3.quadratic function

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1

Additional Mathematics SPM Chapter 3

© Penerbitan Pelangi Sdn. Bhd.

3Quadratic Functions

1. (b), (c), (d), (e) and (h)

2. f ( x ) = 4 x 2 – 8 x + 6

When x = –2,

f (–2) = 4(–2)2 – 8(–2) + 6

= 16 + 16 + 6= 38

3. f ( x ) = x 2 – 3 x + 2

When f ( x ) = 0,

x 2 – 3 x + 2 = 0

( x – 1)( x – 2) = 0

x – 1 = 0 or x – 2 = 0

x = 1 or x = 2

4. f ( x ) = –3 x 2 + 5 x – 1

When f ( x ) = 1,

–3 x 2 + 5 x – 1 = 1

–3 x 2 + 5 x – 1 – 1 = 03 x 2 – 5 x + 2 = 0

(3 x – 2)( x – 1) = 0

3 x – 2 = 0 or x – 1 = 0

x =2

— 3

or x = 1

5. (a) (b)

(c) (d)

(e) (f)

6. (a) Two different real roots

(b) One real root or two similar real roots

(c) No real roots

7. (a) The minimum value is 3.

(b) The maximum value is 4.

(c) The minimum value is –10.

(d) f ( x ) = 2΄ 1 — 2

( x – 3)2 +1

— 4 ΅

= ( x – 3)2 +1

— 2

Therefore, the minimum value is1

— 2

.

(e) f ( x ) =1

— 3

[6 – ( x + 1)2] + 5

= 2 – 1

— 3

( x + 1)2 + 5

= – 1

— 3

( x + 1)2 + 7

Therefore, the maximum value is 7.

(f) The minimum value is 3.

8. (a) f ( x ) = x 2 – 4 x + 2

= x 2 – 4 x + 4 — 2

2

– 4 —

2

2

+ 2

= ( x – 2)2 – 4 + 2

= ( x – 2)2 – 2Hence, the minimum value is –2.

(b) f ( x ) = 2 x 2 + 6 x – 5

= 2( x 2 + 3 x ) – 5

= 2 ́x 2 + 3 x + 3 — 2

2

– 3 — 2

2

΅ – 5

= 2΄ x + 3 — 2

2

– 9

— 4

΅ – 5

= 2 x + 3 — 2

2

– 9

— 2

– 5

= 2 x + 3 — 2

2

– 19

—– 2

Hence, the minimum value is – 19

—– 2

.

(c) f ( x ) = x 2 + 5 x

= x 2 + 5 x + 5 — 2

2

– 5 — 2

2

= x + 5 — 2

2

– 25

—– 4


—– 4

.



2



(d) f ( x ) = 6 x – x 2

= –( x 2 – 6 x )

= –

΄ x 2 – 6 x + 6

—

2

2

– 6 —

2

2

΅= –[( x – 3)2 – 9]

= – ( x – 3)2 + 9

Hence, the maximum value is 9.

(e) f ( x ) = 3 – 4 x – x 2

= – x 2 – 4 x + 3

= –( x 2 + 4 x ) + 3

= – ́ x 2 + 4 x + 4 — 2

2

– 4 — 2

2

΅ + 3

= –[( x + 2)2 – 4] + 3

= –( x + 2)2 + 4 + 3

= –( x + 2)2 + 7


(f) f ( x ) = 4 x – 2 x 2

= –2 x 2 + 4 x

= –2( x 2 – 2 x )

= –2 ́x 2 – 2 x + 2 — 2

2

– 2 — 2

2

΅= –2[( x – 1)2 – 1]

= –2( x – 1)2 + 2


(g) f ( x ) = 10 + 5 x – 3 x 2

= –3 x 2 + 5 x + 10

= –3 x 2 – 5

— 3 x + 10

= –3΄ x 2

–

5

— 3 x + 5

— 6 2

– 5

— 6 2

΅ + 10

= –3΄ x – 5

— 6

2

– 25

—– 36 ΅ + 10

= –3 x – 5

— 6

2

+25

—– 12

+ 10

= –3 x – 5

— 6

2

+145

—–– 12

Hence, the maximum value is145

—–– 12

.

