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9.2: QUADRATIC FUNCTIONS: Quadratic Function: A function that can be written in the form y = ax2+bx+c where a ≠ 0. Standard Form of a Quadratic: A function written in descending degree order, that is ax2+bx+c. Quadratic Parent Graph: The simplest quadratic function f(x) = x2. Parabola: The graph of the function f(x) = x2. Axis of Symmetry: The line that divide the parabola into two identical halves Vertex: The highest or lowest point of the parabola. Minimum: The lowest point of the parabola. Maximum: The highest point of the Line of Symmetry parabola. Vertex = Minimum GOAL: FINDING THE VERTEX OF 2 ax +bx+c: To find the vertex of a quadratic equation where a ≠ 1, we must know the following: 1) − 2) − 3) y( = axis of symmetry . = the x value of the vertex of the parabola. − ) = the y value of the vertex of the parabola, that is; min or max Ex: Provide the graph, vertex, domain and range of: y= 2 3x -9x+2 SOLUTION: To graph we can create a table or we can find the vertex with other two points Faster. y = 3x2-9x+2 1) − = axis of symmetry . a = 3, b = -9 −(−) − () 1.5 Thus x = 1.5 = axis of symmetry SOLUTION: (Continue) 2) − 3) y( y= = . = x value of the vertex ( 1.5, ) − ) = the y value of the vertex of the parabola y = 3x2-9x+2 − 2 − 3( ) -9( )+2 y = 3(1.5)2-9(1.5)+2 y = 3(2.25)-9(1.5)+2 y = 6.75-13.5+2 y = -4.75 Vertex = ( 1.5, -4.75) SOLUTION: (Continue) Vertex = ( 1.5, -4.75) Now: choose one value of x on the left of 1.5: x=0 x = 0 3(0)2-9(0)+2 y=2 (0, 2) choose a value of x on the right of 1.5: x=3 x = 3 3(3)2-9(3)+2 y=2 (3, 2) We not plot our three point: (0,2), vertex(1.5, -4.75) and (3,0) SOLUTION: Vertex: ( 1.5, -4.75) (0, 2) (3, 2) Domain: (-∞, ∞) Range: (-4.75, ∞) REAL-WORLD: During a basketball game halftime, the heat uses a sling shot to launch T-shirts at the crowd. The T-shirt is launched with an initial velocity of 72 ft/sec. The T-shirt is caught 35 ft above the court. How long will it take the T-shirt to reach its maximum height? What is the maximum height? What is the range of the function that models the height of the T-shirt over time? SOLUTION: To solve vertical motion problems we must know the following: An object projected into the air reaches the following maximum height: h= -16 Where 2 t + vt + c t = time, v = initial upward velocity c = initial height. SOLUTION: During ….The T-shirt is launched with an initial velocity of 72 ft/sec. The T-shirt is caught 35 ft above the court. From the picture we can see that initial height is 5ft. − Thus t = unknow, v = initial upward velocity = 72 ft/sec c = initial height= 5 ft h= -16 Now: 2 t + vt + c h= -16 − t= − t= − 2 t + 72t + 5 a = -16, b = 72 t = t = 2.25 −() (−) SOLUTION: (Continue) 2) − 3) y( y= = 2.25 = x value of the vertex: ( 2.25, − ) ) = the y value of the vertex: y = -16t2+72t+5 − 2 − -16( ) +72( )+5 y = -16(2.25)2+72(2.25)+5 y = -16(5.06)+72(2.25)+5 y = -81+162+5 y = 86 Vertex = ( 2.25, 86) SOLUTION: (Continue) Vertex = ( 2.25, 86) Now: choose one value of x on the left of 1.5: x=0 x=0 -16(0)2+72(0)+5 y = 5 (0, 5) choose a value of x on the right of 1.5: x=3 x = 4.5 -16(4.5)2+72(4.5)+5 y = (4.5, 5) We not plot our three point: (0,5), vertex(2.25, 86) and (4.5, 5) SOLUTION: Vertex: ( 2.25, -86) (0, 5) (4.5, 5) Domain: (-∞, ∞) Range: (-∞, 86) Maximum : 86 ft. VIDEOS: Quadratic Graphs and Their Properties Graphing Quadratics: http://www.khanacademy.org/math/algebra/quadratics/ graphing_quadratics/v/graphing-a-quadratic-function Interpreting Quadratics: http://www.khanacademy.org/math/algebra/quadratics/q uadratic_odds_ends/v/algebra-ii--shifting-quadraticgraphs VIDEOS: Quadratic Graphs and Their Properties Graphing Quadratics: http://www.khanacademy.org/math/algebra/quadratics/g raphing_quadratics/v/quadratic-functions-1 CLASSWORK: Page 544-545: Problems: 1, 2, 3, 4, 5, 8, 10, 13, 1619, 23, 25, 26.