Name: ______________________________________________

Creating and Solving EquationsLESSON

2-2

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 8 18

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 VariableReteach

2n is the same as “a number times 2.”

8 is the same as “the sum of...and 8.”

18 is the same as “is 18.”

LESSON

2-3

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 If you know the circumference, you could find the diameter by using a formula for d.

ExamplesSolve for d. Since d is multiplied by use division to undo this.

Divide both sides by

or

Solve for C. What has been done to C? First undo adding 32.

Subtract 32 from both sides.

Multiply both sides by the reciprocal of

Simplify.

Solve each formula for the indicated variable.

1. for b 2. A lw, for l 3. for s

_________________________ _________________________ _________________________

4. P a b c, for c 5. for t 6. for H

_________________________ _________________________ _________________________

Solve each equation for the indicated variable.

7. m n p q 360, for n 8. for s 9. for a

_______________ _________________________ _________________________

Creating and Solving InequalitiesReteach

An inequality, such as 2x 8, has infinitely many numbers in its

LESSON

2-4

solution. For example, 2, 0, 1 ,2

and 350 are all values of x that make

this 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

3x15 Note that 6 and 8 make this true, but it is not true for 6 or 8.

3 153 3x

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 2.2c

8 11

_________________________________________ __________________________________________________________________________________ __________________________________________________________________________________ _________________________________________

Solve each inequality.

3. 15 3 4s 4.34

j 1 4

_________________________________________ _________________________________________

5. 8c 4 4(c 3) 6. 5(x 1) 3x 10 8x_________________________________________ _________________________________________

7. 8 4a 12 2a 10 8. 0.6t 3 15

Graphing FunctionsReteach

To check whether an equation is a function, isolate the y variable on one side of the equation and simplify.

4x 2y 82y 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.

LESSON

3-4

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) 4f(x) 4 4 f(x) 6 4 f(x) 8 4f(x) 8 f(x) 10 f(x) 12

Use a table to find the ordered pairs. Graph the ordered pairs.

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. 4.

function? ____________ function? ____________ function? _____________

Using InterceptsReteach

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

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.

LESSON

5-2

is the graph of 4x 2y 12.

To find the y-intercept, substitute x 0. Solve for y. 4(0) 2y 12

He can buy 6 boxes of popcorn (y) if he buys 0 peanuts. y 6To find the x-intercept, substitute y 0. Solve for x. 4x 2(0) 12

He can buy 3 bags of peanuts (x) if he buys 0 popcorn. 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

Interpreting Rate of Change and SlopeReteach

LESSON

5-3

Find the rate of change, or slope, for the graph of a straight line by finding

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.

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. 3.

_________________________ _________________________ _________________________

4. 5. 6.

_______________ _________________________ _________________________

Direct VariationReteach

When a situation can be described by an equation of the form y kx, where k is a constant,

LESSON

5-4

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

x 2 3 5 10

y 8 12 ? ?

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. 5. 6.

_________________________ _________________________ _________________________

For each table, y varies directly with x. Determine the missing value.7. 8. 9.

_________________________ _________________________ _________________________

Slope-Intercept FormReteach

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.

x 1 3 5

y 6 18 30

x 2 4 6

y 8 16 ?

x 2 4 8

y 6 12 24

x 3 7 11

y 12 28 ?

x 5 7 9

y 10 14 ?

x 1 4 9

y 5 20 45

LESSON

6-1

An equation is in slope-intercept form if it is written as:

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 1: Solve for y.

2x 6y 12 Subtract 2x from both sides.

2x 2x

Divide both sides by 6.

1 23

y x Simplify.

Step 2: Find the slope and y-intercept.

slope: 1 13 3

m

y-intercept: b 2

Step 3: Graph the line.

Plot (0, 2).

Then count 1 down (because the rise is negative) and 3 right (because the run is positive) and plot another point.

Draw a line connecting the points.

Find each slope and y-intercept. Then graph each equation.

1. 1 32

y x 2. 3x y 2 3. 2x y 3

slope:_______________ slope:_____________ slope:___________________

y-intercept:___________ y-intercept:_________ y-intercept:______________

Point-Slope FormReteach

m is the slope.b is the y-intercept.