(h) f ( x ) = (2 x – 1)( x + 3)

= 2 x 2 + 5 x – 3

= 2

x 2 +

5 —

2

x

– 3

= 2 ́x 2 +5

— 2 x + 5

— 4

2

– 5 — 4

2

΅ – 3

= 2΄ x + 5 — 4

2

– 25

—– 16 ΅ – 3

= 2 x + 5 — 4

2

– 25

—– 8

– 3

= 2 x + 5 — 4

2

– 49

—– 8


—– 8

.

(i) f ( x ) = (1 – 4 x )( x + 2)

= x + 2 – 4 x 2 – 8 x

= – 4 x 2 – 7 x + 2

= – 4 x 2 + 7 —

4 x + 2

= – 4́ x 2 +7

— 4 x + 7

— 8

2

– 7 — 8

2

΅ + 2

= – 4́ x + 7 —

8

2

– 49

—– 64 ΅ + 2

= – 4 x + 7 — 8

2

+49

—– 16

+ 2

= – 4 x + 7 — 8

2

+81

—– 16

Hence, the maximum value is81

—– 16

.

9. (a) f ( x ) = x 2 – 4

Therefore, the minimum point is (0, – 4).

x –2 2 4

f ( x) 0 0 12

f (x )

x

–4 –2 2 4

12

0

(b) f ( x ) = 3 x 2

+ 5Therefore, the minimum point is (0, 5).

x –1 3

f ( x) 8 32

f (x )

x –1 3

5

8

32

0

(c) f ( x ) = 8 – x 2

Therefore, the maximum point is (0, 8).

x –3 ±ͱ⒓8 3

f ( x) –1 0 –1

f (x )

x

–1

8

–3 3 – 8ͱ ⒓⒓⒓ 8 ⒓⒓⒓0



3



(d) f ( x ) = 10 – 2 x 2


x –3 ±ͱ⒓5 4

f ( x) –8 0 –22

f (x )

x

–8

10

–3 4 – 5ͱ ⒓⒓⒓ 5 ⒓⒓⒓0

–22

(e) f ( x ) = x ( x + 2)

= x 2 + 2 x

= x 2 + 2 x + 12 – 12

= ( x + 1)2 – 1

Therefore, the minimum point is (–1, –1).

x – 4 –2 0 2

f ( x) 8 0 0 8

f (x )

x

(–1,–1) –4

8

–2 20

(f) f ( x ) = ( x – 1)(2 x + 1)= 2 x 2 – x – 1

= 2 x 2 – x

— 2

– 1

= 2 ́x 2 – x

— 2

+ 1 — 4

2

– 1 — 4

2

΅ – 1

= 2΄ x – 1

— 4

2

– 1

—– 16

΅ – 1

= 2 x – 1

— 4

2

– 1

— 8

– 1

= 2 x – 1

— 4

2

– 9

— 8

Therefore, the minimum point is (1

— 4

, – 9

— 8

).

x –1 – 1

— 2

0 1 2

f ( x) 2 0 –1 0 5

f (x )

x 1 2 –1 1

2 – –

9

8 – –)1

4 ( –,

2

0

5

–1

(g) f ( x ) = –( x – 3)2 + 5


x –2 0 3 – ͱ⒓5 3 + ͱ⒓5 6

f ( x) –20 – 4 0 0 –4

f (x )

x –2 –4

–20

(3, 5)

63 + 5ͱ⒓⒓3 – 5ͱ⒓⒓0

(h) f ( x ) = x 2 + 4 x + 5

= x 2 + 4 x + 22 – 22 + 5

= ( x + 2)2 + 1

Therefore, the minimum point is (–2, 1).

x –3 0 1

f ( x) 2 5 10

f (x )

x

(–2, 1)

–3 1

2

5

10

0

(i) f ( x ) = 2 x 2

+ 6 x – 8= 2( x 2 + 3 x ) – 8

= 2 ́x 2 + 3 x + 3 —

2

2

– 3 — 2

2

΅ – 8

= 2΄ x + 3 — 2

2

– 9

— 4 ΅ – 8

= 2 x + 3 — 2

2

– 9

— 2

– 8

= 2 x + 3 — 2

2

– 25

—– 2

Therefore, the minimum point is (– 3

— 2

, – 25

—– 2

).

