Real Numbers and Algebra Chapter 1 Real Numbers and Algebra

1.1 Describing Data with Set of numbers Natural Numbers are counting numbers and can be expressed as N = { 1, 2, 3, 4, 5, 6, …. }  Set braces { }, are used to enclose the elements of a set. A whole numbers is a set of numbers, is given by W = { 0, 1, 2, 3, 4, 5, ……}  

…Continued The set of integers include both natural and the whole numbers and is given by   I = { …, -3, -2, -1, 0, 1, 2, 3, ….}   A rational number is any number can be written as the ratio of two integers p/q, where q = 0. Rational numbers can be written as fractions and include all integers. Some examples of rational numbers are  8/1, 2/3, -3/5, -7/2, 22/7, 1.2, and 0.

…continued Rational numbers may be expressed in decimal form that either repeats or terminates.  The fraction 1/3 may be expressed as 0.3, a repeating decimal, and the fraction ¼ may be expressed as 0.25, a terminating decimal. The overbar indicates that 0.3 = 0.3333333…. Some real numbers cannot be expressed by fractions. They are called irrational numbers. 2, 15, and  are examples of irrational numbers.

0 is called the additive identity and a . 1 = 1 . a = a, Identity Properties  For any real number a,   a + 0 = 0 + a = a, 0 is called the additive identity and a . 1 = 1 . a = a, The number 1 is called the multiplicative identity. Commutative Properties For any real numbers a and b, a + b = b + a (Commutative Properties of addition) a.b = b.a (Commutative Properties of multiplication)  

(a.b) . c = a . (b . c) (Associative Properties for multiplication) …Continued Associative Properties  For any real numbers a, b, c, (a + b) + c = a + (b + c) (Associative Properties of addition) (a.b) . c = a . (b . c) (Associative Properties for multiplication) Distributive Properties a(b + c) = ab + ac and a(b- c) = ab - ac

1.2 Operation on Real Numbers The Real Number Line -3 -2 -1 0 1 2 3 Origin -2 2 -2 = 2 Absolute value cannot be negative 2 = 2 -3 -2 -1 0 1 2 3 Origin

…Continued If a real number a is located to the left of a real number b on the number line, we say that a is less than b and write a<b. Similarly, if a real number a is located to the right of a real number b, we say that a is greater than b and write a>b. Absolute value of a real number a, written a , is equal to its distance from the origin on the number line. Distance may be either positive number or zero, but it cannot be a negative number.

Arithmetic Operations Addition of Real Numbers To add two numbers that are either both positive or both negative, add their absolute values. Their sum has the same sign as the two numbers. Subtraction of real numbers For any real numbers a and b, a-b = a + (-b). Multiplication of Real Numbers The product of two numbers with like signs is positive. The product of two numbers with unlike signs is negative. Division of Real Numbers For real numbers a and b, with b = 0, a/b = a . 1/b That is, to divide a by b, multiply a by the reciprocal of b.

1.3 Bases and Positive Exponents Squared 4 Cubed 4 4 4 4 4 4 . 4 = 42 4 . 4. 4 = 43 Exponent Base

Powers of Ten Power of 10 Value 103 1000 102 100 101 10 1 10-1 1/10 = 0.1 10-2 1/ 100 = 0.01 10-3 1/ 1000 = 0.001

1.3 Integer Exponents Let a be a nonzero real number and n be a positive integer. Then   an = a. a. a. a……a (n factors of a )  a0 = 1, and  a –n = 1/an a -n b m b -m = a n a -n b n b = a  

… cont The Product Rule am . an = a m+n   For any non zero number a and integers m and n, am . an = a m+n The Quotient Rule For any nonzero number a and integers m and n, am = a m – n a n

For nonzero numbers a and b and any integer n. Raising Products To Powers For any real numbers a and b and integer n, (ab) n = a n b n Raising Powers to Powers For any real number a and integers m and n, (am)n = a mn Raising Quotients to Powers For nonzero numbers a and b and any integer n. a n = an b bn

…Continued A positive number a is in scientific notation when a is written as b x 10n, where 1 < b < 10 and n is an integer.  Scientific Notation Example : 52,600 = 5.26 x 104 and 0.0068 = 6.8 x 10 -3  

1.4 Variables, Equations , and Formulas A variable is a symbol, such as x, y, t, used to represent any unknown number or quantity. An algebraic expression consists of numbers, variables, arithmetic symbols, paranthesis, brackets, square roots. Example 6, x + 2, 4(t – 1)+ 1, X + 1

…cont An equation is a statement that says two mathematical expressions are equal.  Examples of equation  3 + 6 = 9, x + 1 = 4,  d = 30t, and x + y = 20   A formula is an equation that can be used to calculate one quantity by using a known value of another quantity. The formula y = x/3 computes the no. of yards in x feet. If x= 15, then y=15/3= 5.

Square roots The number b is a square root of a number a if b2 = a. Example - One square root of 9 is 3 because 32 = 9. The other square root of 9 is –3 because (-3)2 = 9. We use the symbol to 9 denote the positive or principal square root of 9. That is, 9 = +3. The following are examples of how to evaluate the square root symbol. A calculator is sometimes needed to approximate square roots, 4 = + 2 - The symbol ‘ + ‘ is read ‘plus or minus’. Note that 2 represents the numbers 2 or –2.  

Cube roots The number b is a cube root of a number a if b3 = a   The number b is a cube root of a number a if b3 = a The cube root of 8 is 2 because 23 = 8, which may be written as 3 8 = 2. Similarly 3 –27 = -3 because (- 3)3 = - 27. Each real number has exactly one cube root.

1.5 Introduction to graphing Relations is a set of Ordered pairs. If we denote the ordered pairs in a relation (x,y), then the set of all x-values is called the domain of the relation and the set of all y values is called the range.

