y  _________(1  ______) ___
2. The population of a town is 4200 and increasing at a rate
of 3% per year; 7 years
y  _________(1  ______) ___
Write an exponential decay function to model the situation. Then
find the value of the function after the given amount of time.
3. Monthly car sales for a certain type of car are $350,000
and are decreasing at a rate of 3% per month; 6 months
_______________________________________
y  _________(1  ______) ___
LESSON
14-1
Understanding Polynomial Expressions
Reteach
Polynomials have special names based on the number of terms.
POLYNOMIALS
No. of
Terms
Name
1
2
Monomial Binomial
3
4 or more
Trinomial
Polynomial
The degree of a monomial is the sum of the exponents in the monomial. The degree of a
polynomial is the degree of the term with the greatest degree.
Examples
Find the degree of 8x 2 y 3 .
Find the degree of 4ab  9a 3 .
8x 2 y 3 The exponents are 2 and 3.
4ab  9a 3
2
3
The degree of
the binomial is 3.
The degree of the monomial
is 2  3  5.
Identify each polynomial. Write the degree of each expression.
1. 7m3n5
_______________________
2. 4x2y3  y4  7
3. x5  x5y
________________________
________________________
You can simplify polynomials by combining like terms.
The following are like terms:
4y and 7y
8x2 and 2x2
7m5 and m5
same variables raised to same power
The following are not like terms:
Examples
Add 3x2  4x  5x2  6x.
3x2  5x2  4x  6x
8x  10x
2
3x2 and 3x
same variable,
different exponent
47 and 7y
one with variable,
one constant
8m and m5
same variable but
different power
Identify and rearrange like terms so they are together.
Combine like terms.
Simplify each expression.
4. 2y2  3y  7y  y2
_______________________
5. 8m4  3m  4m4
________________________
6. 12x5  10x4  8x4
________________________
LESSON
14-2
Adding Polynomial Expressions
Reteach
You can add polynomials by combining like terms.
These are examples of like terms:
4y and 7y
8x2 and 2x2
m5 and 7m5
These are like terms because they
have the same variables and same
exponent.
These are not like terms:
3x2 and 3x
4y and 7
same variable
but different
exponent
one with a
variable, one
is a constant
8m and 8n
different
variables
Add (5y2  7y  2)  (4y2  y  8).
(5y2  7y  2 )  (4y2  y  8 )
Identify like terms.
(5y2  4y2)  ( 7y  y )  ( 2  8 )
Rearrange terms so that like terms are
together.
9y2  8y  10
Combine like terms.
Add (5y2  7y  2)  (4y2  y  8).
(5y2  7y  2 )  (4y2  y  8 )
Identify like terms.
(5y2  4y2)  ( 7y  y )  ( 2  8 )
Rearrange terms so that like terms are
together.
9y2  8y  10
Combine like terms.
Add.
1. (6x2  3x)  (2x2  6x)
____________________________________________________
2. (m2  10m  5)  (8m  2)
____________________________________________________
3. (6x3  5x)  (4x3  x2  2x  9)
____________________________________________________
4. (2y5  6y3  1)  (y5  8y4  2y3  1) ____________________________________________________
Subtracting Polynomial Expressions
LESSON
14-3
Reteach
To subtract polynomials, you must remember to add the opposites.
Find the opposite of (5m3  m  4).
(5m3  m  4)
(5m3  m  4)
Write the opposite of the polynomial.
5m3  m  4
Write the opposite of each term in the polynomial.
Subtract (4x3  x2  7)  (2x3).
(4x3  x2  7)  (2x3)
Rewrite subtraction as addition of the opposite.
(4x3  x2  7)  (2x3)
Identify like terms.
(4x3  2x3)  x2  7
Rearrange terms so that like terms are together.
2x3  x2  7
Combine like terms.
Subtract (6y4  3y2  7)  (2y4  y2  5).
(6y4  3y2  7)  (2y4  y2  5)
Rewrite subtraction as addition of the opposite.
(6y4  3y2  7 )  (2y4  y2  5)
Identify like terms.
(6y4  2y4 )  (3y2  y2)  (7 5)
Rearrange terms so that like terms are together.
4y4  4y2  12
Combine like terms.
Subtract.
1. (9x 3  5x)  (3x)
_______________________________________
2. (6t 4  3)  (2t 4  2)
_______________________________________
3. (2x3  4x  2)  (4x3  6)
_______________________________________
4. (t 3  2t)  (t 2  2t  6)
_______________________________________
5. (4c5  8c2  2c  2)  (c3  2c  5)
LESSON
15-1
Multiplying Polynomial Expressions by Monomials
Reteach
To multiply monomial expressions, multiply the constants, and then multiply
variables with the same base.
Example
Multiply (3a2b) (4ab3).
(3a2b) (4ab3)
(3  4) (a2  a) (b  b3)
Rearrange so that the constants and the variables with the same
bases are together.
12a3b4
Multiply.
To multiply a polynomial expression by a monomial, distribute the monomial to each term
in the polynomial.
Example
Multiply 2x(x2  3x  7).
2x(x2  3x  7)
(2x)x2  (2x)3x  (2x)7
Distribute.
2x  6x  14x
Multiply.
3
2
Multiply.
1. (5x 2 y 3 ) (2xy)
2. (2xyz) (4x 2 yz)
___________________________
3. (3x) (x 2 y 3 )
___________________________
___________________________
Fill in the blanks below. Then complete the multiplication.
4. 4(x  5)

