This document summarizes a book titled "The Development of the Thai Language Teaching Materials for Grade 3-4 Students" by Dr. Somchai Srisa-an.
The book was published in 2001 to provide Thai language teaching materials for grades 3-4. It includes 4 chapters, with each chapter focusing on a different grade level (grade 3, chapter 1 and grade 4, chapter 4).
The summary highlights that the book aims to develop Thai language skills for students in grades 3-4 and provides teaching materials tailored to each grade level. It also seeks to appropriately introduce students to the Thai language in order to enhance their language abilities and prepare them for further study.
Han Geurdes' slides of his talk at the Växjö conference “Quantum Theory: Advances and Problems”, Monday 10 June, 2013.
Reference: doi:10.1016/j.rinp.2014.06.002
Results in Physics Volume 4, 2014, pages 81–82
“A probability loophole in the CHSH” by Han Geurdes.
I analyse this alleged disproof of Bell's theorem in my own paper http://arxiv.org/abs/1506.00223
Solutions manual for calculus an applied approach brief international metric ...Larson612
Solutions Manual for Calculus An Applied Approach Brief International Metric Edition 10th Edition by Larson IBSN 9781337290579
Full download: https://goo.gl/RtxZKH
Han Geurdes' slides of his talk at the Växjö conference “Quantum Theory: Advances and Problems”, Monday 10 June, 2013.
Reference: doi:10.1016/j.rinp.2014.06.002
Results in Physics Volume 4, 2014, pages 81–82
“A probability loophole in the CHSH” by Han Geurdes.
I analyse this alleged disproof of Bell's theorem in my own paper http://arxiv.org/abs/1506.00223
Solutions manual for calculus an applied approach brief international metric ...Larson612
Solutions Manual for Calculus An Applied Approach Brief International Metric Edition 10th Edition by Larson IBSN 9781337290579
Full download: https://goo.gl/RtxZKH
Operation “Blue Star” is the only event in the history of Independent India where the state went into war with its own people. Even after about 40 years it is not clear if it was culmination of states anger over people of the region, a political game of power or start of dictatorial chapter in the democratic setup.
The people of Punjab felt alienated from main stream due to denial of their just demands during a long democratic struggle since independence. As it happen all over the word, it led to militant struggle with great loss of lives of military, police and civilian personnel. Killing of Indira Gandhi and massacre of innocent Sikhs in Delhi and other India cities was also associated with this movement.
Unit 8 - Information and Communication Technology (Paper I).pdfThiyagu K
This slides describes the basic concepts of ICT, basics of Email, Emerging Technology and Digital Initiatives in Education. This presentations aligns with the UGC Paper I syllabus.
Palestine last event orientationfvgnh .pptxRaedMohamed3
An EFL lesson about the current events in Palestine. It is intended to be for intermediate students who wish to increase their listening skills through a short lesson in power point.
Read| The latest issue of The Challenger is here! We are thrilled to announce that our school paper has qualified for the NATIONAL SCHOOLS PRESS CONFERENCE (NSPC) 2024. Thank you for your unwavering support and trust. Dive into the stories that made us stand out!
Instructions for Submissions thorugh G- Classroom.pptxJheel Barad
This presentation provides a briefing on how to upload submissions and documents in Google Classroom. It was prepared as part of an orientation for new Sainik School in-service teacher trainees. As a training officer, my goal is to ensure that you are comfortable and proficient with this essential tool for managing assignments and fostering student engagement.
Synthetic Fiber Construction in lab .pptxPavel ( NSTU)
Synthetic fiber production is a fascinating and complex field that blends chemistry, engineering, and environmental science. By understanding these aspects, students can gain a comprehensive view of synthetic fiber production, its impact on society and the environment, and the potential for future innovations. Synthetic fibers play a crucial role in modern society, impacting various aspects of daily life, industry, and the environment. ynthetic fibers are integral to modern life, offering a range of benefits from cost-effectiveness and versatility to innovative applications and performance characteristics. While they pose environmental challenges, ongoing research and development aim to create more sustainable and eco-friendly alternatives. Understanding the importance of synthetic fibers helps in appreciating their role in the economy, industry, and daily life, while also emphasizing the need for sustainable practices and innovation.
The Roman Empire A Historical Colossus.pdfkaushalkr1407
The Roman Empire, a vast and enduring power, stands as one of history's most remarkable civilizations, leaving an indelible imprint on the world. It emerged from the Roman Republic, transitioning into an imperial powerhouse under the leadership of Augustus Caesar in 27 BCE. This transformation marked the beginning of an era defined by unprecedented territorial expansion, architectural marvels, and profound cultural influence.
The empire's roots lie in the city of Rome, founded, according to legend, by Romulus in 753 BCE. Over centuries, Rome evolved from a small settlement to a formidable republic, characterized by a complex political system with elected officials and checks on power. However, internal strife, class conflicts, and military ambitions paved the way for the end of the Republic. Julius Caesar’s dictatorship and subsequent assassination in 44 BCE created a power vacuum, leading to a civil war. Octavian, later Augustus, emerged victorious, heralding the Roman Empire’s birth.
Under Augustus, the empire experienced the Pax Romana, a 200-year period of relative peace and stability. Augustus reformed the military, established efficient administrative systems, and initiated grand construction projects. The empire's borders expanded, encompassing territories from Britain to Egypt and from Spain to the Euphrates. Roman legions, renowned for their discipline and engineering prowess, secured and maintained these vast territories, building roads, fortifications, and cities that facilitated control and integration.
The Roman Empire’s society was hierarchical, with a rigid class system. At the top were the patricians, wealthy elites who held significant political power. Below them were the plebeians, free citizens with limited political influence, and the vast numbers of slaves who formed the backbone of the economy. The family unit was central, governed by the paterfamilias, the male head who held absolute authority.
Culturally, the Romans were eclectic, absorbing and adapting elements from the civilizations they encountered, particularly the Greeks. Roman art, literature, and philosophy reflected this synthesis, creating a rich cultural tapestry. Latin, the Roman language, became the lingua franca of the Western world, influencing numerous modern languages.
Roman architecture and engineering achievements were monumental. They perfected the arch, vault, and dome, constructing enduring structures like the Colosseum, Pantheon, and aqueducts. These engineering marvels not only showcased Roman ingenuity but also served practical purposes, from public entertainment to water supply.
The Indian economy is classified into different sectors to simplify the analysis and understanding of economic activities. For Class 10, it's essential to grasp the sectors of the Indian economy, understand their characteristics, and recognize their importance. This guide will provide detailed notes on the Sectors of the Indian Economy Class 10, using specific long-tail keywords to enhance comprehension.
For more information, visit-www.vavaclasses.com
How to Split Bills in the Odoo 17 POS ModuleCeline George
Bills have a main role in point of sale procedure. It will help to track sales, handling payments and giving receipts to customers. Bill splitting also has an important role in POS. For example, If some friends come together for dinner and if they want to divide the bill then it is possible by POS bill splitting. This slide will show how to split bills in odoo 17 POS.
The Art Pastor's Guide to Sabbath | Steve ThomasonSteve Thomason
What is the purpose of the Sabbath Law in the Torah. It is interesting to compare how the context of the law shifts from Exodus to Deuteronomy. Who gets to rest, and why?
