Algebra 1 2nd Quarter - Review

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

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

Functions and Linear Equations

Functions

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).

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

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

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

Vertical Line Test This IS NOT the graph of a Function

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

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

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.

Linear Equations

Linear Equations Graphing Linear Equations Graphing Linear Inequalities Writing Linear Equations

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.

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

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

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

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.

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

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

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

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

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

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.

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.

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) ? √

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

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

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

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 ⅔ 9 Slope = ⅔ y-intercept = 9

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

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

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)

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 = -9 + b +9 +9 3) Solve for b (y-intercept) 17 = b y = -3x + b 17 4) Substitute for b

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 = 4 + b 3) Solve for b (y-intercept) -4 -4 4) Substitute for b -1 = b y = ⅖ x + b - 1

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 ⅔ 2 Point (2, 5) Slope = ⅔

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

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

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

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

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

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

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

Systems of Linear Equations1

Linear Systems Solving by Graphing Solving Algebraically Substitution Linear Combination

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) √

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.

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.

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 = -10+9 = -1 2x +10x -45 = 15 (5, -1) 12x = 60 x = 5 5

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.

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)

} 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

} 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 + 20 = 29 3x = 9 x = 3

} 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.

Quadratics and Polynomials

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

( ) 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 = =

( ) 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

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

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

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

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

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

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)

Multiplying Polynomials Use the distributive property and Laws of Exponents -5x2 (3x2 + 4x - 11) = -15x4 - 20x3 + 55x2 2x (3x - 11) +3(x -5) = 6x2 - 22x + 3x -15 = 6x2 - 19x -15 (x+ 3)(x - 8) = x2 - 8x + 3x - 24 = x2 - 5x - 24

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

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

Factoring

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

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 + 16 = (x __ )(x __ ) – – 4 4 Factors of 16 1 16 2 8 4 Both Negative Factors have the same sign

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

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

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

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.

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.

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 )

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 = 5 or x = -7

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 = 6 or x = 2

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 = 8 or x = -3

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

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

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

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 = 5 or -1

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

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

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

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

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?

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

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 + 8 = 7 + and - square roots - 8 - 8 Solve for x x = -8 + 7 = -1 x = -8 - 7 = -15

The Quadratic Formula and Discriminant

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

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

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

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 = -6 -2i = -3 - i 2

Graphing Quadratic Equations

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

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)2 +7 Vertex=(4, 7) y = (x + 5)2 - 9 Vertex=(-5, -9) y = -2(x + 1)2 + 6 Vertex=(-1, 6) Note: the value of a doesn’t effect the vertex.

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

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 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.

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.

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 .

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

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

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

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

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 = 9 ­ 18 +4 y = -5 Vertex=(-3, -5)

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.

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

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

Exponential Functions and Sequences

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

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.

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.

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

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)

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

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

Scatter Plots, Trends and Statistics

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.

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

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

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 0 1 2 3 4 5 6 7 8

Education Write the Equation of the Line of Best Fit. Slope = 30 15 4 2 95 90 85 80 75 70 65 60 Run = 4 Intercept = 60 Rise = 30 15 2 y = x + 60 0 1 2 3 4 5 6 7 8

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 = 45 + 60 y = 105 No. You’ll never get more than 100% Or maybe 100? 0 1 2 3 4 5 6 7 8

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

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) + 16.45 y = 10.35 + 16.45 y = 26.80 This is extrapolation. It uses a point outside The data.

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