Curriculum for the high school grades (9-12) for the state of North Carolina (US) : Curriculum for the high school grades (9-12) for the state of North Carolina (US)
Objectives : Objectives This module explains the curriculum for the high school grades (9-12) for the state of North Carolina (US) Country: United States
State: North Carolina
Time Zone: Eastern Standard Time (GMT-5) till 11th March 2007
Subject: Mathematics
Topic: Algebra 2
Language: English
Grade: Grade 9-12
Introduction : Students will be expected to
Describe and translate among graphic, algebraic, numeric, and verbal representations of relations and use those representations to solve problems
Extend their use of symbols to include vectors and matrices
Use technology to assist in developing models and analytical solutions. use appropriate terminology and notation to define function, domain, range, composition, and inverses of functions
Expand their understanding of functions to include power, polynomial, exponential, periodic, piece-wise, and recursively defined functions
Solve equations, inequalities, and systems using algebraic, tabular, numerical, and graphical methods Introduction
Contents : Algebra 2 includes:
Study of advanced algebraic concepts including functions, polynomials, and rational expressions, systems of functions and inequalities, and matrices.
Study of how to describe and translate among graphic, algebraic, numeric, tabular, and verbal representations of relations and use those representations to solve problems.
Emphasis should be placed on practical applications and modeling. Contents
Pre-requisites :
What your student must already know?
Operate with matrices to solve problems.
Create linear models, for sets of data, to solve problems.
Use linear functions and inequalities to model and solve problems.
Use quadratic functions to model problems and solve by factoring and graphing.
Use systems of linear equations or inequalities to model and solve problems.
Graph and evaluate exponential functions to solve problems. Pre-requisites
Competency Goal 1 : Competency Goal 1 1.01 Simplify and perform operations with rational exponents and logarithms (common and natural) to solve problems. Sample Problems: Simplify expression. 1/12 1/4
1331 . 1331
1/12 1/4 1/12 + 1/4
1331 . 1331 = 1331 Multiply powers.
1/3
= 1331 Add exponents.
= 11 3 1331 = 11
Competency Goal 1 : Competency Goal 1 Solve log 1 a = 2
8 3
log 1 a = 2 Original Equation
8 3
a = 1 2/3 Definition of logarithm
8
a = 1 3 2/3 1/8= (1/2)3
2
a = 1 2 Power of a power
2
a = 1 Simplify
4
Competency Goal 1 : 1.02 Define and compute with complex numbers. Sample Problems: Competency Goal 1 Simplify
(–7 + 5i) + (12 + 3i)
(–7 + 5i) + (12 + 3i)
= (–7 + 12) + (5 + 3)i
= 5 + 8i Simplify
(1 + 2i)(3 – 4i)
(1 + 2i)(3 – 4i)
= 1(3) + 1(–4i) + 2i (3) + 2i (–4i)
= 3 – 4i + 6i – 8i 2
= 3 + 2i – 8(–1) i 2 = –1
= 11 + 2i
Competency Goal 1 : Sample Problem: 1.03 Operate with algebraic expressions (polynomial, rational, complex fractions) to solve problems. Competency Goal 1 Find (m + n)(2m2 – 5mn + 8n)
(m + n)(2m2 – 5mn + 8n)
= m(2m2 – 5mn + 8n) + n(2m2 – 5mn + 8n) Distributive Property
= m(2m2) + m(– 5mn) + m(8n) + n(2m2) + n(– 5mn) + n(8n) Distributive Property
= 2m3 – 5m2n + 8mn + 2m2n – 5mn2 + 8n2 Multiply monomials.
= 2m3 – 3m2n + 8mn - 5mn2 + 8n2 Combine like terms.
Competency Goal 1 : Sample Problem: 1.04 Operate with matrices to model and solve problems. Competency Goal 1 The table below shows the total number of gold, silver, and bronze medals won by 3 countries in the Olympics.
Country Gold Silver Bronze
Norway 83 87 69
Soviet Union 78 57 59
United States 59 59 41
How many more gold than bronze medals did each country win?
The data about the gold and bronze medals can be organized in two matrices. Then you can find the difference between the two matrices to answer the question. 83 69 83-69
78 - 59 = 78-59 Subtract corresponding elements
59 41 59-41
= 14 Norway
19 Soviet Union
18 United States
Competency Goal 1 : Sample Problem: 1.05 Model and solve problems using direct, inverse, combined and joint variation. Competency Goal 1 The volume of any cylinder can be found by using the formula V = πr2h, where r is the radius of the cylinder and h is its height.
Solve the formula for h. V = πr2h Original formula
h = V Divide each side by πr 2.
װr2 If the radius of a cylinder is 2 in. and the volume is 6π in. , what is the height? h = V Formula from part a
װr2
= 6 π r = 2, V= 6
π (2) 2
= 1.5 Simplify
The height of the cylinder is 1.5 in.
