HW#1: Two Problems on graph paper SWBAT… analyze the characteristics of the graphs of quadratic functions 6/2/10 Agenda 1. WU (15 min) 2. Notes on graphing quadratics & properties of quadratics (30 min) WARM-UP Write the hw in your planners Review tests Path of a baseball (back of agenda) HW#1: Two Problems on graph paper

Mean = 80% Range = 27% - 113%

Mean = 85% Range = 60% - 100%

Mean = 81% Range = 40% - 107%

Mean = 82% Range = 27% - 113%

To get us warmed up and ready for quadratics… Complete the path of a baseball on the back of this week’s agenda

Solving Quadratic Equations by Using the Quadratic Formula Monday, 5/31 Tuesday, 6/1 Wednesday, 6/2 Thursday, 6/3 Friday, 6/4 NO CLASSES MEMORIAL DAY Path of a baseball Graphing & Properties of Quadratic Functions HW#1 HW#2 Activity on big graph paper: Graphing Quadratics HW#3 (quiz) TI-84 Graphing Calculator Investigation Activity: Transformations of Quadratics HW#4 Monday, 6/7 Tuesday, 6/8 Wednesday, 6/9 Thursday, 6/10 Friday, 6/11 Solving Quadratic Equations by Using the Quadratic Formula HW#5 Review for Final ½ Day: A FINAL EXAM!!!! Monday, 6/14 Tuesday, 6/15 Wednesday, 6/16 Thursday, 6/17 Friday, 6/18 TBA Last Day of Class!

Properties of Quadratic Functions Agenda: Standard form of a quadratic (1 slide) Graphing quadratics (1 slide) Finding solutions to quadratics (1 slide) Characteristics of quadratic functions (3 slides) Quadratic graphs examples (1 slide) HW Problem

Standard form of a quadratic y = ax2 + bx + c When the power of an equation is 2, then the function is called a quadratic

Graphs of Quadratics The graph of any quadratic equation is a parabola To graph a quadratic, set up a table and plot points Example: y = x2 x y -2 4 -1 1 0 0 1 1 2 4 . x y y = x2

HOMEWORK #1 – On Graph Paper (Warm Up – will be collected) 1.) Graph y = -x2 + 1 using a table of values (answer on the front side) 2.) How are the graphs of y = -x2 + 1 and y = -x + 1 different? (answer on the back side)

Characteristics of Quadratic Functions Axis of symmetry . x-intercept vertex y-intercept x y To find the solutions graphically, look for the x-intercepts of the graph (Since these are the points where y = 0)

Characteristics of Quadratic Functions The shape of a graph of a quadratic function is called a parabola. Parabolas are symmetric about a central line called the axis of symmetry. The axis of symmetry intersects a parabola at only one point, called the vertex (ordered pair). The lowest point on the graph is the minimum. The highest point on the graph is the maximum. The maximum or minimum is the vertex

Finding the solutions of a quadratic 1. Set y or f(x) equal to zero: 0 = ax2 + bx + c 2. Factor 3. Set each factor = 0 4. Solve for each variable 1)Algebraically (last week and next slide to review) 2)Graphically (today  in a few slides) In general equations have roots, Functions haves zeros, and Graphs of functions have x-intercepts

Ex: f(x) = x2 – 8x + 12 0 = (x – 2)(x – 6) x – 2 = 0 or x – 6 = 0 Directions: Find the zeros. Ex: f(x) = x2 – 8x + 12 0 = (x – 2)(x – 6) x – 2 = 0 or x – 6 = 0 x = 2 or x = 6 Factors of 12 Sum of Factors, -8 1, 12 13 2, 6 8 3, 4 7 -1, -12 -13 -2, -6 -8 -3, -4 -7

Key Concept: Quadratic Functions Parent Function f(x) = x2 Standard Form f(x) = ax2 + bx + c Type of Graph Parabola Axis of Symmetry y-intercept c

Axis of symmetry examples http://www.mathwarehouse.com/geometry/parabola/axis-of-symmetry.php

Example: y = x2 – 4 1. What is the axis of symmetry? 2. What is the vertex: (0, -4) x 3. What is the y-intercept: (0, -4) x y -2 0 -1 -3 0 -4 1 -3 2 0 1. (0, -4) 2. x = 2 or x = -2 3. -4 4. x = 0 4. What are the solutions: (x-intercepts) x = -2 or x = 2

HOMEWORK #1 – On Graph Paper 1.) For y = -x2 + 1 Graph using a table of values The axis of symmetry The vertex The y-intercept The solutions (x-intercepts) 2.) How are the graphs of y = -x2 + 1 and y = -x + 1 different?

Example: y = -x2 + 1 1. Axis of symmetry: 2. Vertex: 3. y-intercept: (0,1) 2. Vertex: y 3. y-intercept: (0,1) 4. x-intercepts: x = 1 or x = -1 y = -x2 + 1 x y -2 -3 -1 0 0 1 1 0 2 -3 x 1. (0, 1) 2. x = 1 or x = -1 3. 1 4. x = 0

Vertex formula x = -b 2a Steps to solve for the vertex: Step 1: Solve for x using x = -b/2a Step 2: Substitute the x-value in the original function to find the y-value Step 3: Write the vertex as an ordered pair ( , )

Example 1 Find the vertex: y = x2 – 4x + 7 x = -b = -(-4) = 4 = 2 a = 1, b = -4 x = -b = -(-4) = 4 = 2 2a 2(1) 2 y = x2 – 4x + 7 y = (2)2 – 4(2) + 7 = 3 The vertex is at (2,3)

Example 2 Find the vertex: y = x2 + 4x + 7 x = -b = -4 = -4 = -2 a = 1, b = 4 x = -b = -4 = -4 = -2 2a 2(1) 2 y = x2 + 4x + 7 y = (-2)2 + 4(-2) + 7 = 3 The vertex is at (-2,3)

HOMEWORK Find the vertex: y = 2(x – 1)2 + 7 y = 2(x – 1)(x – 1) + 7 a = 2, b = -4 x = -(-4)/(2(2)) = 1 y = 2(1 – 1)2 + 7 y = 2(0)2 + 7 Answer: (1, 7)

Example 3: (HW1 Prob #9) Find the vertex: y = 4x2 + 20x + 5 a = 4, b = 20 x = -b = -20 = -20 = -2.5 2a 2(4) 8 y = 4x2 + 20x + 5 y = 4(-2.5)2 + 20(-2.5) + 5 = -20 The vertex is at (-2.5,-20)

Example 4: (HW1 Prob #12) Find the vertex: y = 5x2 + 30x – 4 a = 5, b = 30 x = -b = -30 = -30 = -3 2a 2(5) 10 y = 5x2 + 30x – 4 y = 5(-3)2 + 30(-3) – 4 = -49 The vertex is at (-3,-49)