Participating Authors: 
Crystal Abren  Quae Johnson  Susan Sztain 
Brian Bastidas  Tim Jorgensen  Rachel Taylor 
Kristine Behen  Monique Joseph  Amanda Thomas 
Jim Bennett  Gino King  Ryan Thurlow 
Matt Berryman  Erin Kunkle  Christine Vargas 
Sussan Campos  Tami Lacey  Kelly Virden 
Ryan Carr  Beverley Martin  Amy Westlake 
Helen Copeland  Jorge Martinez  Dr. Louis Wildman 
John Craft  Steven F. Martinez  
Janet Dibbini  Justin Mclelland  Educational 
Sally Dibbini  Brian Moss  Administration 
Joshua Evans  Pamela Nero  Program 
Anne Falcon  Margaret Newton  School of 
Martina Fisher  Kim Paz  Education 
Alfred Guerrero  Shari Peters  CSUBakersfield 
Daniel Hansford  Lorrie Robinson  July 2009 
Anne Harvey  Desiree See  
Mary Hommowun  Imelda SimosValdez  
Kari Humecky  C. L. Stockdale  
Mattie Jacobs  Raymond StrasserKing  
Kathi Johnson  Cecil Swetland 
Introduction
Dr. Louis Wildman
California State UniversityBakersfield
As Stanford University Education Professor William Damon says, schools need to give students a better understanding of why they are in school in the first place. In particular, students need to know why they are learning what is being taught. They need to understand how the knowledge and skills they are learning can help them accomplish their life goals. That is the only way to motivate students in a lasting way. (Tully, Susannah, “Helping Students Find a Sense of Purpose,” The Chronicle Review, March 13, 2009, p. B14B15.)
The “No Child Left Behind” Act has focused attention on reading and mathematics, assuming that students would understand that these subjects are important. But, as we have found in writing this booklet, neither students nor, surprisingly, their teachers are able to cite simple practical applications of elementary algebra. Hence it is no wonder that student motivation is so weak.
We believe that by presenting simple practical applications of algebra, students will gain a clearer understanding of why they are studying this subject.
As Professor Damon elaborates:
Students learn bits of knowledge that they may see little use for; and from time to time someone at a school assembly urges them to go and do great things in the world. When it comes to drawing connections between the two—that is, showing students how a math formula or a history lesson could be important for some purpose that a student may wish to pursue—schools too often leave their students flat.
If you visit a typical classroom and listen for the teacher’s reasons for why the students should do their schoolwork, you will hear a host of narrow, instrumental goals, such as doing well in the course, getting good grades, and avoiding failure, or perhaps—if the students are lucky—the value of learning a specific skill for its own sake. But rarely (if ever) will you hear the teacher discuss with students broader purposes that any of these goals might lead to . . . How can schools expect that young people will find meaning in what they are doing if they so rarely draw their attention to considerations of the personal meaning and purpose of the work others do?
. . . most pervasive is a sense of emptiness that has ensnared many young people in long periods of drift during a time in their lives when they should be defining aspirations and making progress toward their fulfillment.
For too many young people today, apathy and anxiety have become the dominant moods, and disengagement or even cynicism has replaced the natural hopefulness of youth. That is not a problem that can be addressed by solutions advanced in the past. The message that young people do best when they are challenged to strive must be expanded to include an answer to the question: For what purpose? (ibid.)
This booklet was suggested by Brianne Blanton during her culminating oral examination for a master’s degrees in educational administration at California State UniversityBakersfield. Her husband, Peter, teaches algebra, and she had noticed that many students wanted to know “why?”
This booklet provides sample answers to first year algebra students who ask “why are we studying each of the California Algebra I Standards. It is written by graduate educational administration students who hope that it will be useful for algebra teachers.
Dr. Louis Wildman
Professor and Coordinator
Educational Administration Program
California State UniversityBakersfield
1.0 Students identify and use the arithmetic properties of subsets of integers and rational, irrational, and real numbers, including closure properties for the four basic arithmetic operations where applicable.
Example:
The rational numbers form a closed system for addition, subtraction, multiplication, and division. Hence, subtracting two rational numbers produces a rational number
3 _ 1 = 9 _ 4 = 5
4 3 12 12 12
Application:
Average velocity:
Vav = X_{2}  X_{1} T_{2}  T_{1} 
A particle moved from an xcoordinate of 12 centimeters to an xcoordinate of 20 centimeters. At time T_{1} = 4 seconds and time T_{2} = 8 seconds. What is the average velocity?
20 – 12 = 8 = 2 cm/sec
8  4 4
1.1 Students use properties of numbers to demonstrate whether assertions are true or false.
Examples:
The sum of two even numbers will always be even.
2n + 2m = 2(n + m), which is always an even number if n and m are integers.
Multiplying two integers, the product is sometimes larger than the factors.
5 times 6 = 30: the product is larger than the factors.
5 times 6 = 30: the product is smaller than either factor.
0 times 0 = 0: the product is equal to the factors.
8 2/3 times 3 1/2 has a product between 24 and 36, because 8 and 3 are less than 8 2/3 and 3 1/2, and 9 and 4 are greater than 8 2/3 and 3 1/2.
Applications:
Being able to quickly estimate a product is useful in bargaining and in recognizing mistakes.
In agriculture a farmer may face an aphid population that can double every three days. If the farmer did not recognize the mathematics of exponential growth, the farmer would fail to quickly respond to the aphid infestation. For example, assume that the farmer spots 20 aphids. Notice how fast this problem can spread over a 30 day period:
Aphids:  20  40  80  160  320  640  1280  2560  5120  10,240  … 
Days:  0  3  6  9  12  15  18  21  24  27  … 
2.0 Students understand and use such operations as taking the opposite, finding the reciprocal, taking a root, and raising to a fractional power. They understand and use the rules of exponents.
Examples:
