Download PPT Chapter 8 Review

Chapter 8
Review
Quadratic
Functions
§ 8.3
Graphing Quadratic
Equations in Two
Variables
Graphs of Quadratic Equations
We spent a lot of time graphing linear equations
in chapter 3.
The graph of a quadratic equation is a parabola.
The highest point or lowest point on the parabola
is the vertex.
Axis of symmetry is the line that runs through
the vertex and through the middle of the
parabola.
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Graphs of Quadratic Equations
Example
y
Graph y = 2x2 – 4.
x
y
2
4
1
–2
0
–4
–1
–2
–2
4
(–2, 4)
(2, 4)
x
(–1, – 2)
(1, –2)
(0, –4)
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Intercepts of the Parabola
Although we can simply plot points, it is helpful
to know some information about the parabola
we will be graphing prior to finding individual
points.
To find x-intercepts of the parabola, let y = 0 and
solve for x.
To find y-intercepts of the parabola, let x = 0 and
solve for y.
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Characteristics of the Parabola
If the quadratic equation is written in standard
form, y = ax2 + bx + c,
1) the parabola opens up when a > 0 and
opens down when a < 0.
b
2) the x-coordinate of the vertex is  .
2a
To find the corresponding y-coordinate, you
substitute the x-coordinate into the equation
and evaluate for y.
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Graphs of Quadratic Equations
Example
Graph y = –2x2 + 4x + 5.
Since a = –2 and b = 4, the
graph opens down and the
x-coordinate of the vertex
4
is 
1
y
(0, 5)
(1, 7)
(2, 5)
2(2)
x
y
3
–1
2
5
1
7
0
5
–1
–1
(–1, –1)
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(3, –1)
x
7
The Graph of a Quadratic Function
The vertical line which passes through the vertex is
called the Axis of Symmetry or the Axis
Recall that the equation of a vertical line is x =c
For some constant c
b
x coordinate of the vertex
2a
The y coordinate of the vertex is
4ac  b 2
2a
The Axis of Symmetry is the x =
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b
2a
8
The Quadratic function

Opens up when a>o
opens down a < 0
Vertex
axis of symmetry
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Identify the Vertex and Axis of
Symmetry of a Quadratic Function
 Vertex
=(x, y). thus
 Vertex =   b , f  


 Axis
2a

b 
 
2a  
of Symmetry: the line x =
b
2a
 Vertex
is minimum point if parabola opens up
 Vertex
is maximum point if parabola opens down
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Identify the Vertex and Axis of
Symmetry
f ( x)  3 x  6 x
2
 Vertex
x=
y=
 Vertex
b
2a
=

6
2(3)
 b 
f     f  1 
 2a 
= (-1, -3)
 Axis of Symmetry is x =
= -1
-3
b
2a
= -1
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8.5 Quadratic Solutions
The number of real solutions is at
most two.
No solutions
One solution
Two solutions
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Identifying Solutions
2
Example f(x) = x - 4
Solutions are -2 and 2.
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Identifying Solutions
Now you try this
problem.
2
f(x) = 2x - x
Solutions are 0 and 2.
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Graphing Quadratic Equations
The graph of a quadratic equation is a
parabola.
The roots or zeros are the x-intercepts.
The vertex is the maximum or
minimum point.
All parabolas have an axis of
symmetry.
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Graphing Quadratic Equations
One method of graphing uses a table with arbitrary
x-values.
Graph y = x2 - 4x
x
y
0
1
2
3
4
0
-3
-4
-3
0
Roots 0 and 4 , Vertex (2, -4) ,
Axis of Symmetry x = 2
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Graphing Quadratic Equations
Try this problem y = x2 - 2x - 8.
x
y
-2
-1
1
3
4
Roots
Vertex
Axis of Symmetry
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8.6 – Solving Quadratic Equations by Factoring
A quadratic equation is written in the Standard
Form,
2
ax  bx  c  0
where a, b, and c are real numbers and a  0.
Examples:
x  7 x  12  0
2
(standard form)
x  x  7  0
3 x  4 x  15
2
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8.6 – Solving Quadratic Equations by Factoring
Zero Factor Property:
If a and b are real numbers and if ab  0 ,
then a  0 or b  0 .
Examples:
x  x  7  0
x0
x7  0
x0
x  7
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8.6 – Solving Quadratic Equations by Factoring
Zero Factor Property:
If a and b are real numbers and if ab  0 ,
then a  0 or b  0 .
Examples:
 x 103x  6  0
x  10  0
3x  6  0
x  10  10  0  10 3x  6  6  0  6
3x 6
x  10

