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Algebra 1 2nd Quarter - Review.

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Presentation on theme: "Algebra 1 2nd Quarter - Review."— Presentation transcript:

1 Algebra 1 2nd Quarter - Review

2 Topics from Semester II
Functions and Linear Equations Systems of Linear Equations Quadratics and Polynomials Exponential Functions and Sequences Scatter Plots, Trends and Statistics

3 Functions and Linear Equations
f(x) = mx + b

4 Functions and Linear Equations

5 Functions

6 An Ordered Pair (x, y) is a pair of numbers that describes a location in the coordinate plane.
A Relation is a set of ordered pairs. { (3, 5), (4, 7), (5, 9) (6, 11) } A Function is a relation where each input (x) has only one output (y).

7 Mapping Functions (and relations) are sometimes presented as mapping one set of numbers (input) to another (output). The arrows connect each input (x) with the output (y) it produces. input output

8 Function: Each x has only one y
Not a Function Function

9 Vertical Line Test A Vertical Line
hits the graph of a function only once. This IS the graph of a Function

10 Vertical Line Test This IS NOT the graph of a Function

11 Domain and Range Domain
The Domain of a Relation is the set of all Input values (x). This is also called the Independent Variable. Notice: Domain Input Independent all contain the word “in” Domain

12 Domain and Range Range The Range of a Relation is the set of all
Output values (y). This is called the Dependent Variable because its value depends on the input. Domain Range Input Output Independent Dependent Range

13 Evaluating a Function Many Functions are defined by algebraic rules (or equations). To evaluate the function for a given input value, substitute the value into the equation and simplify. f(x) = 3x – 5 {f is the name of the function and x is the input } If x=4, then f(4) = 3(4) – 5 = 7 So f(4) = 7 and the point (4,7) is on the graph of the function.

14 Linear Equations

15 Linear Equations Graphing Linear Equations
Graphing Linear Inequalities Writing Linear Equations

16 Graphing Linear Equations
Graph: y = 2x -5 y = 2(0) – 5 y = 2(1) – 5 From a table: X Y – 5 1 – 3 Pick values for x That will be easy to use. And evaluate for y. x = zero is usually easiest Pick another x Close by.

17 Graphing Linear Equations
Graph: y = 2x -5 From a table: X Y – 5 1 – 3 Plot each Point And draw the Line through them.

18 Graphing Linear Equations
Graph: y = -⅖x +5 y = -⅖(5) + 5 y = -⅖(0) + 5 From a table: -2 + 5 X Y 5 5 3 Pick values for x That will be easy to use. And evaluate for y. x = zero is usually easiest Pick denominator As second x

19 Graphing Linear Equations
Graph: y = -⅖x +5 From a table: X Y 5 5 3 Plot each Point And draw the Line through them.

20 Graphing Linear Equations
Graph: y = ⅔x -3 Using y-intercept and slope. Plot the y-intercept on the y-axis. run = 3 Use slope to find A second point. rise = 2 Draw line through the 2 points.

21 Graphing Linear Equations
Graph: 4x - 3y = 24 -3y = 24 4x = 24 From standard form: X Y -8 6 Because 24 is a multiple of both 4 and 3, it’s easy to use x and y intercepts. x = zero for y-intercept y = zero For x-intercept

22 Graphing Linear Equations
Graph: 4x - 3y = 24 From standard form: X Y -8 x-intercept on x-axis 6 Plot each Point And draw the Line through them. y-intercept on y-axis

23 Graphing Linear Equations
Graph: 5x - 3y = 24 From standard form: Because 24 is not a multiple of 5. Convert to y = mx + b

24 Convert from Standard form to Slope-Intercept
Standard Slope-Intercept ax + by = c y = mx + b 5x - 3y = 24 -5x -5x Move x-term to other side by adding the opposite -3y = -5x + 24 Solve for y -3 -3 -3 y = 5/3x - 8

25 Graphing Linear Equations
Graph: 5x - 3y = 24 y = 5/3x - 8 y = 5/3(0) - 8 y = 5/3(3) - 8 X Y -8 3 -3

26 Graphing Linear Inequalities
Graph the Line If < or > use a dashed line (not part of solution). If ≤ or ≥ use a solid line. Identify a point in the solution set: Usually, check the origin (0,0) If point is a solution, shade that side of the line Otherwise, shade the other side of the line.