15. 5
r
y x
x y y = 2x
1
x = 2 y ( 0) x x 2 y
2
{x x 2} x R
y = 2x
1
y 0 ( )
0
{y y 0}
x y y
x
y = 2x
1
x – 2 = y
1
x = 2
y
1
5
16. 6
y 0 x
{y y 0}
y x (x, y)
(1) r = {(x, y) R R y = x} (x R, y R)
(2) r = {(x, y) R R y = –3} (x R, y = –3)
(3) r = {(x, y) R R y = 5x – 2} (x R, y = R)
(4) r = {(x, y) R R y = 7 – 2x} (x R, y = R)
(5) r = {(x, y) R R y = x2
} (x R, y 0)
(6) r = {(x, y) R R y = (x – 3)2
} (x R, y 0)
(7) r = {(x, y) R R y = x2
– 8} (x R, y –8)
(8) r = {(x, y) R R y = x } (x R, y 0)
(9) r = {(x, y) R R y = x + 5} (x R, y 5)
(10) r = {(x, y) R R y = x – 7} (x R, y –7)
(11) r = {(x, y) R R y = x
1
} (x 0, y 0)
(12) r = {(x, y) R R y = 5x
1
} (x 5, y 0)
(13) r = {(x, y) R R y = 5x } (x –5, y 0)
(14) r = {(x, y) R R y = x7 } (x 7, y 0)
(15) r = {(x, y) R R y = 2
x } (x R, y 0)
(16) r = {(x, y) R R y = 5x2
} (x R, y 5 )
(17) r = {(x, y) R R y = 2
x7 } ( 7 < x < 7 , 0 < y < 7 )
(18) r = {(x, y) R R y = 2x5 } (x –2, y 5)
6
17. 7
y = x x
1) x
2) x
3) x
x
{x x 0}
y x (x, y) r
r = {(x, y) y = x2
, –2 x 2}
r {y 0 y 4}
4 ( .4 – .6)
1. (linear function)
1) X Y
(1) (x, y) 1
x y (x, y) = (1, 1) 1
(x, y) 1 x > 0 y > 0
7
18. 8
(2) 2, 3 4
(x, y)
1
2
3
4
x > 0, y > 0
x < 0, y > 0
x < 0, y < 0
x > 0, y < 0
(3) X Y
X y (x, 0)
Y x (0, y)
2) y = ax + c
(1) y = ax + c y = x y = ax
a 0 a 1 y = ax
y = x
Y
X
(1, 1)
Y
X
y = ax, a > 0
0
Y
X
y = ax, a < 0
0
2
2–2
–2
8
19. 9
(2) y = x y = x + c
1
y = x y = x + c c > 0
y = x
y = x + 1
y = x + 2
y = x + 3
2
y = x y = x + c c < 0
y = x
y = x – 1
y = x – 2
y = x – 3
Y X
y = x + c c 0 y = x
3) y = ax + c a, c 0
y = ax + c y = x
y = 3x + 1 y
(1) y = x
0
Y
X
y = x
1
2
3
Y
X0
–3
–2
–1
Y
X0
9
20. 10
(2) y = 3x
(3) y = 3x + 1
y = 3x + 1 y = 3x X (
3
1
, 0) Y
(0, 1)
4) y = ax + c
Y X y = 3x + 1
y = 3x + 1 y = 3x, y = 3x + 2, y = 3x – 1
0
Y
X
y = 3x
y = 3x + 1
1
–1
y = x
0
Y
X
y = 3x
10
21. 11
2. (Quadratic function)
y = x2
y = x2
1. y = ax2
1) a > 1 y = 2x2
, y = 3x2
, y = 5x2
y = ax2
a > 1
y = x2
2) 0 < a < 1 y = 2
x
2
1
, y = 2
x
3
1
, y = 2
x
5
1
y = ax2
0 < a < 1
y = x2
Y
X0
Y
X0
Y
X0
11
22. 12
3) y = ax2
, a < 0 y = –x2
y = ax2
a > 0 a < 0 2
2. y = x2
+ c, c > 0
y = x2
+ c, c > 0 y = x2
y = x2
y = x2
+ 1
y = x2
+ 2
y = x2
+ 3
3. y = x2
– c, c > 0
y = x2
– c, c > 0 y = x2
y = x2
y = x2
– 1
y = x2
– 2
y = x2
– 3
4. y = (x + c)2
, c > 0
y = (x + c)2
, c > 0 y = x2
y = x2
y = (x + 1)2
y = (x + 2)2
y = (x + 3)2
Y
X0
–3 –2 –1
Y
X
0
3
2
1
X
Y
0
–3
–2
–1
Y
X0
12
23. 13
5. y = (x – c)2
, c > 0
y = (x – c)2
, c > 0 y = x2
y = x2
y = (x – 1)2
y = (x – 2)2
y = (x – 3)2
y = ax2
, y = (x – a)2
y = x2
+ c
y = a(x – h)2
+ k
y = 2(x – 1)2
+ 2 y = x2
1) y = x2
2) y = (x – 1)2
y = x2
1 2 3
Y
X0
Y
X1
y = (x – 1)2
Y
X
y = x2
0
0
13
24. 14
3) y = 2(x – 1)2
y = (x – 1)2
4) y = 2(x – 1)2
+ 1 y = 2(x – 1)2
y = a(x – h)2
+ k
1) a > 0, h > 0, k > 0 y = 3)1x(
2
1 2
a > 0, h > 0, k < 0 y = 3(x – 3)2
– 3
a > 0, h < 0, k > 0 y = 2(x + 1)2
+ 1
a > 0, h < 0, k < 0 y = (x + 2)2
– 2
2) a < 0, h > 0, k > 0 y = – 3)1x(
2
1 2
a < 0, h > 0, k < 0 y = –3(x – 3)2
– 3
a < 0, h < 0, k > 0 y = –2(x + 1)2
+ 1
a < 0, h < 0, k < 0 y = –(x + 2)2
– 2
Y
X
Y
X
1
1
y = 2(x – 1)2
y = 2(x – 1)2
+ 1
10
0
14
25. 15
ax2
+ bx + c = 0 a 0
a
ax2
+ bx + c = 0 a(x + h)2
+ k = 0 x = a2
b
y
1. y = ax2
ax2
= 0
a > 0 a < 0
y = 0 x = 0
ax2
= 0 x = 0
2. y = a(x – c)2
y = a(x + c)2
c > 0
a(x – c)2
= 0 a(x + c)2
= 0
y = a(x – c)2
, c > 0 y = a(x + c)2
, c > 0
c1 c2 c3
Y
X0 –c3 –c2 –c1
Y
X
0
Y
X0
Y
X
0
a < 0
Y
X
0
15
26. 16
a(x – c)2
= 0 x = c
a(x + c)2
= 0 x = –c
3. ax2
+ bx + c = 0, a 0
x = a2
ac4bb 2
y = ax2
+ bx + c a(x – h)2
+ k
(h, k) a
a > 0 (–h, –k) (0, –k) (h, –k)
(–h, k) (0, k) (h, k) a < 0
a
X
(–h, –k) (h, –k)
Y
a > 0
0
(h, k)
Y
X
(–h, k)
(h, –k)(–h, –k)
a > 0 a > 0
a < 0 a < 0
0
Y
X
(–h, k) (h, k)
a < 0
0
16
27. 17
x + a < 0 x + a > 0
1 x + 3 8
1 x + 3 8
x – 5 0
y = x – 5 y
y x – 5 0 x 5
2 y1 = x + 3
y2 = 8
y1 y2
y1 = y2 y1 y2 x = 5
y1 x + 3 y2 8 x 5
Y
X
–5
50
y = x – 5
y2 = 8
Y
X
3
5–3
y1 = x + 3
0
17
28. 18
2 x2
– 2x – 3 < 0
1) x2
x2
x2
– 2x – 3
x2
-2x-3 (x-c)2
X
2) X
X
x2
– 2x – 3 = 0
(x – 3)(x + 1) = 0 x = –1, 3
1) X (–1, 0) (3, 0)
x2
– 2x – 3 < 0 –1 < x < 3
x, y (x, y)
–1
Y
X30
18
29. 19
1.
1) {(0, 1), (1, –2), (2, 0), (3, 2)}
2) {(0, –1), (1, –2), (1, 1), (2, 2), (3, 0)}
2.
1) 3y = 2x + 4
2) y = 4 – x2
3.
1) f(1) f(0) f(x) = –2x – 7
2) f(0) f(4) f(x) = 3 – x
3) f(–2) f(0) f(x) = x + 4
4) f(–1) f(2) f(x) =
4.
1) 2)
3) 4)
2x , x 0
2x + 1 , x < 0
Y
X
y3
0
Y
X0
y4
y1
Y
X
1
0–1
Y
X0
y2
–2–1
19
30. 20
5.
1) f = {(–3, 0), (–1, 4), (0, 2), (2, 2), (4, –1)}
2) y = 5x
1
6. A = {–1, 0, 1}
y
1) y = x
2) y = 1x
7.
1) y = –x2
– 2 2) y = x2
+ 2x + 3 3) y = 2 x
8.