x –3 0 1 2

f ( x) –8 –8 0 12

–25

2 – – )3

2( –,

f (x )

x –3

12

–8

0 1 2



4



(j) f ( x ) = ( x – 4)2

Therefore, the minimum point is (4, 0).

x 0 4 5

f ( x) 16 0 1

f (x )

x 4 50

1

16

(k) f ( x ) = – x 2 + 6 x – 9

= –( x 2 – 6 x ) – 9

= –( x 2 – 6 x + 32 – 32) – 9

= –[( x – 3)2 – 9] – 9

= –( x – 3)2


x 0 3 4

f ( x) –9 0 –1

f (x )

x 0

–1

–9

(3, 0)

4

10. (a) x ( x – 2) у 0

0 2x

f (x )

The range of values of x is x р 0 or x у 2.

(b) ( x – 3)( x – 4) р 0

0 43x

f (x )

The range of values of x is 3 р x р 4.

(c) x 2 – 3 x – 4 Ͼ 0

( x – 4)( x + 1) Ͼ 0

4 –10

x

f (x )

The range of values of x is x Ͻ –1 or x Ͼ 4.

(d) 2 x 2 + 5 x – 3 Ͻ 0

(2 x – 1)( x + 3) Ͻ 0

1 –2

–3 0x

f (x )

The range of values of x is –3 Ͻ x Ͻ 1

— 2

.

(e) ( x – 3)( x + 2) р –4

x 2 – x – 6 р –4

x 2 – x – 2 р 0

( x – 2)( x + 1) р 0

–1 2

0x

f (x )

The range of values of x is –1 р x р 2.

(f) (2 x – 1)( x – 3) р 4( x – 3)

2 x 2 – 6 x – x + 3 р 4 x – 12

2 x 2 – 11 x + 15 р 0

( x – 3)(2 x – 5) р 0

5 –

230

x

f (x )

The range of values of x is5

— 2

р x р 3.

(g) x 2 + 4

——— 5

р 2 x – 1

x 2 + 4 р 5(2 x – 1)

x 2 + 4 р 10 x – 5

x 2 – 10 x + 9 р 0

( x – 1)( x – 9) р 0



5



910x

f (x )

The range of values of x is 1 р x р 9.

(h) x (1 – 4 x ) Ͻ 5 x – 8

x – 4 x 2 – 5 x + 8 Ͻ 0

– 4 x 2 – 4 x + 8 Ͻ 0

x 2 + x – 2 Ͼ 0

( x + 2)( x – 1) Ͼ 0

1 –2 0x

f (x )

The range of values of x is x Ͻ –2 or x Ͼ 1.

1. (a) x -coordinate of the maximum part = 1 + 7 ——–

2= 4

Therefore, the equation of the axis of symmetry

is x = 4.

(b) f ( x ) = p – ( x + q)2

= 5 – ( x – 4)2

2. f ( x ) = 2 x 2 – 16 x + k 2 + 2k + 1

= 2( x 2 – 8 x ) + k 2 + 2k + 1

= 2 ́x 2 – 8 x + 8 — 2

2

– 8 — 2

2

΅ + k 2 + 2k + 1

= 2[( x – 4)2 – 16] + k 2 + 2k + 1

= 2( x – 4)2 – 32 + k 2 + 2k + 1

= 2( x – 4)2 + k 2 + 2k – 31

Given minimum value = –28,

∴ k 2 + 2k – 31 = –28

k 2 + 2k – 3 = 0

(k + 3)(k – 1) = 0

k = –3, 1

3. (a) f ( x ) = 2( x – 3)2 + k

p is the x -coordinate of the minimum point.

Therefore, p = 3.

(b) k is the minimum value of f ( x ).

Therefore, k = –4.

(c) The equation of the axis of symmetry is x = 3.

4. x ( x + 2) у 3

x 2 + 2 x – 3 у 0

( x + 3)( x – 1) у 0

1 –3 0x

f (x )

The range of values of x is x р –3 or x у 1.