Example 1. Find the domain and range for the relation given by   Find the domain and range for the relation given by S = {( -1, 5), (0,1), (2, 4), (4,2), (5,1)} Solution The domain D is determined by the first element in each ordered pair, or D ={-1, 0, 2, 4,5} The range R is determined by the second element in each ordered pair, or R = {1,2,4,5}

The Cartesian Coordinate System Quadrant II y Quadrant I y (1, 2) 2 1 -1 -2 Origin 2 1 -1 -2 x x -2 -1 1 2 -2 -1 1 2 Quadrant III Quadrant IV The xy – plane Plotting a point

Scatterplots and Line Graphs If distinct points are plotted in the xy- plane, the resulting graph is called a scatterplot. Y 7 6 5 4 3 2 1 (4, 6) (3, 5) (5, 4) (2, 4) (6, 3) (1, 2) X 1 2 3 4 5 6 7

Using Technology

Ex 9 page 44 Make a table for y = x 2 / 9, starting at x = 10 and incrementing by 10 and compare The table for example 4 ( pg 41) Go to Y= and enter x 2 /9 Go to 2nd then table set and enter Go to 2nd then table

Viewing Rectangle ( Page 57 ) Ymax }Ysc1 Xmax Xmin Xsc1 Ymin [ -2, 3, 0.5] by [-100, 200, 50]

Making a scatterplot with a graphing calculator Plot the points (-2, -2), (-1, 3), (1, 2) and (2, -3) in [ -4, 4, 1] by [-4, 4, 1] (Example 10, page 58) Go to 2nd then stat plot Go to Stat Edit then enter points Scatter plot [ -4, 4, 1] by [-4, 4, 1]

Example 11 Cordless Phone Sales Year 1987 1990 1993 1996 2000 Phones (millions) 6.2 9.9 18.7 22.8 33.3 Go to Stat edit and enter data Enter line graph Hit graph Enter datas in window [1985, 2002, 5] by [0, 40, 10]

Linear Functions and Models Chapter 2 Linear Functions and Models

Ch 2.1 Functions and Their Representations A function is a set of ordered pairs (x, y), where each x-value corresponds to exactly one y-value.   Input x Output y Function f (x, y) Input Output

…continued y is a function of x because the output y is determined by and depends on the input x. As a result, y is called the dependent variable and x is the independent variable To emphasize that y is a function of x, we use the notation y = f(x) and is called a function notation.   A function f forms a relation between inputs x and outputs y that can be represented verbally (Words) , numerically (Table of values) , symbolically (Formula), and graphically (Graph).  

Representation of Function y Table of Values Graph 20 16 12 8 4 x (yards 1 2 3 4 5 6 7 y(feet) 9 12 15 18 21 y= 3x x 0 4 8 12 16 20 24 Numerically Graphically

Diagrammatic Representation Function Not a function x y x y 3 6 9 (1, 3), (2, 6), (3, 9) 1 2 3 4 5 6 1 2 1 2 3 4 5 (1, 4), (2, 5), (2, 6) (1,4), (2, 4), (3, 5) Example 1, pg - 77

Domain and Range Graphically The domain of f is the set of all x- values, and the range of f is the set of all y-values y 3 2 1 Range R includes all y – values satisfying 0 < y < 3 x Domain D includes all x values Satisfying –3 < x < 3 Range -3 -2 -1 0 1 2 3 Domain Pg 79 Ex-5

Vertical Line Test If each vertical line intersects the graph at most once, then it is a graph of a function 5 4 3 2 1 -1 -2 -3 -4 -5 Not a function (-1, 1) -4 -3 -2 -1 0 1 2 3 4 (-1, -1)

…Continued Not a function Example 9 page - 81 (-1, 1) (1, -1) 4 3 2 1 -2 -3 (-1, 1) -3 -2 -1 0 1 2 3 (1, -1) Example 9 page - 81

Using Technology Graph of y = 2x - 1 Hit Y and enter 2x - 1 -1 -3 1 2 3 Graph of y = 2x - 1 Hit Y and enter 2x - 1 Hit 2nd and hit table and enter data [ - 10, 10, 1] by [ - 10, 10, 1]

2.2 Linear Function A function f represented by f(x) = ax + b, where a and b are constants, is a linear function. 100 90 80 70 60 100 90 80 70 60 f(x) = 2x + 80 0 1 2 3 4 5 6 0 1 2 3 4 5 6 Scatter Plot A Linear Function Ex- 1, 2, 3, 4, 5 Pg 91

Modeling data with Linear Functions Example 7 1500 1250 1000 750 500 250 Cost (dollars) 8 12 16 20 x Credits Symbolic Representation f(x) = 80x + 50 Numerical representation 4 8 12 16 $ 370 $ 690 $1010 $1330

Using a graphing calculator Example 5 Give a numerical and graphical representation f(x) = ½ x - 2 Numerical representation – Y1 = .5x – 2 starting x = -3 Graphical representation – [ -10, 10, 1] by [-10, 10, 1]

2.3 The Slope of a line Y 8 7 6 5 4 3 2 1 Cost (dollars) Rise = 3 Slope = Rise = 3 Run 2 Run = 2 1 2 3 4 5 6 x Gasoline (gallons ) Cost of Gasoline Every 2 gallons purchased the cost increases by $3

Where x1 = x2. That is, slope equals rise over run. rise y2 - y1 m = run = x2 - x1   y2 (x2, y2) y2 –y1 y1 (x1, y1) x2 –x1 Rise The Slope m of the line passing through the points (x1 y1 ) and (x2, y2) is m= y2 –y1/x2 –x1 Where x1 = x2. That is, slope equals rise over run.   Run Ex- 1, 2 pg 104

-1 m = - ½ < 0 m = 2 > 0 2 Positive slope Negative slope m = 0 4 3 2 1 -1 -2 4 3 2 1 -1 -2 -3 2 -1 m = 2 > 0 m = - ½ < 0 2 -4 -2 1 2 3 4 - 4 -2 1 2 Positive slope Negative slope m = 0 m is undefined Zero slope Undefined slope

(0, 4) Example 2 - Sketch a line passing through the point (0, 4) and having slope - 2/3 y - values decrease 2 units each times x- values increase by 3 (0 + 3, 4 – 2) = (3, 2) 4 3 2 1 ( 0, 4) Rise = -2 ( 3, 2) - 4 - 3 - 2 1 0 1 2 3 4

The line with slope m and y = intercept b is given by Slope-Intercept Form   The line with slope m and y = intercept b is given by   y= mx + b The slope- Intercept form of a line Example – 3, and 4, 5, 6, 7 pg - 106