Competency Goal 2 : Competency Goal 2 Sample Problem: Determine whether f(x) = –7x + 3 and g(x) = –x – 3 are inverse functions.
7
Check to see if the compositions of f(x) and g(x) are identity functions. The functions are not inverses since neither [ f 0 g](x) nor [ g 0 f](x) equals x
Competency Goal 2 : Competency Goal 2 Sample Problem: 2.02 Use quadratic functions and inequalities to model and solve problems; justify results.
a) Solve using tables, graphs, and algebraic properties.
b) Interpret the constants and coefficients in the context of the problem. Solve x2 + 9 = –6x by graphing.
Write the equation in the ax2 + bx + c = 0 form.
x2 + 9 = –6x x2 + 6x + 9 = 0 Add 6x to each side.
Graph the related quadratic function
f(x) = x2 + 6x + 9.
x –5 –4 –3 –2 –1
f(x) 4 1 0 1 4
Notice that the graph has only one x–intercept, –3.
Thus, the equation’s only solution is –3.
Competency Goal 2 : Competency Goal 2 Sample Problem Write an exponential function whose graph passes through the given points. (0, –1) and (2, –9)
Since the graph of the function contains the point (0, –1), the y-intercept, and value of a, is –1.
Substitute the values for x and y from the point (2, –9) into an exponential function to find the value of b.
y = abx Exponential function
–9 = (–1)b2 Replace x with 2, y with –9, and a with –1.
9 = b2 Divide each side by –1
3 = b Take the square root of each side.
An exponential function whose graph passes through the points (0, –1) and (2, –9) is y = (–1)(3) x .
Competency Goal 2 : Competency Goal 2 Sample Problem: 2.04 Create and use best-fit mathematical models of linear, exponential, and quadratic functions to solve problems involving sets of data.
a) Interpret the constants, coefficients, and bases in the context of the data.
b) Check the model for goodness-of-fit and use the model, where appropriate, to draw conclusions or make predictions. The scatter plot shows the relationship between average monthly temperature during the winter and the corresponding heating bill for that month. Write an equation in slope-intercept form for the best-fit line. Step 1
First, select two points on the line and find the slope. m = y2 - y1 Definition of slope
x2 - x1
= 175 - 100 (x1, y1) = (40, 100); (x2, y2) = (20, 175)
20 - 40
= -3.75 Simplify.
Competency Goal 2 : Competency Goal 2 Step 2
Next, find the y-intercept.
y = mx + b Slope-intercept form
100 = -3.75(40) + b Replace (x, y) with (40, 100) and m with –3.75.
250 = b Simplify.
Step 3
Write the equation.
y = mx + b Slope-intercept form
y = -3.75x + 250 Replace m with –3.75 and b with 40.
b. Predict the monthly heating bill for a winter month having average temperature of 10 o F.
y = -3.75x + 250 Write the equation of the best-fit line.
y = -3.75(10) + 250 Replace x with 10.
y = 212.5 Simplify.
A prediction for the monthly heating bill for a winter month in which the average temperature is 10 o F is $212.50.
Competency Goal 2 : Competency Goal 2 Sample Problem: 2.05 Use rational equations to model and solve problems; justify results.
a) Solve using tables, graphs, and algebraic properties.
b) Interpret the constants and coefficients in the context of the problem.
c) Identify the asymptotes and intercepts graphically and algebraically. The Belgrade Intermediate School needs a new roof. Since the weather is turning cold quickly, the school board decides to hire two construction companies to complete the project. Rick’s Roofing claims to be able to complete the job in 8.5 days. Phil’s Fix–It Inc. claims to be able to complete the job in 11 days. If the time estimates of both companies are accurate, how long will it take for the two companies to complete the job working together? In 1 day, Rick’s Roofing can complete 1/8.5 of the roof.
In 2 days, Rick’s Roofing can complete 1 . 2 or 2 of the roof.
8.5 8.5
In d days, Rick’s Roofing can complete 1 . d or d of the roof.
8.5 8.5
Competency Goal 2 : Competency Goal 2 Likewise, in d days, Phil’s Fix–It Inc. can complete 1 . d or d of the roof.
11 11
Together, they will complete the whole roof.
Part completed plus part completed equals entire job
by Rick s by Phil s
d + d = 1
8.5 11
Solve the equation.
d + d = 1
8.5 11
93.5 (d + d ) = 93.5 (1) Multiply each side by 93.5.
8.5 11
11d + 8.5d = 93.5 Simplify.
19.5d = 93.5 Simplify.
d = 4.8 Divide each side by 19.5.
It will take about 4.8 days to complete the roof.
