Slide 1

Functions

Quadratics and Polynomials

Adding PolynomialsGroup and Combine Like TermsLike 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

Special ProductsSquare of a Binomial= (x+ 3)(x + 3)= x2 + 3x +3x + 9= x2 + 6x + 9(x+ 3)2Or use pattern(a+ b)2 = a2 + 2ab + b2(a - b)2 = a2 - 2ab + b2Rewrite and distribute(x+ 8)2 = x2 + 2x(8) + 82= x2 + 16x + 64(x- 5)2 = x2 - 2x(5) + 52= x2 - 10x + 25

Factoring

Completing the Square

(x + b)2 = x2 + 2bx + b2

The Quadratic Formulaand Discriminant

Graphing Quadratic Equations

Vertical Line Test

A Vertical Linehits the graph of a function only once.This IS the graph of a FunctionVertical Line Test

This IS NOT the graph of a FunctionDomain and RangeThe Domain of a Relation is the set of all Input values (x).This is also called the Independent Variable.Notice: DomainInput Independent all contain the word in

DomainDomain and RangeThe 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 RangeInput Output Independent Dependent

RangeEvaluating a FunctionMany 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 = 7So f(4) = 7 and the point (4,7) is on the graph of the function.

Linear Equations

Linear EquationsGraphing Linear EquationsGraphing Linear InequalitiesWriting Linear EquationsGraphing Linear Equations

Graph: y = 2x -5From a table:XYPick values for xThat will be easy to use.And evaluate for y.x = zero is usually easiest0y = 2(0) 5 5Pick another xClose by.1y = 2(1) 5 3Graphing Linear Equations

Graph: y = 2x -5From a table:XYPlot each PointAnd draw the Line through them.0 51 3Graphing Linear Equations

From a table:XYPick values for xThat will be easy to use.And evaluate for y.x = zero is usually easiest0y = -(0) + 5 5Pick denominatorAs second x5y = -(5) + 5 3Graph: y = -x +5 -2 + 5Graphing Linear Equations

Graph: y = -x +5From a table:XYPlot each PointAnd draw the Line through them.0 55 3Graphing Linear Equations

Graph: y = x -3Using y-interceptand slope.Plot the y-intercepton the y-axis.Use slope to findA second point.rise = 2run = 3Draw line through the 2 points.Graphing Linear Equations

From standard form:XYBecause 24 is a multipleof both 4 and 3, its easyto use x and y intercepts.x = zero for y-intercept0 -3y = 24-8y = zeroFor x-intercept04x = 24 6Graph: 4x - 3y = 24Graphing Linear Equations

From standard form:XYPlot each PointAnd draw the Line through them.0 -86 0y-intercept on y-axisx-intercept on x-axisGraph: 4x - 3y = 24Graphing Linear Equations

From standard form:Because 24 is not a multiple of 5.Convert to y = mx + bGraph: 5x - 3y = 24Convert from Standard form to Slope-Intercept Standard Slope-Intercept y = 5/3x - 8Move x-term to other sideby adding the opposite-5xax + by = c y = mx + b-3y = -5x + 24Solve for y5x - 3y = 24-5x-3-3-3Graphing Linear Equations

Graph: 5x - 3y = 24 y = 5/3x - 8XY0 -83-3 y = 5/3(0) - 8 y = 5/3(3) - 8Graphing Linear Inequalities

Graph: y 1, graph has a vertical stretch

y=x2Wider than y=x2More narrow than y=x2Graphing from Vertex Form

y=x2If a < 0, graph is Reflected over the x-axis.y= -x2y= -x2

Graphing from Standard FormStandard Form: y = ax2 +bx + c1) Axis of Symmetry: x=-b2aThis is the x value of the VertexQuadratic FormulaGraphing from Standard FormDetermine Coordinates of the VertexStandard Form: y = ax2 +bx + c y = x2 +6x +4Vertex=(-3, -5)1) Axis of Symmetry: x=-b2a y = x2 +6x +4a = 1, b = 6, c = 4x= - 6 2(1)= -3This is the x value of the Vertex2) Substitute and calculate y y = (-3)2 +6(-3) +4 y = 9 18 +4 y = -5

