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 parabola. Line of Symmetry

Vertex = Minimum

GOAL:

FINDING THE VERTEX OF ax2+bx+c: To find the vertex of a quadratic equation where a ≠ 1, we must know the following:1) = axis of symmetry .

2) = the x value of the vertex of the parabola.

3) y( ) = the y value of the vertex of the parabola, that is; min or max

Ex: Provide the graph, vertex, domain and range of:

y = 3x2-9x+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

−𝒃𝟐𝒂

1.5

Thus x = 1.5 = axis of symmetry

SOLUTION: (Continue)

2) = x value of the vertex ( 1.5, )

3) y( ) = the y value of the vertex of the parabola y = 3x2-9x+2

y = 3()2-9()+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=0x = 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)

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 t2 + vt + cWhere t = time,

v = initial upward velocity c = initial height.

SOLUTION:

h= -16 t2 + vt + c

Thus t = unknow, v = initial upward velocity = 72 ft/sec c = initial height= 5 ft

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 t2 + 72t + 5Now: t = a = -16, b = 72 t =

t = t = 2.25

SOLUTION: (Continue)

2) = 2.25 = x value of the vertex: ( 2.25, )

3) y( ) = the y value of the vertex:y = -16t2+72t+5

y = -16()2+72()+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

SOLUTION: (Continue) Vertex = ( 2.25, 86)

Now: choose one value of x on the left of 1.5: x=0x = 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)

SOLUTION: Vertex: ( 2.25, -86)

(0, 5) (4.5, 5)

Domain: (-∞, ∞)

Range: (-∞, 86)

Maximum : 86 ft.

VIDEOS: Quadratic Graphs and Their Properties

Interpreting Quadratics:http://www.khanacademy.org/math/algebra/quadratics/quadratic_odds_ends/v/algebra-ii--shifting-quadratic-graphs

http://www.khanacademy.org/math/algebra/quadratics/graphing_quadratics/v/graphing-a-quadratic-function

Graphing Quadratics:

VIDEOS: Quadratic Graphs and Their Properties

Graphing Quadratics:http://www.khanacademy.org/math/algebra/quadratics/graphing_quadratics/v/quadratic-functions-1

CLASSWORK:

Page 544-545:

Problems: 1, 2, 3, 4, 5, 8, 10, 13, 1619, 23, 25, 26.