5. f ( x ) = 2 x 2 – 12 x + 5

= 2( x 2 – 6 x ) + 5

= 2[( x – 3)2 – 32] + 5

= 2( x – 3)2 – 18 + 5

= 2( x – 3)2 – 13

∴ p = 2, q = –3 , – r + 1 = –13

r = 14

6. (a) f ( x ) = – x 2 + 6 px + 1 – 4 p2

= –( x 2 – 6 px ) + 1 – 4 p2

= – ́ x 2 – 6 px + 6 p —–

2

2

– 6 p —–

2

2

΅ + 1 – 4 p2

= –[( x – 3 p)2 – 9 p2] + 1 – 4 p2

= – ( x – 3 p)2 + 9 p2 + 1 – 4 p2

= – ( x – 3 p)2 + 1 + 5 p2

The maximum value given is q2 – p.

Therefore, q2 – p = 1 + 5 p2

5 p2 + p + 1 = q2

(b) x = 3 is symmetrical axis

3 p = 3

p = 1

Substitute p = 1 into 5 p2 + p + 1 = q2,

5(1)2 + 1 + 1 = q2

q2 = 7

q = ±ͱ⒓7Hence, p = 1, q = ±ͱ⒓7

1. x -coordinate of maximum point = – 4 + 0 ——––

2= –2

Equation of the axis of symmetry is x = –2



6



2. Let x be the x -coordinate of A

0 + x ——–

2= 3

x = 6The coordinates of A are (6, 4).

3. Let x be the x -coordinate of A

x + 6 ——–

2= 2

x = 4 – 6

x = –2

The coordinates of A are (–2, 0).

4. x -coordinate of A = 0 + 8 ——–

2= 4

Let C be the centre of OB,

4

5

A

C O

AC2 = OA2 – OC2

= 52 – 42

= 9

AC = 3

The coordinates of A are (4, 3).

5. x -coordinate of minimum point = 0 + 4 ——–

2= 2

x -coordinate of minimum point for the image is –2.

6. (a) y = ( x – p)2 + q and minimum point is (2, –1)

Hence, p = 2 and q = –1

(b) y = ( x – 2)2 – 1

When y = 0,

( x – 2)2 – 1 = 0

( x – 2)2 = 1

x – 2 = ±1

x = ±1 + 2

= 1, 3

Hence, A is (1, 0) and B is (3, 0).

7. f ( x ) = 2k + 1 – x + 1 — 2 p

2

Given (–1, k ) is the maximum point.

Therefore, 2k + 1 = k

k = –1

x +1

— 2 p = 0 when x = –1,

–1 +1

—

2

p = 0

1

— 2 p = 1

p = 2

8. Given ( p, 2q) is the minimum point of

y = 2 x 2 – 4 x + 5

= 2( x 2 – 2 x ) + 5

= 2( x 2 – 2 x + 12 – 12) + 5

= 2[( x – 1)2 – 1] + 5

= 2( x – 1)2 – 2 + 5

= 2( x – 1)2 + 3

2q = 3

q = 3 — 2

p – 1 = 0

p = 1

9. (a) Since (1, 4) is the point on y = x 2 – 2kx + 1,

substitute x = 1, y = 4 into the equation,

4 = 12 – 2k (1) + 1

2k = –2

k = –1

(b) y = x 2 – 2(–1) x + 1

= x 2 + 2 x + 1

= ( x + 1)2

Minimum value of y is 0.

10. f ( x ) = – x 2 – 8 x + k – 1

= –( x 2 + 8 x ) + k – 1

= –( x 2 + 8 x + 42 – 42) + k – 1

= –[( x + 4)2 – 16] + k – 1

= –( x + 4)2 + 16 + k – 1

= –( x + 4)2 + 15 + k

Since 13 is the maximum value,

then 15 + k = 13

k = –2

11. f ( x ) = 2 x 2

– 6 x + 7= 2( x 2 – 3 x ) + 7

= 2 ́x 2 – 3 x + 3 — 2

2

– 3 —

2

2

΅ + 7

= 2΄ x – 3

— 2

2

– 9

— 4

΅ + 7

= 2 x – 3

— 2

2

– 9

— 2

+ 7

= 2 x – 3

— 2

2

+5

— 2



7



The minimum point is (3

— 2

,5

— 2

).

x –1 0 3

f ( x) 15 7 7

–, –

3

5322

0

7

15

x

f (x )

–1

The range is5

— 2

р f ( x ) р 15.