2^{2}= 2 times 2 = 4 3^{2}= 3 times 3 = 9
2^{3}= 2 times 2 times 2 = 8 3^{3}= 3 times 3 times 3 = 27
2^{4}= 2 times 2 times 2 times 2 = 16 3^{4}= 3 times 3 times 3 times 3 = 81
16^{1/2}= the square root of 16 = 4 81^{1/2}= the square root of 81 = 9
8^{1/2} = √ 8 = approximately 2.8284 27^{1/2}= √ 27 = approximately 5.1962
4^{1/2} = √ 4 = 2 9^{1/2}= √ 9 = 3
Application
When one borrows money to purchase a house or a car, the total amount (At) actually borrowed is determined the by the following formula:
A_{t}= P(1 + r/n)^{nt} where P is the amount borrowed (called the principal), r is the interest rate per year, n is the number of conversion periods per year, and t is the number of years of the loan.
Let’s say that you borrow $20,000 to help pay for a new car. Then, $20,000 = P, the principal.
Further, let’s say that you pay off this loan, which will be compounded semiannually, in 3 years.
Then, n =2 , and t = 3.
Finally, let’s assume that the interest rate is 8%. Then r = .08
Substituting in the formula, we see that:
A_{3}=20,000(1 + .08/2)^{(2)(3)}
A_{3}=20,000(1 + .04)^{(6)}
A_{3}=20,000(1.04)^{6}
A_{3}=20,000(1.2653) = $25,306.38
How much would it have cost to borrow the $20,000 had the interest of 8% only been compounded annually?
A_{3}=20,000(1 + .08/1)^{(1)(3)} = 20,000(1.08)^{3} = 20,000(1.2597) = $25,194
3.0 Students solve equations and inequalities involving absolute values.
Examples:
Find the values of x that satisfy:  x – 2 = 3 Answer: 5 and 1
Find the values of x that satisfy:  x – 2 < 3 Answer: 1 < x < 5
Application:
A dog food manufacturer has a tolerance of 0.25 pound per bag of dog food advertised as weighing 5 pounds. Write and solve an absolute value inequality that describes acceptable weight for “5 pound” bags.
Verbal Actual weight  Ideal weight ≤ Tolerance
Model
Labels: Actual weight = x (pounds)
Ideal weight = 5 (pounds)
Tolerance = 0.25 (pounds)
Algebraic
Model  x  5  ≤ 0.25
Equivalent
compound
inequality 0.25 ≤ x  5 ≤ 0.25
Add 5 to
each
expression: 4.75 ≤ x ≤ 5.25
Therefore the actual weight can range between 4.75 pounds and 5.25 pounds.
4.0 Students simplify expressions before solving linear equations and inequalities in one variable, such as 3(2x5) + 4(x2) = 12.
Example:
3(2x – 5) + 4(x – 2) 
= 
12  
6x – 15 + 4x – 8 
= 
12  Distributive law 
6x + 4x – 15 – 8 
= 
12  Commutative law 
10x  23 
= 
12  Combining like terms 
10x 
= 
12 + 23  Adding 23 to both sides 
10x 
= 
35  Combining like terms 
x 
= 
35/10 = 7/2  Multiplying both sides by 1/10 
Applications:
(1) Let’s say you only have $20.00 to spend on gas, and gas is $3.50 a gallon. How many gallons can you buy?
3.50x = 20.00, where x is the number of gallons one can purchase.
x = 5.71 gallons
(2) Let’s say you have to move 2,500 miles across the country for a new job in Buffalo, New York. How much money do you need to save for gas, if the national average cost per gallon is $3.23 and your car gets 20 miles to the gallon?
(2,500/20)($3.23) = x, the amount of money needed.
Notice that (2,500/20) gives you the number of gallons needed.
(125) ($3.23) = x
$403.75 = x
(3) The oil refinery is equipped with a main pump that can fill the tank of an oil truck in 20 minutes, and an auxiliary pump which alone can fill the oil truck in 30 minutes. Sometimes it is necessary to fill the oil trucks faster to get them on the road, and so both pumps are used simultaneously. How much will that reduce the filling time?
If x is the number of minutes it takes to fill the tank with both pumps operating, then in one minute, both pumps can fill 1/x of the tank.
In one minute, the main pump can fill 1/20 of the tank. Similarly, in one minute, the auxiliary pump can fill 1/30 of the tank. With the two pumps running simultaneously, in one minute, the combined contribution of each pump will be
1/20 + 1/30 of the tank. This means that 1/20 + 1/30 = 1/x. Multiplying both sides by 60x, we get 3x + 2x = 60, or 5x = 60, or x = 12.
This means that there can be an 8 minute reduction over the 20 minute time it takes the main pump to fill an oil truck, if both pumps operate simultaneously.
5.0 Students solve multistep problems, including word problems, involving linear equations and linear inequalities in one variable and provide justification for each step.
Example:
5x – 2y 
= 
7  
x + 3y 
= 
15  
6x + y 
= 
22  Adding the equations 
y 
= 
22 – 6x  Subtracting 6x from both sides 
5x – 2(226x) 
= 
7  Substituting y = 22 – 6x in the first equation. 
5x – 44 + 12x 
= 
7  Using the distributive axiom. 
17x 
= 
7 + 44  Using the commutative axiom, adding, and adding 44 to both sides 
17x 
= 
51  Adding 
x 
= 
3  Dividing both side by 17 
y 
= 
22 – (6)(3) = 22 – 18 = 4  Substituting x = 3 in y = 22 – 6x. 
Application:
A gas station charges $3.33 per gallon, and $5 for a car wash. If drivers fillup their cars and get a car wash, the gas station reduces the charge per gallon of gas to $3.23 The gas station manager has noticed that this incentive motivates more drivers to get a car wash. Assume that the average motorist buys 12 gallons of gas per visit to the gas station, and that five out of a hundred motorists would also get a car wash without the advertised incentive. (A) Under those normal conditions, how much money does the gas station take in for every hundred motorists? (B) How many additional car washes would be necessary to at least break even, if gas is lowered by 10 cents a gallon?
Under normal conditions, this is how much money the gas station takes in per 100 motorists:
$3.33(12)(100) + $5(5) = $3,996 + $25 = $4,021
To calculate how many additional car washes it would take to make up the difference:
$3.23(12)(100) + $5(5 + x) 
= 
$4,021 
$3,876 + 25 + 5x 
= 
$4,021 
$3,901 + 5x 
= 
$4,021 
5x 
= 
$120 
x 
= 
$24 
In other words, the number of car washes would have to increase from 5 to 24+5=29 out of every 100 motorists who utilized the gas station in order for this incentive to break even. More than 29 motorists would have to purchase car washes in order for this gas station to justify reducing the price of gasoline. Do you think that this incentive would work?
Justin Mclelland
6.0 Students graph a linear equation and compute the x and yintercepts (e.g., graph 2x + 6y = 4). They are also able to sketch the region defined by linear inequality (e.g., they sketch the region defined by 2x + 6y < 4).
Example:
Graph 2x + 6y 
= 
4  
6y 
= 
2x+4  
y 
= 
(1/3)x + (2/3)  
Make a table: 
x 
y 