Graphing from Standard Formxy-3-5Plot two points on the same side of the vertex.-20-4 4Vertex=(-3, -5) y = x2 +6x +4 y = (-2)2 +6(-2) +4= 4 12 +4 y = (0)2 +6(0) +4= 4

Graphing from Standard Formxy-3-5-20-4 4 y = x2 +6x +4Copy points to the other side of the vertex.

Graphing from Standard Formxy-3-5-20-4 4 y = x2 +6x +4Draw a smooth curve to match the plotted pointsExponential Functions and Sequences

Exponential Functions

y = bx

If the base, b > 1Exponential GrowthIf the base, 0 < b < 1Exponential Decayy = 2xy = xExponential Functions

y = bx

If the base, b > 1Exponential GrowthIf the base, 0 < b < 1Exponential Decayy = 2xy = xIn either Growth or Decay there is an Asymptotewhere the curve approaches a horizontal line.Exponential Functions

y = bx +k

If the base, b > 1Exponential GrowthIf the base, 0 < b < 1Exponential Decayy = 2xy = xThe Asymptote is always the x-axis unless the graph is Translated up or down.y = 2x+3y = x -2Asymptote: y = 3Asymptote: y = -2Exponential Functionsy = bx

If the base, b > 1Exponential Growthy = 2xy = 3x(0,1)Any number raised to the zero power equals one.So a key point on the graph is where the base is raised to the zero power.Exponential Functionsy = b(x-h)

If the base, b > 1Exponential Growthy = 2xy = 2(x-2)(0,1)A key point on the graph is where the base is raised to the zero power.With a horizontal shiftthis isnt the y-intercept.(2,1)Shift 2 units to right.Exponential Functionsy = bx

If the base, b > 1Exponential Growthy = 2xy = 3x(1,2)Any number raised to the first power equals that number.So that is another key point on the graph.(1,3)x2x011224x3x011329Exponential Functions

y = bxIf the base, 0 < b < 1Exponential Decayy = xxx01-12-24b0 = 1 is still a key pointOther key points may usenegative exponentsScatter Plots, Trends and Statistics

Scatter Plots, Trends & Correlation

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

The table shows Predicted cost fora middle income family to raise a child.Draw a scatter plotAnd describe the relationship.There is a strong positive correlationLine of Best Fit

Draw a scatter plot of the data and a line of best fitWrite equation of the line.Slope of the line = -1y-intercept = 1Equation of the line y = -x + 1EducationThe table gives hours studying and grade.Draw the scatter plot and line of best fit.

0 1 2 3 4 5 6 7 8HoursGrade3842775921700604903759590858075706560EducationWrite the Equation of the Line of Best Fit.

0 1 2 3 4 5 6 7 89590858075706560Rise = 30Run = 4Slope = 30 15 4 2Intercept = 6015 2y = x + 60EducationPredict the grade for aStudent who studied 6 hrs.

0 1 2 3 4 5 6 7 8959085807570656015 2y = x + 6015 2y = (6) + 60y = 45 + 60y = 105Or maybe 100?Could this line go on forever?No. Youll never get more than 100%Baseball

Slope = 18.75 17.60 = 1.15y = 1.15x + b17.60 = 1.15(1) + bUse 2000 as starting point.Then 2001 would be x=116.45 = by = 1.15x + 16.45Where x is the numberof years since 2000Baseball

y = 1.15x + 16.45Where x is the numberof years since 2000Use Equation to tell the Price of a ticket in 2009.y = 1.15(9) + 16.45y = 10.35 + 16.45y = 26.80This is extrapolation.It uses a point outside The data. StatisticsMean: Average = Sum/number of instancesMedian: Middle Number (when sorted)Mode:Most Common NumberRange:Highest - Lowest