12. f ( x ) = 5 – 4 x – 2 x

2

= –2 x 2 – 4 x + 5

= –2( x 2 + 2 x ) + 5

= –2( x 2 + 2 x + 12 – 12) + 5

= –2[( x + 1)2 – 1] + 5

= –2( x + 1)2 + 2 + 5

= –2( x + 1)2 + 7

When f ( x ) = –1,

–2( x + 1)2 + 7 = –1

–2( x + 1)2 = –8

( x + 1)2 = 4

x + 1 = ±2

x = ±2 – 1

= –3 or 1

13. y = ( x – 3)2 – 4

Minimum point is (3, –4).

x –1 0 1 5 6

y 12 5 0 0 5

x 0 1 5 6

5

12

–1

(3, –4)

y

The range is – 4 р y р 12.

14. y = – x 2 + 4 x – 5

= –( x 2 – 4 x ) – 5

= –( x 2 – 4 x + 22 – 22) – 5

= – [( x – 2)2 – 4] – 5

= –( x – 2)2 + 4 – 5

= –( x – 2)2 – 1

Maximum point is (2, –1).

x –1 0 3

y –10 –5 –2

x 0 3

(2, –1) –1

–5

–10

y

The range is –10 р y р –1.

15. y = Ȋ9 – ( x – 3)2ȊMaximum point is (3, 9).

x –1 0 6 7

y 7 0 0 7

x

y

0 –1 6

(3, 9)

7

7

The range is 0 р y р 9.

16. 3 x 2 Ͻ x 3 x 2 – x Ͻ 0

x (3 x – 1) Ͻ 0

–

13

0x

f (x )

The range is 0 Ͻ x Ͻ 1

— 3

.

17. 3 x – x 2

——— 2 Ͻ 13 x – x 2 Ͻ 2

– x 2 + 3 x – 2 Ͻ 0

x 2 – 3 x + 2 Ͼ 0

( x – 1)( x – 2) Ͼ 0

20x

f (x )

1

The range is x Ͻ 1 or x Ͼ 2.



8



18. Given that x – 2 y = 1,

∴ x = 1 + 2 y .......................................ᕃ

Substitute ᕃ into y + 3 у 2 xy,

y + 3 у 2(1 + 2 y) y y + 3 у 2 y + 4 y2

0у 4 y2 + y – 3

0у (4 y – 3)( y + 1)

3 –

4

0y

f (y )

–1

The range is –1 р y р 3

— 4

.

19. f ( x ) Ͻ 0

5 x 2 – 4 x – 1 Ͻ 0

(5 x + 1)( x – 1) Ͻ 0

1 – –

5

0x

f (x )

1

The range is – 1

— 5

Ͻ x Ͻ 1.

20. g( x ) Ͼ 04 x 2 – 9 Ͼ 0

(2 x + 3)(2 x – 3) Ͼ 0

3 – –

23 –

2

0x

g (x )

The range is x Ͻ – 3

— 2

or x Ͼ 3

— 2

.

21. (a) Since y = 3 x 2 – 9 x + t Ͼ 0 for all values of x and

it does not have root when y = 0.

Then, b2 – 4ac Ͻ 0 for 3 x 2 – 9 x + t = 0

(–9)2 – 4(3)(t )Ͻ 0

81 – 12t Ͻ 0

–12t Ͻ –81

t Ͼ –81 —— –12

t Ͼ 27

—– 4

(b) Let f ( x ) = a( x – b)2 + c

f ( x ) = a( x – 2)2 + 0

f ( x ) = a( x – 2)2

Substitute x = 0, f ( x ) = –3 into the equation,–3 = a(0 – 2)2

= 4a

a = – 3

— 4

Hence, the quadratic function is

f ( x ) = – 3

— 4

( x – 2)2.

22. (a) Given 2 x 2 – 3 y + 2 = 0

3 y = 2 x 2 + 2

y =2 x 2

—– 3

+2

— 3

................ᕃ

Substitute ᕃ into y Ͻ 10,

2 x 2

—– 3

+2

— 3

Ͻ 10

2 x 2 + 2 Ͻ 30

2 x 2 – 28 Ͻ 0

x 2 – 14 Ͻ 0

( x + ͱ⒓⒓14)( x – ͱ⒓⒓14 ) Ͻ 0

ͱ⒓⒓14 –ͱ⒓⒓14

0x

f (x )

The range is – ͱ⒓⒓14 Ͻ x Ͻ ͱ⒓⒓14 .