Example - 3 3 2 Y = ½ x 1 Y = ½ x + 2 -1 -2 -3 -3 -2 -1 1 2 -3 -2 -1 1 2 Y = ½ x - 2

Analyzing Growth in Walmart Example 9 Year 1997 1999 2002 2007 Employees 0.7 1.1 1.4 2.2 3.0 2.5 2.0 1.5 1.0 0.5 Employees (millions) m3 m2 m1 Years 0 1999 2003 2007 m1 = 1.1 – 0.7 = 0.2 m2 = 1.4 - 1.1 = 0.1 and 1999 – 1997 2002 – 1999 m3 = 2.2 - 1.4 = 0.16 2007 - 2002 Average increase rate

The point- slope form of a line The line with slope m passing through the point (x1 , y1 ) is given by y = m ( x - x1 )+ y1 Or equivalently, y – y1 = m (x –x1) The point- slope form of a line (x, y) y – y1 (x1, y1) x – x1 m = y – y1 / x – x1 Ex 1, 2,3 pg 117

Horizontal and Vertical Lines Equation of vertical line y y x = h x x b h y= b Equation of Horizontal Line

…Continued Parallel Lines Two lines with the same slope are parallel. m1 = m2 Perpendicular Lines Two lines with nonzero slopes m1 and m2 are perpendicular if m1 m2 = -1 Examples – 7, 8 Page - 123

m2 = -1 m1 = 1 m2 = - 1/m1 m2 = - 1/2 m1 = 2 m1

Linear Equations and Inequalities Chapter 3 Linear Equations and Inequalities

3.1 Linear Equation in One Variable A Linear Equation in one variable is an equation that can be written in the form     ax + b = 0  Where a = 0 A linear function can be written as f(x) = ax + b Examples of Linear equation. 2x –1 = 0, -5x = 10 + x, and 3x + 8 = 5  

Properties of Equality Addition Property of Equality If a, b, c are real numbers, then a = b is equivalent to a+ c = b + c. Multiplication Property of Equality If a, b, c are real numbers with c = 0, then a = b is equivalent to ac = bc. Example 1,2,3,4 pg 146

Example 7 Solving a Linear equation graphically using technology [ -6, 6, 1] by [-4, 4, 1]

Example 8 In 1987 CD and LP record sales were both 107 million [ 1984, 1991, 1] by [0, 350, 50] in 1987 In 1987 CD and LP record sales were both 107 million

Standard form of a line An equation for a line is in standard form when it is written as ax + b = c, where a, b, c are constants With a, b and c are constants with a and b not both 0 To find x-intercept of a line, let y = 0 in the equation and solve for x To find y-intercept of a line, let x = 0 in the equation and solve for x

3.2 Linear Inequality in One Variable A linear inequality in one variable is an inequality that can be written in the form   ax + b > 0, where a = 0. ( The symbol > may be replaced with >, <, or > ) There are similarities among linear functions, equations, and inequalities. A linear function is given by f(x) = ax + b, a linear equation by ax + b = 0, and a linear inequality by ax + b > 0.

Examples of linear inequalities are 2x + 1< 0, 1-x > 6, and 5x + 1 < 3 – 2x A solution to an inequality is a value of the variable that makes the statement true. The set of all solutions is called the solution set. Two inequalities are equivalent if they have the same solution set. Inequalities frequently have infinitely many solutions. For example, the solution set to the inequality x- 5> 0 includes all real numbers greater than 5, which can be written as x > 5. Using set builder notation, we can write the solution set as { x x > 5 }. Meaning This expression is read as “ the set of all real numbers x such that x is greater than 5. “

3.2 Solving a problem Using a variable to an unknown quantity Number problem n is the smallest even integers Three consecutive three even integers n, n + 2, n +4 n is the smallest odd integers Three consecutive odd integers n, n + 2, n + 4

Mixing acid ( Ex 8 pg 163) 20 % 2 liters 60% x liters + = 50% Step 1 x: liters of 60% sulphuric acid x + 2: Liters of 50% sulphuric acid Step 2 Concentration Solution Amount Pure Acid 0.20 (20%) 2 0.20(2) 0.60 (60%) x 0.60x 0.50(50%) x + 2 0.50(x + 2) Equation 0.20(2) + 0.60x = 0.50(x + 2) (pure acid in 20% sol.) (pure acid in 60% sol.) (pure acid in 50% sol.) Step 3 Solve for x 0.20(2) + 0.60x = 0.50(x + 2) 2 (2) + 6x = 5(x + 2) Multiply by 10 4 + 6x = 5x + 10 (Distributive Property) Subtract 5x and 4 from each side x = 6 Six liters of the 60% acid solution should be added to the 2 liters of 20% acid solution. Step 4 If 6 liters of 60% acid solution are added to 2 liters of 20% solution, then there will be 8 liters of acid solution containing 0.60(6) + 0.20(2) = 4 liters of pure acid. Check The mixture represents a 4/8 = 0.50 or 50% mixture

Ex 46 (Pg 166) Anti freeze mixture A radiator holds 4 gallons of fluid x represents the amount of antifreeze that is drained and replaced The remaining amount is 4 – x 20 % of + 70 % of solution of = 50% of solution 4 – x gallons x gallons 4 gallons 0.20(4-x) + 0.70x = 0.50(4) 0.8 – 0.2x + 0.7x = 2 0.5x = 1.2 x = 2.4 The amount of antifreeze to be drained and replaces is 2.4 gallons

Geometric Formulas Perimeter of triangle P = a+b+c unit h Perimeter of triangle P = a+b+c unit Area of triangle A = ½ bh sq.unit Area of Rectangle =LW sq.unit Perimeter P = 2(L + W) unit Area of Parallelogram A = bh sq.unit Area & Volume of cylinder A = 2 rh sq.unit V = r2 h cu.unit Area & Volume of a cube A = 6a2 sq.unit V = a3 cu.unit a c b W L h b h r a a

3.3 Properties of Inequalities Let a, b, c be real numbers.  a < b and a+c < b+c are equivalent ( The same number may be added to or subtracted from both sides of an inequality.) If c > 0, then a < b and ac < bc are equivalent. (Both sides of an inequality may be multiplied or divided by the same positive number) If c< 0, then a < b and ac > bc are equivalent. Each side of an inequality may be multiplied or divided by the same negative number provided the inequality symbol is reversed.  