1) y = (x – 4)2
– 3 2) y = –(x – 4)2
+ 3
3) y = (x + 4)2
– 3 4) y = –(x + 4)2
+ 3
( ) ( )
( ) ( )
Y
X
Y
X
Y
X
Y
X
–3
3
–3
3
– 4
4
– 4
4
0 0
0 0
20
34. 24
4. 1)
Y y1 1
y1
y1
y1 {y y –1}
2) Y
y2 1
x y
y2
y2
y2
3)
Y y3 1
y3
y3 {x x R, x –1}
y3 {y y = 2}
4)
Y y4 1 (0, 0)
(0, –2)
y4 x
0 y
y4 {x x 0}
y4 {y y 0 y = –2}
Y
X0
y4
Y
X0
y2
Y
X
y3
0
y1
Y
X
1
0–1
2
–1
–2
24
35. 25
5. 1) f = {(–3, 0), (–1, 4), (0, 2), (2, 2), (4, –1)}
{–3, –1, 0, 2, 4}
{–1, 0, 2, 4}
2) f(x) = 5x
1
{x x R x –5}
{y y R y 0}
6. 1) y = x
r = {(x, y) y = x x A}
r = {(–1, 1), (0, 0), (1, 1)}
2) y = 1x
r = {(x, y) y = 1x x A}
r = {(–1, 0), (0, 1), (1, 2 )}
7. 1) y = –x2
– 2
{y y R y –2}
2) y = x2
+ 2x + 3
= (x2
+ 2x + 1) + 2
= (x + 1)2
+ 2
{y y R y 2}
Y
X0
–2
Y
X
0
2
–1
25
36. 26
3) y = 2 x
{y y R y 0}
8. 1) ( ) 2) ( ) 3) ( ) (4) ( )
9. ABCD 120
f(x) ABCD
f(x) = x(60 – x)
f(x) = 60x – x2
y = 60x – x2
x =
a2
b
x =
2
60
= 30 y = 30(60-30) = 900
x = 30 y = 900
120 900
A B
D C
x x
60 – x
60 – x
Y
X0
Y
X
0
900
30 60
(30, 900)
26
37. 27
10. x y
x y 2
x – y = 2 y = x – 2
1 xy = –3
x(x – 2) = –3
x2
– 2x + 3 = 0
(x2
– 2x + 1) + 2 = 0
(x – 1)2
+ 2 = 0
y = (x – 1)2
+ 2
y X
2
–3
2 x y –3
xy = –3 y = x
3
y1 = x – 2 y2 = x
3
Y
X0
2
1
x y1 = x – 2 y2 = x
3
–3 –5 1
–2 – 4 2
3
–1 –3 3
0 –2
1 –1 –3
2 0
2
3
3 1 –1
Y
X0
3
2
–2
–3
y1 y2
2 –3
27
38. 28
11. y
x
y = 500 + 125x
12. x
10 , 1 x 2
f(x) = 20 , 3 x 4
30 , 5 x 7
13. 1) y = –x2
– 2 y
(0, –2)
2) y = x2
– 4x y = (x2
– 4x + 4) – 4 = (x – 2)2
– 4
y
(2, – 4)
Y
X0
(0, –2)
Y
X0
– 4
2
(2, – 4)
28
39. 29
14. 1) x + 1 = 0
y = x + 1
y
x y = 0
x + 1 = 0
2) –x2
+ 4 = 0
y = –x2
+ 4 y
y x
x
–x2
+ 4 = 0
x2
= 4
x = –2, 2
–x2
+ 4 = 0 x = –2, 2
15. 1) 2x + 1 < 3
2x – 2 < 0 2(x – 1) < 0
y1 = 2(x – 1)
y1
Y
X0
1
y = x + 1
Y
X0
4
y1
2–2
y2
29
40. 30
y1 < 0 x < 1
2x + 1 < 3 x < 1
2) x2
+ 4x – 5 < 0
y = x2
+ 4x – 5
= (x2
+ 4x + 4) – 5 – 4
= (x + 2)2
– 9
y = (x + 2)2
– 9
X
x2
+ 4x – 5 = 0
(x + 5) (x – 1) = 0
x = –5, 1
x2
+ 4x – 5 < 0 –5 < x < 1
3) x – 2 > 0
y = x – 2
y
y > 0 x 2
1
Y
X0
Y
X0–5 1
Y
X
0
y1
2
y1 = 2x – 2
–2
–9
30
41. 31
1.1
1. 1) x y
x y
1
2
3
4
5
6
7
6.00 .
6.03 .
6.01 .
6.05 .
6.06 .
6.02 .
0.01 .
2) 155
. . . . 155 .
155 . ( .)
( .)
3) 2
1
A h
A = h10
2
1
A = 5h
.
.
.
.
40
42
38
31
42. 32
2. 1) 35
1.5
x ( )
y ( )
y = 35 + 1.5x
2) a
ABCD a
ABCD x
x2
= a2
+ a2
x2
= 2a2
x = a2
x = a2
3) 200
3
x ( )
y ( )
y = 200 + 3x
A Ba
a
CD
x
32
43. 33
3. A B
1)
2)
3)
b
1
4)
4. 1)
2) a b 1
3)
4) a 1
a
b
c
1
2
3
A B
a
b
c
1
2
3
A B
a
b
c
d
A B
1
2
3
4
a
b
c
d
A B
1
2
3
4
33
44. 34
5. 1) {(2, 10), (3, 15), (4, 20)}
2) {(–7, 3), (–2, 1), (–2, 4), (0, 7)}
1
3) {(–2, 1), (0, 1), (2, 1), (4, 1), (–3, 1)}
4) {(5, 0), (3, –1), (0, 0), (5, –1), (3, –2)}
1
2
3
4
10
15
20
A B
–7
–2
0
1
3
4
7
A B
–3
–2
0
2
4
1
A
B
0
3
5
–2
–1
0
A B
34
45. 35
6. A = {a, b, c} B = {1, 2}
A B
(1) (2)
(3) (4)
7. A B 3 A
B
A = {a, b, c} B = {1, 2, 3}
(1) (2)
(3) (4)
(5) (6)
a
b
c
1
2
A B
a
b
c
1
2
3
A B
a
b
c
1
2
3
A B
a
b
c
1
2
3
A B
a
b
c
1
2
3
A B
a
b
c
1
2
3
A B
a
b
c
1
2
3
A B
a
b
c
1
2
A B
a
b
c
1
2
A B
a
b
c
1
2
A B
35
46. 36
8. Df = {–2, –1, 0, 1, 2} f
1) f(x) = x2
f = {(–2, 4), (–1, 1), (0, 0), (1, 1), (2, 4)}
2) f(x) =
1x
x2
2
f = {(–2, 5
4
), (–1, –1), (0, 0), (1, 1) (2,
5
4
)}
3) f(x) = 2x
f = {(–2, 0), (–1, 1), (0, 2 ), (1, 3 ), (2, 2)}
4) f(x) = x + 1
f = {(–2, 1), (–1, 0), (0, 1), (1, 2), (2, 3)}
9. 1)
f(a) = 2
f(b) = 4
f(c) = 3
f(d) = 1
2)
f(a) = 1
f(b) = 4
f(c) = 1
f(d) = 3
10. 1) Y 1
2) Y 1
3) Y 1
4) Y 1
a
b
c
d
1
2
3
4
a
b
c
d
1
2
3
4
36
47. 37
11. 1)
{y y –2}
2)
{y y 0}
3) {x x 1}
{y y 0}
4)
{y y 3}
12. 1) f(–1) = 3
f(0) = 0
f(1) = – 3
2) g(–2) = 0
g(0) = – 4
13. 1) g(x) = x2
– 2x
(1) g(2) = 0 (2) g(–3) = 15
2) f(s) =
1s
1
(1) f(4) =
5
1
(2) f(0) = 1
14. 1) {x x R} {y y R}
2) {x x R} {y y R}
3) {x x R} {y y =
2
1
}
4) {x x 2} {y y 0}
5) {x x –2} {y y 0}
6) {x x R} {y y 0}
7) {x x R} {y y –1}
8) {x x R} {y y 0}
9) {x x R} {y y 0}
10) {x x R} {y y 0}
37
48. 38
1.2
1. 1) y1 = 5x + 3 y2 = 5x – 3
2) y1 = –x + 3 y2 = –x – 3
3) y1 = 5 – x y2 = 5 + x
4) y1 = x + 2 y2 = –x – 2
Y
X
y2
3
y1 –3
Y
X
3
–3
y2
y1
Y
X
y1 y2
5
Y
X
y2 y1
2
–2
0
–3 3
–5 0 5
–2 0
38
49. 39
2. 1) ( ) 2) ( ) 3) ( )
3. 1) (3, 5) y
y = 1x
7
2
y x = 3
y = 1)3(
7
2
y =
7
6
1
x = 3 y =
7
6
1
(3,
7
6
1 ) y (3, 5)
2) (– 4, –5) y
y = –7 – 2x
y x = – 4
y = –7 – 2(– 4)
y = 1
x – 4 y = 1
(– 4, 1) (– 4, –5)
(– 4, 1)
0
(– 4, –5)
Y
X– 4
–5
1
(3, 5)
0
Y
X
2
3
(3,
7
6
1 )
39
50. 40
3) y = –1
(x, –1) x
(4, –5) y = –1
4. 1) x
f(x)
150 5
f(x) = 150 + 5x
x f(x)
0
1
2
3
4
150
155
160
165
170
0 1 2 3 4 5
170
x
f(x)
165
160
155
150
(4, –5)
Y
Xy = –1
–5
40
40
51. 41
2) x
f(x)
6,000 5%
f(x) = 6,000 + 0.05x
x f(x)
0
1,000
2,000
3,000
4,000
6,000
6,050
6,100
6,150
6,200
3) 2.54
x
y
y = 2.54x
x y
0
1
2
3
4
0
2.54
5.08
7.62
10.16
x
1,000 2,000 3,000 4,000
f(x)
6,000
6,100
6,200
Y
0 1 2 3 4 5
X
2
4
6
8
10
41
52. 42
5. 1)
( )
34,000
200,000 28,000 150,000
A
b
c
x ( )
A = c + bx
34,000 200,000
34,000 = c + 200,000 x --------------- (1)
28,000 150,000
28,000 = c + 150,000 x --------------- (2)
(1) – (2) 6,000 = 50,000 x
x = 100
12
100
12
12
2) x = 100
12
(1)
34,000 = c + 200,000 100
12
34,000 = c + 24,000
c = 10,000
10,000
3)
s
f(s) s
f(s) = 10,000 + 100
12
(s)
42
53. 43
6. 1) 12,000 /
10%
x
f(x) x
f(x) = 12,000 + 12,000(100
10
)x
= 12,000 + 1,200x
= 12,000(1 +
10
x
)
2) x = 5
f(5) = 12,000(1 +
10
5
)
= 12,000
2
3
= 18,000
1.3.1
1.
