Algebra II

Unit 2

Preview

1. Write the equation of the line passing through (-2, 3) and (4, 0)

2. Create a scatter plot for thedata at right.

3. Draw a trend line (line of best fit). 4. For the scatter plot,

what kind of relationship do you see?(positive, negative or no corelation?)

131415151820Percent

15129630Years Since 1985

Source: U.S. News & World Report

Percent of students sending applications to two colleges

Preview

1. 0 – 3 = - ½

4 – (-2)

0 = - ½ (4) + b

y = - ½ x + 2 OR

y – 3 = - ½ ( x - - 2)

y – 3 = - ½ x – 1

y = - ½ x + 2 4. negative

1) Find slo

pe

2) Find b and/or

equatio

n

Modeling Real World Data

Unit 2.1

(Glencoe Chapter 2.5)

Algebra II

I. Write a Prediction Equation

A. Prediction equation:

1. also known as a regression equation

2. use two points and write a slope- intercept equation

Use points what best fit the data (are solid with the line of best fit)

I. Write a Prediction Equation B. Example:

1. Use points (3, 18) and (15, 13) because they are solid points

2. Find slope: 13 – 18 = -5 or - 0.42 15 – 3 12

3. Find y-intercept point (“b) use one point: 18 = -0.42(3) + b

so 19.26 = “b” 4. Write an equation in y = mx + b form:

y = -0.42x + 19.26

131415151820Percent

15129630Years Since 1985

Source: U.S. News & World Report

Percent of students sending applications to two colleges

I. Write a Prediction EquationC. Example:

y = -0.42x + 19.261. What is the real meaning of the –

0.42?

2. Does the 19.26 fit with the line of best fit on our graph?

Your equation will always be in y = mx + b form

II. Make a Prediction Equation

A. Example:

y = -0.42x + 19.26 Use the regression equation (Prediction

Equation) to make a prediction: 1. What will the percent be in year 18?

y = -0.42(18) + 19.26

= 11.7 2. What will the percent be in 2010?

(HINT: what number year will that be?)

Let’s Review the process: The Prediction Equation

Then use two good points to write an equation in y = mx + b form.

Then predict by substituting an “x” value.

This is real data: You may round!

Practice Today:

You may work in your working groups:Unit 2.1 (Practice 2.5) Handout#1 (use (1.4, 24) and (2.4, 15)and # 2 use (7500, 61) and (9700, 50)

1.Create a scatter plot

2.Write a prediction equation

3.Predict the missing values

Check It Out!1. b) Prediction equation:

y = - 9x + 36.6 c) Prediction: 18.6 miles per gallon2. b) Prediction equation:

y = - 0.005x + 98.5 c) Prediction: 38.5º F

III. Use Technology

Handout Follow instructions in your groupsComplete all parts of the handout.Every student turns in a handout!

Systems of Equations

Unit 2.2(Glencoe Chapter 3.1-3.3)

Algebra II

I. Review Systems of Two Equations

Solve. Use any method.1. x + 2y = 10

x + y = 6

2. 2x + 3y = 12

5x – 2y = 11

(2, 4)

(3, 2)

I. You Try: Take Five

Solve using any method: 1. 2p + 4q = 18 3p – 6q = 3

2. 3x – y = 4 y = 4 + x

3. y = 3x y = - 2x + 5

II. Solve Systems of Two Equations by Graphing

1. y = -2x + 5

y = x – 1

2. x – 2y = 0

x + y = 6

Check it out: Calculator Tips

In y= menu, enter equation 1 as y1 and equation 2 as y2

Graph (remember zoom ► 6)Find intersection: 2nd trace (calc) ►5

(intersect) enter enter enter enterWrite as ordered pair ( x, y)

II. You Try: Take Five

1. 2x + 3y = 12

2x – y = 4

2. 2x – y = 6

x – 8y = - 12

II. How did you do?

1. 2x + 3y = 12

2x – y = 4

2. 2x – y = 6

x – 8y = - 12

For Your Practice

Page 122 Practice Quiz 1#1 – 4

Bellwork 2-04-10

Page 120 #14, 22Word problem:

Practice Quiz 1 Page 122 #5!

Page 120 #14 (2, 7)

#22 (3, -1)

Page 122 #5 Hartsfield 78 million

O Hare 72.5 million

Systems of Equations: Part 2

Unit 2.2Glencoe Chapter 3.5

Algebra II

II. Systems of Three Equations (Three Variables)

A. One solution (x, y, z) 1. 5x + 3y + 2z = 2

2x + y – z = 5x + 4y + 2z = 16

Use elimination or substitution to create a system of 2 variables:

5x + 3y + 2z = 2 2(2x + y – z = 5) 2(2x + y – z = 5 ) x + 4y + 2z = 16

5x + 3y + 2z = 2 4x + 2y – 2z = 10 4x + 2y – 2z = 10 x + 4y + 2z = 16 9x + 5y = 12 5x + 6y = 26

Make a plan

before doing

anything!

Use 1st and 2

nd and

Eliminate z!

Use 2 nd and 3 rd and

Eliminate z!

III. Systems of Three Equations (Three Variables)

A. One solution (x, y, z) 1. 9x + 5y = 12 5x + 6y = 26 Now solve the new system:

6(9x + 5y = 12) -5(5x + 6y = 26)

54x + 30y = 72-25x – 30y = -13029x = - 58

so, x = - 2

Find y: 9(-2) + 5y = 12

-18 + 5y = 12

so, y = 6

Find z: -2 + 4(6) + 2z = 16

22 + 2z = 16

so, z = - 3 (-2, 6, -3)

III. Systems of Three Equations

x + 2y = 12

3y – 4z = 25

X + 6y + z = 20

(6, 3, -4)

You Try: Take Ten

Textbook Page 142# 5, 7

How did you do?