x2
3x  6
3 3
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8.6 – Solving Quadratic Equations by Factoring
Solving Quadratic Equations:
1) Write the equation in standard form.
2) Factor the equation completely.
3) Set each factor equal to 0.
4) Solve each equation.
5) Check the solutions (in original equation).
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8.6 – Solving Quadratic Equations by Factoring
x  3 x  18
2
x  3x  18  0
Factors of 18 :
1, 18 2, 9 3, 6
 6
 x  3  x  6   0
 3
2
x3  0
x  3
x6  0
x6
2
 3  6   18
36 18  18
18  18
2
 3  3  18
9  9  18
18  18
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8.6 – Solving Quadratic Equations by Factoring
If the Zero Factor
Property is not used,
then the solutions will
be incorrect
x  3 x  18
x  x  3  18
x  18
x  3  18
18
2
 3 18   18
324  54  18
270  18
2
x  3  3  18  3
x  21
 21
2
 3  21  18
441  63  18
378  18
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8.6 – Solving Quadratic Equations by Factoring
x  x  4  5
x  4x  5
2
x  4x  5  0
2
 x  1 x  5  0
x 1  0
x 5  0
x  1
x 5
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8.6 – Solving Quadratic Equations by Factoring
x  3x  7   6
 x  33x  2  0
3x  7 x  6
x  3  0 3x  2  0
x  3
3x  2
x2
3
2
3x  7 x  6  0
Factors of 3 :
1, 3
Factors of 6 :
1, 6 2, 3
2
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8.6 – Solving Quadratic Equations by Factoring
9 x  24 x  16
2
9 x  24 x  16  0
2
9 and 16 are perfect squares 
3x  43x  4  0
3x  4  0
3x  4
4
x
3
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8.6 – Solving Quadratic Equations by Factoring
2 x  18 x  0
3
2x  x  9   0
2
2x  x  3  x  3  0
2x  0 x  3  0 x  3  0
x 3
x0
x  3
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8.6 – Solving Quadratic Equations by Factoring
 x  3  3 x
 20 x  7   0
Factors of 3 : 1, 3 Factors of 7 : 1, 7
2
 x  3  x  7   3x  1  0
x3  0
x  3
x7  0
x7
3x  1  0
3x  1
1
x
3
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8.6 – Quadratic Equations and Problem Solving
A cliff diver is 64 feet above the surface of the
water. The formula for calculating the height (h)
2
of the diver after t seconds is: h  16t  64.
How long does it take for the diver to hit the surface
of the water?
2
0  16t  64
2
0  16  t  4 
0  16  t  2   t  2 
t 20
t  2
t 2  0
t  2 seconds
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8.6 – Quadratic Equations and Problem Solving
The square of a number minus twice the number is
63. Find the number.
x is the number.
x 2x  63
2
x  2 x  63  0
Factors of 63 : 1, 63 3, 21 7, 9
2
 x  7  x  9  0
x7  0
x 9  0
x  7
x 9
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8.6 – Quadratic Equations and Problem Solving
The length of a rectangular garden is 5 feet more than
its width. The area of the garden is 176 square feet.
What are the length and the width of the garden?
l  w  A The width is w. The length is w+5.
 w  5 w  176
 w  11  w  16  0
w  5w  176
w 11  0
w  11
2
w  5w  176  0
Factors of 176 :
1, 176 2, 88 4, 44
8, 22 11, 16
2
w  11 feet
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w 16  0
w  16
l  11  5
l  16 feet
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8.6 – Quadratic Equations and Problem Solving
Find two consecutive odd numbers whose product is
23 more than their sum?
x  2.
Consecutive odd numbers: x
x  x  2   x  x  2  23 x  5  0
2
x  2 x  2 x  25
x  5
2
x  2 x  2 x  2 x  25  2 x 5  2  3
2
x  25