1) y = 2x2
2) y = –2x2
Y
X0
Y
X
0
43
54. 44
3) y = 2x2
+ 1
4) y = 2x2
– 1
5) y = –2x2
+ 1
6) y = –2x2
– 1
Y
X0
Y
X
0
1
–1
Y
X0
1
–1
Y
X0
44
55. 45
7) y = (x – 1)2
8) y = (x + 1)2
9) y = (x – 1)2
– 1
10) y = (x + 1)2
+ 1
1
–1
–1
1
Y
X0
Y
X0
Y
X
Y
X0–1
1
0
45
56. 46
2. 1) y1 = x2
y2 = 2x2
y3 = 5x2
y4 = 11x2
2) y1 = x2
y2 = 2
x
2
1
y3 = 2
x
5
1
3) y1 = 2x2
y2 = –2x2
4) y1 = 0.5x2
y2 = –0.5x2
5) y1 = (x – 3)2
y2 = (x – 4)2
y3 = (x – 5)2
Y
X
Y
X
y1
y2
y3
Y
X
y1
y2
Y
X
y1
y2
Y
X
y1 y2y3
3 4 50
y1
y2
y3
y4
0
0
0
0
46
59. 49
X
Y
0
16) y = (x + 4)2
– 7
17) y = 3(x – 3)2
+ 3
18) y = –2(x + 2)2
+ 1
3. 1) y = (x – 4)2
– 3
( )
–3
4
Y
X
3
3
Y
X–2
1
Y
X–4
–7
0
0
0
49
60. 50
X
Y
0
X
Y
0
2) y = –(x – 4)2
+ 3
( )
3) y = (x + 4)2
– 3
( )
4) y = –(x + 4)2
+ 3
( )
3
4
–3
– 4 X
Y
0
3
– 4
50
61. 51
5) y = 2(x – 2)2
( )
6) y = (x + 3)2
– 4
( )
7) y = 3)1x(
2
1 2
( )
Y
X0
Y
X–3 3
-4
0
–3
–1 X
Y
0
2
51
62. 52
8) y = –2(x + 3)2
+ 2
( )
9) y = x2
– 2x + 3
= (x2
– 2x + 1) + 2
= (x – 1)2
+ 2
( )
10) y = 2x2
– 4x + 5
= 2(x2
– 2x) + 5
= 2(x2
– 2x + 1) +5 – 2
= 2(x – 1)2
+ 3
( )
2
–3 X
Y
0
2
0 1
Y
X
Y
X
3
-1 10
52
63. 53
4. 1) y = x2
– 2x – 3
y = (x2
– 2x + 1) – 3 – 1
y = (x – 1)2
– 4
h = 1 k = – 4
(1, – 4)
x2
y
y
2) y = x2
– 4x + 8
y = (x2
– 4x + 4) + 8 – 4
y = (x – 2)2
+ 4
h = 2 k = 4
(2, 4)
x2
y
y
Y
X0
– 4
(1, – 4)
1
Y
X0
(2, 4)
4
2
53
64. 54
3) y = 2x2
+ 4x + 8
y = 2(x2
+ 2x + 4)
= 2[(x2
+ 2x + 1) + 3]
= 2[(x + 1)2
+ 3]
= 2(x + 1)2
+ 6
h = –1 k = 6
(–1, 6)
x2
y
y
4) y = 3x2
+ 12x + 3
y = 3x2
+ 12x + 3
= 3(x2
+ 4x + 1)
= 3[(x2
+ 4x + 4) + 1 – 4]
= 3[(x + 2)2
– 3]
= 3(x + 2)2
– 9
h = –2 y = –9
(–2, –9)
x2
y
y
Y
X
(–1, 6)
–1 0
6
Y
X
(–2, –9)
0
–2
–9
54
65. 55
5) y = –x2
+ 2x + 1
y = –(x2
– 2x – 1)
= –[(x2
– 2x + 1) – 1 – 1]
= –[(x – 1)2
– 2]
= –(x – 1)2
+ 2
h = 1 k = 2
(1, 2) x2
y y
5. 1) y = –3x2
+ 6x + 3
y = ax2
+ bx + c , a 0 x =
a2
b
y = –3x2
+ 6x + 3 a = –3 b = 6
x =
)3(2
)6(
= 1
y = –3(1)2
+ 6(1) + 3
= –3 + 6 + 3
= 6
(1, 6)
x2
y
Y
X0
(1, 2)
2
1
Y
X0 1
(1, 6)
6
55
66. 56
2) y = 2x2
– 4x
y = ax2
+ bx + c , a 0 x
a2
b
y = 2x2
– 4x a = 2 b = – 4
x =
)2(2
)4(
= 1
y = 2(1)2
– 4(1)
= –2
(1, –2)
x2
y
3) y = 2x2
+ 4x + 2
y = ax2
+ bx + c, a 0 x
a2
b
y = 2x2
+ 4x + 2 a = 2 b = 4
x =
)2(2
)4(
= –1
y = 2(–1)2
+ 4(–1) + 2
= 2 – 4 + 2
= 0
(–1, 0)
x2
y
Y
X0
(1, –2)
–2
1
56
67. 57
4) y = 2x2
– 2x – 24
y = ax2
+ bx + c, a 0 x =
a2
b
y = 2x2
– 2x – 24 a = 2 b = –2
x =
)2(2
)2(
=
4
2
=
2
1
y = 24)
2
1
(2)
2
1
(2 2
= 241
2
1
= –24 2
1
)
2
1
24,
2
1
(
x2
y
Y
X0
)
2
1
24,
2
1
(
1
Y
X
(–1, 0) 0
57
68. 58
1.3.2 (1)
1. 1) x2
= 16
x2
– 16 = 0
y = x2
– 16 y
y X y = 0 , x = – 4, 4
x2
– 16 = 0 x2
= 16 x = – 4, 4
2) 3x2
= 27
3x2
– 27 = 0
y = 3x2
– 27 y
y X y = 0 , x = –3, 3
3x2
– 27 = 0 3x2
= 27 x = –3, 3
3) 2x2
= 8
2x2
– 8 = 0
y = 2x2
– 8 y
X
Y
0– 4 4
(0, –16)
X
Y
0–3 3
(0, –27)
58
69. 59
y = 0 x = –2, 2
2x2
– 8 = 0 2x2
= 8 x = –2, 2
4) x2
= 0
y = x2
y
y X (0, 0) y = 0 , x = 0
x2
= 0 x = 0
5) x2
= –8
x2
+ 8 = 0
y = x2
+ 8 y
X
Y
0–2 2
(0, –8)
X
Y
0 (0, 0)
X
Y
0
(0, 8)
–8
8
59
70. 60
y 8
x y = 0
y x x2
+ 8 = 0
x2
+ 8 = 0 x2
= –8
2. 1) x2
+ 8x + 16 = 0
(x + 4)2
= 0
y = (x + 4)2
y
y X (– 4, 0)
x2
+ 8x + 16 = 0 x = – 4
2) 8x2
= 16x – 3
8x2
– 16x + 3 = 0
8(x2
– 2x) + 3 = 0
8(x2
– 2x + 1) + 3 – 8 = 0
8(x – 1)2
– 5 = 0
y = 8(x – 1)2
– 5 y
y X
8x2
+ 8x + 16
X
Y
(– 4, 0)
X
Y
1
–5
0
(1, –5)
60
71. 61
3) 6x2
= 4x + 3
6x2
– 4x – 3 = 0
y = ax2
+ bx + c, a 0 x
a2
b
y = 6x2
– 4x – 3 a = 6 b = – 4
x =
)6(2
)4(
=
3
1
y = 3)
3
1
(4)
3
1
(6 2
= 3
3
4
3
2
=
3
11
x2
y )
3
11
,
3
1
(
y X
6x2
= 4x + 3
4) 2x2