Plot (0, 2).

Count 3 right.

Count 1 down.

LESSON

6-2

An equation is in slope-intercept form if it is written as:

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.

Slope m between any two points on the line.

At the y-intercept, x 0 and y b.

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. slope y-intercept 3 __________________________________

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

m is the slope.b is the y-intercept.

Transforming Linear FunctionsReteach

For a linear function f(x) mx b, changing the value of b moves the graph up or down.

Description Equation y -intercept or b

Parent function f(x) x 0Translate up 2 g(x) x 2 2Translate down 4 h(x) x 4 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

6-4

Parallel and Perpendicular LinesReteach

The chart shows how parallel and perpendicular lines differ.

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)

Find each slope and y-intercept.

1. Are the lines whose equations are y 2x 4 and parallel or perpendicular?

Explain.__________________________________________________________________________________________________________________________________________________________________________________

2. Are the lines whose equations are y 4 7(x 3) and y 2 7(x 5) parallel or perpendicular? Explain.__________________________________________________________________________________________________________________________________________________________________________________

Using Functions to Solve One-Variable EquationsReteach

LESSON

7-1

Lines Slopes y-intercept

parallel same different

perpendicular negative reciprocals of each other

can be the same or different

LESSON

7-2

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.

For f(x)

slope and y-intercept 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 andare usually easiest for integers.

Solve with a table.

f(x) 3 2x and g(x) 12 3x

1. In the algebraic example above, if x 2, what does f(x) equal? _____ What does g(x)

equal? ______

x f (x) 3 2x g(x) 12 3x

0 3 12

1 1 9

2 1 6

3 3 3

4 5 0

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? _____________________________________________________

Linear Inequalities in Two VariablesReteach

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.

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

7-3

Scatter Plots and Trend LinesReteach

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

8-1

_________________________ _________________________ _________________________

Solving Linear Systems by GraphingReteach

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. 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.

1. 2.

____________

____________

3. 4.

LESSON

9-1

____________

____________

Solving Linear Systems by SubstitutionReteach

You can use substitution to solve a system of equations if one of the equations is already solved for a variable.

Solve

23 10y x

x y

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.

3x y 103x (x 2) 10 Substitute x 2 for

y.4x 2 10 Combine like terms.

2 24 84 84 4

2

xx

x

Step 3: Now substitute x 2 back into one of the original equations to find the value of y.

y x 2y 2 2y 4

The solution is (2, 4).

Check:Substitute (2, 4) into both equations.y x2 3xy104 22 3(2)4 10

4 4 64 ? 10

10 ? 10

You may need to solve one of the equations for a variable before solving with substitution.

Solve each system by substitution.

1. 2.

LESSON

9-2

x 2

_________________________________________ _________________________________________

3. 4.

_______________________________ _________________________________________

Solving Linear Systems by Adding or SubtractingReteach

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.

8x 8 Solve for x.x 1

Substitute x 1 into 3x 2y 7 and 3x 2y 7solve for y: 3(1) 2y 7

2y 4y 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.

1. 2.

__________________________________ ________________________________

LESSON

9-3

Use the elimination method when the coefficients of one of the variables are the same or opposite.

The y-terms have opposite coefficients, so add.

Add the equations.

3. 4.

__________________________________ ________________________________

Solving Linear Systems by Multiplying FirstReteach

To solve a system by elimination, you may first need to multiply one of the equations to make the coefficients match.

Solve for y: 11y 11 Substitute 1 for y in x 3y 10.11 11 x 3(1) 10

y 1 x 3 10 3 3

x 7The solution to the system is the ordered pair (7, 1).

You may need to multiply both of the equations to make the coefficients match.

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.

1. 2.

LESSON

9-4

Multiply bottom equation by 2.

Multiply the top by 2 and the bottom by 3.

__________________________________ ________________________________

3. 4.

Graphing Systems of Linear InequalitiesReteach

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

Solve each system of linear inequalities by graphing. Check your answer by testing an ordered pair from each region of your graph.