27 Graphing Linear Inequalities
Graph: y <⅔x -3 Using y-intercept and slope. Plot the y-intercept on the y-axis. Use slope to find A second point. Draw line through the 2 points.

28 Graphing Linear Inequalities
Graph: y < ⅔x -3 Shade below the line? Check a point. Is (3, -4) a solution? Shade ? y < ⅔x -3 -4 < ⅔(3) -3 ? -4 < (2 -3) ?

29 Graphing Linear Inequalities
Graph: y < ⅔x -3 Shade below the line!

30 Writing Linear Equations
Linear Equations in Three Forms Point-Slope Form y - y1 = m( x – x1) Slope-Intercept Form y = mx + b Standard Form ax + by = c

31 Writing Linear Equations
Point-Slope Form y - y1 = m( x – x1) Point (x1, y1) Slope Slope-Intercept Form y = mx + b y-intercept

32 Slope Intercept ⅔ Slope-Intercept Form y = mx + b
Write the equation of a line with a slope of ⅔ And a y-intercept of 9 Slope-Intercept Form y = mx + b 9 Slope = ⅔ y-intercept = 9

33 Slope Intercept Slope-Intercept Form y = m x + b
Write the equation of a line Parallel to y = -3x +5 with a y-intercept of -6 Parallel lines have the same slope. Slope-Intercept Form y = m x + b -3 - 6 Slope = -3 y-intercept = -6

34 Slope Intercept Write the equation of a line
Perpendicular to y = ½x - 13 Through the point (0, 5) The slopes of Perpendicular lines are negative inverses. Negative inverse of ½ is -2 Slope-Intercept Form y = mx + b -2 The y-intercept… is where x = 0. Slope = -2 the point (0, 5) 5 y = -2x + b

35 Slope Intercept Write the equation of a line with a slope of -3
Through the point (3, 8) Slope-Intercept Form y = mx + b -3 We don’t know the y-intercept… Slope = -3 8 y = -3 x + b (3) We do know one point on the graph. The point (3, 8)

36 Slope Intercept Write the equation of a line with a slope of -3
Through the point (3, 8) Slope-Intercept Form y = mx + b y = -3x + b 1) Substitute in slope 8 = -3(3) + b 2) Substitute x and y 8 = b 3) Solve for b (y-intercept) 17 = b y = -3x + b 17 4) Substitute for b

37 Slope Intercept Write the equation of a line with a slope of ⅖
Through the point (10, 3) Slope-Intercept Form y = mx + b y = ⅖x + b 1) Substitute in slope 3 = ⅖(10) + b 2) Substitute x and y 3 = b 3) Solve for b (y-intercept) 4) Substitute for b -1 = b y = ⅖ x + b - 1

38 Point-Slope Write the equation of a line with a slope of ⅔
through the point (2, 5) Point-Slope Form y - y1 = m( x – x1) 5 2 Point (2, 5) Slope = ⅔

39 Point-Slope Write the equation of a line with a slope of 4
through the point (-2, -3) Point-Slope Form y - y1 = m( x – x1 ) + 3 4 + 2 Point (-2, -3) Slope = 4

40 Point-Slope Write the equation of a line parallel to y= 2x - 9
through the point (-5, 7) Point-Slope Form y - y1 = m( x – x1) 7 + 5 2 Point (-5, 7) Slope = 2 Parallel Lines have the Same Slope

41 Convert from Point-Slope to Slope-Intercept
Point-Slope Slope-Intercept y - y1 = m( x – x1) y = mx + b y - 5 = 3( x – 2) Distribute y - 5 = 3x – 6 +5 +5 Solve for y y = 3x – 1 skip