– 4x + 1 = 0
2(x2
– 2x) + 1 = 0
2(x2
– 2x + 1) + 1 – 2 = 0
2(x – 1)2
– 1 = 0
y = 2(x – 1)2
– 1 x2
y (1, –1)
X
Y
–5
)
3
11
,
3
1
(
0
61
72. 62
X
2x2
– 4x + 1 = 0
5) –8x2
– 24 = 0
y = –8x2
– 24
y
y = –8x2
– 24 X
–8x2
– 24 = 0
3. 1) –(x + 1)2
+ 1 = 0
y = –(x + 1)2
+ 1 y
y X
–(x + 1)2
+ 1 = 0
X
Y
–1
1
X
Y
–24
X
Y
–1
1
(1, –1)
(–1, 1)
0
0
0
62
73. 63
2) 7(x + 2)2
= 0
y = 7(x + 2)2
y
y X
7(x + 2)2
= 0
3) (x – 4)2
= – 4 (x – 4)2
+ 4 = 0
y = (x – 4)2
+ 4 y
y X
(x – 4)2
= – 4
4) (x + 7)2
= 3 (x + 7)2
– 3 = 0
y = (x + 7)2
– 3 y
y X
(x + 7)2
= 3
X
Y
(–2, 0)
X
Y
4
4
X
Y
–7
–3(–7, –3)
0
0
0
63
74. 64
4. 1) (1) {x x R}
{y y 0}
(2) (– 4, 0)
(3) y = 0
2) (1) x x R
{y y – 4}
(2) (–3 , – 4)
(3) y = – 4
3) (1) {x x R}
{y y 2}
(2) (–3, 2)
(3) y = 2
4) (1) x x R
{y y –3}
(2) (–1, –3)
(3) y = –3
5) (1) x x R
{y y –1}
(2) (2, –1)
(3) y = –1
5. 1) y = x2
– 8x + 15 y a(x – h)2
+ k
(x2
– 8x + 15) = (x2
– 8x + 16) + 15 – 16
= (x – 4)2
– 1
a = 1, h = 4 k = –1
a > 0 y (4, –1)
y = (x – 4)2
– 1
64
75. 65
1) Df = {x x R}
Rf = {y y –1}
2) (4, –1)
3) y –1
4) X X
x2
– 8x + 15 = 0
(x – 3)(x – 5) = 0
x = 3, 5
X (3, 0) (5, 0)
2) y = x2
– 2x – 4 y a(x – h)2
+ k
x2
– 2x – 4 = (x2
– 2x + 1) – 4 – 1
= (x – 1)2
– 5
a = 1 , h = 1 k = –5
a > 0 f (1, –5)
y = (x – 1)2
– 5
1) Df = {x x R}
Rf = {y y –5}
2) (1, –5)
Y
X
0–2 1
–5
(1, –5)
Y
X0
(4, –1)
65
76. 66
3) y –5
4) X X x = a2
ac4bb 2
x2
– 2x – 4 = 0 a = 1 b = –2 c = –4
x =
)1(2
)4)(1(4)2()2( 2
= 2
1642
= 51
X (1 – 5 , 0) (1 + 5 , 0)
3) y = x2
+ 8x + 13 y a(x – h)2
+ k
x2
+ 8x + 13 = (x2
+ 8x + 16) + 13 – 16
= (x + 4)2
– 3
a = 0, h = – 4 k = –3
a > 0 y (– 4, –3)
y = (x + 4)2
– 3
1) Df = {x x R}
Rf = {y y R, y –3}
2) (– 4, –3)
3) y –3
4) X X x = a2
ac4bb 2
x2
+ 8x + 13 = 0 a = 1, b = 8, c = 13
x = )1(2
)13)(1(488 2
= 2
52648
Y
X0
(– 4, – 3)
– 4
–3
66
77. 67
= 2
128
= 2
328
= 34
X (– 4 – 3 , 0) (– 4 + 3 , 0)
4) y = 2x2
+ 4x + 4 y a(x – h)2
+ k
2x2
+ 4x + 4 = 2(x2
+ 2x + 2)
= 2[(x2
+ 2x + 1) + 2 – 1]
= 2[(x + 1)2
+ 1]
= 2(x + 1)2
+ 2
a = 2, h = –1 k = 2
a > 0 y (–1, 2)
y = 2(x + 1)2
+ 2
1) Df = {x x R}
Rf = {y y R, y 2}
2) (–1, 2)
3) y 2
4) X
5) y = 3x2
– 12x + 6 y a(x – h)2
+ k
3x2
– 12x + 6 = 3(x2
– 4x + 2)
= 3[(x2
– 4x + 4) + 2 – 4]
= 3[(x – 2)2
– 2]
= 3(x – 2)2
– 6
Y
X0
(–1, 2)
2
–1
67
78. 68
a = 3, h = 2 k = –6
a > 0 y (2, –6)
y = 3(x – 2)2
– 6
1) Df = {x x R}
Rf = {y y R, y –6}
2) (2, –6)
3) y y –6
4) X X x = a2
ac4bb 2
3x2
– 12x + 6 = 0 a = 3, b = –12, c = 6
x = )3(2
)6)(3(4)12()12( 2
= 6
7214412
= 6
72
2
= 6
26
2
= 22
X (2 + 2 , 0) (2 – 2 , 0)
6) y = x(x – 1) – 1 y a(x – h)2
+ k
x(x – 1) – 1 = x2
– x – 1
= (x2
– x + 4
1
) – 1 – 4
1
= 4
5
)
2
1
x( 2
a = 1, h = 2
1
k = 4
5
Y
X0
(2, –6)
2
–6
68
79. 69
a > 0 y )
4
5
,
2
1
(
y = x(x – 1) – 1
1) Df = {x x R}
Rf = {y y R, y
4
5
}
2) (
4
5
,
2
1
)
3) y
4
5
4) X X x = a2
ac4bb 2
x2
– x – 1 = 0 a = 1, b = –1, c = –1
x =
)1(2
)1)(1(4)1()1( 2
=
2
411
=
2
51
X (
2
51
, 0) (
2
51
, 0)
7) y = x2
– 4x – 7 y a(x – h)2
+ k
x2
– 4x – 7 = (x2
– 4x + 4) – 7 – 4
= (x – 2)2
– 11 = 0
a = 1, h = 2 k = –11
a > 0 y (2, –11)
Y
X0
(
4
5
,
2
1
)
–2 2
–2
69
80. 70
y = (x – 2)2
– 11
1) Df = {x x R}
Rf = {y y R, y –11}
2) (2, –11)
3) y –11
4) X X x = a2
ac4bb 2
x2
– 4x – 7 = 0 a = 1, b = – 4, c = –7
x =
)(
))((4)()( 2
12
7144
=
2
28164
=
2
1124
= 112
X (2 + 11 , 0) (2 – 11 , 0)
8) y = x2
– 2x + 5 = 0 y a(x – h)2
+ k
x2
– 2x + 5 = (x2
– 2x + 1) + 5 – 1
= (x – 1)2
+ 4
a = 1, h = 1 k = 4
a > 0 y (1, 4)
y = (x – 1)2
+ 4
Y
X0
(2, –11)
2
–11
70
81. 71
1) Df = {x x R}
Rf = {y y R, y 4}
2) (1, 4)
3) y 4
4) X
6.