1. 2. 3.

LESSON

10-2

Graph of the system

Two ordered pairs that are solutions: (3, 4) and (5, 2)

All solutions are in this double shaded area.

_________________________ _________________________ _________________________

Modeling with Linear SystemsReteach

LESSON

10-3

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?

Write two equations. Use the information in each row of the chart.

Write each equation in slope-intercept form.

Set the equations equal to each other and solve.

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: ____________________________________

Understanding Geometric SequencesReteach

Sticky Notes Pens TotalNumber of Items n p 12Cost 4n 2p 40

Organize the information.

n p 12n p 12

4n 2p 404n 2p 40n 10

n p 12n 4 12n 8She bought 8 sticky notes.

p 12 10

12 10

2

4 pShe bought 4 pens.

LESSON

12-1

Cost 4n 2p 40

Number of Items n p 12

4n 2p 40

n p 12

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

common ratio is , the geometric sequence is: 72, 24, 8, , …

Completeeachtable.

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 a1 a2 a3 a4 a5

Value

Modeling Exponential Growth and DecayReteach

LESSON

13-2

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.

Step 1: Identify the variables.

a 236,000——r 0.04

Step 2: Substitute for a and r.

y a (1 r)t

y 236,000 (1 0.04)t

The exponential growth function is y 236,000 (1.04)t.

Step 3: Substitute for t.

y 236,000 (1.04)9

335,902

The 2009 population was about 335,902 people.

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.

a 35,000——r 0.06

Step 2: Substitute for a and r.

y a (1 r)t

y 35,000 (1 0.06)t

The exponential decay function is y 35,000 (0.94)t.

Step 3: Substitute for t.

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

2. The population of a town is 4200 and increasing at a rate of 3% per year; 7 years

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_________________________________________

Growth;the growth factor is greater than 1.Decay; the growth factor is less than

1 and greater than 0.

y _________(1 ______) ___

y _________(1 ______) ___

y _________(1 ______) ___

Understanding Polynomial ExpressionsReteach

Polynomials have special names based on the number of terms.

POLYNOMIALS

No. of Terms 1 2 3 4 or more

Name Monomial Binomial 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 8x2y3. Find the degree of 4ab 9a3.

8x2y3 The exponents are 2 and 3.

The degree of the monomialis 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:

The following are not like terms:

Examples

Add 3x2 4x 5x2 6x.3x2 5x2 Identify and rearrange like terms so they are together.8x2 10x Combine like terms.

Simplify each expression.4. 2y2 3y 7y y2 5. 8m4 3m 4m4 6. 12x5 10x4 8x4

_________________________ _________________________ _________________________

LESSON

14-1

The degree of the binomial is 3.

same variable,different exponent

one with variable,one constant

same variable but different power

3x2 and 3x 47 and 7y 8m and m5

4y and 7y 8x2 and 2x2 7m5 and m5

same variables raised to same power

Adding Polynomial ExpressionsReteach

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 not like terms: 3x2 and 3x 4y and 7 8m and 8n

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

14-2

different variables

LESSON

14-3

one with a variable, one is a constant

These are like terms because they have the same variables and same exponent.

same variable but different exponent

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)

Multiplying Polynomial Expressions by MonomialsReteach

To multiply monomial expressions, multiply the constants, and then multiply

LESSON

15-1

variables with the same base.

ExampleMultiply (3a2b) (4ab3).

(3a2b) (4ab3)

(3 4) (a2 a) (b b3) Rearrange so that the constants and the variables with the samebases are together.

12a3b4 Multiply.

To multiply a polynomial expression by a monomial, distribute the monomial to each term in the polynomial.

ExampleMultiply 2x(x2 3x 7).

2x(x2 3x 7)

(2x)x2 (2x)3x (2x)7 Distribute.

2x3 6x2 14x Multiply.