0 
2/3 
(0,2/3) is the yintercept  
1 
1/3 

2 
0 
(2,0) is the xintercept 
Graph 2x + 6y < 4
Start by graphing 2x + 6y = 4, as this graph represents all points on the line.
Now the graph of 2x + 6y < 4 will be on one side of the line or the other.
To determine which side, simply check out a point, such as (0,0) which is not on the line. Since (2)(0) + (6)(0) is less than 4, the graph of 2x + 6y < 4 will be on the “(0,0)” side of the line.
Application:
Greg’s allowance is $5.00 a week if he does all his chores. Assuming Greg does his chores, graph how much allowance money Greg will have as the weeks go by, if he doesn’t spend any of his allowance.
[Graph of y = 5x]
7.0 Students verify that a point lies on a line, given an equation of the line. Students are able to derive linear equations by using the pointslope formula.
Examples:
Given the following ordered pairs:
x 
y 
0 
2 
1 
5 
2 
8 
3 
11 
Which of the following equations includes all of the above points?
A.  y 
= 
2x +1  
B.  y 
= 
3x+ 2  (correct) 
C.  y 
= 
1x + 2 
Which points lies on the line defined by 2y = x + 4 ?
A.  (2,0) 
B.  (0,2) (correct) 
C.  (2,2) 
D.  (2,3) (correct) 
Application:
A carpenter wants to construct a simple cabin with a pitched roof. The inside height of this “Aframe” cabin next to the wall must be 8 feet high. The owner wants the inside height of the roof in the middle of the room (at the tallest point) to be 12 feet high. The owner wants the slope of the roof to be 1/3. How wide then must the cabin be?
Answer:
We know that the equation of a line, given the slope (m) and yintercept (b), is:
y = mx + b. If we use the xaxis to represent the floor of the cabin, and the yaxis to represent a wall, then the yintercept is (0, 8). The equation of the line then becomes:
y = (1/3)x + 8, and we want to know what x will be when y = 12.
Substituting in our equation:
14  = 
(1/3)x +8 
6  = 
(1/3)x 
18  = 
x 
This demonstrates that the roof will rise from 8 feet (next to the wall) to 14 feet in the center of room, which is 14 feet from the wall. The total width of the room would then be 28 feet—a very wide room.
How wide would the room be, if the highest point in the middle was 12 feet ? Experiment by changing the slope of the roof.
8.0 Students understand the concepts of parallel lines and perpendicular lines and how those slopes are related. Students are able to find the equation of a line perpendicular to a given line that passes through a given point.
Example:
For two lines to be perpendicular, the slope of one must be the negative reciprocal of the other. For example, consider the following two linear equations:
y=2x + 1 and y= (1/2)x + 5
The slope of this line is 2. The slope of this line is –(1/2), the negative reciprocal of the first line.
Application: Why would anyone care if two lines were parallel ?
Gas mileage is affected by the tow of the car. Wheels that are not parallel will lower gas mileage and increase tire wear. Car owners who upgrade their suspension or lower their vehicle can use a simple, inexpensive technique to correct the toe of the wheels.
Start by running a string around the entire vehicle, making sure that the string crosses the vertical center of each wheel. The string should touch the front and back parts of all wheels. If not, the car is out of toe and the wheels are not parallel. If the string only touches the front or back portion of a wheel, adjustments can be made, depending on the make of the car.
The “camber” angle is the angle between the vertical reference (perpendicular to the road surface) and the line that runs through the center of the wheel. Zero camber indicates a wheel perpendicular to the road. Positive or negative camber negatively affects tire wear and car handling, causing the vehicle to pull left or right.
Specialized tools are needed to measure camber, although directions for homemade tools can be found on the internet. Camber can be adjusted following manufacturer’s directions.
Other practical applications using parallel and perpendicular lines include road construction (most American towns are laid out with parallel and perpendicular lines), architectural design, railroad tracks, building frameworks, window panes and blinds, power lines, and the goal posts on a football field. The cross bar is perpendicular to the two outside posts.
9.0 Students solve a system of two linear equations in two variables algebraically and are able to interpret the answer graphically. Students are able to solve a system of two linear inequalities in two variables and to sketch the solution sets.
Example:
Solve:
x + y = 7 and x + 4y = 8
y = x 7 —> x + 4(x – 7) = 8
x + 4x 28 = 8
5x  28 = 8
5x = 20
x = 4 —> y = 4 – 7
y = 3
The two lines intersect at (4,3).
Application:
All of us are occasionally faced with a situation where we must choose between two payment methods for some service: Payment Method A requires an initial payment plus so much per month, versus Payment Method B requires no initial payment but a higher monthly rate. In such situations, Payment Method A is usually the better choice if one is going to continue purchasing the service for a long period of time. Payment B is better if one is only going to use the service for a short amount of time.
Let’s examine a specific purchase, and determine the breakeven point.
Businesses have website contracts with website hosting services to store and maintain computer files that make up their websites. Larry’s Internet Provider Service charges a $10.00 set up fee plus $21.95 per month. Ajax Website Hosting, Inc. charges $22.45 per month, but without a set up fee. Let’s find the breakeven point.
Solution:
Let y be the total cost after x months.
Larry’s Internet Provide Service costs = y = 10 + 21.95 x
Ajax Website Hosting, Inc. costs = y = 22.45 x
To solve these two simultaneous linear equations, we substitute 22.45x for y in the first equation:
22.45 x = 10 + 21.95 x Subtracting 21.95x from both sides, we get:
.5 x = 10 or x = 20. This means that by month 20, the total cost one would have spent on the two services over 20 months would have been the same.
In month 18:  y = 10 + (21.95)(18) = 10 + 395.10 = 405.10  Larry’s Internet 
y = (22.45)(18) = 404.10  Ajax Website Hosting  
In month 19:  y = 10 + (21.95)(19) = 10 + 417.05 = 427.05  Larry’s Internet 
y = (22.45)(19) = 426.55  Ajax Website Hosting  
In month 20:  y = 10 + (21.95)(20) = 10 + 439 = 449.  Larry’s Internet 
y = (22.45)(20) = 449.  Ajax Website Hosting  
In month 21:  y = 10 + (21.95)(21) = 10 + 460.95 = 470.95  Larry’s Internet 
y = (22.45)(21) = 471.45  Ajax Website Hosting  
In month 22:  y = 10 + (21.95)(22) = 10 + 482.90 = 492.90  Larry’s Internet 
y = (22.45)(22) = 493.90  Ajax Website Hosting 
Now one might conclude that it doesn’t make much difference which payment method is chosen, and it only makes a dollar difference by month 22. However, consider the difference after five years or 60 months:
y = 10 + (21.95)(60) = 10 + 1317 = 1327.00
y = (22.45)(60) = 1347
Now there is a $20. difference.
How much difference would there be after ten years?
y = 10 + (21.95)(120) = 10 + 2634 = 2644
y = (22.45)(120) = 2694
There would be a $50 difference in the amount spent.
10.0 Students add, subtract, multiply, and divide monomials and polynomials. Students solve multistep problems, including word problems, by using these techniques.
Example:
(x^{2} + 13x + 46) + (x^{2} + 7x + 10) = 2x^{2} + 20x +56
Applications:
You are putting a stone border (of the same width) along the front and an adjacent side of a rectangular garden that measures six yards by fifteen yards. You have only enough stone to cover 46 square yards. How wide should the border be?
Solution:
Area of Border = Total Area – Area of Garden
46  = 
(x + 15)(x + 6) – (15)(6) 
46  = 
(x^{2} + 6x + 15x + 90) – 90 
46  = 
(x^{2} + 21x + 90) – 90 
46  = 
x^{2} + 21x 
0  = 
x^{2} + 21x – 46 
0 
= 
(x + 23) (x – 2) 
x + 23 = 0 
or 
x – 2 = 0 
x = 23 
or 
x = 2 
While the solutions are 23 and 2, only x = 2 is a reasonable solution, because negative values for dimension don’t make sense. Hence the border should be 2 yards wide.
How many miles would someone who purchased a $45,000 car that gets 45 miles to the gallon have to drive to come out even with the individual who purchased a car for $35,000 which gets 35 miles to the gallon? Assume that gas costs $3 per gallon.
$45,000 + 3(d/45) = 
$35,000 + 3(d/35)  We want to know what d will be when the two sides of the equation will be equal 
10,000 = 
(3d/35)  (3d)/45)  
10,000 = 
135d  105d 1575 