x 
5. 3x(x  8)
5
___________________________

x 
6. 2x(x 2  6x  3)
8
___________________________

 x2  
 6x  
___________________________
Multiply.
7. 5(x  9)
___________________________
10. 3(5  x 2  2)
8. 4x(x 2  8)
___________________________
11. (5a 3 b) (2ab)
9. 3x 2 (2x 2  5x  4)
___________________________
12. 5y(y 2  7y  2)
3
LESSON
15-2
Multiplying Polynomial Expressions
Reteach
Use the Distributive Property to multiply binomial and polynomial expressions.
Examples
Multiply (x  3) (x  7).
(x  3) (x  7)
x(x  7)  3(x  7)
Distribute.
(x)x  (x)7  (3)x  (3)7
Distribute again.
x  7x  3x  21
Multiply.
x2  4x  21
Combine like terms.
2
Multiply (x  5) (x2  3x  4).
(x  5) (x2  3x  4)
x(x2  3x  4)  5 (x2  3x  4)
Distribute.
(x)x2  (x)3x  (x)4  (5)x2  (5)3x (5)4
x3  3x2  4x  5x2  15x  20
x3  8x2  19x  20
Distribute again.
Multiply.
Combine like terms.
Fill in the blanks below. Then finish multiplying.
1. (x  4) (x  5)
__ (x  5)  __ (x  5)
_______________________
2. (x  2) (x  8)
3. (x  3) (x  6)
__ (x  8)  __ (x  8)
_______________________
__ (x  6)  __ (x  6)
________________________
Multiply.
4. (x  2) (x  3)
_______________________
5. (x  7) (x  7)
6. (x  2) (x  1)
_______________________
________________________
Fill in the blanks below. Then finish multiplying.
7. (x  3) (2x2  4x  8)
__ (2x2  4x  8)  __ (2x2  4x  8)
_______________________________________
8. (x  2) (6x2  4x  5)
__ (6x2  4x  5)  __ (6x2  4x  5)
________________________________________
LESSON
16-1
Characteristics of Quadratic Functions
Reteach
To analyze a function of the form y  ax2, where a is not 0, you can take
notes about the equation.
Look.
Think.
Look.
variable squared
quadratic function
U shaped graph
y  3x2
y  3x 2
Think.
number times variables squared
Point (0, 0) is the highest or
lowest point on the U.
Remember:
The line x  0 divides the U into left and right parts that are identical.
The U is symmetric with x  0 as the line of symmetry.
Now look at the number 3, the coefficient of x2.
Look.
3 is greater
than 0.
Look.
3 is greater
than 1.
vertical stretch
Think.
The U opens upward.
(0, 0) is the lowest
point on the U.
Think.
The U is narrower than
the U that represents
the graph of y  x2.
In this example, the coefficient of x2 is positive. Use similar thinking when the
coefficient is negative. The U will flip over the x-axis of the one shown here.
Answer each question about y  3x2.
1. Does the graph open up or down? __________________________
2. Is (0, 0) the highest (maximum) or lowest (minimum) point on the graph?
__________________
Answer each question about y  0.1x2.
3. Is the graph wider or more narrow than the graph of y  x2? __________________________
4. What is an equation of the axis of symmetry of the graph? __________________________
Answer each question about y  0.1x2.
5.
Does the graph open up or down? __________________________
LESSON
16-2
Transforming Quadratic Functions
Reteach
A parabola has the equation f(x)  a(x  h)2  k. Identify:
a. a, a stretch if a  1 or compression if 0 a  1
b. h, the horizontal translation
c. k, the vertical translation
The vertex is (h, k) and the parabola opens up if a  0 and opens down if a  0.
In parabola f(x)  4 (x  3)2  5, the stretch is 4, the horizontal translation
is 3 to the right, and the vertical translation is up 5. The vertex is (3, 5),
and the parabola opens up.
Complete 1–4 for parabola f(x)  2(x  7)2  9.
1. Stretch or shrink? _______________
2. Open up or down? ______________
3. Horizontal translation? ____________
4. Vertical translation? ______________
Complete 5–8 for parabola f ( x ) 
1
( x  4)2  8.
2
5. Stretch or shrink? _______________
6. Open up or down? ______________
7. Horizontal translation? ____________
8. Vertical translation? ______________
For a parabola that opens up, the vertex represents the minimum
point. For a parabola that opens down, the vertex represents the
maximum point.
The following graph is a translation of y  x2.
9. The vertex is (_____, ______).
10. Is the vertex a maximum or a minimum?
_______________________________________
11. The quadratic equation for the graph is
_______________________________________ .
LESSON
16-3
Interpreting Vertex Form and Standard Form
Reteach
The equation of a parabola can be written in either vertex or standard form.
Vertex Form
Standard Form
y  a( x  h)  k
y  ax 2  bx  c
2
Find the vertex of the quadratic equation y  2( x  1)2  4.
The vertex is the lowest point of a parabola when the parabola opens up.
When the equation is written in vertex form, the coordinates of the vertex are (h, k).
y  a( x  h)2  k
h  1, k  4
y  2( x  1)2  4
The vertex is the ordered pair (1, 4) and the line of symmetry is x  1.
To change the equation from vertex form to standard form, do the following:
y  2( x  1)2  4
y  2( x  1)( x  1)  4
Show the factors.
y  2( x 2  2 x  1)  4
Expand.
y  2x  4 x  2  4
Multiply.
y  2x 2  4 x  2
Simplify.
2
 ( b )  ( b )  
, f
When written in standard form, the coordinates of the vertex are 
 .
 2a  
 2a
y  ax 2  bx  c
(b ) ( 4)