42 Convert from Point-Slope to Slope-Intercept
Point-Slope Slope-Intercept y + 3 = ⅔( x – 6) Distribute y + 3 = ⅔x – 4 Solve for y - 3 - 3 y = ⅔ x – 7

43 Convert from Slope-Intercept to Standard form
Slope-Intercept Standard y = mx + b ax + by = c y = 3x + 7 Move x-term to other side by adding the opposite - 3x - 3x -3x + y = 7

44 Convert from Slope-Intercept to Standard form
Slope-Intercept Standard y = mx + b ax + by = c y = ⅓x + 5 Move x-term to other side by adding the opposite - ⅓x - ⅓x 3( - ⅓x + y = 5 ) Clear Fractions -x + 3y = 15

45 Convert from Standard form to Slope-Intercept
Standard Slope-Intercept ax + by = c y = mx + b 3x + 2y = 18 Move x-term to other side by adding the opposite -3x -3x 2y = -3x + 18 2 2 2 Solve for y y = - x + 9 3 2

46 Systems of Linear Equations1

47 Linear Systems Solving by Graphing Solving Algebraically Substitution
Linear Combination

48 Linear Systems √ y = 2x – 5 1 = 2(3) – 5 y = -x + 4 1 = -(3) + 4
Solving by Graphing Graph Each Equation. Solution is the point where the graphs Intersect. Check Algebraically (3, 1)

49 Linear Systems 2x + 3y = 6 y = -⅔x + 4 Solving by Graphing
Graph Each Equation. Solution is the point where the graphs Intersect. Parallel Lines Don’t Intersect. There is no solution.

50 Linear Systems 2x + 5y = 15 y = -⅖x + 3 Solving by Graphing
Graph Each Equation. Both Equations are For the same line. There are infinitely many solutions.

51 Linear Systems By Substitution: (5, -1) 4x +2y = 18 2x -5y = 15
Solve for one variable in terms of the other Substitute in other equation 4x +2y = 18 2x -5y = 15 2y = -4x +18 y = -2x +9 -2x +9 y = -2( )+9 2x -5( ) = 15 y = = -1 2x +10x -45 = 15 (5, -1) 12x = 60 x = 5 5

52 Substitution (5, 7) = y = 3x – 8 y=3x - 8 + 2x -2x +17 + 2x y=3(5) - 8
Set the two equations equal (since they both equal y). y = 3x – 8 = y=3x - 8 + 2x -2x +17 + 2x y=3(5) - 8 5x – 8 = 17 5x = 25 x = 5 y= 7 (5, 7) Substitute in either of the equations.

53 Linear Combination + (7, 5) 3x – y = 16 2x + y = 19 5x = 35 x = 7 7
Also called Elimination because the goal is to eliminate one of the variables. 3x – y = 16 2x + y = 19 Add the two equations. + Opposites in the two equations will cancel out, eliminating the variable. 5x = 35 x = 7 7 Solve for the remaining variable. 2( ) + y = 19 Substitute in any of the equations. 14 + y =19 y =5 (7, 5)

54 } Linear Combination (8, 3) 2( 2x – y = 13 3x +2y = 30 ) 4x –2y = 26
Need Opposites in the two equations 2( 2x – y = 13 3x +2y = 30 ) So multiply one (or both) equation(s) } Add these two equations. 4x –2y = 26 7x = 56 x = 8 3x +2y = 30 3( 8) +2y = 30 24 +2y = 30 (8, 3) 2y = 6 y = 3

55 } Linear Combination (3, 4) 3( 2x –3y = -6 3x +5y = 29 ) ) -2(
Need Opposites in the two equations 3( 2x –3y = -6 3x +5y = 29 ) So multiply one (or both) equation(s) ) -2( } 6x –9y = -18 Add these two equations. -6x –10y = -58 3x +5y = 29 -19y = -76 3x +5(4) = 29 y = 4 (3, 4) 3x = 29 3x = 9 x = 3

56 } Linear Combination 2( 2x – y = 13 -4x +2y = 30 ) 4x –2y = 26 0 = 56
Need Opposites in the two equations 2( 2x – y = 13 -4x +2y = 30 ) So multiply one (or both) equation(s) } Add these two equations. 4x –2y = 26 0 = 56 Since this is false, there is no solution. These are parallel lines. If both variables are eliminated and the resulting statement is true then there are infinitely many solutions.

