bell ringer 4/2/15 find the axis of symmetry, vertex, and solve the quadratic eqn. 1. f(x) = x 2 +...
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Bell Ringer 4/2/15Bell Ringer 4/2/15
Find the Axis of symmetry, vertex, and solve the quadratic eqn.
1. f(x) = x2 + 4x + 4
2. f(x) = x2+ 2x - 3
Bell Ringer 4/3/15Bell Ringer 4/3/15
Find the Axis of symmetry, vertex, and solve the quadratic eqn.
1. f(x) = x2+ 2x - 3
Bell Ringer 4/6/15Bell Ringer 4/6/15
#1 & 2 Find the axis of symmetry (AoS), and vertex of the following functions.
1.f(x) = X2 – 4
2.f(x) = -2X2 – 8x + 10
3.If a< 0 which way will the parabola open?
4.Graph the function f(x)= 2x2 – 4x – 1 by solving for the AoS, vertex, and using x = 2 & x = 3.
7.4 & 7.5 Graphing Quadratic 7.4 & 7.5 Graphing Quadratic FunctionsFunctions
• DefinitionsDefinitions
• 3 forms for a quad. function3 forms for a quad. function
• Steps for graphing each formSteps for graphing each form
• ExamplesExamples
• Changing between eqn. formsChanging between eqn. forms
Quadratic FunctionQuadratic Function•A function of the form A function of the form
y=axy=ax22+bx+c where a+bx+c where a≠0 making a ≠0 making a u-shaped graph called a u-shaped graph called a parabolaparabola..
Example quadratic equation:
Vertex-Vertex-
• The lowest or highest pointThe lowest or highest point
of a parabola.of a parabola.
VertexVertex
Axis of symmetry-Axis of symmetry-
• The vertical line through the vertex of the The vertical line through the vertex of the parabola.parabola.
Axis ofSymmetry
Standard Form EquationStandard Form Equationy=axy=ax22 + bx + c + bx + c
• If a is If a is positivepositive, u opens , u opens upupIf a is If a is negativenegative, u opens , u opens downdown
• The x-coordinate of the vertex is atThe x-coordinate of the vertex is at• To find the y-coordinate of the vertex, plug the To find the y-coordinate of the vertex, plug the
x-coordinate into the given eqn.x-coordinate into the given eqn.• The axis of symmetry is the vertical line x=The axis of symmetry is the vertical line x=• Choose 2 x-values on either side of the vertex x-Choose 2 x-values on either side of the vertex x-
coordinate. Use the eqn to find the coordinate. Use the eqn to find the corresponding y-values. corresponding y-values.
• Graph and label the 5 points and axis of Graph and label the 5 points and axis of symmetry on a coordinate plane. Connect the symmetry on a coordinate plane. Connect the points with a smooth curve.points with a smooth curve.
a
b
2
a
b
2
Graph
y = 2x2 - 8x + 6
Bell Ringer 4/7/15Bell Ringer 4/7/15
Standard Form: Transformations
f(x) = x2 g(x) = x2 + 4z(x) = x2 - 2
z(x)
g(x)
f(x)
5
y
x5-5
The simplest quadratic functions are of the form f (x) = ax2 (a 0) These are most easily graphed by comparing them with the graph of y = x2.
Example: Compare the graphs of
, and2xy 2
2
1)( xxf 22)( xxg
2
2
1)( xxf
22)( xxg
2xy
Standard Form: Transformations
Transformations (Cont.)
Graph of Transformations (cont.)
Vertex Form EquationVertex Form Equation
• If a > 0, parabola opens upIf a > 0, parabola opens up
If a < 0, parabola opens down.If a < 0, parabola opens down.
• The vertex is the point (h , k).The vertex is the point (h , k).
• The axis of symmetry is the vertical line x = h.The axis of symmetry is the vertical line x = h.
• If h>0 then parent function y = xIf h>0 then parent function y = x2 2 , h units to , h units to the right.the right.
• If h<0 then parent function y = xIf h<0 then parent function y = x2 2 , h units to , h units to the left.the left.
• K shift same as in standard formK shift same as in standard form
The Vertex form for the equation of a quadratic function is:
f (x) = a(x – h)2 + k (a 0)
Example: Graph f (x) = (x – 3)2 + 2 and find the vertex and axis.
f (x) = (x – 3)2 + 2 is the same shape as the graph ofg (x) = (x – 3)2 shifted upwards two units. g (x) = (x – 3)2 is the same shape as y = x2 shifted to the right three units.
f (x) = (x – 3)2 + 2
g (x) = (x – 3)2y = x 2
- 4x
y
4
4
vertex (3, 2)
x
y
4
4
Example: Graph and find the vertex and x-intercepts of f (x) = – ( x – 3)2 + 16
a < 0 parabola opens downward.
h = 3, k = 16 axis x = 3, vertex (3, 16).
Find the x-intercepts by solving
x = 7, x = –1 x-intercepts (7, 0), (–1, 0)
x = 3
(7, 0)(–1, 0)
(3, 16)
Example 3: GraphExample 3: Graphy=-.5(x+3)y=-.5(x+3)22+4+4• a is negative (a = -.5), so parabola opens down.a is negative (a = -.5), so parabola opens down.• Vertex is (h,k) or (-3,4)Vertex is (h,k) or (-3,4)• Axis of symmetry is the vertical line x = -3Axis of symmetry is the vertical line x = -3• Table of values Table of values x y x y
-1 2-1 2 -2 3.5 -2 3.5
-3 4-3 4 -4 3.5-4 3.5 -5 2-5 2
Vertex (-3,4)
(-4,3.5)
(-5,2)
(-2,3.5)
(-1,2)
x=-3
Now you try one! Ex. 4Now you try one! Ex. 4
y=2(x-1)y=2(x-1)22+3+3
•Open up or down?Open up or down?
•Vertex?Vertex?
•Axis of symmetry?Axis of symmetry?
•Table of values with 5 points?Table of values with 5 points?
(-1, 11)
(0,5)
(1,3)
(2,5)
(3,11)
X = 1
Copyright © by Houghton Mifflin Copyright © by Houghton Mifflin Company, Inc. All rights reserved.Company, Inc. All rights reserved. 1919
Example: A basketball is thrown from the free throw line from a height of six feet. What is the maximum height of the ball if the path of the ball is: 21
2 6.9
y x x
The path is a parabola opening downward. The maximum height occurs at the vertex.
2 ,9
162
9
1 2
baxxy
.92
a
bxAt the vertex,
1592
fa
bf
So, the vertex is (9, 15). The maximum height of the ball is 15 feet.
Copyright © by Houghton Mifflin Copyright © by Houghton Mifflin Company, Inc. All rights reserved.Company, Inc. All rights reserved. 2020
Example: A fence is to be built to form a rectangular corral along the side of a barn 65 feet long. If 120 feet of fencing are available, what are the dimensions of the corral of maximum area?
barn
corralx x
120 – 2xLet x represent the width of the corral and 120 – 2x the length.
Area = A(x) = (120 – 2x) x = –2x2 + 120 x
The graph is a parabola and opens downward.The maximum occurs at the vertex where ,
2a
bx
a = –2 and b = 120 .304
120
2
a
bx
120 – 2x = 120 – 2(30) = 60The maximum area occurs when the width is 30 feet and the length is 60 feet.