x1 x2 X
x1, x2 y = 0
x2
– 2x – 8 = 0
(x + 2)(x – 4) = 0
x = –2, 4
x1 = –2 x2 = 4
y = x2
– 2x – 8 y a(x – h)2
+ k
= (x2
– 2x + 1) – 8 – 1
= (x – 1)2
– 9
a = 1, h = 1, k = –9
(1, –9)
y1 –9
Y
X0
(1, 4)
4
1
x1 x2
Y
X0
y1
71
82. 72
7. 1) y = (x – 3)(x – 6)
(x – 3)(x – 6) = 0
x = 3, 6
2) y = (x – 6)(x + 4)
(x – 6)(x + 4) = 0
x = – 4, 6
3) y = x(5 – x)
x(5 – x) = 0
x = 0, 5
4) y = x2
+ 2
Y
X
3
6
Y
X–4 6
Y
X50
Y
X
0
0
0
2
72
83. 73
5) y = x2
+ 4x + 12
= (x2
+ 4x + 4) + 8
= (x + 2)2
+ 8
6) y = 2x2
– 12x + 6
2x2
– 12x + 6 = 0
x2
– 6x + 3 = 0
x =
)(
))(()()( 2
12
31466
=
2
246 = 63
x = 63 63
7) y = –x2
– 2x – 1
y = –(x2
+ 2x + 1)
= –(x + 1)2
y = 0
x = –1
8) y = 15 + 2x – x2
15 + 2x – x2
= 0
(x – 5)(x + 3) = 0
x = –3, 5
Y
X
8
4
0–2
Y
X0
–6
–12
Y
X
(–1, 0)
Y
X
(1, 16)
–3 0 1 5
3
73
84. 74
2
1
1.3.2 (2)
1. 1) x2
1
y1 = x2
y2 = 1
y1 y2
x2
1 x 1 x –1
2) 4x2
1
y1 = 4x2
y2 = 1
y1 y2
4x2
< 1 2
1
< x <
2
1
Y
X
y1
y2 = 1
–1 0 1
Y
X
y1
2
1
2
1
y2 = 1
1
–1 1
74
85. 75
3) 5 – x2
1
5 – x2
– 1 > 0
4 – x2
> 0
y = 4 – x2
y
y > 0 –2 < x < 2
y1 = 5 – x2
y2 = 1 y1 y2
y1 > y2 –2 < x < 2
4) –(x – 1)(x + 5) 0
y = –(x – 1)(x + 5)
x2
y
X (–5, 0) (1, 0)
y 0 x –5 x 1
Y
X–2 0 2
4
Y
X–5 0 1
Y
X–2 0 2
1
5
y2
75
86. 76
2. 1) x2
– x – 2 0
(x – 2)(x + 1) 0
x2
y X (–1, 0) (2, 0)
y
x2
– x – 2 0 x –1 x 2
2) x2
– 3x – 1 < 3
x2
– 3x – 4 < 0
(x – 4)(x + 1) < 0
y = (x – 4)(x + 1)
x2
y
X (4, 0) (–1, 0) y
y < 0 –1 < x < 4
3) x2
+ 2x 3
x(x + 2) 3
y1 = x(x + 2) y2 = 3
x2
y1 X (0, 0) (–2, 0)
y1 y2
Y
X0–1 2
Y
X0–1 4
76
88. 78
1.3.3
1. 1) x 45
y
x + y = 45
y = 45 – x
2) xy
xy = x(45 – x)
= 45x – x2
3) 45x – x2
= 164
x2
– 45x + 164 = 0
(x – 4)(x – 41) = 0
45 164 4 41
4) x y 45
x y 1 44
x 44 y = 1
2. xy2
x + y2
= 6
x + y2
= 6
y2
= 6 – x
xy2
x y2
= 6 – x
xy2
= x(6 – x)
g = x(6 – x)
x(6 – x) x2
g
X (0, 0) (6, 0)
78
89. 79
g g = x(6 – x)
g x = a2
b
x =
)1(2
6
= 3
x = 3 y = x(6 – x)
= 3(6 – 3)
= 9
g (3, 9)
xy2
9
3. 1) x
y x
y = [(100 – 0.1(x)] x
= 100x – 0.1x2
2) y = 100x – 0.1x2
x2
y y
y = ax2
+ bx + c x = a2
b
y = 100x – 0.1x2
a = –0.1 b = 100
x =
)1.0(2
)100(
= 500
y = 100(500) – 0.1(500)2
= 25,000
9
(0, 0)
3
(6, 0)
Y
X
79
90. 80
y
x = 500
500
3) 500 25,000
4. 1) x 200
y
4000 + 200x 80 – x
y = (80 – x)(4,000 + 200x)
= 320,000 + 16,000x – 4,000x – 200x2
= –200x2
+ 12,000x + 320,000
2) y = 375,000
375,000 = – 2002
x + 12,000x + 320,000
200x2
– 12,000x + 55,000 = 0
200(x2
– 60x + 275) = 0
200(x – 55)(x – 5) = 0
x = 5, 55
375,000
2
1 4,000 + 200(5) = 5,000 5
2 4,000 + 200(55) = 15,000 55
25000
(500, 25000)
5000
Y
X
80
91. 81
3)
y = –200x2
+ 12,000x + 320,000
x2
y
y y x = a2
b
y = –200x2
+ 12,000x + 320,000
a = –200, b = 12,000
x =
)200(2
000,12
=
400
000,12
= 30
4000 + 200(30) 10,000
4)
y = –200(x)2
+ 12,000x + 320,000 x = 30
= –200(30)2
+ 12,000(30) + 320,000
= 500,000
500,000
10,000 30 80
5. 1) x 5 75
75 + x
475 – 5x
y
y = (75 + x)(475 – 5x)
= –5x2
+ 100x + 35,625
= –5(x2
– 20x – 7,125)
y = –5x2
+ 100x + 35,625 x2
y y
x =
a2
b
y = –5x2
+ 100x + 35,625 a = –5 b = 100
x =
)5(2
100
= 10
y = –5(10)2
+ 100(10) + 35,625 36,125
81
92. 82
y
(10, 36125)
y 36,125 x = 10
(75 + 10) 85
2) 36,125
1.4
1. 1) y = 3x
(R)
{y y > 0}
(10, 36125)
Y
X
y = 3x
1
0
0
82
93. 83
2) y =
x
3
1
{y y > 0}
3) y = 2x
+ 1
{y y > 1}
4) y = 3x
– 1
{y y > –1}
y =
x
3
1
Y
X
y = 2x
+ 1
0
2
1
Y
X
y = 3x
– 1
–1 0
83
+
-
95. 85
5. v(t) = 0.78C(0.8)t-1
t
800,000 3
v(3)
C = 800,000 t = 3
V(3) = 0.78(800,000)(0.83-1
)
= 0.78(800,000)(0.82
)
= 399,360 400,000
5 t = 5
V(5) = 0.78(800,000)(0.85-1
)
= 0.78(800,000)(0.84
)
= 255,590.4 256,000
10 t = 10
V(10) = 0.78(800,000)(0.810-1
)
= 0.78(800,000)(0.89
)
83,751 84,000
6. S = P(1 + i)n
S
P
i
n
P = 100,000, i =
4
03.0
, n = 3 4 12
S = 43
)
4
03.0
1(000,100
= 12
)0075.1(000,100
100,000(1.09380)
109,380
100,000 3 3%
3 109,380 – 100,000 9,380
85
96. 86
7. S = P(1 + i)n
P = 50,000
i =
12
12.0
= 0.01
n = 6
S = 50,000(1 + 0.01)6
= 50,000(1.01)6
50,000(1.06152)
53,076
53,076
8. S = P(1 + i)n
P = 10,000
i = 0.01
n = 12
S = 10,000(1 + 0.01)12
= 10,000(1.01)12
10,000(1.1268)
11,268
11,268
86
97. 87
1.5
1. 1) y = x + c
c = – 3 y = x – 3 y
{y y –3}
c = 1 y = x + 1 y
{y y 1}
Y
X0
–3
Y
X0
1
87
98. 88
2) y = x – c
c = –3 y = x + 3 y
{y y 0}
c = 1 y = x – 1 y
{y y 0}
3) y = x + c – 2
c = –3 y = x – 3 – 2 y
{y y –2}
Y
X0–3
3
Y
X0 1
Y
X0
–2
88
99. 89
c = 1 y = x + 1 – 2 y
{y y –2}
2. 1) 3 + x = 0 y = 3 + x y
y = 0, x = –3
3 + x = 0 x = –3
2) x – 5 = 0 y = x – 5 y
y = 0 , x = 5
x – 5 = 0 x = 5
Y
X–1
–2
Y
X0
Y
X0 5
–3
89
100. 90
3) 5 – x = 0
y = 5 – x y
y = 0 , x = 5
5 – x = 0 x = 5
4) x – 1 = 0 y = x – 1 y
y = 0 x = –1, 1
x – 1 = 0 x = 1 x = –1
5) x + 7 = 7 x + 7 – 7 = 0
y = x + 7 – 7 y
y = 0 , x = 0 –14
x + 7 – 7 = 0 x + 7 = 7 x = 0 x = –14
Y
X0
Y
X
–1
Y
X0
–7
–14
5
90
101. 91
3. 1) y = x + 7 y
x –7 , x + 7 0
2) y = 5 – x y
x 5 , 5 – x 0
3) y = 2 – x y
x < 2 , 2 – x > 0
–7
Y
X
5
Y
X
2
Y
X0
0
0
91
102. 92
1.6
1. f Y
0 50 100 250 500 1,000 1,500 2,000
2
4
6
8
10
12
14
16
18
20
Y ( )
X
( )
92
103. 93
2.