5,

3
2
x  25  25  25
2
x  25  0
 x  5  x  5  0
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x 5  0
x 5
5 2  7
5, 7
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8.6 – Quadratic Equations and Problem Solving
The length of one leg of a right triangle is 7 meters less than
the length of the other leg. The length of the hypotenuse is 13
meters. What are the lengths of the legs?  Pythagorean Th.
a b  c
a  x b  x  7 c  13 2  x  5  x 12  0
2
2
2
x   x  7   13
x 12  0
x5  0
2
2
x  x  14 x  49  169
x  5
x  12
2
2 x  14 x  120  0
a  12 meters
2
2  x  7 x  60   0
meters
b

12

7

5
Factors of 60 : 1, 60 2, 30
3, 20 4, 15 5, 12 6, 10
2
2
2
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§ 8.7
Solving Quadratic
Equations by the Square
Root Property
Square Root Property
We previously have used factoring to solve
quadratic equations.
This chapter will introduce additional
methods for solving quadratic equations.
Square Root Property
If b is a real number and a2 = b, then
a b
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Square Root Property
Example
Solve x2 = 49
x   49  7
Solve 2x2 = 4
x2 = 2
x 2
Solve (y – 3)2 = 4
y  3   4  2
y=32
y = 1 or 5
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Square Root Property
Example
Solve x2 + 4 = 0
x2 = 4
There is no real solution because the square root
of 4 is not a real number.
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Square Root Property
Example
Solve (x + 2)2 = 25
x  2   25  5
x = 2 ± 5
x = 2 + 5 or x = 2 – 5
x = 3 or x = 7
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Square Root Property
Example
Solve (3x – 17)2 = 28
3x – 17 =  28   2 7
3x  17  2 7
17  2 7
x
3
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§ 8.8
Solving Quadratic
Equations by Completing
the Square
Completing the Square
In all four of the previous examples, the constant in the
square on the right side, is half the coefficient of the x
term on the left.
Also, the constant on the left is the square of the
constant on the right.
So, to find the constant term of a perfect square
trinomial, we need to take the square of half the
coefficient of the x term in the trinomial (as long as the
coefficient of the x2 term is 1, as in our previous
examples).
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Completing the Square
Example
What constant term should be added to the following
expressions to create a perfect square trinomial?
x2 – 10x
add 52 = 25
x2 + 16x
add 82 = 64
x2 – 7x
2
49
7
add   
4
2
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Completing the Square
Example
We now look at a method for solving
quadratics that involves a technique called
completing the square.
It involves creating a trinomial that is a perfect
square, setting the factored trinomial equal to a
constant, then using the square root property
from the previous section.
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Completing the Square
Solving a Quadratic Equation by Completing
a Square
1) If the coefficient of x2 is NOT 1, divide both
sides of the equation by the coefficient.
2) Isolate all variable terms on one side of the
equation.
3) Complete the square (half the coefficient of the
x term squared, added to both sides of the
equation).
4) Factor the resulting trinomial.
5) Use the square root property.
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Solving Equations
Example
Solve by completing the square.
y2 + 6y = 8
y2 + 6y + 9 = 8 + 9
(y + 3)2 = 1
±1
y+3=±1 =
y = 3 ± 1
y = 4 or 2
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Solving Equations
Example
Solve by completing the square.
y2 + y – 7 = 0
y2 + y = 7
y2 + y + ¼ = 7 + ¼
(y +
½)2
=
29
4
1
29
29
y 