Competency Goal 2 : Competency Goal 2 Sample Problem: 2.06 Use cubic equations to model and solve problems.
a) Solve using tables and graphs.
b) Interpret constants and coefficients in the context of the problem. y= 2x3
Graph a Cubic Function (2,16) 2(2)3= 16 2 (1,2) 2(1)3= 2 1 (0,0) 2(0)3= 0 0 (-1, -2) 2(-1) 3= -16 -1 (-2, -16) 2(-2)3= -16 -2 (x, y) y = 2x3 x
Competency Goal 2 : Competency Goal 2 Sample Problem Solve 6 3 2x – 1 – 25 = 5.
In order to remove the cube root, you must first isolate it and then raise each side of the equation to the third power.
6 3 2x – 1 – 25 = 5 Original equation
6 3 2 x – 1 = 30 Add 25 to each side.
3 2 x – 1 = 5 Divide each side by 6.
( 3 2 x – 1 )3 = (5)3 Cube each side.
2x – 1 = 125 Evaluate the cubes.
2x = 126 Add 1 to each side.
x = 63 Divide each side by 2.
Competency Goal 2 : Competency Goal 2 Check:
6 3 2x – 1 – 25 = 5 Original equation
6 3 2(63) – 1 – 25 = 5 Replace x with 63.
6 3 125 – 25 = 5 Simplify.
6(5) – 25 = 5 The cube root of 125 is 5.
5 = 5 Subtract.
The solution is 63.
Competency Goal 2 : Competency Goal 2 Sample Problem 2.08 Use equations and inequalities with absolute value to model and solve problems; justify results.
a) Solve using tables, graphs, and algebraic properties.
b) Interpret the constants and coefficients in the context of the problem. Solve –10 |b + 3| = –40. Check your solutions.
First, divide both sides by –10.
|b + 3| = 4
Case 1 a = b Case 2 a = –b
b + 3 = 4 b + 3 = –4
b + 3 – 3 = 4 – 3 b + 3 – 3 = –4 – 3
b = 1 b = –7
Competency Goal 2 : Competency Goal 2 CHECK:
-10 |b + 3| = -40 -10 |b + 3| = -40
-10 |1 + 3| = -40 -10 |-7 + 3| = -40
-10 |4| = -40 -10 |-4| = -40
-40 = -40 - 10(4)= -40
- 40 = -40
The solutions are -7 or 1. Thus, the solution set is {-7, 1}.
On the number line, we can see that each answer is 4 units away from -3.
Competency Goal 2 : Competency Goal 2 Sample Problem 2.09 Use the equations of parabolas and circles to model and solve problems; justify results.
a) Solve using tables, graphs, and algebraic properties.
b) Interpret the constants and coefficients in the context of the problem. Write an equation for each parabola described. Then draw the graph.
a. vertex (–2, 1), focus (0, 1) Plot these two points on a coordinate plane. You can see that
the parabola will have a horizontal axis of symmetry at y = 1. It will open to the right and the equation will then be of the form x = a(y – k)2 + h. Since the focus is two units to the right of the vertex, the directrix is 2 units to the left of the vertex. The equation of the directrix is x = –4.
Competency Goal 2 : Competency Goal 2 Let (x, y) be any point on this parabola. The distance from this point to the focus must be the same as the distance from this point to the directrix. The distance from a point to a line is measured along the perpendicular from the point to the line. distance from (x, y) to (0, 1) = distance from (x, y) to (–4, y) Graph the equation x = 1 (y – 1)2 – 2 using the information
8
found above and several more points. The length of the latus rectum is 1 or 8 units, so the graph must pass through
1
8
(0, 5) and (0, –3).
(x - 0)2 + (y - 1)2 = [x - (-4) ]2 + (y - y)2
(x – 0)2 + (y – 1)2 = (x + 4)2 + 02 Square each side.
x2 + (y – 1) 2 = x2 + 8x + 16 Square x + 4.
(y – 1)2 – 16 = 8x Isolate the x–term.
1 (y – 1)2 – 2 = x Divide each side by 8.
8
Competency Goal 2 : Competency Goal 2 Sample Problem 2.10 Use systems of two or more equations or inequalities to model and solve problems; justify results. Solve using tables, graphs, matrix operations, and algebraic properties. The U.S. Postal Service allows packages up to 70 pounds with a combined length and girth not over 108 inches to be mailed under the classification of Priority Mail. Write and graph a system of inequalities that represents the range of weights and combined length and girth measures for Priority Mail. Let w represent the weight of packages in pounds. The acceptable weights
are 0 to 70 pounds. We can write this information as two inequalities. 0 ≤ w and w ≤ 70
Let m represent the combined length and girth of a package.
The acceptable measures can also be written as two inequalities.
0 ≤ m and m ≤ 108
Graph all of the inequalities. Any ordered pair in the intersection
of the graphs is a solution of the system.