57 Quadratics and Polynomials

58 Quadratics and Polynomials
Laws of Exponents Operations with Polynomials Special Products Factoring Completing the Square Quadratic Formula Graphing

59 ( ) Laws of Exponents = x2+3 = x5 = x7-3 = x4 = x 4∙3 = x12 = 27a3b6 =
Product property x2(x3) Quotient Property Power of a Power (x4)3 Power of a Product (3ab2)3 Power of a Quotient = x2+3 = x5 x7 x3 = x7-3 = x4 = x 4∙3 = x12 = 27a3b6 ( ) x7 y3 2 x7∙2 y3∙2 x14 y6 = =

60 ( ) Laws of Exponents = x5-5 = x0 = 1 = x5-7 = x-2 = x5 Zero Exponents
Negative Exponents = x5-5 = x0 = 1 ( ) 2x5y3 7x3z-3 = 1 1 x2 x5 x7 = x5-7 = x-2 = -2x-5 y7z-3 -2 In fractions x5 Terms with negative exponent move

61 Laws of Exponents = x5-5 = x0 = 1 = x5-7 = x-2 = x5 Zero Exponents
Negative Exponents = x5-5 = x0 = 1 1 x2 x5 x7 = x5-7 = x-2 = -2 y7 -2 y7z-3 z3 In fractions x5 Terms with negative exponent move

62 Simplifying Expressions
Each variable base appears only once There are no powers of powers All fractions are in simplest form (2gh4)3[(-2g4h)3]2 [-8g12h3]2 (8g3h12) (64g24h6) = 512g27h18

63 Simplifying Expressions
Each variable base appears only once There are no powers of powers All fractions are in simplest form a7b4c5 a2b7c a5 1 1 b3 c4 1 a5c4 b3 a7b4c5 b-3c5 a7 b7 c0 = a7b7

64 Simplifying Expressions
( ) 3 3 ( ) 2x5y6z3 3x7y2 2y4z3 3x2 8y12z9 27x6 Parentheses first Then exponent Negative Exponent means the “inverse” ( ) ( ) -2 2 ( ) a3b4 a8b2c3 2 a8b2c3 a3b4 a10c6 b4 a5c3 b2 So you can flip the fraction first

65 Adding Polynomials (3x3 – 5x2 +x -11) + (4x3 + 7x2 -6x +23) (7x3 +2x2
Group and Combine Like Terms Like Terms have the same variable raised to the same power (3x3 – 5x2 +x -11) + (4x3 + 7x2 -6x +23) (7x3 +2x2 - 5x + 12) (x3 – 5x2 + 4x) + (4x3 -6x +23) (5x3 -5x2 - 2x + 23) Include any terms that are only in one polynomial

66 Subtracting Polynomials
Add the Opposite (3x3 – 5x2 +x -11) - (4x3 + 7x2 -6x +23) (3x3 – 5x2 +x -11) + (-4x3 -7x2 +6x -23) (-x3 -12x2 + 7x - 34) (x3 – 5x2 + 4x) - (4x3 -6x +23) (x3 – 5x2 + 4x) + (-4x3 +6x -23) (-3x3 -5x2 + 10x - 23)

67 Multiplying Polynomials
Use the distributive property and Laws of Exponents -5x2 (3x2 + 4x - 11) = -15x x3 + 55x2 2x (3x - 11) +3(x -5) = 6x x + 3x -15 = 6x x -15 (x+ 3)(x - 8) = x2 - 8x + 3x - 24 = x2 - 5x - 24

