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Solutions to Exercise 2C (Joseph Yeo, Teh Keng Seng, Loh Cheng Yee and Ivy Chew’s New Syllabus Additional Mathematics, 9th Edition ‒ ISBN 9789812374998)

Solved by: Dr Lee Chu Keong (ascklee@gmail.com)

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Exercise 2C (p. 62)

Question 1(a)

𝑎 = +𝑣𝑒 (𝑚𝑖𝑛𝑖𝑚𝑢𝑚)

𝑦 = −4 𝑎𝑡 𝑥 =2

5

Question 1(b)

𝑎 = −𝑣𝑒 (𝑚𝑎𝑥𝑖𝑚𝑢𝑚)

𝑦 = 2 𝑎𝑡 𝑥 =1

2

Question 1(c)

𝑎 = −𝑣𝑒 (𝑚𝑎𝑥𝑖𝑚𝑢𝑚)

𝑦 = −2 𝑎𝑡 𝑥 = −5

2

Question 1(d)

𝑎 = +𝑣𝑒 (𝑚𝑖𝑛𝑖𝑚𝑢𝑚)

𝑦 = −3 𝑎𝑡 𝑥 = −4

3

Question 3(a)

𝑦 = 2𝑥2 − 4𝑥 + 7

= 2 [𝑥2 − 2𝑥 +7

2]

= 2 [(𝑥 − 1)2 − 1 +7

2]

= 2 [(𝑥 − 1)2 +5

2]

= 2(𝑥 − 1)2 + 5

Question 2(b)

𝑦 = 2 [𝑥2 −5

2𝑥−

1

2]

= 2 [(𝑥 −5

4)

2

− (5

4)

2

−1

2]

= 2 [(𝑥 −5

4)

2

−25

16−

1

2]

= 2 [(𝑥 −5

4)

2

− 21

16]

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2

= 2 (𝑥 −5

4)

2

− 41

8

Question 2(c)

𝑦 = 3 [𝑥2 +7

3𝑥 + 3]

= 3 [(𝑥 +7

6)

2

− (7

6)

2

+ 3]

= 3 [(𝑥 +7

6)

2

+ 123

36]

= 3 (𝑥 +7

6)

2

+ 411

12

Question 2(d)

𝑦 = −[𝑥2 − 4𝑥 + 1]

= −[(𝑥 − 2)2 − 4 + 1]

= −[(𝑥 − 2)2 − 3]

= −(𝑥 − 2)2 + 3

Question 2(a)

𝑎 = +𝑣𝑒 (𝑠𝑚𝑖𝑙𝑖𝑛𝑔 𝑐𝑢𝑟𝑣𝑒 − 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑝𝑜𝑖𝑛𝑡)

𝑥-intercepts (substitute 𝑦 = 0):

(1,0) 𝑎𝑛𝑑 (−7,0)

y-intercepts (substitute 𝑥 = 0):

𝑦 = (−1)(7)

= −7

(0, −7)

𝑥 =1 − 7

2

=−6

2

= −3

𝑦𝑚𝑖𝑛𝑖𝑚𝑢𝑚 = (−3 − 1)(−3 + 7)

= −4 × 4

= −16

(−3, −16)

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Question 2(b)

𝑎 = +𝑣𝑒 (𝑠𝑚𝑖𝑙𝑖𝑛𝑔 𝑐𝑢𝑟𝑣𝑒 − 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑝𝑜𝑖𝑛𝑡)

𝑥-intercepts (substitute 𝑦 = 0):

(1

2, 0) 𝑎𝑛𝑑 (−

7

2, 0)

y-intercepts (substitute 𝑥 = 0):

𝑦 = (−1)(7)

= −7

(0, −7)

𝑥 =

12 −

72

2

=−

62

2

= −3

2

𝑦𝑚𝑖𝑛𝑖𝑚𝑢𝑚 = [2 (−3

2) − 1] [2 (−

3

2) + 7]

= [−3 − 1][−3 + 7]

= [−4][4]

= −16

(−3

2, −16)

6 4 2x

15

10

5

y

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Question 2(c)

𝑎 = +𝑣𝑒 (𝑠𝑚𝑖𝑙𝑖𝑛𝑔 𝑐𝑢𝑟𝑣𝑒 − 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑝𝑜𝑖𝑛𝑡)

𝑥-intercepts (substitute 𝑦 = 0):

(−2,0) 𝑎𝑛𝑑 (5,0)

y-intercepts (substitute 𝑥 = 0):

𝑦 = (2)(−5)

= −10

(0, −10)

𝑥 =−2 + 5

2

=3

2

𝑦𝑚𝑖𝑛𝑖𝑚𝑢𝑚 = (3

2+ 2) (

3

2− 5)

= (7

2) (−

7

2)

= −49

2

(3

2, −

49

2)

4 3 2 1 1x

15

10

5

5

10

y

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Question 2(d)

𝑎 = −𝑣𝑒 (𝑓𝑟𝑜𝑤𝑛𝑖𝑛𝑔 𝑐𝑢𝑟𝑣𝑒 − 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑝𝑜𝑖𝑛𝑡)

𝑥-intercepts (substitute 𝑦 = 0):

(−3,0) 𝑎𝑛𝑑 (2,0)

y-intercepts (substitute 𝑥 = 0):

𝑦 = (3)(2)

= 6

(0,6)