15,750,000 = 
30d  
525,000 miles = 
d 
11.0 Students apply basic factoring techniques to second and simple thirddegree polynomials. These techniques include finding a common factor for all terms in a polynomial, recognizing the difference of two squares, and recognizing perfect squares of binomials.
Example:
x^{2} + 2x – 15 = 0 (Notice that the first term is of the 2nd degree, and the second term is of the first degree.
This a 2nd degree equation.)
(x – 3)(x + 5) = 0 (Factoring)
x – 3 = 0 and/or x + 5 = 0
x = 3 x = 5
Application:
A ball is thrown upward with an initial velocity of 40 feet per second, and the distance from the starting point is given by the formula d = 40t 16t^{2}.
Notice that at a halfsecond, the ball will be 40(.5)16(.5)^{2}=204=16 feet high.
After one second, the ball will be 40(1)16(1)^{2}=4016=24 feet high.
After one and a quarter seconds, the ball will be 40(1.25)16(1.25)^{2}=25 feet high.
After one and a half seconds, the ball will be 40(1.5)16(1.5)^{2}=24 feet high.
After 2 seconds, the ball will be (40)(2) – 16 (2)^{2 }= 80 – 64 = 16 feet high.
Apparently, the ball went up to about 25 feet, and then started to come down.
To determine the time the ball will be back to where it started, we need to solve the equation,
asking when will d = 0 ? We can do that by factoring:
0 = 40t – 16t^{2}
0 = (40 – 16t)t
0 = 40 – 16t and t = 0
40 = 16t
5 = 2t
2.5 = t
In other words, the ball will be at the start when t = 0, and back to the start when
t = 2.5 seconds.
12.0 Students simplify fractions with polynomials in the numerator and denominator by factoring both and reducing them to the lowest terms.
Example:
A polynomial is the sum of monomials.
2x^{2} + x – 5 and 4bc^{3} + 4b^{2}c are polynomials.
Simplify: 6x^{2} + 32x + 10
9x^{2} – 1
6x^{2} + 32x + 10 = 2 (x + 5)(3x + 1) = 2 (x + 5)
9x^{2} – 1 (3x 1)(3x + 1) (3x – 1)
Application:
Consider the following fraction:
(x^{2} – 9) (x – 5) = (x + 3)(x – 3) (x – 5) = (x – 3)
(x^{2} – 6x + 5)(x + 3) (x – 5)(x – 1) (x + 3) (x – 1)
x can equal any number except 5 and 1 and 3, because if x = 5,
x^{2} – 6x + 5 = 5^{2} – (6)(5) + 5 = 25 – 30 + 5 = 0, and division by zero is undefined.
The same is true for x = 1.
It is the same for x = 3, because
x + 3 = 3 + 3 = 0. So once again, since division by zero is undefined, x can not equal 3.
Otherwise, if we wish to evaluate
(x^{2} – 9) (x – 5) for any x (except 3, 1, and 5), we can simply use x  3
(x^{2} – 6x + 5)(x + 3) x – 1
For example, if x = 6, then x – 3 is 6 – 3 which is 3
x – 1 6 – 1 5
13.0 Students add, subtract, multiply, and divide rational expressions and functions. Students solve both computationally and conceptually challenging problems by using these techniques.
Example:
Divide 8a^{2} – 15ab + 7b^{2} by a – b.
8a^{2} – 15ab + 7b^{2} = (8a – 7b)(a  b)
Hence, the answer is 8a – 7b.
Applications:
Much of the world measures temperature using the Centigrade thermometer, rather than the Fahrenheit scale. One can use the following conversion formula to change Fahrenheit to Centigrade: T_{C} = (5/9) (T_{F} 32)
When the Fahrenheit temperature in Los Angeles, California is 77 degrees, what is the equivalent Centigrade temperature?
T_{C} = (5/9) (T_{F} 32)
T_{C} = (5/9) (77 – 32) = (5/9) (45) = 25 degree Centigrade
Two girls, Oralia and Teresa, are balanced on a seesaw. The girls find that they must sit at distances from the support such that the products obtained by multiplying the weight of each by her distance from the fulcrum, or point of support, are equal.
Oralia weighs 85 pounds, and Teresa weighs 100 pounds. Oralia is seated 8 feet from the fulcrum. How far is Teresa sitting from the fulcrum?
W_{1}d_{1} = W_{2}d_{2}
(85)(8) = (100)d_{2}
6.8 feet= d_{2}
14.0 Students solve a quadratic equation by factoring or completing the square.
Example:
x^{2}  5x + 24 
= 
0 