 1, and y  2(1)2  4(1)  2  4
a  2, b  4 , so x 
2
2a
2(2)
y  2x  4 x  2
The vertex is the ordered pair (1, 4) and the line of symmetry is x  1.
Find the vertex and axis of symmetry of each quadratic equation.
1. y  ( x  5)2  2
_______________________
4. y  2x 2  12x  24
_______________________
2. y  x 2  6x  8
________________________
5. y  8( x  9)2  5
________________________
3. y  2( x  4)2  1
________________________
6. y  4x 2  16x  1
________________________
LESSON
17-3
Applying the Zero Product Property to Solve Equations
Reteach
Quadratic equations in factored form can be solved by using the Zero Product Property.
If the product of two quantities equals zero, at least one of the quantities must equal zero.
If (x) (y)  0, then
If (x  3) (x  2)  0, then
x  0 or
x  3  0 or
y0
x20
You can use the Zero Product Property to solve any quadratic equation written in
factored form, such as (a  b)(a  b)  0.
Examples
Find the zeros of (x  5)(x  1)  0.
x  5  0 or x  1  0
Set each factor equal to 0.
x  5 or x  1
Solve each equation for x.
Solve (x  7)(x  2)  0.
x  7  0 or x  2  0
Set each factor equal to 0.
x  7 or x  2
Solve each equation for x.
Use the Zero Product Property to solve each equation by filling in the
blanks below. Then find the solutions. Check your answer.
1. (x  6) (x  3)  0
2. (x  8) (x  5)  0
x  ________ or x  ________
x  ________ or x  _________
_______________________________________
________________________________________
Use the Zero Product Property to solve each equation.
3. (y  7)(y  3)  0
_______________________
6. (t  9)(t  3)  0
_______________________
9. (z  6)(z  4)  0
_______________________
4. 0  (x  6)(x  3)
________________________
7. (n  5)(n  3)  0
________________________
10. 0  (x  4)(x  2)
________________________
5. (x  4)(x  3)  0
________________________
8. (a  10)(a  3)  0
________________________
11. 0  (g  3)(g  3)
________________________
LESSON
18-1
Solving Equations by Factoring x2  bx  c
Reteach
To find the factors for a trinomial in the form x2  bx  c, answer these
2 questions.
1. What numbers have a product equal to c?
2. What numbers have a sum equal to b?
Find numbers for which the answer to both is yes.
Factor x2  5x  6.
What numbers have a product equal to c, 6?
1 and 6
1 and 6
2 and 3
2 and 3
What numbers have a sum equal to b, 5?
1 and 6
1 and 6
2 and 3
2 and 3
The factors of x2  5x  6 are (x  2) and (x  3).
Solve the trinomial by setting it equal to 0. Factor and use the Zero
Product Property to solve.
Example
Solve x2  5x  6  0.
x2  5x  6  0
(x  2)(x 3)  0
Factor x2  5x 6.
x  2  0 or x 3  0
Set each factor equal to 0.
x  2 or x  3
Solve each equation for x.
Complete the factoring.
1. x2  x  2
What numbers have a product equal to c, _____?
What numbers have a sum equal to b, _____?
Factors:
Factor.
2. x2  4x  4
_______________________
3. x2  4x  3
________________________
4. x2  3x  10
________________________
Solve.
5. x2  12x  35  0
_______________________
6. x2  9x  18  0