2
4
2
1
29  1  29
y 

2
2
2
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Solving Equations
Example
Solve by completing the square.
2x2 + 14x – 1 = 0
2x2 + 14x = 1
x2 + 7x = ½
x2 + 7x +
49
4
=½+
49
4
=
51
4
7 2
51
(x + ) =
2
4
x
7
51
51


2
4
2
7
51  7  51
x 

2
2
2
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§ 8.9
Solving Quadratic
Equations by the
Quadratic Formula
The Quadratic Formula
Another technique for solving quadratic
equations is to use the quadratic formula.
The formula is derived from completing the
square of a general quadratic equation.
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The Quadratic Formula
A quadratic equation written in standard
form, ax2 + bx + c = 0, has the solutions.
 b  b  4ac
x
2a
2
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The Quadratic Formula
Example
Solve 11n2 – 9n = 1 by the quadratic formula.
11n2 – 9n – 1 = 0, so
a = 11, b = -9, c = -1
9  (9)  4(11)( 1) 9  81  44 9  125
n



22
22
2(11)
2
95 5
22
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The Quadratic Formula
Example
1 2
5
Solve x + x – = 0 by the quadratic formula.
8
2
x2 + 8x – 20 = 0 (multiply both sides by 8)
a = 1, b = 8, c = 20
 8  (8) 2  4(1)( 20)  8  64  80  8  144
x



2(1)
2
2
 8  12 20
4

or ,  10 or 2
2
2
2
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The Quadratic Formula
Example
Solve x(x + 6) = 30 by the quadratic formula.
x2 + 6x + 30 = 0
a = 1, b = 6, c = 30
 6  (6)  4(1)(30)  6  36  120  6   84
x


2
2
2(1)
2
So there is no real solution.
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The Discriminant
The expression under the radical sign in the
formula (b2 – 4ac) is called the discriminant.
The discriminant will take on a value that is
positive, 0, or negative.
The value of the discriminant indicates two
distinct real solutions, one real solution, or no
real solutions, respectively.
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The Discriminant
Example
Use the discriminant to determine the number and
type of solutions for the following equation.
5 – 4x + 12x2 = 0
a = 12, b = –4, and c = 5
b2 – 4ac = (–4)2 – 4(12)(5)
= 16 – 240
= –224
There are no real solutions.
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Solving Quadratic Equations
Steps in Solving Quadratic Equations
1) If the equation is in the form (ax+b)2 = c, use
the square root property to solve.
2) If not solved in step 1, write the equation in
standard form.
3) Try to solve by factoring.
4) If you haven’t solved it yet, use the quadratic
formula.
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Solving Equations
Example
Solve 12x = 4x2 + 4.
0 = 4x2 – 12x + 4
0 = 4(x2 – 3x + 1)
Let a = 1, b = -3, c = 1
3  (3)  4(1)(1) 3  9  4 3  5
x


2
2
2(1)
2
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Solving Equations
Example
Solve the following quadratic equation.
5 2
1
m m 0
8
2
5m 2  8m  4  0
(5m  2)( m  2)  0
5m  2  0 or m  2  0
2
m  or m  2
5
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The Quadratic Formula
Solve for x by completing the square.
 b  b  4ac
x
2a
2
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Yes, you can remember this formula
Pop goes the Weasel
http://www.youtube.com/watch?v=2lbABbfU6Zc&featu
re=related
Gilligan’s Island
http://www.youtube.com/watch?v=3CWTt9QFioY&feat
ure=related
This one I can’t explain
http://www.youtube.com/watch?v=haq6kpWdEMs&feat
ure=related
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How does it work
Equation:
3x 2  5 x  1  0
a3
b5
c 1
 b  b 2  4ac
x
2a
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How does it work
Equation:
3x 2  5 x  1  0
a3
b5
c 1
x
 5 
52  431
23
 5  25  12
x
6
 b  b 2  4ac
x
2a
x
 5  13  5
13