(x  8)(x +3) 
= 
0 
(factoring) 
x  8 = 0 
and 
x + 3 = 0 

x = 8 
and 
x = 3 
Application:
A manager of an apartment house complex is deciding how much monthly rent to charge for each of the 60 newly built units. He has been charging $950 per month, and all 60 apartments have been occupied. He has found that for every additional $25 per month he charges, one apartment is likely to remain vacant. However, since service and maintenance costs are higher for occupied apartments than for vacant ones, he figures he can increase his profit margin by raising the monthly rental and still collect the same total rental of $47,000. The question he must answer is, how much should he charge and how many apartment units would then likely remain vacant at the higher rent?
He starts by letting x represent the number of vacant apartment units, so that 60x will represent the number of occupied apartment units.
The amount of total rent he wants to obtain is $57,000. Hence, the number of occupied apartment units times the amount he will be charging must equal $57,000.
Hence, $57,000 = (60 – x)($950 +$25x) Explantion: remember that for
every vacant unit he has raised the rent $25.
Hence, if he raises the rent $150 = (3)(25),
he would only have 60 – 3 = 57 rented apartments.)
57,000 = 57 – 950x + 1500x – 25x^{2} (Notice that we must arrange terms
so that one side equals 0. No other number will work.)
0 = 550x – 25x^{2}
0 = 22x – x^{2}
0 = (22 – x)(x)
0 = 22 – x and x = 0
X = 22
Hence, when all of the apartments are rented (x = 0), the total rent is $57,000. However, the apartment owner can realize the same total and increase his profit margin by raising the monthly rental per unit to $950 + (22)($25) = $1500, with 22 vacant units.
15.0 Students apply algebraic techniques to solve rate problems, work problems, and percent mixture problems.
Mixture Example:
What quantities of silver 60% and 82% pure must be mixed together to give 12 ounces of silver 70% pure?
Let x = the number of ounces of 60% pure silver.
Let y = the number of ounces of 82% pure silver.
The total amount of x + y = 12 ounces.
Now .60x will be the actual amount of silver in the “60% type,” and
.82y will be the actual amount of silver in the “82% type.”
Hence, .60x + .82y = (.70)(12), since (.70)(12) will be the actual amount of silver in the total mixture.
With this information we can solve the two simultaneous equations:
.60x + .82y = (.70)(12)
x + y = 12
.60x + .82y = 8.4
x + y = 12
.60x + .82y = 8.4
.60x + .60y = 7.2 (Multiplying both sides by .60)
.22y = 1.2 (Subtracting )
y = 5 5/11
x + 5 5/11 = 12
x = 6 6/11
16. Students understand the concepts of a relation and a function, determine whether a given relation defines a function, and give pertinent information about given relations and functions.
Any set of ordered pairs is a relation. Thus, (2, 5), (2, 6) and (2, 7) could define a relation. However, that relation is not a function, because a function is a singlevalued relation. For every first value, a function specifies a specific second value—not three, as in the above mentioned relation.
Example:
This is an important distinction. For example, suppose that we wish to purchase three hamburgers, each of which costs $1.50. Multiplying 3 times $1.50 = $4.50 The result can be written as (3, $4.50). Here, if the customer knows how many hamburgers he or she is purchasing, there is one total purchase price. We say that the total purchase price, P, is a function of the number of hamburgers purchased f(h). This can also be written: P =f(h) In this example, f(h) = $1.50 times h, or $1.50 h.
Application:
A rental company charges $50 plus $42 per day to rent 100 chairs for a wedding. Write a function to represent this situation, and calculate how much it will cost to rent the chairs for threedays.
In this situation, the total cost, C, of renting the chairs will depend upon how many days, d, they will be needed, plus $50. Hence, C is a function of d plus $50. This can be written: C = f(d) + 50. In this case, f(d) = $42 times d, or $42 d.
If we wish to rent the chairs for three days, f(d) = $42 times 3 = $126. But then we need to remember that there is a $50 fee, regardless of how many days we rent the chairs. Hence, C = f(d) + $50 = $126 + 50 = $176.
Application:
Sometimes costs will depend upon two functions. For example, when renting a truck, the cost might depend upon the number of days rented and the number of miles driven.
Let’s say a rental company charges $34 per day and 23 cents per mile. The cost (C) will then be a function of both the number of days (d) the truck is rented, and the number of miles (m) the truck is driven. We can write that as: C = f_{1}(d) + f_{2}(m)
In this case f1(d) = $34d and f2(m) = .23m
Let’s assume that we will need to truck for 2 days and we drove it 250 miles. We can then calculate our total cost as:
C = f_{1}(d) + f_{2}(m) = $34d + .23m = $34(2) + .23(250) = $68 + $57.50 = $125.50
Application:
Newly hatched channel catfish typically weigh about 0.06 grams. During the first six weeks of life their weight increases by about 10% each day. The following exponential mathematical model describes their growth:
y = C(1 + r)^{t} where y is the weight of the catfish after so many days (t); C is the initial weight in grams; and r is the growth rate.
After one week, or 7 days, their weight is:
y = 0.06(1 +0.10)7
y = 0.12 grams
Assume that you purchased a $25,000 car. It is expected to lose value or depreciate at the rate of 12% per year. The following exponential mathematical model describes what
the car will be worth after t years:
y = C (1 – r)^{t} where y is the value of the car after t years; C is the initial value of
the car; and r is the rate of depreciation.
Let’s examine what the car will be worth after 10 years:
y = 25,000(1  .12)10
y = 25,000(.2785) = $6,962.50
17.0 Students determine the domain of independent variables and the range of dependent variables defined by a graph, a set of ordered pairs, or a symbolic expression.
Example:
Find the domain of y = 2√x
Use several values in the domain to make a table of values for the function.
Solution:
Since the square root is defined only when x, in this case, is nonnegative, the domain of y = 2√x consists of all nonnegative numbers.
Table:
x 
y 
0 
2√0 = 0 
1 
2√1 = 2 
2 
2√2 = 2.8 
3 
2√3 = 3.5 
4 
2√4 = 4.5 
Application:
The production of corn is sometimes a profitable business, depending upon the weather and how much it costs to bring the corn to market. According to Iowa State University, the amount of corn grown per acre depends upon the seeding rate.
Based upon this research, Iowa State believes that the best seeding rate for corn is about 35,000 seeds per acre.
What is the domain and range represented by the above graph?
{Domain: 20,000 to 45,000 plants per acre}
{Range: 77% to 100% of maximum yield}
Corn seed prices are about $1.30 per 1,000 seed. The corn grain market price is about $3.00 per bushel. (A bushel is 56 pounds.) Corn farmers produce about 200 bushels of corn per acre.
How much would 35,000 seeds cost? {$45.50} How much does 200 bushels of corn sell for? {$600.00} What other costs will a corn farmer face? {planting, weed control, pest management, harvesting, transportation to market.}
18.0 Students determine whether a relation defined by a graph, a set of ordered pairs, or a symbolic expression is a function and justify the conclusion.
If no two ordered pairs in a relation have the same first component and different second componets, then the relation is called a function. Graphically, this means that no vertical line will interest the graph at more than one point.
Examples of Graphs of Functions:
Graph: y = x+1
x 
y 