Multiply.1. (5x2y3) (2xy) 2. (2xyz) (4x2yz) 3. (3x) (x2y3)

___________________________ ___________________________ ____________________________

Fill in the blanks below. Then complete the multiplication.4. 4(x 5) 5. 3x(x 8) 6. 2x(x2 6x 3)

                  5x                   8x 2                 6           3x x___________________________ ___________________________ ____________________________

Multiply.7. 5(x 9) 8. 4x(x2 8) 9. 3x2(2x2 5x 4)

___________________________ ___________________________ ____________________________

10. 3(5 x2 2) 11. (5a3b) (2ab) 12. 5y(y2 7y 2)

Multiplying Polynomial ExpressionsReteach

Use the Distributive Property to multiply binomial and polynomial expressions.

ExamplesMultiply (x 3) (x 7).

LESSON

15-2

(x 3) (x 7)

x(x 7) 3(x 7) Distribute.(x)x (x)7 (3)x (3)7 Distribute again.x2 7 x 3 x 21 Multiply.x2 4x 21 Combine like terms.

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 Distribute again.x3 3 x 2 4 x 5 x 2 15 x 20 Multiply.x3 8x2 19x 20 Combine like terms.

Fill in the blanks below. Then finish multiplying.1. (x 4) (x 5) 2. (x 2) (x 8) 3. (x 3) (x 6)

__ (x 5) __ (x 5) __ (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) 8. (x 2) (6x2 4x 5)

__ (2x2 4x 8) __ (2x2 4x 8) __ (6x2 4x 5) __ (6x2 4x 5)_________________________________________ _________________________________________

Characteristics of Quadratic FunctionsReteach

To analyze a function of the form y ax2, where a is not 0, you can take notes about the equation.

Look. Think. Look. Think.

LESSON

16-1

number times variables squared Point (0, 0) is the highest or

lowest point on the U.

y 3x2

Remember:

Now look at the number 3, the coefficient of x2.

Look. Think.

Look. Think.

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? __________________________

Transforming Quadratic FunctionsReteach

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

3 is greaterthan 1.vertical stretch

variable squared quadratic function U shaped graph

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.

LESSON

16-2

The U opens upward. (0, 0) is the lowest

point on the U.

The U is narrower than the U that represents the graph of y x2.

3 is greaterthan 0.

b. h, the horizontal translationc. 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

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

________________________________________.

Interpreting Vertex Form and Standard FormReteach

The equation of a parabola can be written in either vertex or standard form.Vertex Form Standard Form

LESSON

16-3

Find the vertex of the quadratic equation 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).

The vertex is the ordered pair and the line of symmetry is x 1.

To change the equation from vertex form to standard form, do the following:

When written in standard form, the coordinates of the vertex are

, so

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. 2. 3.

_________________________ _________________________ _________________________

4. 5. 6._______________ _________________________ _________________________

Applying the Zero Product Property to Solve Equations

Reteach

LESSON

17-3

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 y 0 x 3 0 or 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.

ExamplesFind 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 4. 0 (x 6)(x 3) 5. (x 4)(x 3) 0

_________________________ _________________________ _________________________

6. (t 9)(t 3) 0 7. (n 5)(n 3) 0 8. (a 10)(a 3) 0_________________________ _________________________ _________________________

9. (z 6)(z 4) 0 10. 0 (x 4)(x 2) 11. 0 (g 3)(g 3)_________________________ _________________________ _________________________

Solving Equations by Factoring x2 bx cReteach

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?

LESSON

18-1

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.

ExampleSolve 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

_________________________ _________________________ _________________________

Solving Equations by Factoring ax2 bx cReteach

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

LESSON

18-2

equation to get 0 on the right side.

ExampleSolve 3x2 4x 8 6x by factoring.

3x2 4x 8 6x3x2 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 or x 4 Solve each equation.

Sometimes you can factor out a common factor.

ExampleSolve 3x2 12x 12 0 by factoring.

3x2 12x 12 0 Factor out a common factor.3 (x2 4x 4) 0 Factor the quadratic expression.3 (x 2) (x 2) 0 Set each factor equal to 0.

3 0 or x 2 0 Solve each equation.

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 5. 6x2 3x 2 4x 6. 7x2 8x 10x 11

_________________________ _________________________ _________________________

7. 18x2 24x 8 0 8. 10x2 25x 15 0 9. 6x2 96_________________________ _________________________ _________________________

Solving Equations by Taking Square RootsReteach

These equations have something in common. They have the same roots.