Fig. 3.14 Graphical Solutions (Pg 172) Distance (miles) y1 y2 y2 y1 x = 2 0 1 2 3 4 Time ( hours ) Distances of two cars 350 300 250 200 150 100 50 Distance ( milesDistance When x = 2 , y1 = y2, ie car 1 and car 2 both are 150 miles From Chicago y1 < y2 when x< 2 car 1 is closer to Chicago than car 2 y1 is below the graph of y2 y1 > y2 when x > 2 Car 1 is farther from Chicago than Car 2 Y1 above the graph of y2

…Continued Ex 5 (Pg 173) Solving an inequality graphically Solve 5 – 3x < x – 3 y1 = 5 – 3x and y2 = x – 3 Intersect at the point (2, -1) 5 4 3 2 1 X = 2 y1 y2 -4 -3 -2 -1 0 1 2 3 4 y1 = y2 when x = 2 -1 -2 -3 (2, -1) y1 < y2 when x > 2 , y1 is below the graph of y2 Combining the above result y1 < y2 when x > 2 Thus 5 – 3x < x – 3 is satisfied when x > 2. The solution set is {x / x> 2 }

Using Technology Hit window Hit Graph Hit Y Enter inequality Enter [ -5, 5, 1] by [-5, 5, 1]

Ex 80 ( Pg 178) Sales of CD and LP records Hit Y , enter equations Enter table set Enter window Hit Table Hit graph 1987 or after CD sales were greater than or equal to LP records

3.4 Compound inequalities A compound inequality consists of two inequalities joined by the words and or or.   The following are two examples of compound inequalities.  2x > -3 and 2x < 5 2(1) > -3 and 2(1) < 5 1 is a solution True True 5 + 2 > 3 or 5 – 1 < -5  5 is a solution x + 2 > 3 or x – 1 < -5 True False Example 1 Example 2

…. Cont. If a compound inequality contains the word and, solution must satisfy both inequalities. For example, x = 1 is a solution of the first compound inequality because  2 (1) > -3 and 2 (1) < 5 True True  are both true statements.   If a compound inequality contains the word or, solution must satisfy atleast one of the two inequalities. Thus x = 5 is a solution to the second compound inequality.   5 + 2 > 3 or 5 – 1 < -5 True False

Symbolic Solutions and Number Lines ] x < 6 x> - 4 x < 6 and x > - 4 -8 -6 -4 -2 0 2 4 6 8 ( -8 -6 -4 -2 0 2 4 6 8 ( ] -8 -6 -4 -2 0 2 4 6 8 - 4< x < 6

Three-part inequality Sometimes compound inequality containing the word and can be combined into a three part inequality. For example, rather than writing x > 5 and x < 10 We could write the three-part inequality 5 < x < 10 ( ] -1 0 1 2 3 4 5 6 7 8 9 10 5 < x < 10

Example 5 Page 184 Solve x + 2 < -1 or x + 2 > 1 x < -3 or x > -1 ( subtract 2 ) The solution set for the compound inequality results from taking the union of the first two number lines. We can write the solution, using builder notation, as { x x < - 3} U { x x > - 1 } or {x x < - 3 or x > - 1} x < - 3 x > - 1 x < - 3 or x > -1 ] ) - 4 -3 -2 -1 0 1 2 3 4 ( - 4 -3 -2 -1 0 1 2 3 4 ) ( - 4 -3 -2 -1 0 1 2 3 4

Interval Notation Table 3.5 (Page 186) Inequality Interval Notation Number line Graph - 1 < x < 3 ( - 1, 3) - 3 < x < 2 ( - 3, 2] - 2 < x < 2 [ - 2, 2 ] x < - 1 or x > 2 ( -  , - 1) U (2,  ) x > - 1 ( - 1,  ) x < 2 ( - , 2 ] ( ) -4 -3 -2 -1 0 1 2 3 4 ( ] -4 -3 -2 -1 0 1 2 3 4 [ ] ) ( ( ]

Example 6 (Page 185) Solving a compound inequality numerically and graphically Using Technology Tution at private colleges and universities from 1980 to 1997 Can be modelled by f(x) = 575(x – 1980) + 3600 Estimate when average tution was between $8200 and $10,500. Hit Y and enter equation Hit Window and enter Hit 2nd and Table [ 1980, 1997, 1] by [3000, 12000, 3000 ] Hit 2nd and calc and go to Intersect and enter 4 times to get intersection

Ex 85 (Page 189) Solve graphically and numerically Ex 85 (Page 189) Solve graphically and numerically. Write your answer in interval notation x + 1 < -1 or x + 1 > 1 Y1 = -1, Y2 = x + 1, Y3 = 1 [ -5, 5, 1] by [ -5, 5, 1] x + 1 < - 1 or x + 1 > 1 Solution in interval notation is ( -  -2) U (0,  )

School Enrollment Example 92 (Pg 190) 70 60 55 50 Enrollment (millions) 1980 1990 2000 Year

Systems of Linear Equations Chapter 4 Systems of Linear Equations

4.1 Systems of Linear Equations in two variables The system of linear equation with two variables. Each equation contains two variables, x and y. Example x + y = 4 x – y = 8 An ordered pair (x, y) is a solution to linear equation if the values for x and y satisfy both equations. The standard form is ax + by = c dx + ey = k Where a, b, c, d, e, k are constants.

Types of Linear Equations in two variables A system of linear equations in two variables can be represented graphically by two lines in the xy-plane   The lines intersect at a single point, which represents a unique solution. Consistent system The equations are called independent equation If the two lines are parallel it is an inconsistent system and no solution If two lines are identical and every point on the line represents solution and give infinitely many solutions. The equations are called dependent equations

..cont. Example The equations x + y = 1 and 2x + 2y = 2 are equivalent. If we divide the second equation by 2 we obtain the first equation. As a result, their graphs are identical and every point on the line represents a solution.Thus there are infinitely many solutions, and the system of equations is a dependent system. y y y x x x Inconsistent dependent Unique

The Substitution Method Consider the following system of equations.  2x + y = 5 3x – 2y = 4  It is convenient to solve the first equation for y to obtain y = 5 – 2x. Now substitute ( 5 – 2x) for y into the second equation.   3x – 2(y) = 4 Second equation 3x – 2(5-2x) = 4, Substitute  to obtain a linear equation in one variable. 3x – 2(5 – 2x) = 4 3x – 10 + 4x = 4 Distributive property 7x – 10 = 4 Combine like terms 7x = 14 Add 10 to both sides x = 2 Divide both sides by 7 To determine y we substitute x = 2 into y = 5 – 2x to obtain y = 5 – 2(2) = 1  The solution is (2, 1).