68 Special Products Square of a Binomial (x+ 3)2 = (x+ 3)(x + 3)
Rewrite and distribute (x+ 3)2 = (x+ 3)(x + 3) = x2 + 3x +3x + 9 = x2 + 6x + 9 Or use pattern (a+ b)2 = a2 + 2ab + b2 (x+ 8)2 = x2 + 2x(8) + 82 = x2 + 16x + 64 (a - b)2 = a2 - 2ab + b2 (x- 5)2 = x2 - 2x(5) + 52 = x2 - 10x + 25

69 Special Products Product of a sum and difference (x+ 3)(x - 3)
= x2 - 3x +3x - 9 = x2 - 9 (4x+ 5)(4x - 5) = 16x2 - 20x +20x - 25 = 16x2 - 25 (a+ b)(a - b) = a2 – b2 Pattern: (x+ 7)(x - 7) = x2 – 72 = x2 – 49

70 Factoring

71 Factoring: x2 + bx + c x2 + bx + c = (x + m)(x + p )
So, we’re looking for two factors of c that add to b. x2 + 7x + 12 = (x +__ )(x +__ ) 3 4 Factors of 12 1 12 2 6 3 4 = 7

72 Factoring: x2 + bx + c x2 + 9x + 18 = (x __ )(x __ ) + +
3 + 6 Factors of 18 1 18 2 9 3 6 Both Positive Factors have the same sign x2 – 8x = (x __ )(x __ ) 4 4 Factors of 16 1 16 2 8 4 Both Negative Factors have the same sign

73 Factoring: x2 + bx – c x2 + 7x – 8 = (x __ )(x __ ) – +
1 Factors of 8 1 8 2 4 Larger Positive Factors have different signs x2 – 4x – 21 = (x __ )(x __ ) + 3 7 Factors of 21 1 21 3 7 Larger Negative Factors have different signs

74 Factoring: x2 – c This is a “difference of squares”
Pattern: (a2 – b2) = (a + b)(a – b) (m – n) = (√m + √ n)(√m– √n) x2 – 16 = (x + __ )(x – __ ) √ 16 = 4 4 4 √ 121 = 11 x2 – 121 = (x + __ )(x –__ ) 11 11 √ 4x2 = 2x 4x2 – 169 = ( x + __ )( x –__ ) 2 13 2 13 √ 169 = 13

75 Factoring: ax2 – ac This “difference of squares”
Is hidden by a common factor. (ax2 – ab2) = a(x2 – b2) = a(x + b)(x – b) 3x2 – 75 = 3(x2 – 25) =3(x + __ )(x – __ ) √ 25 = 5 5 5 2x3 – 98x = 2x(x2 – 49) = √ 49 = 7 2x(x + __ )(x –__ ) 7 7 4x2 – 324 = 4(x2 – 81 ) √ 81 = 9 4(x + __ )(x –__ ) 9 9

76 Factoring by Grouping = ac + ad + bc + bd a(c + d) +b (c + d)
(a + b)(c + d) = ac + ad + bc + bd a(c + d) +b (c + d) (a + b)(c + d) To factor by “grouping” Split the polynomial into two “groups” Pull out a common factor on Left See if the resulting binomial is a factor on right. Find common factor on Right Regroup into binomial factor s.

77 Factoring by Grouping x3 + 3x2 + 5x + 15 x2(x + 3) +5 (x + 3)
To factor by “grouping” Split the polynomial into two “groups” Pull out a common factor on Left See if the resulting binomial is a factor on right. Find common factor on Right Regroup into binomial factors.