2 
1 

1 
0 

0 
1 

1 
2 

2 
3 

Graph: y = x^{2}
x 
y 

2 
4 

1 
1 

0 
0 

1 
1 

2 
4 

Graph: y = x^{3}
x 
y 

2 
8 

1 
1 

(1/2) 
(1/8) 

0 
0 

1/2 
1/8 

1 
1 

2 
8 

Graph: y = 8/(x^{2} + 4) (Called "The Witch of Agnessi")
x 
y 

5 
8/29 

3 
8/13 

2 
1 

1 
8/5 

0 
2 

1 
8/5 

2 
1 

3 
8/13 

5 
8/29 

Examples of Graphs of Relations:
Graph:
x^{2} + y^{2} = 25
y^{2} = 25  x^{2}
y = √(25  x^{2})
x 
y 

5 
0


4 
±3


3 
±4


2 
±4.58


1 
±4.89


0 
±5


1 
±4.89


2 
±4.58


3 
±4


4 
±3 

5 
0 

Notice that if x > 5 or x < 5 y is imaginary.
Graph: y^{2} = x^{3}/(1  x) (Called "The Cissoid of Diocles.")
y = √( x^{3})/√(1  x^{})
x 
y 

0 
0 

1/8 
±.05(approximately) 

1/4 
±.14(approximately) 

1/2 
±.5(approximately) 

3/4 
±1.3(approximately) 

.9 
±2.7(approximately) 

.99 
±9.85(approximately) 