Note that adding the two equations eliminates the variable y Elimination Method The elimination method is the second way to solve linear systems symbolically. This method is based on the property that ‘ equals added to equals are equal.’ That is, if a = b and c = d Then a + c = b + d Note that adding the two equations eliminates the variable y Example 2x – y = 4 x + y = 1 3x = 5 or x = 5/3 and solve for x Substituting x = 5/3 into the second equation gives 5/3 + y = 1 or y = - 2/3 The solution is (5/3, - 2/3)

Solve the system of equations using Technology 4.1 Pg 226 Ex 43 Ex 44 No solution Infinitely many solutions

Burning calories Ex 74 pg 239 During strenuous exercise an athelete can burn on Rowing machine Stair climber 10 calories per minute 11.5 calories per minute In 60 minute an athelete burns 633 calories by using both exercise machines Let x minute in rowing machine, y minute in stair climber The equations are x + y =60 10x + 11.5 y = 633 Find x and y -10x - 10 y = -600 (Multiply by -10) Add 1.5y = 33, y = 33/1.5= 22 minute in stair climber and x = 38 minute in rowing machine

Mixing acids Ex 76 ( Pg 239 ) x represents the amount of 10% solution of Sulphuric acid y represents the amount of 25 % solution of Sulphuric acid According to statement x + y = 20 10% of x + 25% of y = 0.18(20) 0.10x + .25 y = 3.6 .1x + .25y = 3.6 Multiply by 10 x + 2.5y = 36 x + y = 20 Subtract 1.5 y = 16 y = 16/1.5 = 10.6 x = 20 – y = 20 – 10.6 = 9.4 Mix 9.4 ml of 10 % acid with 10.6 ml of 25% acid

River current ( Ex 84, pg 240 ) x Speed of tugboat y Speed of current Distance = Speed x Time 165 = (x – y) 33 Upstream x – y = 5 165 = (x + y) 15 Downstream x + y = 11 By elimination method 2x = 16, x = 8 y = 3 The tugboat travels at a rate of 8 mph and river flows at a rate of 3 mph

Solving Linear inequalities in Two variables x < 1 x > 1 - 2 -1 1 2 3 - 2 -1 1 2 -2 -1 0 1 2 3 4 1 - 2 -1 x – 2y < 4 y < 2x - 1 x –intercept = 4 (4, 0) y –intercept = -2 (0, -2) Choose a test point Let x = 0, y = 0 x - 2y < 4 0 - 2(0) < 4 0 < 4 which is true statement Shade containing (0, 0)

Solving System of Linear Inequalities ( Pg 244) x + y < 4 y > x Testing point x + y < 4 x = 1, y = 2 1 + 2 < 4 ( True ) 4 3 2 1 Shaded region Testing point 2 > 1 ( True) (1, 2) (1, 2) y > x -4 -3 -2 -1 0 1 2 3 4 Shaded region Shaded region To solve inequalities ( 1, 2)

Modeling target heart rates (Ex 5 Pg 246 ) For Aerobic Fitness A person’s Maximum heart rate ( MHR) = 220 - A 200 175 150 125 100 75 50 25 Heart Rate Beats Per minute T = - 0.8A + 196 ( Upper Line ) T = - 0.7 A + 154 (Lower Line ) 0 20 30 40 50 60 70 80 Age in years (30, 150) is a solution 0.8 (40) + 196 <= 165 When A = 40 yrs - 0.7 (40) + 196> = 125 150 < - 0.8 (30) + 196 = 172 True 150 > - 0.7 (30) + 196 = 133 True

Solving a system of linear inequalities with technology Ex 6 ( Pg 247 ) Shade a solution set for the system of inequalities, using graphing calculator - 2x + y > 1 or y> 2x + 1 2x + y < 5 or y < 5 – 2x [ - 15, 15, 5 ] by [ - 10, 10, 5 ] [ - 15, 15, 5 ] by [ - 10, 10, 5 ] Hit 2nd and Draw then go to Shade

Linear Functions and Polynomial Functions   Every linear function can be written as f(x) = ax +b and is an example of a polynomial function. However, polynomial functions of degree 2 or higher are nonlinear functions. To model nonlinear data we use polynomial functions of degree 2 or higher.

Polynomial Expressions and Functions Chapter 5 Polynomial Expressions and Functions

The following are examples of polynomial functions.   f(x) = 3 Degree = 0 Constant f(x) = 5x – 3 Degree = 1 Linear f(x) = x2 – 2x –1 Degree = 2 Quadratic f(x) = 3x3 + 2x2 –6 Degree = 3 Cubic

Polynomial Expressions A term is a number, a variable, or a product of numbers and variables raised to powers. Examples of terms include 15, y, x4, 3x3 y, x-1/2 y-2 , and 6 x-1 y3 If the variables in a term have only nonnegative integer exponents, the term is called a monomial. Examples of monomials include 4, 5y, x2, 5x 2 z4, - x y4 and 6xy4

Monomials x x x x x x y y y Volume x3 Total Area xy + xy + xy = 3xy

Modeling AIDS cases in the United States Ex 11 ( Pg 308 ) 600 500 400 300 200 100 Aids Cases (Thousands) 0 4 6 8 10 12 14 Year ( 1984 1994 ) f(x) = 4.1x2 - 25x + 46 f(7) = 4.1(7) 2- 25.7 + 46 = 71.9 f(17) = 4.1 (17) 2 – 25.7 + 46 = 805.9

