Transcript ReviewChapter3

MAT 171 Chapter 3 Review
The following is a brief review of Chapter 3 for Test 2
that covers Chapters 2 & 3 and Section 10.7. This does
NOT cover all the material that may be on the test.
Click on Slide Show and View Slide Show.
Read and note your answer to the question.
Advance the slide to see the answer.
Dr. Claude Moore, Math Instructor, CFCC
Slide 3 - 1
Active Learning Lecture Slides
For use with Classroom Response Systems
Chapter 3
© 2009 Pearson Education, Inc.
Slide 3 - 2
Chapter 3:
Quadratic Functions and
Equations; Inequalities
3.1
3.2
3.3
3.4
3.5
The Complex Numbers
Quadratic Equations, Functions, Zeros, and Models
Analyzing Graphs of Quadratic Functions
Solving Rational Equations and Radical Equations
Solving Equations and Inequalities with Absolute
Values
Copyright © 2009 Pearson Education, Inc.
Slide 3 - 3
Express
12
in terms of i.
a. 2 3
b. 2 3i
c. 4 3i
d. 2 3i
Copyright © 2009 Pearson Education, Inc.
Slide 3 - 4
Express
12
in terms of i.
a. 2 3
b. 2 3i
c. 4 3i
d. 2 3i
Copyright © 2009 Pearson Education, Inc.
Slide 3 - 5
Simplify: (4 – 2i) – (8 + 3i).
a. –4 – 5i
b. –4 + i
c. 12 – 5i
d. –3i
Copyright © 2009 Pearson Education, Inc.
Slide 3 - 6
Simplify: (4 – 2i) – (8 + 3i).
a. –4 – 5i
b. –4 + i
c. 12 – 5i
d. –3i
Copyright © 2009 Pearson Education, Inc.
Slide 3 - 7
Simplify: (3 + 2i)(4 – i).
a. 14 + 5i
b. 10 + 5i
c. 12 – 2i
d. 7 + i
Copyright © 2009 Pearson Education, Inc.
Slide 3 - 8
Simplify: (3 + 2i)(4 – i).
a. 14 + 5i
b. 10 + 5i
c. 12 – 2i
d. 7 + i
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Slide 3 - 9
Simplify: i17.
a. i
b. 1
c. i
d. 1
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Slide 3 - 10
Simplify: i17.
a. i
b. 1
c. i
d. 1
Copyright © 2009 Pearson Education, Inc.
Slide 3 - 11
Simplify: i23.
a. i
b. 1
c. i
d. 1
Copyright © 2009 Pearson Education, Inc.
Slide 3 - 12
Simplify: i23.
a. i
b. 1
c. i
d. 1
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Slide 3 - 13
Solve: 5x2 + 17x + 6 = 0.
a.
b.
2
3, 
5
5
 ,3
2
c.
2
 ,3
5
d.
5
3,
2
Copyright © 2009 Pearson Education, Inc.
Slide 3 - 14
Solve: 5x2 + 17x + 6 = 0.
a.
b.
2
3, 
5
5
 ,3
2
c.
2
 ,3
5
d.
5
3,
2
Copyright © 2009 Pearson Education, Inc.
Slide 3 - 15
Solve: 5t2 – 2t + 6 = 0.
a. 1  29i
b.
c.
d.
1  31
5
1  5i
5
1  29i
5
Copyright © 2009 Pearson Education, Inc.
Slide 3 - 16
Solve: 5t2 – 2t + 6 = 0.
a. 1  29i
b.
c.
d.
1  31
5
1  5i
5
1  29i
5
Copyright © 2009 Pearson Education, Inc.
Slide 3 - 17
Solve x2 + 6x = 2 by completing the
square.
a. 3  11
b. 3  2 11
c. 3  7
d. 3  2 7
Copyright © 2009 Pearson Education, Inc.
Slide 3 - 18
Solve x2 + 6x = 2 by completing the
square.
a. 3  11
b. 3  2 11
c. 3  7
d. 3  2 7
Copyright © 2009 Pearson Education, Inc.
Slide 3 - 19
Solve x2 + 8x = 12 by completing the
square.
a. 4  2 7
b.
2, 6
c.
4  8 7
2
d. 4  2 7
Copyright © 2009 Pearson Education, Inc.
Slide 3 - 20
Solve x2 + 8x = 12 by completing the
square.
a. 4  2 7
b.
2, 6
c.
4  8 7
2
d. 4  2 7
Copyright © 2009 Pearson Education, Inc.
Slide 3 - 21
Find the zeros of f(x) = 6x2 + 13x – 15.
a.
15
b.
5
 ,3
6
c.
d.
5
3,
6
6
3,
5
Copyright © 2009 Pearson Education, Inc.
Slide 3 - 22
Find the zeros of f(x) = 6x2 + 13x – 15.
a.
15
b.
5
 ,3
6
c.
d.
5
3,
6
6
3,
5
Copyright © 2009 Pearson Education, Inc.
Slide 3 - 23
The sum of the base and height of a
triangle is 56 in. Find the height for
which the area is maximum.
a. h = 14 in.
b. h = 28 in.
c. h = 42 in.
d. h = 112 in.
Copyright © 2009 Pearson Education, Inc.
Slide 3 - 24
The sum of the base and height of a
triangle is 56 in. Find the height for
which the area is maximum.
a. h = 14 in.
b. h = 28 in.
c. h = 42 in.
d. h = 112 in.
Copyright © 2009 Pearson Education, Inc.
Slide 3 - 25
For the graph of the function
f(x) = 3x2 – 12x + 16 find the vertex.
a. (2, 4)
b. (2, 52)
c.
(2, 4)
d.
(6, 4)
Copyright © 2009 Pearson Education, Inc.
Slide 3 - 26
For the graph of the function
f(x) = 3x2 – 12x + 16 find the vertex.
a. (2, 4)
b. (2, 52)
c.
(2, 4)
d.
(6, 4)
Copyright © 2009 Pearson Education, Inc.
Slide 3 - 27
For the graph of the function
f(x) = 4x2 – 8x – 1 find the axis of
symmetry.
a. x = 2
b. x = 1
1
c. x 
4
d. x = 5
Copyright © 2009 Pearson Education, Inc.
Slide 3 - 28
For the graph of the function
f(x) = 4x2 – 8x – 1 find the axis of
symmetry.
a. x = 2
b. x = 1
1
c. x 
4
d. x = 5
Copyright © 2009 Pearson Education, Inc.
Slide 3 - 29
For the graph of the function
f(x) = 6x2 – 6x – 3 find the axis of
symmetry.
a. x = 2
1
b. x 
2
1
c. x 
4
d. x = 2
Copyright © 2009 Pearson Education, Inc.
Slide 3 - 30
For the graph of the function
f(x) = 6x2 – 6x – 3 find the axis of
symmetry.
a. x = 2
1
b. x 
2
1
c. x 
4
d. x = 2
Copyright © 2009 Pearson Education, Inc.
Slide 3 - 31
The Cotes have 30 feet of picket fence with
which to enclose a flower garden. What
dimensions should the garden have in order
to maximize area?
a. 2 ft by 13 ft
b. 4 ft by 11 ft
c. 7.5 ft by 7.5 ft
d. 10 ft by 5 ft
Copyright © 2009 Pearson Education, Inc.
Slide 3 - 32
The Cotes have 30 feet of picket fence with
which to enclose a flower garden. What
dimensions should the garden have in order
to maximize area?
a. 2 ft by 13 ft
b. 4 ft by 11 ft
c. 7.5 ft by 7.5 ft
d. 10 ft by 5 ft
Copyright © 2009 Pearson Education, Inc.
Slide 3 - 33
3
11
Solve:

 1.
x  5 3x  1
a. 3
b. 3, 7
c. 1
d. 7, 3
Copyright © 2009 Pearson Education, Inc.
Slide 3 - 34
3
11
Solve:

 1.
x  5 3x  1
a. 3
b. 3, 7
c. 1
d. 7, 3
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Slide 3 - 35
Solve: 2x  8  6  10.
a. 2
b. 4
c. 14
d. 16
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Slide 3 - 36
Solve: 2x  8  6  10.
a. 2
b. 4
c. 14
d. 16
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Slide 3 - 37
Solve:
a.
6  2n  2  2.
25
b. 5, 69
c. 5
1
d. 
3
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Slide 3 - 38
Solve:
a.
6  2n  2  2.
25
b. 5, 69
c. 5
1
d. 
3
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Slide 3 - 39
Solve: 3x  2  11.
13
a. 
3
13
b.  , 3
3
3 1
c.  ,
13 3
d.
3
Copyright © 2009 Pearson Education, Inc.
Slide 3 - 40
Solve: 3x  2  11.
13
a. 
3
13
b.  , 3
3
3 1
c.  ,
13 3
d.
3
Copyright © 2009 Pearson Education, Inc.
Slide 3 - 41
Solve: 8m  3  7.
a.
b.
c.
d.
1 5
 ,
2 4
1

2
1 5
,
2 4
5
4
Copyright © 2009 Pearson Education, Inc.
Slide 3 - 42
Solve: 8m  3  7.
a.
b.
c.
d.
1 5
 ,
2 4
1

2
1 5
,
2 4
5
4
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Slide 3 - 43
Solve: x  6  5.
a.  , 11 1, 
b.
11, 1
c.  ,1  11, 
d.
1,11
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Slide 3 - 44
Solve: x  6  5.
a.  , 11 1, 
b.
11, 1
c.  ,1  11, 
d.
1,11
Copyright © 2009 Pearson Education, Inc.
Slide 3 - 45
Solve: x  4  12.
a.  , 8 16, 
b.
8,16
c.  , 16 8, 
d. 8,16
Copyright © 2009 Pearson Education, Inc.
Slide 3 - 46
Solve: x  4  12.
a.  , 8 16, 
b.
8,16
c.  , 16 8, 
d. [8, 16]
Copyright © 2009 Pearson Education, Inc.
Slide 3 - 47