________________________
7. x2  x  20  0
________________________
LESSON
18-2
Solving Equations by Factoring ax2  bx  c
Reteach
When a factorable quadratic expression is written in standard form,
ax2  bx  c  0, you can use the Zero Product Property to solve the
equation.
To solve a quadratic equation, move all terms to the left side of the
equation to get 0 on the right side.
Example
Solve 3x2  4x  8  6x by factoring.
3x2  4x  8  6x
3x2  4x  6x  8  0
Subtract 8 and add 6x to both sides.
3x2  10x  8  0
Simplify.
(3x  2)(x  4)  0
Factor the quadratic expression.
3x  2  0 or x  4  0
Set each factor equal to 0.
x
2
or x  4
3
Solve each equation.
Sometimes you can factor out a common factor.
Example
Solve 3x2  12x  12  0 by factoring.
3x2  12x  12  0
3 (x2  4x  4)  0
3 (x  2) (x  2)  0
Factor out a common factor.
Factor the quadratic expression.
Set each factor equal to 0.
30
Solve each equation.
or
x20
x2
Use the Zero Product Property to find the solutions.
1. (2x  3)(x  9)  0
2. (5x  1)(x  2)  0
_______________________
________________________
3. 2(3x  1)(3x  1)  0
________________________
Solve the equations by factoring.
4. 2x2  5x  3  2x
_______________________
7. 18x2  24x  8  0
_______________________
5. 6x2  3x  2  4x
________________________
8. 10x2  25x  15  0
________________________
6. 7x2  8x  10x  11
________________________
9. 6x2  96
________________________
LESSON
19-1
Solving Equations by Taking Square Roots
Reteach
These equations have something in common. They have the
same roots.
This comes from adding
2x2  5  13
2x2  18
5 to each side of 2x 2  5  13.
But 2x2  18 is easier to read and solve.
2x 2  18
x2  9
This comes from dividing each
side of 2x2  18 by 2.
Now x2  9 is very easy to solve.
x2  9
x 9
x  3
This comes from taking the
square roots of 9.
Here is another example.
Given
3x  7  13
2
Simpler
3x2  6
Simpler still
x2  2
Done
x 2
Identify the reason for each step in the solution.
1. 4x2  1  15
®
4x2  16
®
x2  4
®
x  2
________________________________________________________________________________________
2. 2x2  3  9
®
2x2  6
®
x2  3
®
x 3
________________________________________________________________________________________
Solve using square roots.
3. x2  9
_______________________
6. x2  400  0
_______________________
9. (x  6)2  144
_______________________
12. (x  3)2  121
_______________________
4. x2  16
________________________
7. x2  49  0
________________________
10. (x  5)2  81
________________________
13. (x  1)2  36
________________________
5. x2  1
________________________
8. x2  64  0
________________________
11. (x  4)2  100
________________________
14. (x  2)2  4
________________________
LESSON
19-2
Solving Equations by Completing the Square
Reteach
To solve a quadratic equation, complete the square. Here is an
example.
Solve x2  10x  24.
Leave room for adding a number to each side of the equation.
¯
¯
x2  10x  _______  24  _______
What number?
2
Answer The square of one half of 10, the coefficient of x
1