78 Factoring: ax2 + bx + c 3x2 + 10x + 8 3(8) = 24 3x2 + 4x+ 6x + 8
Find the product ac 3(8) = 24 3x2 + 4x+ 6x + 8 We’re looking for factors of 24 that add to 10 Split middle term. x(3x + 4) +2 (3x + 4) 1 24 2 12 3 8 4 6 Factor by grouping. (x +2 )(3x +4 )

79 Solving Quadratic Equations by Factoring
Zero Product Rule: If the product of two numbers is zero, then at least one of the numbers is zero. (x – 5)(x +7) = 0 (x – 5)= 0 or (x +7) = 0 x = or x = -7

80 Solving Quadratic Equations by Factoring
Factors of 12 x2 - 8x + 12 = 0 1 12 2 6 3 4 (x – 6)(x - 2) = 0 (x – 6)= 0 or (x - 2) = 0 x = or x = 2

81 Solving Quadratic Equations by Factoring
Factors of 24 x2 - 5x - 24 = 0 1 24 2 12 3 8 4 6 (x – 8)(x +3) = 0 (x – 8)= 0 or (x + 3) = 0 x = or x = -3

82 Solving: x2 – c = 0 This is a “difference of squares”
Pattern: (a2 – b2) = (a + b)(a – b) x2 – 25 = 0 √ 25 = 5 (x + __ )(x – __ ) = 0 5 5 (x + 5)= 0 or (x - 5) = 0 x = -5 or x = 5

83 Solving: x2 – c = 0 Or take the square root x2 – 49 = 0 x2 = 49
x = 7 or -7

84 Solving: take the square root
2x2 – 47 = 25 First solve for x2 2x2 = 72 Then take square root On both sides x2 = 36 √x2 = ±√ 36 = ± 6 x = 6 or -6

85 Solving: take the square root
2(x - 2)2 – 4 = 14 First solve for (x-2)2 2(x - 2)2 = 18 Then take square root On both sides (x – 2)2 = 9 √(x – 2)2 = ±√ 9 x – 2 = ± 3 x = 2+3 or 2-3 x = or -1

86 Solving: Perfect Square
Cannot take the square root if there is an x term, Unless it’s a perfect square. x2 -10x +25 = 12 (x – 5)2 = 12 Rewrite as (x-5)2 √(x – 5)2 = ±√ 12 Then take square root On both sides x – 5 = ± √12 x = 5 ± √12

87 Completing the Square (x + b)2 = x2 + 2bx + b2

88 Completing The Square (x +n)2= x2 + 2nx + n2 x2 + bx + c 2n = b n = ½b
Square of a binomial x2 + bx + c Standard form 2n = b n = ½b c = n2

89 Completing The Square (x +n)2= x2 + 2nx + n2 x2 + 12x + 36 2n =12
Square of a binomial x2 + 12x + 36 Standard form 2n =12 n = ½(12) = 6 c = 62 = 36

90 Completing The Square (x +n)2= x2 + 2nx + n2 x2 + 14x + c 2n =14
Square of a binomial x2 + 14x + c Standard form 2n =14 n = ½(14) = 7 c = 72 = 49 49 What value must c have to represent a binomial square?

91 Completing The Square x2 + 16x + 24 = 9 x2 + 16x + = -15 x2 + 16x + 64
Subtract 24 to create a space for c x2 + 16x = -15 n = ½b = ½(16) = 8 8 +64 c = n2 = 82 = 64 64 x2 + 16x + 64 = 49 = 49 Add c to both sides of equation (x + n )2

92 Completing The Square √ x2 + 16x + 24 = 9 x2 + 16x + = -15
64 +64 x2 + 16x + 64 = 49 = 49 (x + 8 )2 = 49 Take square root on both sides of equation x = 7 + and - square roots - 8 - 8 Solve for x x = = -1 x = = -15

93 The Quadratic Formula and Discriminant

94 Quadratic Formula ax2 + bx + c = 0 x2 + 16x + 15 = 0
Standard form x2 + 16x + 15 = 0 a=1, b = 16, c = 15

95 Quadratic Formula x2 + 16x + 15 = 0 x = -16 + 14 = -1 2
a=1, b = 16, c = 15 x = = -1 2 x = -16 – 14 = -15 2

96 Quadratic Formula x2 + 8x + 16 = 0 x = -8 = -4 2 a=1, b = 8, c = 16
discriminant x = -8 = -4 2 When b2-4ac = 0 there is only 1 solution