Modeling heart rate of an athelete ( ex 12 ) 250 200 150 100 50 Heart Rate (bpm) 0 1 2 3 4 5 6 7 8 Time (minutes ) P(t) = 1.875 t2 – 30t + 200 Where 0 < t < 8 P(0) = 1.875(0) 2 – 30(0) + 200 = 200 (Initial Heart Rate) P(8) = 1.875(8) 2– 30(80 )+ 200= 80 beaqts per minute (After 8 minutes) The heart rate does not drop at a constant rate; rather, it drops rapidly at first and then gradually begins to level off

A PC for all (Ex 102 Pg 312) x = 0 corresponds to 1997 200 150 100 50 Worldwide computer 0 1998 2000 2002 2004 Year f(x) = 0.7868 x2 + 12x + 79.5 x = 0 corresponds to 1997 x = 1 corresponds to 1998 and so on x = 6 corresponds to 2003 f(6) = 0.7868 x 6 2 + 12 x 6 + 79.5 = 179.8248 = 180 million(approx)

5.2 Review of Basic Properties Using distributive properties Multiply 4 (5 + x) = 4.5 + 4. x = 20 + 4x 20 4x 4 x Area : 20 + 4x

Using Technology [-6, 6, 1] by [ -4, 4, 1 ] [-6, 6, 1] by [ -4, 4, 1 ]

….Cont Multiplying Binomials Multiply (x+ 1)(x + 3) Geometrically Symbolically a) Geometrically x + 1 1 x 3 x2 3x x + 3 x 3 Area: (x + 1)(x + 3) Area: x2 + 4x + 3 b) Symbolically apply distributive property (x + 1)(x + 3) = (x + 1)(x) + (x+ 1)(3) = x.x + 1.x + x.3 + 1.3 = x 2 + x + 3x + 3 = x2 + 4x + 3 x

Difference Some Special products (a + b) (a - b) = a 2 - b 2 Sum (a + b) 2 = a 2 + 2ab + b 2 Difference (a – b) 2 = a 2 - 2ab + b 2

(a + b) 2 = a2 + ab + ab + b2 = a2 + 2ab + b2 Squaring a binomial (a + b) 2 = a2 + ab + ab + b2 = a2 + 2ab + b2 a a2 ab b ba b2 a b (a + b)2 = a2 + 2ab + b2

5.3 Factoring Polynomials Common Factors Factoring by Grouping Factoring and Equations ( Zero Product Property)

Zero- Product Property   For all real numbers a and b, if ab = 0, then a = 0 or b = 0 ( or both)

Ex – 5 Solving the equation 4x – x2 = 0 graphically and symbolically Numerically x y -1 -5 0 0 3 4 5 -5 Symbolically 4x – x 2 = 0 x(4 – x) = 0 (Factor out x ) x = 0 or 4 – x = 0 ( Zero Product property) x = 0 or x = 4 4 3 2 1 -3 -2 -1 1 2 3 4 5 6

5.4 Factoring Trinomials Factoring x2 + bx + c x2 + bx + c , find integers m and n that satisfy m.n = c and m + n = b x2 + bx + c = (x + m)(x + n)

5.4 Factoring Trinomials ax 2 + bx + c by grouping Factoring Trinomials with Foil 3x 2+ 7x + 2 = ( 3x + 1 ) ( x + 2 ) x + 6x 7x If interchange 1 and 2 (3x + 2)(x + 1) = 3x2 + 5x + 2 which is incorrect 2x + 3x = 5x

5.5 Special types of Factoring Difference of Two Squares (a –b) (a + b) = a2 - b2 (a + b) 2 = a2 + 2ab + b2 (a –b) 2 = a2 - 2ab + b2 Sum and Difference of Two Cubes (a + b)(a2 – ab + b2) = a3 + b3 (a –b)(a2 + ab + b2 ) = a3 – b3 Verify ( a + b) (a2 – ab + b2) = a. a2 – a . ab + a. b2 + b. a2 -b . ab + b. b2 = a3 - a2 b + a b2 + a2 b - a b2 + b3 = a3 + b3 Perfect Square

5.6 Polynomial Equations Solving Quadratic Equations Higher Degree Equations

Using Technology [ -4.7, 4.7, 1] by [-100, 100, 25] Y = 16x^4 – 64x^3 + 64x^2 x = 0 , y = 0, x = 2 y = 0

Rational Expressions and Functions Chapter 6 Rational Expressions and Functions

6.1 Rational Expressions and Functions Rational Function   Let p(x) and q(x) be polynomials. Then a rational function is given by f(x) = p(x)/ q(x)   The domain of f includes all x-values such that q(x) = 0 Examples - 4 , x , 3x2 –6x + 1 x x – 5 3x - 7

Identify the domain of rational function (Ex 2 pg 376) g(x) = 2x x2 - 3x + 2 Denominator = x2 - 3x + 2 = 0 (x – 1)(x –2) = 0 Factor x = 1 or x = 2 Zero product property Thus D = { x / x is any real number except 1 and 2 }

Using technology ( ex 57, pg 384 ) [ -4.7 , 4.7 , 1 ] by [ -3.1, 3.1, 1 ] (ex 64, pg 384)

Highway curve ( ex 72, page 384 ) R(m) = 1600/(15m + 2) 500 400 300 200 100 Radius 0 0.2 0.4 0.6 0.8 slope a) R(0.1) = 1600 / (15(0.1) + 2) = 457 About 457 : a safe curve with a slope of 0.1 will have a minimum radius of 457 ft b) As the slope of banking increases , the radius of the curve decreases c) 320 = 1600/(15m + 2) , 320( 15m + 2) = 1600 , 4800m +640 = 1600 4800m = 960, m =960/4800 = 0.2

Evaluating a rational function Ex 4, Pg -377 Evaluate f(-1), f(1), f(2) Numerical value x -3 -2 -1 0 1 2 3 y 3/2 4/3 1 0 __ 4 3 f(x) = 2x x - 1 4 3 2 1 -4 -3 -2 -1 1 2 3 Vertical asymptote f(-1) = 1 f(1) = undefined and f(2) = 4

A/B – C/D = AC/BD B, C, and D are nonzero 6.2 Products and Quotients of Rational Expression  To multiply two rational expressions, multiply numerators and multiply denominators.  A/B. C/D = AC /BD B and D are nonzero. Example 2/3 . 5/7 = 10/21  To divide two rational expressions, multiply by the reciprocal of the divisor.  A/B – C/D = AC/BD B, C, and D are nonzero Example 3/4 - 2/4 = 3.4/4.5 = 3/5

