1. 9.2: QUADRATIC FUNCTIONS: Quadratic Function: A function that can be written in the form y = ax 2 +bx+c where a ≠ 0. Standard Form of a Quadratic: A function written in descending degree order, that is ax 2 +bx+c.

2. Quadratic Parent Graph: The simplest quadratic function f(x) = x 2. Parabola: The graph of the function f(x) = x 2. Axis of Symmetry: The line that divide the parabola into two identical halves Vertex: The highest or lowest point of the parabola.

3. Minimum: The lowest point of the parabola. Maximum: The highest point of the parabola. Line of Symmetry Vertex = Minimum

4. GOAL:

5. FINDING THE VERTEX OF ax 2 +bx+c: To find the vertex of a quadratic equation where a ≠ 1, we must know the following:

6. Ex: Provide the graph, vertex, domain and range of: y = 3x 2 -9x+2

7. SOLUTION: To graph we can create a table or we can find the vertex with other two points  Faster. y = 3x 2 -9x+2  a = 3, b = -9  1.5 Thus x = 1.5 = axis of symmetry

8. SOLUTION: (Continue) y = 3x 2 -9x+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)

9. 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) x = 3  3(3) 2 -9(3)+2  y = 2  (3, 2) choose a value of x on the right of 1.5:  x=3 We not plot our three point: (0,2), vertex(1.5, -4.75) and (3,0)

10. SOLUTION: Vertex: ( 1.5, -4.75) (0, 2) (3, 2) Domain: (-∞, ∞) Range: (-4.75, ∞)

11. 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?

12. SOLUTION: To solve vertical motion problems we must know the following: An object projected into the air reaches the following maximum height: h= -16 t 2 + vt + c Where t = time, v = initial upward velocity c = initial height.

13. SOLUTION: h= -16 t 2 + vt + c 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.  h= -16 t 2 + 72t + 5  a = -16, b = 72  t = 2.25

14. SOLUTION: (Continue) y = -16t 2 +72t+5  y = -16(2.25) 2 +72(2.25)+5  y = -81+162+5  y = 86 Vertex = ( 2.25, 86)  y = -16(5.06)+72(2.25)+5

15. 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) x = 4.5  -16(4.5) 2 +72(4.5)+5  y =  (4.5, 5) choose a value of x on the right of 1.5:  x=3 We not plot our three point: (0,5), vertex(2.25, 86) and (4.5, 5)

16. SOLUTION: Vertex: ( 2.25, -86) (0, 5) (4.5, 5) Domain: (-∞, ∞) Range: (-∞, 86) Maximum : 86 ft.

17. VIDEOS: Quadratic Graphs and Their Properties Interpreting Quadratics: http://www.khanacademy.org/math/algebra/quadratics/q uadratic_odds_ends/v/algebra-ii--shifting-quadratic- graphs http://www.khanacademy.org/math/algebra/quadratics/ graphing_quadratics/v/graphing-a-quadratic-function Graphing Quadratics:

18. VIDEOS: Quadratic Graphs and Their Properties Graphing Quadratics: http://www.khanacademy.org/math/algebra/quadratics/gr aphing_quadratics/v/quadratic-functions-1

19. CLASSWORK: Page 544-545: Problems: 1, 2, 3, 4, 5, 8, 10, 13, 16  19, 23, 25, 26.