6
6
6
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The Discriminant
The number in the square root of the quadratic
formula.
b  4ac
2
Given x  5 x  6  0
2
 5
 416 
25  24  1
2
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The Discriminant
The Discriminant can be negative, positive or zero
If the Discriminant is positive,
there are 2 real answers.
If the square root is not a prefect
square
25
( for example
),
then there will be 2 irrational
roots
2 5
( for example
).
Martin-Gay, Developmental Mathematics
64
The Discriminant
The Discriminant can be negative, positive or zero
If the Discriminant is positive,
there are 2 real
answers.
If the Discriminant is zero,
there is 1 real answer.
If the Discriminant is negative,
there are 2 complex
answers.
Martin-Gay, Developmental
Mathematics
complex
answer have i.
65
Solve using the Quadratic formula
x  8 x  33
2
Martin-Gay, Developmental Mathematics
66
Solve using the Quadratic formula
x  8 x  33
2
x  8 x  33  0
2
x
  8 
 8  41 33
21
2
Martin-Gay, Developmental Mathematics
67
Solve using the Quadratic formula
x 2  8 x  33
x 2  8 x  33  0
x
  8 
 82  41 33
21
8  196 8  14
x

2
2
8  14 22
x

 11
2
2
8  14  6
x

 3
2
2
Martin-Gay, Developmental Mathematics
68
Solve using the Quadratic formula
x  34 x  289  0
2
Martin-Gay, Developmental Mathematics
69
Solve using the Quadratic formula
x  34 x  289  0
2
x
  34  
 34
21
2
 41289 
Martin-Gay, Developmental Mathematics
70
Solve using the Quadratic formula
x  34 x  289  0
2
x
  34 
 342  41289
21
34  1156  1156
x
2
34  0 34
x

 17
2
2
Martin-Gay, Developmental Mathematics
71
Solve using the Quadratic formula
x  6x  2  0
2
Martin-Gay, Developmental Mathematics
72
Solve using the Quadratic formula
x  6x  2  0
2
x
  6 
 6
21
2
 412
6  36  8 6  28
x

2
2
6 2 7
x 
 3 7
2
2
Martin-Gay, Developmental Mathematics
73
Solve using the Quadratic formula
x 2  13  6 x
x 2  6 x  13  0
x
  6 
 62  4113
21
Martin-Gay, Developmental Mathematics
74
Solve using the Quadratic formula
x 2  13  6 x
x 2  6 x  13  0
x
  6  
 62  4113
21
6  36  52 6   16
x

2
2
Martin-Gay, Developmental Mathematics
75
Solve using the Quadratic formula
x 2  13  6 x
x 2  6 x  13  0
x
  6  
 62  4113
21
6  36  52 6   16
x

2
2
6  4i 6 4
x
  i
2
2 2
x  3  2i
Martin-Gay, Developmental Mathematics
76
Describe the roots
Tell
2 me the Discriminant and the type of roots
x  6x  9  0
Martin-Gay, Developmental Mathematics
77
Describe the roots
Tell
2 me the Discriminant and the type of roots
x  6x  9  0
0, One rational root
Martin-Gay, Developmental Mathematics
78
Describe the roots
Tell
2 me the Discriminant and the type of roots
x  6x  9  0
0,x 2One
 3xrational
 5  0 root
Martin-Gay, Developmental Mathematics
79
Describe the roots
Tell
2 me the Discriminant and the type of roots
x  6x  9  0
0,x 2One
 3xrational
 5  0 root
-11, Two complex roots
Martin-Gay, Developmental Mathematics
80
Describe the roots
Tell
2 me the Discriminant and the type of roots
x  6x  9  0
0,x 2One
 3xrational
 5  0 root
-11,
Two complex roots
2
x  8x  4  0
Martin-Gay, Developmental Mathematics
81
Describe the roots
Tell
2 me the Discriminant and the type of roots
x  6x  9  0
0,x 2One
 3xrational
 5  0 root
-11,
Two complex roots
2
x  8x  4  0
80, Two irrational roots
Martin-Gay, Developmental Mathematics
82