5. (-1, -3, 7)7. (5, 2, -1)

BELLWORK 2-05-10

Page 142#12, 14#10, 11

Check Page 14212. (3, 4, 7)14. (2, -3, 6)10. 6c + 3s + r = 42 (cost)

c + s + r = 13 ½ (pounds)

r = 2s (rice/sausage)11. (4 ½ , 3, 6)

4 ½ lb. chicken, 3 lb. sausage, 6 lb. rice

Rice is Bigger!

IV. Systems of Three Equations Special Cases

1. 2x + y – 3z = 5x + 2y – 4z = 76x + 3y – 9z = 15

Eliminate x : 2x + y – 3z = 5 -6(x + 2y – 4z = 7)-2(x + 2y – 4z = 7) 6x + 3y – 9z = 15

2x + y – 3z = 5 - 6x – 12y + 24z = -42 -2x – 4y + 8z = - 14 6x + 3y – 9z = 15

- 3y + 5z = - 9 -9 y + 15z = - 27

Are these the same line????

Infinitely many solutions

GCF: 3; Solve for 0 = 0

And prove same line

See page 140

EX. 2

Special Cases 2. 3x – y – 2z = 4

6x + 4y + 8z = 11 9x + 6y + 12z = - 3

Eliminate y:4(3x – y – 2z = 4) -3(6x + 4y + 8z = 11)

6x + 4y + 8z = 11 2(9x + 6y + 12z = -3)

12x – 4y – 8z = 16 -18x – 12y – 24z = - 336x + 4y + 8z = 11 18x + 12y + 24z = - 6 18x = 27 0 = - 39

not a true statement!

IV. Systems of Three Equations

No solution

Practice 2.1 – 2.3

Worksheet for Extra Practice

Dividing Polynomials

Unit 2.3(Chapter 5.3)

Algebra II Honors

I. Divide Polynomials

(x4 – 2x3 + x2 – 3x + 2) ÷ (x – 2)Is the same as:

x4 – 2x3 + x2 – 3x + 2x – 2

and

(x4 – 2x3 + x2 – 3x + 2)(x – 2) – 1

I. A. Divide Polynomials:No Remainder

1. (x4 – 4x3 + x2 + 7x – 2 ) ÷ (x – 2)

I. A. You Try:

(2x4 + 7x3 – 14x2 – x – 30) (x + 5) ÷÷

I. B. Divide Polynomials:with placeholder

2. (x4 – 5x3 – 13x2 + 7) ÷ (x +1)

You must use a placeholder when a variable is missing!

I. B. Divide Polynomials:(a > 1) ) with placeholder

3. (4x4 – 17x2 + 14x – 3)(2x – 3)-1

You must use a placeholder when a variable is missing!

Take Ten: You Try

Page 238 Practice Quiz 1#9, 10

II.A. Divide Polynomials (Monomial)

1. 4x3y2 + 8xy2 – 12x2y3

4xy

2. 5a2b – 15ab3 + 10a3b4

5ab

For Your Practice

TEXTBOOK PAGE 236 – 237 #17,18, 22, 24, 32Copy problems in form set up for

division and divide.Write all remainders as fractions.

II. B. Using Synthetic Division (divisor in form of x – r!

1. (5x3 – 13x2 + 10x – 8) ÷ (x – 2)

5 - 13 10 - 8

2

5

(you do not subtract, you add because you changed the sign of the divisor…)

10

-3

-6

4

8

05x2 – 3x + 4

You Try:

Use Synthetic Division:

(x3 + 5x2 – x – 6) ÷ (x + 2)

II. B. Using Synthetic Division (divisor in form of x – r!

2. (4x4 – 5x2 + 2x + 4) ÷ (2x – 1)

2 0 - 5/2 1 2

½ ½

-2

-1

0

0

2

2x3 + x2 – 2x + 2

x - ½

Divide everything by 2!

1 2

1

1 2Simplify!

Change sign back!

2x3 + x2 – 2x + 4

2x - 1

You Try:

Use Synthetic Division:

(2x4 + 3x3 – 2x2 – 3x – 6) ÷ (2x + 3)

Harder Factoring

Unit 2.4(Chapter 5.4)

Algebra II Honors

I. Review:Factor ax2 + bx + c, a ≠ 1

Factor: Use Guess and Check or Product/Sum

1. 2a2 + 3a + 1

2. 6x2 + 13x + 6

II. Factor by Grouping (4 terms)

Look for GCF in pairs of terms1. 8xy + 20x + 6y + 15

2. 18x2 – 30x – 3x + 5

II. Factor by Grouping (4 terms)

Look for GCF in pairs of terms: You try.

3. 8ax – 6x – 12a + 9

4. 10x2 – 14xy – 15x + 21y

III. Sum and Difference of Two cubes

Sum of 2 Cubesa3 + b3 = (a + b)(a2 – ab + b2)Difference of 2 cubesa3 – b3 = (a – b)(a2 + ab + b2)Factor:8a3 + 27 = (2a + 3)(4a2 – 6a + 9)

III. Sum and Difference of Two cubes

You Try:8m3 + 27

c3 – 216

Take Ten:

Factor .1. x3 + 5x2 – 2x – 10 2. 64x3 – y3

3. ab – 5a + 3b – 15 4. x4 – 81

5. x3y3 + 8 6. 2x2 + x - 15

For Your Practice:

Textbook Page 242 – 243 # 6, 19, 20, 24, 28, 35, 36

Factor by Grouping: > 4 terms

Factor by grouping (group by 3 terms):1. 7ac2 + 2bc2 – 7ad2 – 2bd2

2. 5a2x + 4aby + 3acz – 5abx – 4b2y – 3bcz

Watch for trinomials or binomials that can be factored again!