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![Page 1: Section 11.1 Quadratic Functions and Their Graphs Copyright © 2013, 2009, and 2005 Pearson Education, Inc.](https://reader035.fdocuments.in/reader035/viewer/2022062717/56649e215503460f94b0e09f/html5/thumbnails/1.jpg)
Section 11.1
Quadratic Functions and Their Graphs
Copyright © 2013, 2009, and 2005 Pearson Education, Inc.
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Objectives
• Graphs of Quadratic Functions
• Min-Max Applications
• Basic Transformations of Graphs
• More About Graphing Quadratic Functions (Optional)
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The graph of any quadratic function is a parabola.
The vertex is the lowest point on the graph of a parabola that opens upward and the highest point on the graph of a parabola that opens downward.
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The graph is symmetric with respect to the y-axis. In this case the y-axis is the axis of symmetry for the graph.
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Example
Use the graph of the quadratic function to identify the vertex, axis of symmetry, and whether the parabola opens upward or downward.a. b.
Vertex (0, 2)Axis of symmetry: x = –2Open: up
Vertex (0, 4)Axis of symmetry: x = 0Open: down
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Example
Find the vertex for the graph of Support your answer graphically.Solutiona = 2 and b = 8
Substitute into the equation to find the y-value.
2( ) 2 8 3.f x x x
82
2(2)x
2
bx
a
2( ) 2( 2) 8( 2) 3
8 16 3
11
f x
The vertex is (2, 11), which is supported by the graph.
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Example
Identify the vertex, and the axis of symmetry on the graph, then graph.SolutionBegin by making a table of values.Plot the points and sketch a smooth curve.
The vertex is (0, –2)axis of symmetry x = 0
2( ) 2f x x
x f(x) = x2 – 2
3 7
2 2
1 1
0 2
1 1
2 2
3 7
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Example
Identify the vertex, and the axis of symmetry on the graph, then graph.SolutionBegin by making a table of values.Plot the points and sketch a smooth curve.
The vertex is (2, 0)axis of symmetry x = 2
2( ) ( 2)g x x
x g(x) = (x – 2)2
0 4
1 1
2 0
3 1
4 4
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Example
Identify the vertex, and the axis of symmetry on the graph, then graph.SolutionBegin by making a table of values.Plot the points and sketch a smooth curve.
The vertex is (2, 0)axis of symmetry x = 2
2( ) 2 3h x x x
x h(x) = x2 – 2x – 3
2 5
1 0
0 3
1 4
2 3
3 0
4 5
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Example
Find the maximum y-value of the graph of
SolutionThe graph is a parabola that opens downward because a < 0. The highest point on the graph is the vertex.a = 1 and b = 2
2( ) 2 3.f x x x
( 2)1
2 2( 1)
bx
a
2( ) ( 1) 2( 1) 3
4
f x
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Example
A baseball is hit into the air and its height h in feet after t seconds can be calculated by a. What is the height of the baseball when it is hit?b. Determine the maximum height of the baseball.Solutiona. The baseball is hit when t = 0.
b. The graph opens downward because a < 0. The maximum height occurs at the vertex. a = –16 and b = 64.
2( ) 16 64 2.h t t t
2( ) 16 64 2h t t t 2(0) 16(0) 64(0) 2
2
h
64 642
2 2( 16) 32
bx
a
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Example (cont)
2( ) 16 64 2 h t t t
2(2) 16(2) 64(2) 2
66
h
The maximum height is 66 feet.
2( ) 16 64 2 h t t t
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Basic Transformations of Graphs
The graph of y = ax2, a > 0.As a increases, the resulting parabola becomes narrower.When a > 0, the graph of y = ax2 never lies below thex-axis.
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Example
Compare the graph of g(x) = –4x2 to the graph of f(x) = x2. Then graph both functions on the same coordinate axes.SolutionBoth graphs are parabolas.The graph of g opens downward and is narrower than the graph of f.
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Section 11.2
Parabolas and Modeling
Copyright © 2013, 2009, and 2005 Pearson Education, Inc.
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Objectives
• Vertical and Horizontal Translations
• Vertex Form
• Modeling with Quadratic Functions (Optional)
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Vertical and Horizontal Translations
The graph of y = x2 is a parabola opening upward with vertex (0, 0).
All three graphs have the same shape.y = x2
y = x2 + 1 shifted upward 1 unity = x2 – 2 shifted downward 2 units
Such shifts are called translations because they do not change the shape of the graph only its position
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Vertical and Horizontal Translations
The graph of y = x2 is a parabola opening upward with vertex (0, 0).y = x2
y = (x – 1)2 Horizontal shift to the right 1 unit
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Vertical and Horizontal Translations
The graph of y = x2 is a parabola opening upward with vertex (0, 0).y = x2
y = (x + 2)2 Horizontal shift to the left 2 units
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Example
Sketch the graph of the equation and identify the vertex.