( )
1,000
1,000 1,000
( 1,000 1,000 )
15.00
10.00
f ( Y
)
5
10
15
1,000 X
( )
Y ( )
25
35
45
55
5,0002,000 3,000 4,000
93
108. 98
1.
1) 1) sin A
2) cos B
3) tan A
4) tan B
2)
1) sin A sin B
2) cos A cos B
3) tan A tan B
3)
1) sin A sin B
2) cos A cos B
3) tan A tan B
A C
B
c
a
b
A C
B
25
20
15
8 15
17B
C
A
109. 99
2. x
1) 2)
3) 4)
5) 6)
3. ABC
1) sin 2) cos
3) tan 4) AC
18
x
30
12
60
x
x
60
A C
B
50 21
42x
50
32
42
25
10
x
x 73 x
110. 100
4. a, b c
1) 2)
3) 4)
5. ABC AB 18 AD A
AD BC AD
6. 1)
C
A B
C
D
18
100
30
30
C
A B8
b a
60
c
b6
B A45 45
b
B
a
C A
45
30
A
BaC
c7
60
112. 102
1.
1) 1) sin A =
c
a
2) cos B =
c
a
3) tan A =
b
a
4) tan B =
a
b
2)
sin A =
25
20
=
5
4
, sin B =
25
15
=
5
3
cos A =
25
15
=
5
3
, cos B =
25
20
=
5
4
tan A =
15
20
=
3
4
, tan B =
20
15
=
4
3
3)
sin A =
17
8
, sin B =
17
15
cos A =
17
15
, cos B =
17
8
tan A =
15
8
, tan B =
8
15
A C
B
c
a
b
A C
B
25
20
15
8 15
17B
C
A
113. 103
2. x
1) sin 30 =
18
x
x = 18 sin 30 = 18 0.50 = 9
2) cos 60 =
12
x
x = 12 cos 60 = 12 0.5 = 6
3) tan 60 =
42
x
x = 42 tan 60 = 42 1.732 = 72.744
4) tan x =
50
32
= 0.64
tan 33 = 0.649
x 33
5) cos x =
7
3
= 0.42
cos 65 = 0.423
x 65
6) sin x =
25
10
= 0.40
sin 24 = 0.407
x 24
18
x
30
12
60
x
x
60
42
X
50
32
73
X
25
X
10
114. 104
3. ABC
1) sin
sin =
50
21
0.42
2) cos
sin 25 = 0.423
25
cos 25 = 0.906
3) tan
tan = tan 25
tan 25 = 0.466
4) AC
cos =
50
AC
AC = 50 cos
= 50 cos 25
= 50 0.906
= 45.3
4. a, b c
1) sin 30 = 8
a
, a = 8 sin 30
= 2
1
8 = 4
sin 60 = 8
b
, b = 8 sin 60
= 8 0.866 = 6.92830
C
A B8
b a
60
A C
B
50 21
115. 105
2) tan 60 =
7
a
a = 7 1.732
= 12.124
B = 90 – 60 = 30
sin B = sin 30 = c
7
c = 7 sin 30
= 7 0.5 = 3.5
3) sin A = c
6
c =
Asin
6
=
45
6
sin
=
0.707
6 = 8.487
ABC
(A = B = 45 )
b = 6
4) sin 45 =
30
b
b = 30 sin 45 = 30 0.707
= 21.21
A = 90 – 45 = 45
A = B = 45
ABC
a = b = 21.21
5. ABC AB 18 AD A
AD BC AD
A B
C
D
18
60
a
bC
B
A
45
30
cB A
b6
C
45 45
116. 106
ABC
AB = AC = BC = 18
BAC = CBA = ACB = 60
AB = AC
ABC
BD = CD = 2
18
9
ABD D
sin B =
18
AD
AD = 18 sin B
= 18 sin 60
sin 60 = 0.866
AD = 18 0.866
= 15.588
6. 1) x
tan 30 =
100
x
x = 100 tan 30
= 100 0.577 57.77
57.77
2) x
tan 60 =
12
x
x = 12 tan 60
= 12 1.732
= 20.784
20.8
18
9
A
D
B
100
30
x
12
60
118. 108
2.1
1. 1) sin A =
13
5
cos A =
13
12
sin B =
13
12
cos B =
13
5
tan A =
12
5
tan B =
5
12
2) sin A =
5
3
cos A =
5
4
sin B =
5
4
cos B =
5
3
tan A =
4
3
tan B =
3
4
3) sin A =
41
4
cos A =
41
5
sin B =
41
5
cos B =
41
4
tan A =
5
4
tan B =
4
5
2. 1) sin A =
AB
BC
2) sin B =
AB
AC
3) cos A =
AB
AC
4) cos B =
AB
BC
5) tan B =
BC
AC
6) tan A =
AC
BC
A
B
C
13
12
5
C
3
5
4
C
4 5
41
B A
B C
A
A
B
119. 109
3.
1) sin A = c
a
2) cos B = c
a
3) cos A = c
b
4) sin B = c
b
4. 1) sin 32 = 0.530 2) sin 4 = 0.070
3) sin 81 = 0.988 4) sin 17 = 0.292
5) cos 29 = 0.875 6) tan 18 = 0.325
7) tan 81 = 6.314 8) sin 51.5 0.7825
9) cos 67.5 0.383 10) tan 42.5 0.9165
5. 1) sin = 0.53
sin 32 = 0.53
= 32
2) sin = 0.899
sin 64 = 0.899
= 64
3) sin =
5
2
0.4
sin 24 = 0.407
= 24 ( )
4) sin =
8
5
0.625
sin 39 = 0.629
= 39 ( )
5) sin =
23
15
0.652
sin 41 = 0.656
= 41 ( )
C
b a
A Bc
120. 110
6. 1) cos = 0.5
cos 60 = 0.500
= 60
2) cos = 0.99
cos 8 = 0.990
= 8
3) cos = 0.75
cos 41 = 0.755
= 41 ( )
4) cos =
5
3
0.6
cos 53 = 0.602
= 53 ( )
5) cos =
13
9
0.692
cos 46 = 0.695
= 46 ( )
7. 1) tan = 0.404
tan 22 = 0.404
= 22
2) tan = 4.011
tan 76 = 4.011
= 76
3) tan =
4
9
2.250
tan 66 = 2.246
= 66 ( )
4) tan =
7
4
0.571
tan 30 = 0.577
= 30 ( )
5) tan =
13
28
2.154
tan 65 = 2.145
= 65 ( )
121. 111
8.
