Graphs of Quadratic Functions Graph the function. Compare the graph with the graph of Example 1.
Consider the function: f(x) = 2|x – 2| + 1 1. Does the graph of the function open up or down? 2....
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Transcript of Consider the function: f(x) = 2|x – 2| + 1 1. Does the graph of the function open up or down? 2....
![Page 1: Consider the function: f(x) = 2|x – 2| + 1 1. Does the graph of the function open up or down? 2. Is the graph of the function wider, narrower, or the same.](https://reader035.fdocuments.in/reader035/viewer/2022062515/56649d005503460f949d1bf6/html5/thumbnails/1.jpg)
Consider the function: f(x) = 2|x – 2| + 1
1.Does the graph of the function open up or down?
2.Is the graph of the function wider, narrower, or the same width as y = |x|?
3.What is the vertex of the graph?
4.What is the line of symmetry for the graph?
5.What numbers would you put in the table to complete the graph?
Algebra II 1
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I
Quadratic Functions
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A quadratic equation has a squared term in it, or a degree of two.
The graph of a quadratic makes a “U” shape called a parabola.
There are 3 forms a quadratic equation can be in…
3
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1. Vertex Form: y = a(x – h)2 + k
2. Standard Form: y = ax2 + bx + c
3. Intercept Form: y = a(x – p)(x – q)
4
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y = a(x – h)2 + kVertex: (opposite of h, same as k)Axis of symmetry (AOS): x = opposite of h. a: determines the direction the graph opens, and the width of the graph
a > 0 opens up
a < 0 opens down
|a| < 1 wider than x2
|a| > 1 narrower than x2
|a| = 1 same width as x2
II 5
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Graph:y = - ½(x + 3)2 +
4
Opens down
Wider than x2
Vertex: (-3,4)
AOS: x = -3
Table
Reflect
x
-2
-1
y
3.5
26
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Graph:y = 2(x – 1)2 + 3
Opens up
Narrower than x2
Vertex: (1,3)
AOS: x = 1
Table
Reflect
x
2
3
y
5
117
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Graph:y = -(x + 5)2 + 2
Opens down
Same width as x2
Vertex: (-5,2)
AOS: x = -5
Table
Reflect
x
-4
-3
y
1
-28
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Graph:y = 2(x – 1)2 + 1
Opens up
Narrower than x2
Vertex: (1,1)
AOS: x = 1
Table
Reflect
x
2
3
y
3
99
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Graph:y = -2(x + 2)2
Opens down
Narrower than x2
Vertex: (-2,0)
AOS: x = -2
Table
Reflect
x
-1
0
y
-2
-8I 10
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y = ax2 + bx + c
AOS: x = – b/(2a)
Vertex: ( – b/(2a), f(-b/2a))
“a” still determines the direction the graph opens and the width of the parabola
Algebra II 11
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Graph:y = x2 – 2x – 5
Opens up, Same width as x2
AOS: x = - b / (2a) x = 2 / (21) x = 1
y = (1)2 – 2(1) – 5Vertex: (1, -6)
Table
Reflect
x
2
3
y
-5
-2I 12
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Graph:y = 2x2 – 4x + 3
Opens up, Narrower than x2
AOS: x = - b / (2a) x = 4 / (22) x = 1
y = 2(1)2 – 4(1) + 3Vertex: (1, 1)
Table
Reflect
x
2
3
y
3
9I 13
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Graph:y = ½x2 + x – 6
Opens up, Wider than x2
AOS: x = - b / (2a) x = -1 / (2½) x = -1
y = ½(-1)2 + (-1) – 6Vertex: (-1, -6.5)
Table
Reflect
x
0
1
y
-6
-4.514
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Graph:y = -2x2 + 1
Opens down, Narrower than x2
AOS: x = - b / (2a) x = -0 / (2-2) x = 0
y = -2(0)2 + 1Vertex: (0, 1)
Table
Reflect
x
1
2
y
-1
-715
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Graph:y = 2x2 – 4x – 1
Opens up, Narrower than x2
AOS: x = - b / (2a) x = 4 / (22) x = 1
y = 2(1)2 – 4(1) – 1Vertex: (1, -3)
Table
Reflect
x
2
3
y
-1
5I 16
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y = a(x – p)(x – q)x-intercepts: (opposite of p, 0) ,
(opposite of q, 0)AOS: x = (opp p + opp q)