SolutionThe graph is similar to y = x2 except it has been translated 3 units down.
The vertex is (0, 3).
2 3y x
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Example
Sketch the graph of the equation and identify the vertex.
SolutionThe graph is similar to y = x2 except it has been translated left 4 units.
The vertex is (4, 0).
2( 4)y x
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Example
Sketch the graph of the equation and identify the vertex.
SolutionThe graph is similar to y = x2 except it has been translated down 2 units and right 1 unit.
The vertex is (1, 2).
2( 1) 2y x
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Example
Compare the graph of y = f(x) to the graph of y = x2. Then sketch a graph of y = f(x) and y = x2 in the same xy-plane.
SolutionThe graph is translated to the right 2 units and upward 3 units.The vertex for f(x) is (2, 3) and the vertex of y = x2 is (0, 0).The graph opens upward and is wider.
21( ) ( 2) 3
4f x x
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Example
Write the vertex form of the parabola with a = 3 and vertex (2, 1). Then express the equation in the form y = ax2 + bx + c.SolutionThe vertex form of the parabola is where the vertex is (h, k).a = 3, h = 2 and k = 1
To write the equation in y = ax2 + bx + c, do the following:
2( ) ,y a x h k
2)3( 12y x
2)3( 12y x 2( 4 43 1)y x x
23 12 12 1y x x 23 12 13y x x
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Example
Write each equation in vertex form. Identify the vertex. a. b.Solutiona. Because , add and subtract 16 on the right.
2 8 13y x x 22 8 7y x x
2 28
162 2
b
2 8 13y x x
2 16 1 68 13y x x
24 3y x
The vertex is (4, 3).
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Example (cont)
b. This equation is slightly different because the leading coefficient is 2 rather than 1. Start by factoring 2 from the first two terms on the right side.
22 8 7y x x
2 24
42 2
b
2
2
2 8 7
2( 4 ) 7
y x x
x x
2 42 4 74y x x
2 42 4 7 8y x x
22 2 1y x The vertex is ( 2, 1).
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Section 11.3
Quadratic Equations
Copyright © 2013, 2009, and 2005 Pearson Education, Inc.
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Objectives
• Basics of Quadratic Equations
• The Square Root Property
• Completing the Square
• Solving an Equation for a Variable
• Applications of Quadratic Equations
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Basics of Quadratic Equations
Any quadratic function f can be represented by f(x) = ax2 + bx + c with a 0.
Examples:
2 2 21( ) 2 1, ( ) 2 , and ( ) 2 1
3f x x g x x x h x x x
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Basics
The different types of solutions to a quadratic equation.
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Example
Solve each quadratic equation. Support your results numerically and graphically. a. b. c. Solutiona. Symbolic: Numerical: Graphical:
23 2 0x 2 9 6x x 2 2 8 0x x
The equation has no real solutions because x2 ≥ 0 for all real numbers x.
x y
1 5
0 2
1 5
2
2
2
3 2 0
3 2
2
3
x
x
x
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Example (cont)
b. 2 9 6x x 2
2
9 6
6 9 0
( 3)( 3) 0
3 0 or 3 0
3 or 3
x x
x x
x x
x x
x x
The equation has one real solution.
x y
5 4
4 1
3 0
2 1
1 4
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Example (cont)
c. 2 2 8 0x x 2 2 8 0
( 2)( 4) 0
2 0 or 4 0
2 or 4
x x
x x
x x
x x
The equation has two real solutions.
x y
4 0
2 8
1 9
0 8
2 0
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The Square Root Property
The square root property is used to solve quadratic equations that have no x-terms.
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Example
Solve each equation. a. b. c.Solutiona.
2 10x 225 16 0x 2( 3) 16x
2 10x
2 10x
10x
b. 225 16 0x 225 16x
2 16
25x
16
25x
4
5x
c. 2( 3) 16x
( 3) 16x
3 4x
3 4x
1 or 7x
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Real World Connection
If an object is dropped from a height of h feet, its distance d above the ground after t seconds is given by
2( ) 16d t h t
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Example
A toy falls 40 feet from a window. How long does the toy take to hit the ground? Solution 2( ) 16d t h t
2( ) 40 16d t t 240 16 0t
216 40t
2 40
16t
40
16t 4 10 2 10
416
10
2 1.6 sec.
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Example
Find the term that should be added to to form a perfect square trinomial.
SolutionCoefficient of x-term is –8, so we let b = –8. To complete the square we divide by 2 and then square the result.
2 8x x
2 28
162 2
b
2 2168 ( 4)x x x
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Example
Solve the equationSolutionWrite the equation in x2 + bx = d form.
2 8 13 0x x
2 28
162 2
b
2 8 13x x 2 168 3 61 1x x
2( 4) 3x
4 3x
4 3x
5.73 or 2.27x
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Example
Solve the equationSolutionWrite the equation in x2 + bx = d form.