(b) 2 x 2 – 8 x – 10 = 2( x 2 – 4 x ) – 10

= 2( x 2 – 4 x + 22 – 22) – 10

= 2[( x – 2)2 – 4] – 10

= 2( x – 2)2 – 8 – 10

= 2( x – 2)2 – 18

Therefore, a = 2, b = –2 and c = –18

Hence, the minimum value of 2 x 2 – 8 x – 10 is

–18.

23. 5 – 2 x р 0 3 x 2 – 4 x Ͼ –1

3 x 2 – 4 x + 1 Ͼ 0

(3 x – 1)( x – 1) Ͼ 0

11 –

3

0x

f (x )

x Ͻ 1

— 3

, x Ͼ 1

–2 x р –5

x у –5 —– –2

x у 5

— 2



9



5 –

21 –

31

x у5—2

x Ͻ x Ͼ 11—3

The range is x у 5

— 2

.

24. 5 Ͻ f ( x ) Ͻ 9

5 Ͻ 5 – 3 x + x 2 Ͻ 9

5 Ͻ 5 – 3 x + x 2 , 5 – 3 x + x 2 Ͻ 9

5 – 3 x + x 2 – 9 Ͻ 0

x 2 – 3 x – 4 Ͻ 0

( x – 4)( x + 1) Ͻ 0

4 –1 0x

f (x )

–1 Ͻ x Ͻ 4

0 Ͻ x 2 – 3 x

0 Ͻ x ( x – 3)

30x

f (x )

x Ͻ 0, x Ͼ 3

0 –1 43

x < 0 x > 3

–1 < x < 4

The range is –1 Ͻ x Ͻ 0 or 3 Ͻ x Ͻ 4.

25. 1 у x 2 + 3 x – 3 Ͼ –3

x 2 + 3 x – 3 Ͼ –3

x 2 + 3 x Ͼ 0

x ( x + 3) Ͼ 0

–3 0x

f (x )

x Ͻ

–3, x Ͼ

0

1 у x 2 + 3 x – 3 ,

0 у x 2 + 3 x – 4

0 у ( x + 4)( x – 1)

1 –4 0x

f (x )

– 4р

x р

1

–4

x < –3 x > 0

–4 ഛ x ഛ 1

–3 0 1

The range is – 4 р x Ͻ –3 or 0 Ͻ x р 1.

26. Ȋ x 2 + 5 x – 10Ȋ Ͼ 4

x 2 + 5 x – 10 Ͻ – 4 , x 2 + 5 x – 10 Ͼ 4

x 2 + 5 x – 14 Ͼ 0

( x + 7)( x – 2) Ͼ 0

–7 20

x

f (x )

x Ͻ –7, x Ͼ 2

x 2 + 5 x – 6 Ͻ 0

( x + 6)( x – 1) Ͻ 0

–6 10

x

f (x )

–6 Ͻ x Ͻ 1

The ranges are –6 < x < 1, x < –7, x > 2.

27. f ( x ) = (r + 1) x 2 + 2rx + r – 3

Given that f ( x ) does not intersect the x -axis,

therefore b2

– 4ac Ͻ

0(2r )2 – 4(r + 1)(r – 3) Ͻ 0

4r 2 – 4(r 2 – 2r – 3) Ͻ 0

4r 2 – 4r 2 + 8r + 12 Ͻ 0

2r + 3 Ͻ 0

r Ͻ – 3

— 2

28. Given that f ( x ) = 9 – 6 x + 2 x 2 does not have real root

when f ( x ) = k ,

∴ 9 – 6 x + 2 x 2 = k

2 x 2 – 6 x + 9 – k = 0

Use b2 – 4ac Ͻ 0,

(– 6)2 – 4(2)(9 – k ) Ͻ 0

36 – 72 + 8k Ͻ 0–36 + 8k Ͻ 0

8k Ͻ 36

k Ͻ 36

—– 8

k Ͻ 9

— 2

29. Ȋ2 x 2 + 10 x – 20Ȋ р 8

–8 р 2 x 2 + 10 x – 20 р 8

–8р 2 x 2 + 10 x – 20 , 2 x 2 + 10 x – 20 р 8

2 x 2 + 10 x – 28 р 0

x 2 + 5 x – 14 р 0

( x + 7)( x – 2) р 0

–7 20

x

f (x )

–7 р x р 2

0р 2 x 2 + 10 x – 12

0р x 2 + 5 x – 6

0р ( x + 6)( x – 1)

–6 10

x

f (x )

x р –6, x у 1



10



x р –6 x у 1

–7 ഛ x ഛ 2

–6 –7 1 2

The range is –7 р x р –6 or 1 р x р 2.