1  undefined  
19.0 Students know the quadratic formula and are familiar with its proof by completing the square.
Example of completing the square:
x^{2} + 2x – 15 = 0 Divide both sides of the equation so that the coefficient of x^{2} equal to 1.
x^{2} + 2x + 1 = 15 + 1 Add to each side the square of half the coefficient of x.
x + 1 = ±4 Take the square root of each side.
x + 1 = 4 and x + 1 = 4 Solve the two resulting equations separately.
x = 3 and x = 5
Application:
Tom (T) can paint a building in 5 days less than the time it takes Jim (J). Working together, they can complete the work in 6 days. How long does it take each to do the job alone?
Of these two solutions, the only one that makes sense is that Tom takes 10 days, since he couldn’t take a negative amount of time. Therefore, Jim would take 15 days.
20.0 Students use the quadratic formula to find the roots of a seconddegree polynomial and to solve quadratic equations.
Example:
Solve, using the quadratic formula: 2x^{2} – 6x – 9 = 0
Using the quadratic formula, a=2, b=6, and c=9.
Hence, x = 6±√(36+72) = 6±10.39 = 4.1 and 1.1
(2)(2) 4
Application:
The distance s a body falls in a vacuum is given by s = v_{0}t + (1/2)gt^{2},
where s is in feet, t is in seconds, v0 is the initial velocity in feet per second,
and g is the constant of acceleration due to gravity (approximately 32 ft/sec/sec).
How long will it take a body to fall 150 feet if v0 is 20 feet per second?
s = v_{0}t + (1/2)gt^{2}
150 = 20t + (1/2)(32)t^{2}
0 = 16t^{2} + 20t – 150
t = 20±√(400 – 9600) = 20 ±100 = 2.5 seconds [3.75 doesn’t fit]
32 32
If the body starts to fall from rest, then 150 = (0)t + (1/2)(32)t^{2} = 16 t^{2}
We then solve for t in the equation: 0 = 16t^{2}  150
t = 0±√[02(4)(16)(150)] = ±√9600 = 3.06 seconds
(2)(16) 32
21.0 Students graph quadratic functions and know that their roots are the xintercepts.
22.0 Students use the quadratic formula or factoring techniques or both to determine whether the graph of a quadratic function will intersect the xaxis in zero, one, or two points.
Example:
Consider the following equations:
(A) x^{2} – 2x – 3 = 0  
(B) x^{2} – 2x + 1 = 0  
(C) x^{2} – 2x + 4 = 0  
Case A:
x^{2} – 2x – 3 = 0
(x +1)(x – 3) = 0
x + 1 = 0 and x 3 = 0
x = 1 and x = 3
In this example, the graph of the equation crosses the xaxis at two points,
represented by the two solutions to the
equation.
Case B:
x^{2} – 2x + 1 = 0
(x – 1)(x – 1) = 0
x = 1 This is an example of a quadratic equation with just one solution. The
graph touches the xaxis at just one point.
Case C:
x^{2} – 2x + 4 = 0
x = 2 ±√(4 – 16) = 2 ±√(– 12) = 2 ±√4√(– 3) = 2 ±2√(– 3)
2 2 2 2
x = 1±√(3), which is an imaginary solution.
Hence the graph of equation (C) does not cross the xaxis.
Notice that the graphs of all three equations form the shape of a parabola, which is same shape as the St. Louis Arch and the McDonald’s arches.
23.0 Students apply quadratic equations to physical problems, such as the motion of an object under the force of gravity.
Example solution of a quadratic equation:
x^{2} + 10x + 16 = 0
(x + 8)(x + 2) = 0
x + 8 = 0 and x + 2 = 0
x = 8 x = 2
Application:
x(t) = x_{0} + v_{0}t + (½)at^{2}
A car with an initial velocity of 30 meters per second begins to accelerate at a constant rate of 2.8 meters per second. How long does it take the car to travel 200 meters?
200 = 0 + 30t + (1/2)(2.8)t^{2}
0 = 1.4t^{2} + 30t – 200
t = 30 ± √ [(30)^{2} – (4)(1.4)(200)]
2(1.4)
t = 5.34 seconds
24.0 Students use and know simple aspects of a logical argument.
Explanation:
Logical arguments involve:
1. Figuring out what is known;
2. Thinking about what is known means, includes, and/or implies; and
3. Determining whether a conclusion can be reached which is a part of what is known.
Application:
Consider offers made by three different cell phone providers. All of the providers advertise their plan for $69.99 a month with 700 minutes of talk time.
The first provider ABC has a family share plan that includes two lines. This plan requires a minimum of two lines, but this number can be increased up to a maximum of five lines, with each additional line costing $9.99 per month. The ABC package also includes:
The second provider Sprung has a family share plan that includes two lines. This plan requires a minimum of two lines, but this number also be increased up to a maximum of five lines. Each additional line costs $9.99 per month. The Sprung package also includes:
The third provider Horizon has a family share plan that includes one line. This plan requires a minimum of two lines and can have up to a maximum of five lines. Each additional line costs $9.99 per month. The Horizon package also includes:
To select the best plan for your family use a logical argument, first figuring out what is known.
At this point, you know what the three companies offer, but don’t forget what you want. Let’s assume that you have budgeted $75.00 a month for cell phones, and need two.
Based upon your past history, you recall that your family uses their cell phones about 700 minutes a month, and frequently talk throughout the day to family members who live elsewhere in California.
With that information about what is known, your family needs to examine the meaning of each plan for your family’s situation.
Since Horizon offers a family share plan with 700 shared minutes that is offered for $69.99 a month, it is within our budget for cell phone coverage. However, Horizon would require us to have two lines, and the plan only covers one line. Since each additional line costs $9.99 a month, the real price for Horizon would be $79.98, which is more than our budget will allow. Hence, since this is more than we want to spend on cell phones, we definitely need to consider the other two plans.
Since Sprung has a plan for $69.99 a month with 700 minutes including two lines, the Sprung plan is within our budget. Their unlimited mobiletomobile and nationwide long distance service is appealing, but their unlimited night and weekend minutes does not start until 7 p.m. This is a decided drawback to us, because we would like to speak with family members at any time, and not have to wait until after 7 p.m. without incurring an additional fee. Hence, while the Sprung plan better meets our needs than the Horizon plan, we proceed to consider the ABC plan.
The ABC plan costs $69.99 a month with 700 minutes including two lines, and that is within our budget. The ABC plan would allow us to share the minutes, and if we don’t use them they would rollover to another month. Further, they have unlimited nights and weekends, without having to wait until after 7 p.m., and free long distance. Therefore, the ABC plan includes everything that we sought in a cell phone service package. Hence, our family concludes that we should select the ABC plan.
24.1 Students explain the difference between inductive and deductive reasoning and identify and provide examples of each.
Example:
Inductive reasoning is the process of reaching a conclusion about the truth of a statement by making use of observations. Theories in science are based upon inductive reasoning. Scientific theories consist of statements about what is expected to happen in cases similar to other observations.
For example, Galileo rolled a ball down an inclined track and measured how fast it traveled. Using his pulse to measure time, he noticed that the ball proceeded down the track in proportional units to the square of the time, as shown in the following table:
Pulse: 1 2 3 4 5 6 …
Distance: 1 4 9 16 25 36 …
While Galileo couldn’t check out all situations, he used inductive reasoning to reach the above conclusion, based upon his observations.
Deductive reasoning begins with statements accepted without proof. These statements are called axioms. When we use deductive reasoning, we must show that the statement we are trying to prove is a subset of the axioms. For example, in ordinary arithmetic with the set of natural numbers, the axiom or law of trichotomy is accepted without proof. The law of trichotomy says that exactly one of the following is true: a = b; or a = b plus a natural number; or a plus a natural number = b.
Using deductive reasoning we can prove that all positive even natural numbers a=b; a=b+r; or a+r=b, where r is a natural number. Since the positive even natural numbers are a subset of all natural numbers, and since that law of trichotomy holds for all natural numbers, the proposition holds for all positive even natural numbers.
Application:
While scientific theories are based upon observations, Goldbach used inductive reasoning to discover a theory in mathematics. Goldbach observed that every even integer greater than 4 is the sum of two odd primes. While no one has been able to find a counterexample, no one has been able to deductively prove that this observation discovered inductively is always true.
Here are some observations:
6 = 3 + 3 12 = 7 + 5 18 = 13 + 5 24 = 19 + 5 30 = 23 + 7 36 = 31 + 5
8 = 5 + 3 14 = 11 + 3 20 = 17 + 3 26 = 19 + 7 32 = 29 + 3 38 = 19 + 19
10 = 7 + 3 16 = 13 + 3 22 = 19 + 3 28 = 23 + 5 34 = 31 + 3 40 = 37 + 3