97 Quadratic Formula x2 + 6x + 10 = 0 x = -6+2i = -3 + i 2
a=1, b = 6, c = 10 discriminant When b2-4ac < 0 2 imaginary solutions x = -6+2i = -3 + i 2 x = i = -3 - i 2

98 Graphing Quadratic Equations

99 Graphing Quadratic Equations
Determine Coordinates of the Vertex Vertex is the point where the graph turns around. It represents either the Minimum or Maximum Plot two points on the same side of the vertex. Copy those points to the other side of vertex The graph is a “u”-shaped parabola, so is symmetric Draw a smooth curve to match the plotted points

100 Graphing Quadratic Equations
Determine Coordinates of the Vertex Vertex is the point where the graph turns around. It represents either the Minimum or Maximum In Vertex Form: y = a(x –h)2 +k Vertex=(h, k) y = (x – 4) Vertex=(4, 7) y = (x + 5) Vertex=(-5, -9) y = -2(x + 1) Vertex=(-1, 6) Note: the value of a doesn’t effect the vertex.

101 Graphing Quadratic Equations
Determine Coordinates of the Vertex Vertex is the point where the graph turns around. So, the vertex occurs where the amount being squared equals zero since squaring any other amount produces a positive value. y = (2x – 5)2 +7 2x – 5 = 0 2x = 5 y = (2(2.5) – 5)2 +7 x = 2.5 y = (5 – 5)2 +7 Vertex=(2.5, 7) y = 0 +7

102 Graphing from Vertex Form
Determine Coordinates of the Vertex In Vertex Form: y = a(x –h)2 +k Vertex=(h, k) y = (x – 4) Vertex=(4, 5) y = (6 – 4)2 +5 y = (5 – 4)2 +5 = (2)2 +5 = 9 = (1)2 +5 = 6 x y 4 5 5 6 6 9 Plot two points on the same side of the vertex.

103 Graphing from Vertex Form
Determine Coordinates of the Vertex In Vertex Form: y = a(x –h)2 +k Vertex=(h, k) y = (x – 4)2 +5 x y 4 5 5 6 6 9 Copy points to the other side of the vertex.

104 Graphing from Vertex Form
Determine Coordinates of the Vertex In Vertex Form: y = a(x –h)2 +k Vertex=(h, k) y = (x – 4)2 +5 x y 4 5 5 6 6 9 Draw a smooth curve to match the plotted points .

105 Graphing from Vertex Form
y = (x – 4)2 +5 Vertex=(4, 5) The graph of this Quadratic Equation is a translation of the equation y = x2. The vertex (4,5) and every other point have been shifted 4 units to the right And 5 units up. y = (x – 4)2 +5 y = x2

106 Graphing from Vertex Form
In Vertex Form: y = a(x –h)2 +k If 0 < a < 1, graph has a vertical shrink If a > 1, graph has a vertical stretch y=x2 More narrow than y=x2 Wider than y=x2

107 Graphing from Vertex Form
y= -x2 If a < 0, graph is Reflected over the x-axis. y=x2 y= -x2

108 Graphing from Standard Form
Standard Form: y = ax2 +bx + c Quadratic Formula -b 2a 1) Axis of Symmetry: x= This is the x value of the Vertex

109 Graphing from Standard Form
Standard Form: y = ax2 +bx + c y = x2 +6x +4 a = 1, b = 6, c = 4 Determine Coordinates of the Vertex This is the x value of the Vertex -b 2a - 6 2(1) 1) Axis of Symmetry: x= x= = -3 2) Substitute and calculate y y = x2 +6x +4 y = (-3)2 +6(-3) +4 y = ­ y = -5 Vertex=(-3, -5)

110 Graphing from Standard Form
y = x2 +6x +4 Vertex=(-3, -5) y = (-2)2 +6(-2) +4 = 4 ­ 12 +4 y = (0)2 +6(0) +4 = 4 x y -3 -5 -2 -4 4 Plot two points on the same side of the vertex.