𝑥 =−3 + 2

2

= −1

2

𝑦𝑚𝑎𝑥𝑖𝑚𝑢𝑚 = (−1

2+ 3) (2 − (−

1

2))

= (21

2) (2

1

2)

= 61

4

(−1

2, 6

1

4)

2 1 1 2 3 4 5x

12

10

8

6

4

2

y

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Question 2(e)

𝑎 = −𝑣𝑒 (𝑓𝑟𝑜𝑤𝑛𝑖𝑛𝑔 𝑐𝑢𝑟𝑣𝑒 − 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑝𝑜𝑖𝑛𝑡)

𝑥-intercepts (substitute 𝑦 = 0):

(0,0) 𝑎𝑛𝑑 (3

4, 0)

y-intercepts (substitute 𝑥 = 0):

(0,0)

𝑥 =0 +

34

2

=3

8

𝑦𝑚𝑎𝑥𝑖𝑚𝑢𝑚 =3

8(3 − 4 (

3

8))

=3

8(3 −

3

2)

=3

8(

3

2)

=9

16

(3

8,

9

16)

4 3 2 1 1 2 3x

6

4

2

2

4

6

y

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Question 2(f)

𝑎 = +𝑣𝑒 (𝑠𝑚𝑖𝑙𝑖𝑛𝑔 𝑐𝑢𝑟𝑣𝑒 − 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑝𝑜𝑖𝑛𝑡)

𝑥-intercepts (substitute 𝑦 = 0):

(0,0) 𝑎𝑛𝑑 (8

3, 0)

y-intercepts (substitute 𝑥 = 0):

(0,0)

𝑥 =0 +

83

2

=8

6

=4

3

𝑦 =4

3(3 (

4

3) − 8)

=4

3(4 − 8)

=4

3(−4)

= −16

3

(4

3, −

16

3)

0.2 0.2 0.4 0.6 0.8x

0.6

0.4

0.2

0.2

0.4

y

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Question 4

𝑥2 − 5𝑥 + 13 = (𝑥 −5

2)

2

− (5

2)

2

+ 13

= (𝑥 −5

2)

2

−25

4+ 13

= (𝑥 −5

2)

2

+27

4

𝑦𝑚𝑖𝑛 =27

4 𝑎𝑡 𝑥 =

5

2

0.5 1.0 1.5 2.0 2.5x

5

4

3

2

1

1

y

0 1 2 3 4 5x

2

4

6

8

10

12

14

y

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Question 5

𝑦 = 4 [𝑥2 −3

2𝑥 +

7

2]

= 4 [(𝑥 −3

4)

2

− (3

4)

2

+7

2]

= 4 [(𝑥 −3

4)

2

−9

16+

7

2]

= 4 [(𝑥 −3

4)

2

+47

16]

= 4 (𝑥 −3

4)

2

+47

4

Question 6 (newly added)

The value of a stock portfolio is given by the function 𝑦 = 3𝑥2 − 4𝑥 + 5, where y is the

value of the portfolio in thousands of dollars and x is the time in years. Find x when the

value of the portfolio is at its lowest.

Question 7

2𝑥2 + 4𝑥 + 17 = 2 [𝑥2 + 2𝑥 +17

2]

= 2 [(𝑥 + 1)2 − 1 +17

2]

= 2 [(𝑥 + 1)2 +15

2]

= 2(𝑥 + 1)2 + 15

𝑦𝑚𝑖𝑛 = 15 𝑎𝑡 𝑥 = −1

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Question 8

𝑦 = −(𝑥2 + 3𝑥 − 14)

= − [(𝑥 +3

2)

2

− (3

2)

2

− 14]

= − [(𝑥 +3

2)

2

−9

4− 14]

= − [(𝑥 +3

2)

2

−65

4]

= − (𝑥 +3

2)

2

+65

4

𝑦𝑚𝑎𝑥 =65

4= 16

1

4 𝑎𝑡 𝑥 = −

3

2

3 2 1 0 1x

5

10

15

20

25

y

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Question 9 (newly added)

The height, y meters, of the water sprayed from a hose, is given as 𝑦 = −𝑥2 + 4𝑥 − 1,

where x is the horizontal distance traveled by the water. Find the greatest height of the

water sprayed and the horizontal distance from the hose when this occurs.

Question 10

𝑦 = 3 [𝑥2 −4

3𝑥 +

2

3]

= 3 [(𝑥 −2

3)

2

− (2

3)

2

+2

3]

= 3 [(𝑥 −2

3)

2

−4

9+

2

3]

= 3 [(𝑥 −2

3)

2

+2

9]

= 3 (𝑥 −2

3)

2

+2

3

Question 11

𝑦 = −[𝑥2 + 4𝑥 − 5]

= −[(𝑥 + 2)2 − 22 − 5]

= −[(𝑥 + 2)2 − 9]

= −(𝑥 + 2)2 + 9

4 2 2x

5

10

15

y

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Dr. Lee is an experienced teacher who has taught at the Singapore and Temasek Polytechnics. He currently teaches at the Nanyang Technological University. A chemical engineer by training, Dr. Lee feels that everyone benefits from a solid grounding in mathematics, and that this grounding can only be obtained through sufficient practice on a variety of question types.

The solutions to each question are developed with care with the idea of deepening the student’s number sense and strengthening his basic technique of solving mathematical problems. To this end, great attention has been paid to the steps and explanations leading to the final answer.

Dr. Lee is the founder of several mathematics-related initiatives, and can be contacted at ascklee@gmail.com.