LESSON

19-1

This comes from adding5 to each side of

2x2 5 13 2x2 18

But 2x2 18 is easier to read and solve.

Now x2 9 is very easy to solve.

Here is another example.

Given Simpler Simpler still Done3x2 7 13 3x2 6 x2 2

Identify the reason for each step in the solution.1. 4x2 1 15 4x2 16 x2 4

_________________________________________________________________________________________

2. 2x2 3 9 2x2 6 x2 3 _________________________________________________________________________________________

Solve using square roots.3. x2 9 4. x2 16 5. x2 1

_________________________ _________________________ _________________________

6. x2 400 0 7. x2 49 0 8. x2 64 0_________________________ _________________________ _________________________

9. (x 6)2 144 10. (x 5)2 81 11. (x 4)2 100_________________________ _________________________ _________________________

12. (x 3)2 121 13. (x 1)2 36 14. (x 2)2 4_________________________ _________________________ _________________________

Solving Equations by Completing the SquareReteach

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.

This comes from taking thesquare roots of 9.

LESSON

19-2

This comes from dividing eachside of 2x2 18 by 2.

x2 10x _______ 24 _______

What number?

Answer The square of one half of 10, the coefficient of x

Now fill in the blanks with this number.

x2 10x _______ 24 _______

x2 10x 25 1Now you have a simpler equation to work with.

(x 5)2 1

2( 5) 15 1

xx

Finish. x 5 1 x 5 1

x 4 x 6

The solutions are 4 and 6.

Solve by completing the square.1. x2 6x 7 2. x2 8x 12 3. x2 2x 63

_________________________ _________________________ _________________________

4. x2 4x 32 5. x2 14x 24 6. x2 6x 9_________________________ _________________________ _________________________

7. The product of two consecutive positive integers is 56. What are they?_________________________________________________________________________________________

Using the Quadratic Formula to Solve EquationsReteach

Write the quadratic equation in standard form 2 0.ax bx c Use the quadratic formula.

x 2 42

b b aca

Solve 2x2 5x 12 0 Step 3: Simplify.using the quadratic formula.

x2 10x 25 is a perfect squaretrinomial. It equals (x 5)2.

Remember There are twosquare roots.

Two equations to solve.

25 25

LESSON

19-3

2x2 5x 12 0 x 2( 5) ( 5) 4(2)( 12)

2(2)

Step 1: Identify a, b, and c. x 5 25 ( 96)4

x 5 1214

x 5 114

Step 2: Substitute Step 4: Write two equations and solve.into the quadratic formula.

x 2( 5) ( 5) 4(2)( 12)

2(2)x 4 or x

32

Solve using the quadratic formula by filling in the blanks below.1. x2 2x 35 0 2. 3x2 7x 2 0

a ____; b ____; c ____ a ____; b ____; c ____

x 2

4

2x

2

4

2

Simplify: Simplify:_________________________________________ _________________________________________

3. x2 x 20 0 4. 2x2 9x 5 0a ____; b ____; c ____ a ____; b ____; c ____

x 2

4

2x

2

4

2

Simplify: Simplify:_________________________________________ _________________________________________

Choosing a Method for Solving Quadratic EquationsReteach

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:

If there are two terms and one term is a constant try taking square roots:

a 2b 5c 12

x or x

LESSON

19-4

If there is a binomial squared equal to a constant try taking square roots:

If there are three terms try factoring into two binomials:

If the equation does not seem to factor try writing the equation in standard form

and use the quadratic formula

Solve each quadratic equation by any means. Identify the method and explain why you chose it.

1. 2. 3._________________________ _________________________ __________________________________________________ _________________________ __________________________________________________ _________________________ _________________________

4. 5. 6._________________________ _________________________ __________________________________________________ _________________________ __________________________________________________ _________________________ _________________________

7. 8. 9._________________________ _________________________ __________________________________________________ _________________________ __________________________________________________ _________________________ _________________________