20 2 8 7x x
2 24
42 2
b
22 8 7x x
2 442
47
x x
2 1( 2)
2x
12
2x
2 74
2x x
22
2x
12
2x
22
2x
1.29 or 2.71x
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Example
Solve the equation for the specified variable.Solution
216 for w x x
216w x
16
wx
4
wx
2
16
wx
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Example
Use of the Internet in Western Europe has increased dramatically. The figure shows a scatter plot of online users in Western Europe, together with a graph of a function f that models the data. The function f is given by: where the output is in millions of users. In this formula x = 6 corresponds to 1996, x = 7 to 1997, and so on, until x = 12 represents 2002.a. Evaluate f(10) and interpret the result.b. Graph f and estimate the year when the number of Internet users reached 85 million.c. Solve part (b) numerically.
2( ) 0.976 4.643 0.238f x x x
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Example (cont)
Solutiona. Evaluate f(10) and interpret the result.
Because x = 10 corresponds to 2000, there were about 51.4 million users in 2000.
2( ) 0.976 4.643 0.238f x x x
2( ) 0.976(10) 4.643(10) .0238
51.4
f x
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Example (cont)
Solutionb. Graph f and estimate the year when the number of Internet users reached 85 million.
c. Solve part (b) numerically.
2( ) 0.976 4.643 0.238f x x x
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Section 11.4
Quadratic Formula
Copyright © 2013, 2009, and 2005 Pearson Education, Inc.
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Objectives
• Solving Quadratic Equations
• The Discriminant
• Quadratic Equations Having Complex Solutions
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, 0 and 1,xf x a a a
The solutions to ax2 + bx + c = 0 with a ≠ 0 are given by
QUADRATIC FORMULA
2 4.
2
b b acx
a
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Example
Solve the equation 4x2 + 3x – 8 = 0. Support your results graphically.SolutionSymbolic SolutionLet a = 4, b = 3 and c = − 8.
2 4
2
b b acx
a
23 3 4 4 8
2 4x
3 137
8x
3 137
8x
or 3 137
8x
1.1x 1.8x or
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Example (cont)
4x2 + 3x – 8 = 0Graphical Solution
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Example
Solve the equation 3x2 − 6x + 3 = 0. Support your result graphically.SolutionLet a = 3, b = −6 and c = 3.
2 4
2
b b acx
a
26 6 4 3 3
2 3x
6 0
6x
1x
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Example
Solve the equation 2x2 + 4x + 5 = 0. Support your result graphically.SolutionLet a = 2, b = 4 and c = 5.
2 4
2
b b acx
a
24 4 4 2 5
2 2x
4 24
4x
There are no real solutions
for this equation because
is not a real number.
24
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, 0 and 1,xf x a a a
To determine the number of solutions to the quadratic equation ax2 + bx + c = 0, evaluate the discriminant b2 – 4ac.
1. If b2 – 4ac > 0, there are two real solutions.
2. If b2 – 4ac = 0, there is one real solution.
3. If b2 – 4ac < 0, there are no real solutions; there are two complex solutions.
THE DISCRIMINANT AND QUADRATIC
EQUATIONS
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Example
Use the discriminant to determine the number of solutions to −2x2 + 5x = 3. Then solve the equation using the quadratic formula.Solution−2x2 + 5x − 3 = 0Let a = −2, b = 5 and c = −3.
Thus, there are two solutions.
b2 – 4ac
= (5)2 – 4(−2)(−3) = 1
2 4
2
b b acx
a
5 1
2 2x
4
4x
1x
or
6
4x
1.5x
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, 0 and 1,xf x a a a
If k > 0, the solution to x2 + k = 0 are given by
THE EQUATION x2 + k = 0
.x i k
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Example
Solve x2 + 17 = 0.
SolutionThe solutions are
17.i
17 or 17.x i i
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Example
Solve 3x2 – 7x + 5 = 0. Write your answer in standard form: a + bi. SolutionLet a = 3, b = −7 and c = 5. 2 4
2
b b acx
a
7 11
6x
and
27 7 4 3 5
2 3x
7 11
6
ix
7 11
6 6x i
7 11
6 6x i
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Example
Solve Write your answer in standard form: a + bi. SolutionBegin by adding 2x to each side of the equation and then multiply by 5 to clear fractions.
Let a = −2, b = 10 and c = −15.
223 2 .
5 x
x
22 10 15 0x x
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Example (cont)
Let a = −2, b = 10 and c = −15.
223 2 .
5 x
x
2 4
2
b b acx
a
10 20
4x
210 10 4 2 15
2 2x
10 2 5
4
ix
5 5
2 2x i
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Example
Solve by completing the square.SolutionAfter applying the distributive property, the equation becomes
Since b = −4 ,add to each side of the equation.
4 5x x
2 4 5.x x
242 4
2 4 4 5 4x x 2
2 1x 2 1x 2x i
2x i
The solutions are 2 + i and 2 − i.