1) cos = 0.616 cos 52 = 0.616 = 52
2) tan = 0.488 tan 26 = 0.488 = 26
3) sin = 0.982 sin 79 = 0.982 = 79
4) tan = 2.356 tan 67 = 2.356 = 67
5) cos = 0.707 cos 45 = 0.707 = 45
9. 1
AB2
+ BC2
= AC2
BC2
= AC2
– AB2
= 152
– 92
= 144
BC = 12
BC BC = 12
sin A =
AC
BC
=
15
12
0.8
tan A =
AB
BC
=
9
12
1.33
cos C =
AC
BC
=
15
12
0.8
tan C =
BC
AB
=
12
9
0.75
2
cos A =
15
9
0.6
cos 53 = 0.602
A = 53 ( )
C = 180 – 90 – 53 = 37 ( )
sin A = sin 53 = 0.799 0.8
tan A = tan 53 = 1.327 1.33
A B
C
15
9
122. 112
cos C = cos 37 = 0.799 0.8
tan C = tan 37 = 0.754 0.75
10. 1) sin 45 =
14
x
x = 14 sin 45
sin 45 = 0.707
x = 14 0.707
= 9.898
2) tan 30 =
35
x
x = 35 tan 30
=
3
3
35
= 5
cos 30 =
y
35
y =
30cos
35
=
2
3
35
= 10
14 x
45
30 x
y
35
123. 113
3) cos 60 =
8
x
x = 8 cos 60
cos 60 = 0.500
x = 8 0.500
= 4
sin 60 =
8
y
y = 8 sin 60
sin 60 = 0.866
y = 8 0.866
= 6.928
tan 45 =
z
y
z =
45tan
y
45tan
928.6
tan 45 = 1.0
z = 6.928
4) sin 60 =
8
x
x = 8 sin 60
sin 60 = 0.866
x = 8 0.866
= 6.928
sin 30 =
8
y
y = 8 sin 30
sin 30 = 0.50
y = 8 0.5
= 4
11. ABC C cos A =
2
1
cos A =
2
1
sin B = cos A
45
y
z
A
B C
2
1
sin B =
2
1
0.5
y 8
60
x
x
30 60
y
8
124. 114
sin 30 = 0.5 B = 30
cos 60 = 0.5 A = 60
1) sin A = sin 60
sin 60 = 0.866
2) tan A = tan 60
tan 60 = 1.732
3) sin B = cos A =
2
1
0.5
4) cos B = sin A = 0.866
5) tan B = tan 30
tan 30 = 0.577
12. 1)
ABD
BD2
= AB2
+ AD2
BD2
= 82
+ 62
BD2
= 64 + 36
BD2
= 100
BD = 10
BD
BD = 10
2) 1ˆsin =
BD
AB
=
10
8
5
4
3) 1ˆcos =
BD
AD
=
10
6
5
3
4) 2ˆsin =
BD
AD
=
10
6
5
3
5) 2ˆcos =
BD
AB
=
10
8
5
4
A B
D C
8
6
2 4
31
126. 116
13.
1) tan A = 2 a = 10
tan A =
b
a
2 =
b
10
b =
2
10
=
2
52
= 25
2) tan A =
4
1
b = 3
tan A =
b
a
4
1
=
3
a
a =
4
3
3) a = 5 b = 12
sin A =
c
a
c c2
= a2
+ b2
c2
= 52
+ 122
c2
= 25 + 144
c2
= 169
c = 13
c
b
a
B
C A
127. 117
c
c = 13
sin A =
13
5
4) cos B =
3
2
a = 5
cos B =
c
a
3
2
=
c
5
c =
2
35
7.5
5) a = 4 c = 6
tan B =
a
b
b c2
= a2
+ b2
62
= 42
+ b2
b2
= 62
– 42
b2
= 36 – 16
b2
= 20
b = 52
b b = 52
tan B =
4
52
2
5
14. 1)
BG = 12 GH = 8
B =
A C
D
G
B
H
E
128. 118
BGH
sin =
BG
GH
=
12
8
= 0.667
sin 42 = 0.669
42
BDE
BE x
cos =
BD
BE
cos 42 =
15
x
cos 42 = 0.743
0.743 =
15
x
x = 0.743 15 11.145
BE 11.145
2) AC = 9 DE = 6 AB = 15
B
G H8
12
x
B
D E
15
C
E
B
D
A
6
9
15
129. 119
BC
AB2
= AC2
+ BC2
152
= 92
+ BC2
BC2
= 152
– 92
BC2
= 225 – 81
BC2
= 144
BC = 12
BC BC = 12
EC = BC – BE = 12 – 11.145 = 0.855
3) A = 50 BH = 4
ABC BGH
Aˆ = Gˆ = 50
tan 50 =
GH
4
tan 50 = 1.192
1.192 =
GH
4
GH =
192.1
4
GH = 3.356
B
G H
4
50
130. 120
4) AB = 210
AC = 56
ABC A =
cos =
210
56
= 0.949
cos 18 = 0.951
18
ABC = 180
CBˆA 180 – 90 – 18 72
15. DEF DG DE 18
DEF
60
DGE E = 60
sin E =
18
DG
sin 60 =
18
DG
DG = 18 sin 60
= 18 0.866
= 15.588
18
E
G
F
D
18
A C
B
210
56
131. 121
16. JKL LM JL =16
JKL J = K
J = K =
180 = 90 + +
= 45
JLM
sin J =
16
LM
LM = 16 sin 45
sin 45 = 0.707
LM = 16 0.707
= 11.312
17. ABC ACBC Aˆ = 27 BC = 10
ABC AB
sin A =
AB
BC
sin 27 =
AB
10
AB =
27sin
10
sin 27 = 0.454
L
M K
16
J
27
A
BC
10
132. 122
AB =
454.0
10
22.03
AC
tan A =
AC
BC
tan 27 =
AC
10
AC =
27tan
10
tan 27 = 0.510
AC =
510.0
10
AC 19.61
ABC = AB + BC + AC
22.03 + 10 + 19.61
51.64
18. 14, 14 18
AD CAˆB ABC
AD CAˆB ABC AD
BC
A
CB D
18
CB
14 14
A
133. 123
BD = DC = 9
B = C =
cos B =
AB
BD
cos =
14
9
cos = 0.643
cos 50 = 0.643
= 50
180 = BAˆCACˆBCBˆA
180 = 50 + 50 + BAˆC
BAˆC = 180 – 100
BAˆC = 80
50CBˆA , 50ACˆB 80BAˆC
134. 124
2.2
1. 40
20
A
B
C
tan B =
BC
AC
AC = 40 tan 20
tan 20 = 0.364
AC = 40 0.364
AC = 14.56
14.56
40 .
20
A
CB 20
40
135. 125
2. 6.5
4
AB
AC 6.5
BC 4
BCˆA BCˆA =
cos =
AC
BC
=
5.6
4
= 0.615
cos 52 = 0.616
52
sin =
AC
AB
AB = 6.5 sin 52
sin 52 = 0.788
AB = 6.5 0.788
= 5.122
5.122 5
C
4
A
B
6.5
136. 126
3. ABCD 6 55
ABC
sin A =
AC
BC
BC = AC sin 55
sin 55 = 0.819
BC = 6 0.819
= 4.914
cos A =
AC
AB
AB = AC cos 55
sin 55 = 0.574
AB = 6 0.574
= 3.444
4.914
3.444
55
A B
D C
6
137. 127
4.
AB
AC 16
cos A =
AB
AC
AB =
39cos
16
cos 39 = 0.777
AB =
777.0
16
= 20.592
20.6 ( )
5.
AB 2
BC
sin A =
AB
BC
BC = AB sin A
= 2 sin 7
sin 7 = 0.122
BC = 2 0.122
= 0.244
= 0.244 1000
= 244
244
C
39
16A
B
A
B
C
7
2 .
138. 128
6.
AB 9.5
BC
sin A =
AB
BC
BC = 9.5 sin 58
sin 58 = 0.848
BC = 9.5 0.848
= 8.056
8.056 + 1.2 9.2 ( )
7. 45 12
1.5
AB
BC ( )
45
12
A
CB
A C
B
58
9.5
1.5 .
12 .
45
139. 129
ABC
tan C =
BC
AB
AB = BC tan 45
AB = 12 1 12
12 + 1.5 13.5
8.
AC 38
BC + CD
tan 13 =
AC
BC
BC = AC tan 13
tan 13 = 0.231
BC = 38 0.231
= 8.778
tan 16 =
AC
CD
CD = AC tan 16
tan 16 = 0.287
CD = 38 0.287
= 10.906
8.778 + 10.906 = 19.684 19.7
9.
AA // CD
CAA = DCA = 40 ( )
CD // BB
BCD = BBC = 55 ( )
D
A
B
C 13
16
40
55
C
A
B
.
.
x
5
D40
55
A
B
140. 130
AB
BD 5
BCD BDC
tan 55 =
CD
BD
CD =
55tan
5
tan 55 = 1.428
CD =
428.1
5
= 3.5
ACD CDA
tan 40 =
CD
AD
AD = 3.5 tan 40
tan 40 = 0.839
AD = 3.5 0.839
= 2.9365 2.9
5 + 2.9 7.9 ( )
10.
OB
OA
15.07
1
2
3
4
567
8
9
10
11 12
AO
9 .
B
141. 131
12 3 15 1
15.07 . 7
7
15
90
42
cos =
OB
OA
OA = OB cos 42
cos 42 = 0.743
OA = 9 0.743
= 6.687
6.7
12
3