111 Graphing from Standard Form
y = x2 +6x +4 x y -3 -5 -2 -4 4 Copy points to the other side of the vertex.

112 Graphing from Standard Form
y = x2 +6x +4 x y -3 -5 -2 -4 4 Draw a smooth curve to match the plotted points

113 Exponential Functions and Sequences

114 Exponential Functions
y = bx If the base, 0 < b < 1 Exponential Decay If the base, b > 1 Exponential Growth y = 2x y = ½x

115 Exponential Functions
y = bx If the base, 0 < b < 1 Exponential Decay If the base, b > 1 Exponential Growth y = 2x y = ½x In either Growth or Decay there is an Asymptote where the curve approaches a horizontal line.

116 Exponential Functions
y = bx +k If the base, 0 < b < 1 Exponential Decay If the base, b > 1 Exponential Growth y = 2x+3 y = 2x y = ½x Asymptote: y = 3 Asymptote: y = -2 y = ½x -2 The Asymptote is always the x-axis unless the graph is Translated up or down.

117 Exponential Functions
y = bx Any number raised to the zero power equals one. If the base, b > 1 Exponential Growth So a key point on the graph is where the base is raised to the zero power. y = 2x (0,1) y = 3x

118 Exponential Functions
y = b(x-h) A key point on the graph is where the base is raised to the zero power. If the base, b > 1 Exponential Growth With a horizontal shift this isn’t the y-intercept. y = 2x y = 2(x-2) Shift 2 units to right. (2,1) (0,1)

119 Exponential Functions
y = bx Any number raised to the first power equals that number. If the base, b > 1 Exponential Growth So that is another key point on the graph. y = 2x x 2x 1 2 4 x 3x 1 3 2 9 (1,3) (1,2) y = 3x

120 Exponential Functions
y = bx If the base, 0 < b < 1 Exponential Decay b0 = 1 is still a key point Other key points may use negative exponents y = ½x x ½x 1 -1 2 -2 4

121 Scatter Plots, Trends and Statistics

122 Scatter Plots, Trends & Correlation
The scatter plot shows the number of CDs sold from 1999 to 2005. There were about 650 million sold. If the trend continued, how many CDs were sold in 2006? Draw a trend line (line of best fit) So about ½ the points Are on either side.

123 Scatter Plots, Trends & Correlation
The table shows Predicted cost for a middle income family to raise a child. Draw a scatter plot And describe the relationship. There is a strong positive correlation

124 Line of Best Fit Write equation of the line. Slope of the line = -1
Draw a scatter plot of the data and a line of best fit Write equation of the line. Slope of the line = -1 y-intercept = 1 Equation of the line y = -x + 1

125 Education The table gives hours studying and grade. Draw the scatter plot and line of best fit. Hours Grade 3 84 2 77 5 92 1 70 60 4 90 75 95 90 85 80 75 70 65 60

126 Education Write the Equation of the Line of Best Fit. Slope =
95 90 85 80 75 70 65 60 Run = 4 Intercept = 60 Rise = 30 15 2 y = x + 60

127 Education No. You’ll never get more than 100%
Predict the grade for a Student who studied 6 hrs. 15 2 y = x + 60 95 90 85 80 75 70 65 60 15 2 y = (6) + 60 Could this line go on forever? y = y = 105 No. You’ll never get more than 100% Or maybe 100?

128 Baseball Slope = 18.75 – 17.60 = 1.15 y = 1.15x + b
Use 2000 as starting point. Then 2001 would be x=1 Slope = – = 1.15 y = 1.15x + b 17.60 = 1.15(1) + b 16.45 = b y = 1.15x Where x is the number of years since 2000

129 Baseball y = 1.15x + 16.45 Use Equation to tell the
Price of a ticket in 2009. Where x is the number of years since 2000 y = 1.15(9) y = y = 26.80 This is extrapolation. It uses a point outside The data.

130 Statistics Mean: Average = Sum/number of instances Median: Middle Number (when sorted) Mode: Most Common Number Range: Highest - Lowest


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