6.3 Sums and Differences of Rational Expressions   To add (or subtract) two rational expressions with like denominators, add (or subtract) their numerators. The denominator does not change.  A/C + B/C = (A + B)/C Example - 1/5 + 2/5 = (1 + 2) / 5 = 3/5 A/C – B/C = (A – B) /C  3/5 - 2/ 5 = (3 – 2)/5 = 1/5

6.4 Solving rational equations graphically and numerically ( Ex- 3 (a) pg 409 ) 1/2 + x/3 = x/5 Solution- The LCD for 2,3, and 5 is their product, 30. 30( 1/2 + x/3) = x/5 . 30 Multiply by the LCD. 30/2 +30x/3 = 30x/5 Distributive property 15 + 10x = 6x Reduce 4x = -15 Subtract 6x and 15 x = -15/4 Solve Graphically Y1 = 1/2 + x/3 Y2 = X/5 [ -9, 9, 1] by [ -6, 6, 1]

Determining the time required to empty a pool ( pg 415, no.68) A pump can empty a pool in in 40 hours. It can empty 1/40 of the pool in 1 hour. In 2 hour, can empty a pool in 2/40 th of the pool Generally in t hours it can empty a pool in t/40 of the pool. Second pump can empty the pool in 70 hours. So it can empty a pool in t/70 of the pool in t hours. Together the pumps can empty t/40 + t/70 of the pool in t hours. The job will complete when the fraction of the pool is empty equals 1. The equation is t/40 + t/ 70 = 1 Multiply (40)(70) (40)(70) t/40 + t/ 70 = 1 (40)(70) 70t + 40t = 2800 110t = 2800 t = 2800/110 = 25.45 hr Two pumps can empty a pool in 25.45 hr

Example 6( pg – 412) x = speed of slower runner x + 2 = the speed of the winner d = rt , t = d/r The time for slower runner = 3/x, ran 3 miles at x miles per hour The winning time is 3/(x + 2 ) = , the winner ran 3 miles at x+ 2 miles per hr Add 3 minutes = 3/60 = 1/20 hr to the winner’s time, as finishes race 3 minutes ahead of another runner which equals the slower runner’s time 3/(x+2) + 1/20 = 3/x Multiply each side by the LCD, which is 20x(x + 2) 20x(x + 2)( 3/x + 2) + 1/20 = 20x(x + 2)3/x 60x + x(x + 2) = 60(x + 2) Distributive property 60x + x 2 + 2x = 60x + 120 ,, x 2 + 2x – 120 = 0 (x + 12)(x – 10) = 0 Factor x = - 12 or x = 10 Zero product property Running speed cannot be negative. The slower runner is running at 10 miles per hour, and the faster runner is running at 12 miles per hour.

Ex 73 Pg 416 A tugboat can travel 15 miles per hour in still water 36 miles upstream ( 15 – x) Total time 5 hours downstream (15 + x) t = d/r So the equation is 36/(15 – x) + 36/ (15 + x) = 5 The LCD is (15-x)(15 + x) Multiply both sides we get (15 – x)(15 + x)[36/(15 – x) + 36/ (15 + x)] = 5 (15 – x)(15 + x) 540 + 36x + 540 - 36x = 1125 – 5x 2 5x2 – 45 = 0 5x2 = 45 x = + 9, x = 3 mph

Modeling electrical resistance (Ex- 8, pg –413) R1 = 120 ohms R2 = 160 ohms R 1/R = 1/R1 + 1/R2 = 1/120 +1/ 160 = 1/120 . 4/4 + 1/160 . 3/3 LCD = 480 = 4/480 + 3/480 = 7/480 R = 480/7 = 69 ohms

6.6 Proportion a c is equivalent to ad = bc b d Example 6 8 5 x or x = 40/6 = 20/3 h/44 = 6/4 h = 6.44/4 = 66 feet 6 feet h feet 4 feet 44 feet

Modeling AIDS cases Y = 1000 (x – 1981)2 [ 1980, 1997, 2] by [-10000, 800000, 100000]

Radical Expressions and Functions Chapter 7 Radical Expressions and Functions

Chapter 7 Square Root The number b is a square root of a if b2 = a Example 100 = 102 = 10 radical sign Under radical sign the expression is called radicand Expression containing a radical sign is called a radical expression. Radical expressions are 6, 5 + x + 1 , and 3x 2x - 1

Cube Root The number b is a cube root of a if b3 = a Example – Find the cube root of 27 3 27 = 3 33 = 3

Estimating a cellular phone transmission distance The circular area A is covered by one transmission tower is A = R2 2 The total area covered by 10 towers are 10 R , which must equal to 50 square miles Now solve R R = 1.26, Each tower must broadcast with a minimum radius of approximately 1.26 miles

Expression For every real number If n is an integer greater than 1, then a1 n = n a  Note : If a < 0 and n is an even positive integer, then a1 n is not a real number. If m and n are positive integer with m/n in lowest terms, then a m n = n a m = ( n a ) m  Note : If a < 0 and n is an even integer, then a m n is not a real number. a - m n = 1/ a m n a = 0

Properties of Exponent Let p and q be rational numbers Properties of Exponent Let p and q be rational numbers. For all real numbers a and b for which the expressions are real numbers the following properties hold. Page -467 a p . a q = a p + q Product rule a - p = 1/ a p Negative exponents   a/b -p = b a p Negative exponents for quotients a p = a p-q Quotient rule for exponents a q a p q = a pq Power rule for exponents ab p= a p b p Power rule for products a p = a p Power rule for products b b p Power rule for quotients 1 2 3 4 5 6 7

Product Rule for Radical Expressions   Let a and b be real numbers, where n a and n b are both defined. Then n a . n b = n a. b   To multiply radical expressions with the same index, multiply the radicands.