Check out some other even numbers.
24.2 Students identify the hypothesis and conclusion in logical deduction.
The hypothesis of a statement includes all that is assumed, in order to reach the conclusion. A hypothesis is often identified by the qualifier “if.” If … the hypothesis … , then …. the conclusion.
Example:
Hypothesis: If two integers are both odd,
Conclusion: Then their product will be odd.
Proof: Let A and B be any two odd integers.
A and B can be written as 2n – 1 and 2m – 1, because this is the definition of an
odd integer, where n and m are integers.
(2n – 1) (2m – 1) = 4nm – 2n – 2m + 1 = 2(2nm – n – m) + 1
Since 2(2nm – n – m) is an even number, based on the definition of even
numbers, adding one to an even number will always produce an odd number.
Therefore, if A and B are odd integers, the product AB will be an odd integer.
Decisionmaking Application:
A man walks up a restaurant counter where there are three seated customers. Though he had never seen any of them before, he had previously learned that one of them was a used car salesman who was unable to make a truthful statement, one was an attorney who shuffled truth and falsehood so thoroughly that one could never be sure that a given statement of his was true or false, and the other invariably told the truth—an inconvenient trait that made him unemployable.
The man hoped by questioning them to identify each with his occupation. He sat down to the left of the group and ordered a milkshake. Then he turned to the man beside him and asked of him the occupation of the man in the middle. He was told that he sold used cars. Then he spoke to the man in the middle and asked him directly about his occupation.
After he thought about the middle man’s reply, he finished his milkshake and left, having determined the trade of each of the men.
What did each do?
(Notice that all of the foregoing information is a hypothesis. We are now asked to determine what conclusions we can logically draw from that information.)
Solution:
Let’s call the used car salesman: “U”
the attorney: “A,” and
the 3rd man: “T” for truthful.
The first man he asked either told the truth or didn’t tell the truth. If he didn’t tell the truth, he was either the used car salesman (U) or the attorney (A). If he did tell the truth, he must have been the third man (T).
Let’s assume he was T, having told the truth. When the middle man was spoken to, if he said he was an attorney (not telling the truth if this scenario is correct), then we would know that the men were: T, U, and A.
Again, assuming that the first man told the truth, and the middle man said he was a used car salesman, we would know that the first man was not T, because T always tells the truth and U doesn’t.
Now, let’s assume that the first man didn’t tell the truth. Then the first man must be U, because A wouldn’t have given any information. If he is U, then the middle man must either be A or T. If the middle man is A, once again, in speaking with him we will see that we can’t get any information, so the men must be: U, A, and T. If the middle man is determined not to be A, he must be T. So the men must be: U, T, and A.
A Criminal Justice Application:
When investigating a crime scene, a detective has to make hypotheses about what has occurred. The detective looks for clues and evidence that either supports or refutes their hypotheses, which change as more information becomes available.
Here is what a detective found at a crime scene:
Lab Results:
The detective’s hypothesis: A murder has been committed.
Supporting evidence:
A person committing suicide would not be able to shoot themselves twice without leaving residue powder on their hands. They also would not have missed one time if they were shooting themselves. A robbery also seems to have occurred, since a wallet and jewelry are missing. A second party was involved because of the blood on the couch which points to a murder suspect.
24.3 Students use counterexamples to show that an assertion is false and recognize that a single counterexample is sufficient to refute an assertion.
Counterexamples:
Here is a list of the first few prime numbers: 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 . . .
Many have sought a formula which would always produce prime numbers.
f(n) = n^{2} + n + 17 yields primes for n < 16. (Legendre’s formula)
f(n) = 2n^{2} + 29 yields primes for n < 29.
f(n) = n^{2} – n +41 yields primes for all n <41. (Euler’s formula)
f(n) = n^{2} 79n + 1601 yields primes for all n<80.
Now suppose that one discovered Legendre’s formula, and tried it out for n=1, n= 2, n=3, n=4, n=5, and for a few more. One might easily conclude that this formula would always produce prime numbers. However, all one needs to show is one counterexample to prove that Legendre’s formula does not always produce prime numbers.
For example, if n=16, n^{2} + n + 17 = 16^{2} + 16 + 17 = 289, and 289 = (17)(17). Hence, 289 is not a prime number.
25.1 Students use properties of numbers to construct simple, valid arguments (direct and indirect) for, or formulate counterexamples to, claimed assertions.
Example:
Prove: If a number (n) is not evenly divisible by 3, then n^{2} – 1 is evenly divisible by 3.
All numbers (integers) are of the form n, n+1, and n+2.
The above proposition pertains to those numbers that are not evenly divisible by
In other words, if n is divisible by 3, then n+1 and n+2 are not, and we want to prove that when these other numbers of the form (n+1) and (n+2) are squared and
1 is subtracted, that the result will always be evenly divisible by 3.
So, we examine what happens when these other numbers are written in the form
n^{2} – 1:
(n+1)^{2} – 1 = n^{2} + 2n +1 – 1 = n^{2} + 2n = n(n + 2) which
is divisible by 3, since n is.
Likewise, if n is divisible by 3, then (n+2)^{2} – 1 = n^{2} + 4n + 4 – 1 = n^{2} + 4n + 3
which is divisible by 3, since n is and all the terms in the expression are
divisible by 3.
Therefore, when any number not divisible by 3 is transformed into the form
n^{2} – 1, the result is always divisible by 3.
25.2 Students judge the validity of an argument according to whether the properties of the real number system and the order of operations have been applied correctly at each step.
The most vital contribution of the classical Greek period from 600 to 300 B.C. to mathematics was their insistence that all mathematical results be established deductively on the basis of axioms.
To judge the validity of an argument, one must show how each step follows from an axiom or definition.
For example, lets see if we can prove that 23 times 15 = 345, based on the field properties, the definition of a base10 number, and the addition and multiplication tables from 2 to 9.
Statement Justification
(23)(15)
(23)(1· 10 + 5) Definition of a base10 number.
(23)(10 + 5) 1 is the multiplicative identity element.
23·10 + 23·5 Distributive axiom.
(2·10 + 3)(10) + (2·10 + 3)(5) Definition of a base10 number.
2·10·10 + 3·10 + 2·10·5 + 3·5 Distributive axiom
2·10·10 + 3·10 + 2·5·10 + 3·5 Commutative axiom for multiplication
2·10·10 + 3·10 + 10·10 + 15 Basic facts of multiplication
2·10·10 + 3·10 + 1·10·10 + 15 1 is the multiplicative identity element
2·10·10 + 1·10·10 + 3·10 +15 Commutative axiom for addition.
2·10·10 + 1·10·10 + 3·10 +1·10 + 5 Definition of a base10 number.
(2+1)10·10 + (3+1)10 + 5 Distributive axiom
3·10·10 + 4·10 + 5 Basic facts of addition.
345_{base10} Definition of a base10 number.
25.3 Given a specific algebraic statement involving linear, quadratic, or absolute value expressions or equations or inequalities, students determine whether the statement is true sometimes, always or never.
Example:
Let’s define the “square of the opposite” of a rational number, x, as (1)x^{2}
Which, if any, of the following statements is true?
Answer:
Consider numbers on the rational number line:
If x=2, then (1)x^{2}=4.
If x=1, then (1)x^{2}=1.
If x=1/2, then (1)x2=(1/4).
If x=0, then (1)x^{2}=0.
If x=(1/2), then (1)x^{2}=(1/4).
If x=1, then (1)x^{2}=1.
If x=2, then (1)x^{2}=4.
Notice that rational numbers larger than 2 will behave like 2, and that rational numbers smaller than 2 will behave like 2.
Hence, if x is greater than 0, (1)x^{2} will be smaller than x.
If x=0, (1)x2 = 0.
If x is less than 0 but greater than 1, (1)x^{2} will be larger than x.
If x=1, then (1)x^{2} will be equal to x.
If x is less than 1, (1)x^{2} will be less than x.
Therefore, answers C and D are correct.
Application:
Graph the equation: y = (1) x^{2}
x 
y 

2 
4 

1 
1 

0 
0 

1 
1 

2 
4 

Note: Underwood Dudley argues in a recent article ("What Is Mathematics For?") in the Notices of the American Mathematical Society (Vol. 57, Num. 5, May 2010, 608613) that mathematics "is needed very, very seldom, and we do not need to train millions of students in it to keep businesses going." He says that while mathematics "is the most glorious creation of the human intellect," "what mathematics education is for ... [is] to teach reasoning, usually through the medium of silly problems."