algebra ii unit 2. preview 1. write the equation of the line passing through (-2, 3) and (4, 0) 2....
TRANSCRIPT
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!