2Vertex: (opp p + opp q , f(opp p + opp q) )
2 2
***Do not have to make a table***
17
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Graph:y = -(x + 2)(x – 4)Opens down, Same width
as x2
Intercepts:(-2, 0), (4, 0)
AOS: x = (- p + - q) / 2
x = (-2 + 4) / 2
x = 1
y = -(1 + 2)(1 – 4)Vertex: (1, 9)
18
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Graph:y = ½(x – 6)(x –
4)Opens up, Wider than x2
Intercepts:(6, 0), (4, 0)
AOS: x = (- p + - q) / 2
x = (6 + 4) / 2
x = 5
y = ½(5 – 6)(5 – 4)Vertex: (5, -½)
19
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Graph:y = -½x(x – 5)
Opens down, Wider than x2
Intercepts:(0, 0), (5, 0)
AOS: x = (- p + - q) / 2
x = (0 + 5) / 2
x = 2.5
y = -½(2.5)(2.5 – 5)Vertex: (2.5, 3.125)
20
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Graph:y = -2(x – 3)(x +
1)Opens down, Narrower
than x2
Intercepts:(3, 0), (-1, 0)
AOS: x = (- p + - q) / 2
x = (3 + -1) / 2
x = 1
y = -2(1 – 3)(1 + 1)Vertex: (1, 8)
I 21
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Graph:y = ⅓(x – 2)(x +
4)Opens up, Wider than x2
Intercepts:(2, 0), (-4, 0)
AOS: x = (- p + - q) / 2
x = (2 + -4) / 2
x = -1
y = ⅓(-1 – 2)(-1 + 4)Vertex: (-1, -3)
22
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23Algebra II
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General Strategy for Problem Solving1. UNDERSTAND the problem.
• Read and reread the problem• Choose a variable to represent the
unknown• Construct a drawing, whenever possible
2. MODEL the problem with an equation.3. SOLVE the equation.4. INTERPRET the result.
• Check proposed solution in original problem.
• State your conclusion.
24
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25
The equation for the percent of test subjects that felt comfortable at a given temperature x is
y = –3.678x2 + 527.3x – 18,807. What temperature made the greatest percent of test subjects comfortable?
At that temperature, what percent of people felt comfortable?
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26
The Golden Gate Bridge in San Francisco has two towers that rise 500 feet above the road and are connected by cables as shown. Each cable forms a parabola with the
equation y = 1/8960(x – 2100)2 + 8. What is the distance between the two towers? What is the height of
the cable above the road at its lowest point?
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The archway that forms the ceiling of a tunnel can be modeled by the equation y = –0.0355x2 + .923x + 10 where x is the horizontal distance in feet and y is the height in feet from the ceiling to the floor. How many feet from the walls does the ceiling reach its maximum height? What is the maximum height?
27
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The length of a rectangle is three more than twice the width. Determine the dimensions that will give a total area of 27 m2.
28II
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The length of a Ping-Pong table is 3 ft more than twice the width. The area of a Ping-Pong table is 90 square feet. What are the dimensions of a Ping-Pong table?
29
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Find two positive whose sum is 32 and whose product is a maximum.
30
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Find two numbers whose sum is 49 and whose product is a maximum.
31Algebra II
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Find two numbers whose product is a maximum if the sum of the first and five times the second is 80.
32Algebra II
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Write the standard form of the equation of the parabola whose vertex is (1,2) and passes through (3, -6)
33Algebra II
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Write the standard form of the equation of the parabola whose vertex is (-4,11) and that passes through the point (-6,15)
34Algebra II
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1. y = -½(x – 3)(x + 1)
2. y = 2(x + 3)2 – 2
3. y = –x2 + 4x – 2
35Algebra II