2
 2  10   5  25


Now fill in the blanks with this number.
¯
¯
25
25
x2  10x  _______
 24  _______
x2  10x  25  1
Now you have a simpler equation to work with.
(x  5)2  1
x2  10x  25 is a perfect square
trinomial. It equals (x  5)2.
( x  5)   1
2
x  5  1
Finish.
¯
x  5 1
x  4
Remember . 
There are two
square roots.
¯
x  5  1
x  6
Two equations to solve.
The solutions are 4 and 6.
Solve by completing the square.
1. x2  6x  7
_______________________
4. x2  4x  32
_______________________
2. x2  8x  12
________________________
5. x2  14x  24
________________________
3. x2  2x  63
________________________
6. x2  6x  9
________________________
7. The product of two consecutive positive integers is 56. What are they?
________________________________________________________________________________________
Using the Quadratic Formula to Solve Equations
LESSON
19-3
Reteach
Write the quadratic equation in standard form ax 2  bx  c  0. Use the quadratic formula.
x
b  b2  4ac
2a
Solve 2x2  5x  12  0
using the quadratic formula.
Step 3: Simplify.
2x2  5x  12  0
x
( 5)  ( 5)2  4(2)( 12)
2(2)
Step 1: Identify a, b, and c.
x
5  25  ( 96)
4
x
5  121
4
x
5  11
4
a2
b  5
c  12
Step 2: Substitute
into the quadratic formula.
x
( 5)  ( 5)2  4(2)( 12)
2(2)
Step 4: Write two equations and solve.
5  11
5  11
x
or x 
4
4
3
x4
or x  
2
Solve using the quadratic formula by filling in the blanks below.
1. x2  2x 35  0
2. 3x2  7x 2  0
a  ____; b  ____; c  ____
x

   
2
4
  
2
Simplify:
3. x2  x 20  0
   
2
4
  
2
________________________________________
4. 2x2  9x 5  0
a  ____; b  ____; c  ____
x
x

Simplify:
_______________________________________

a  ____; b  ____; c  ____
   
2

4
  
2
Simplify:
_______________________________________
a  ____; b  ____; c  ____
x

     4  
2

2
Simplify:
________________________________________
LESSON
19-4
Choosing a Method for Solving Quadratic Equations
Reteach
The method you choose to solve a quadratic equation depends on the form of the equation.
 If there are two terms and both have the same variable try factoring:
3x 2  6x  0; 3x( x  2)  0, x  0 of x  2
 If there are two terms and one term is a constant try taking square roots:
x 2  25  0; x 2  25,
x 2   25, x  5 of x  5
 If there is a binomial squared equal to a constant try taking square roots:
( x  5)2  9;
( x  5)2   9, x  5  3, x  2 or x  8
 If there are three terms try factoring into two binomials:
x 2  7x  6  0; ( x  1)( x  6)  0, x  1 of x  6
 If the equation does not seem to factor try writing the equation in standard form
ax 2  bx  c  0 and use the quadratic formula
2 x 2  5 x  6  0;
b  b2  4ac
:
2a
( 5)  ( 5)2  4(2)( 6) 5  25  48 5  73


2(2)
4
4
Solve each quadratic equation by any means. Identify the method and
explain why you chose it.
1. 4 x 2  12x  0
2. 5 x 2  30 x  0
3. x 2  49  0
_______________________
________________________
________________________
_______________________
________________________
________________________
_______________________
________________________
________________________
4. 4 x 2  64
5. (2x  5)2  49
6. ( x  5)2  81
_______________________
________________________
________________________
_______________________
________________________
________________________
_______________________
________________________
________________________
7. x 2  14x  49  0
8. 2x 2  14 x  20
9. 3x 2  4 x  20
_______________________
________________________
________________________
_______________________
________________________
________________________
_______________________
________________________
________________________