30. y = x 2 + 5 x – 6

= x 2 + 5 x + 5 — 2

2

– 5 — 2

2

– 6

= x + 5 — 2

2

– 25

—– 4

– 6

= x + 5 — 2

2

– 49

—– 4

The minimum point is (– 5

—

2

, – 49

—–

4

).

x –13 – 6 0 1 3

y 98 0 – 6 0 18

x

0

–6 –13 –6

18

98

1 3

) ) – – , – –

52

494

y

The range is – 49

—– 4

Ͻ y Ͻ 98.

31. y = – x 2 + 2 x – nx + 16

= – x 2 + (2 – n) x + 16

= –[ x 2 – (2 – n) x ] + 16

= – ́ x 2 – (2 – n) x + 2 – n ——–

2

2

– 2 – n ——–

2

2

΅ + 16

= – ́ x – 2 – n

——–

2

2

– 2 – n ——–

2

2

΅ + 16

= – x – 2 – n

——– 2

2

+ 2 – n ——–

2

2

+ 16

Since y = x 2 – 3k and y = – x 2 + 2 x – nx + 16 have the

same axis of symmetry that is x = 0.

then – 2 – n

——– 2

= 0

2 – n = 0

n = 2

The equation y = – x 2 + 2 x – nx + 16

= – x 2 + 2 x – 2 x + 16

= –

x –

2 – 2 ——–

2

2

+

2 – 2

——– 2

2

+ 16

= – x 2 + 16

When y = 0,

– x 2 + 16 = 0

x 2 = 16

x = ±ͱ⒓⒓16

= ±4

Therefore, B = (4, 0)

Substitute x = 4, y = 0 into y = x 2 – 3k ,

0 = 42 – 3k

0 = 16 – 3k

k =16

—– 3

32. (a) y = –2[(3k – x )2 + n] – 10 has a maximum point

(4, 11).

y = –2(3k – x )2 – 2n – 10

∴ 3k – 4 = 0 and –2n – 10 = 11

k =4

— 3

–2n = 21

n = – 21

—– 2

(b) Substitute k =4

— 3

and n = – 21

—– 2

into

y = –2[(3k – x )2 + n] – 10,

y = –2΄(4 – x )2 – 21

—– 2

΅ – 10

= –2(4 – x )2

+ 21 – 10= –2(4 – x )2 + 11

When y = 0,

–2(4 – x )2 + 11 = 0

2(4 – x )2 = 11

(4 – x )2 =11

—– 2

4 – x = ±ͱ⒓⒓⒓11 —–

2

x = 4 ± ͱ⒓⒓⒓11 —–

2

= 4 – ͱ⒓⒓⒓11 —–

2, 4 + ͱ⒓⒓⒓11

—– 2

= 1.655, 6.345

(c) y = –2(4 – x )2 + 11

The maximum point is (4, 11).

x –1 0 5

y –39 –21 9



11



(–1, –39)

–21

(5, 9)

(4, 11)

0x

y

The range is –39 р y р 11.

33. (a) 1 + x ——–

2= 3

x = 6 – 1

= 5

Therefore, B(5, 0)

(b) 3 – p = 0

p = 3

q – 1 = –6

q = –5

(c) The turning point is (3, –6).

34. (a) f ( x ) = –2 x 2 + 12 x + p

= –2( x 2 – 6 x ) + p

= –2( x 2 – 6 x + 32 – 32) + p

= –2[( x – 3)2 – 9] + p

= –2( x – 3)2 + 18 + p

Given f ( x ) = a( x + b)2 + 18By comparison,

a = –2, b = –3, p = 0

(b) The turning point is (3, 18).