Quotient Rule for Radical Expressions Let a and b be real numbers, where a and b are both defined and b = 0.   a/b = a/b The Expression –a If a>0, then –a = a

Square Root Property x2 = k Let k be a nonnegative number. Then the solutions to the equation. x2 = k are x = + k. If k < 0. Then this equation has no real solutions. Ex 2, 3, 4, 5, 6

Technology [ 5, 13, 1] by [0, 100, 10]

To find cube root technologically

Technologically Y1 = x2 [ -6, 6, 1] by [-4, 4, 1]

7.2 Let a and b be real numbers where are both defined , Product rule for radical expression (Pg – 472) = Quotient rule for radical expression where b = 0 (Pg 475) =

Pg -476 Rationalizing Denominators having square roots =

7.3 Operations on Radical Expressions Addition 10 + 4 = (10 + 4) = 14 Subtraction 10 - 4 = (10 - 4) = 6 Rationalize the denominator (Pg 484)

7.6 Complex Numbers Pg 513 x 2 + 1 = 0 x 2 = -1 x =+ Square root property - 1 Now we define a number called the imaginary unit, denoted by i Properties of the imaginary unit i i = - 1 A complex number can be written in standard form, as a + bi, where a and b are real numbers. The real part is a and imaginary part is b

Pg 513 a + ib Complex Number -3 + 2i 5 -3i -1 + 7i - 5 – 2i 4 + 6i Real part a - 3 5 0 -1 -5 4 Imaginary Part b 2 0 -3 7 -2 6

Complex numbers contains the set of real numbers a +bi a and b real Real numbers a +bi b=0 Imaginary Numbers a +bi b =0 Rational Numbers -3, 2/3, 0 and –1/2 Irrational numbers 3 And - 11

Sum or Difference of Complex Numbers Let a + bi and c + di be two complex numbers. Then   Sum ( a + bi ) + (c +di) = (a + c) + (b + d)i Difference (a + bi) – (c + di) = (a - c) + (b – d)i

Quadratic Functions and Equations Chapter 8 Quadratic Functions and Equations

QuadraticFunction A quadratic equation is an equation that can be written as  f(x) = ax2 + bx + c , where a, b, c are real numbers, with a = 0. Axis of symmetry (0, 2) -2 1 -1 2 1 (0, 0) -2 -1 0 1 2 (2, -1) Vertex x = 2

The x-coordinate of the vertex of the graph of Vertex Formula The x-coordinate of the vertex of the graph of y = ax2 +bx +c, a = 0, is given by   x = -b/2a  To find the y-coordinate of the vertex, substitute this x-value into the equation

Example 1 (pg 531) Graph the equation f(x) = x2 -1 whether it is increasing or decreasing and Identify the vertex and axis of symmetry x y = x2 -1 -2 3 -1 0 0 -1 2 3 vertex Equal 3 2 1 -1 Vertex The graph is decreasing when x < 0 And Increasing when x > 0

Example 1(c) pg 532 x y = x2 + 4x + 3 -5 8 -4 3 -3 0 -2 -1 -1 0 0 3 -5 8 -4 3 -3 0 -2 -1 -1 0 0 3 1 8 Axis of symmetry Vertex Equal x = -2 Vertex (-2, -1)

Find the vertex of a parabola f(x) = 2x2 - 4x + 1 Symbolically f(x) = 2x2 – 4x + 1 a = 2 , b = - 4 x = -b/2a = - (-4)/2.2 = 4/4 = 1 To find the y-value of the vertex, Substitute x = 1 in the given formula f(1) = 2. 12 - 4.1 + 1= -1 The vertex is (1, -1) Graphically [ -4.7, 4.7, 1] , [-3.1, 3.1, 1]

Example 5 (Pg 535) Maximizing Revenue The regular price of a hotel room is $ 80, Each room rented the price decreases by $2 900 800 700 600 500 400 (20, 800) Maximum revenue 0 5 10 15 20 25 30 35 40 If x rooms are rented then the price of each room is 80 – 2x The revenue equals the number of rooms rented times the price of each room. Thus f(x) = x(80 – 2x) = 80x - 2x2 = -2x2 + 80x The x-coordinate of the vertex x = - b/2a = - 80/ 2(-2) = 20 Y coordinate f(20) = -2(20)2 + 80 (20) = 800

8.2 Vertical and Horizontal Translations Translated upward and downward y2 = x2 + 1 y1= x2 y3 = x2 - 2 y1= x2 y1= (x-1)2 Translated horizontally to the right 1 unit y1= x2 y2= (x + 2 )2 Translated horizontally to the left 2 units

Vertical and Horizontal Translations Of Parabolas (pg 542) Let h , k be positive numbers.   To graph shift the graph of y = x2 by k units y = x2 + k upward y = x2 – k downward y = (x – k)2 right y = (x +k)2 left Ex 1 pg 543

Vertex Form of a Parabola (Pg 543) The vertex form of a parabola with vertex (h, k) is y = a (x – h)2 + k, where a = 0 is a constant. If a > 0, the parabola opens upward; if a < 0, the parabola opens downward. Ex- 2, 3, 4

Quadratic Equation A quadratic equation is an equation that can be written as ax2 +bx +c= 0, where a, b, c are real numbers with a = 0

Quadratic Equations and Solutions y = x2 + 25 y = 4x2 – 20x + 25 y = 3x2 + 11x - 20 No Solution One Solution Two Solutions

The solutions of the quadratic equation Quadratic Formula The solutions of the quadratic equation ax2 + bx + c = 0, where a, b, c are real numbers with a = 0 No x intercepts One x – intercepts Two x - intercepts Ex 1

Modeling Internet Users Use of the Internet in Western Europe has increased dramatically shows a scatter plot of online users in Western Europe with function f given by f(x) = 0.976 x2 - 4.643x + 0.238 x = 6 corresponds to 1996 and so on until x = 12 represents 2002 90 80 70 60 50 40 30 20 10 f(10) = 0.976(10) 2 - 4.643(10) + 0.238 = 51.4 6 7 8 9 10 11 12 13

8.4 Quadratic Formula The solutions to ax 2 + bx + c = 0 with a = 0 are given by - b + b2 – 4ac X = 2a Ex – 1, 2, 3, 4

The Discriminant and Quadratic Equation To determine the number of solutions to ax2 + bx + c = 0 , evaluate the discriminant b 2 – 4ac > 0, If b 2 – 4ac > 0, there are two real solutions If b 2 – 4ac = 0, there is one real solution If b 2 – 4ac < 0, there are no real solutions , but two complex solution Ex 5, 6, 7, 8, 9