35. (a) y = 4 x 2 – 2kx + 6k has a minimum value of 20

y = 4 x 2 – 2k

—– 4 x + 6k

= 4 ́x 2 – k

— 2 x + k —

4

2

– k — 4

2

΅ + 6k

= 4΄ x – k

— 4

2

– k 2

—– 16

΅ + 6k

= 4 x – k — 4

2

– k 2 — 4

+ 6k

Given minimum value of y = 20.

Therefore, – k 2

— 4

+ 6k = 20

– k 2 + 24k = 80

k 2 – 24k + 80 = 0

(k – 20)(k – 4) = 0

k = 4 or 20

(b) Substitute k = 20 into y = 4 x 2 – 2kx + 6k ,

y = 4 x 2 – 40 x + 120

The equation of the curve after reflection about

the y-axis is y = 4 x 2 + 40 x + 120.

(c) Substitute k = 4 into y = 4 x 2 – 2kx + 6k ,

y = 4 x 2 – 8 x + 24

= 4( x 2 – 2 x ) + 24

= 4[ x 2 – 2 x + (1)2 – (1)2] + 24

= 4[( x – 1)2 – 4]

= 4[( x – 1)2 – 1] + 24

= 4( x – 1)2 – 4 + 24

= 4( x – 1)2 + 20

Hence, the axis of symmetry is x = 1.

36. f ( x ) = 2 x 2 + kx – 4r

= 2 x 2 + k —

2 x – 4r

= 2 ́x 2 +k

— 2 x + k —

4

2

– k — 4

2

΅ – 4r

= 2΄ x + k — 4

2

– k 2

–— 16

΅ – 4r

= 2 x + k — 4

2

– k 2

–— 8

– 4r

x +k

— 4

= 0 when x = k 2 – 3

— 4

,

Therefore, k 2 – 3

— 4

+k

— 4

= 0

4k 2 – 3 + k = 0

4k 2 + k – 3 = 0

(4k – 3)(k + 1) = 0

k = –1 or 3

— 4

– k 2

–— 8

– 4r = k

When k = –1, – 1

— 8

– 4r = –1

4r = 1 – 1

— 8

=7

— 8

r =7

—– 32

When k =3

—

4

, –

9 —– 16

——

8

– 4r =3

—

4

– 9

—–– 128

– 4r =3

— 4

4r = – 9

—–– 128

– 3

— 4

= – 9 + 96 ———–

128

= – 105

—–– 128

r = – 105

—–– 512



12



1. y = 3(2 x – 1)( x + 1) – x (4 x – 5) + 2

= 3(2 x 2 + x – 1) – 4 x 2 + 5 x + 2= 6 x 2 + 3 x – 3 – 4 x 2 + 5 x + 2

= 2 x 2 + 8 x – 1

= 2( x 2 + 4 x ) – 1

= 2( x 2 + 4 x + 22 – 22) – 1

= 2[( x + 2)2 – 4] – 1

= 2( x + 2)2 – 8 – 1

= 2( x + 2)2 – 9

Since a = 2 Ͼ 0, therefore the minimum value of

y is –9.

When y = 0, 2( x + 2)2 – 9 = 0

( x + 2)2 = 9 — 2

x + 2 = ±ͱ⒓⒓9 — 2

x = ±ͱ⒓⒓9 — 2

– 2

= ͱ⒓⒓9 — 2

– 2 or – ͱ⒓⒓9 — 2

– 2

= 0.1213 or – 4.121

When x = 0, y = 2(2)2 – 9

= –1

The minimum point is (–2, –9).

–1 –4.121

(–2, –9)

0.12130

y

x

2. 5 Ͻ Area of rectangle ABCD Ͻ 21

5 Ͻ ( x + 3)( x – 1) Ͻ 21

5 Ͻ ( x + 3)( x – 1)

5 Ͻ x 2 + 2 x – 3

0 Ͻ x 2 + 2 x – 8

0 Ͻ ( x + 4)( x – 2)

0 –4 2

f (x )

x

x Ͻ – 4, x > 2

( x + 3)( x – 1) Ͻ 21

x 2 + 2 x – 3 Ͻ 21

x 2 + 2 x – 24 Ͻ 0

( x – 4)( x + 6) Ͻ 0

0 –6 4

f (x )

x

– 6 Ͻ x Ͻ 4

–4

–6 < x < 4

x < –4 x > 2

2x

–6 4

The range is – 6 Ͻ x Ͻ – 4 or 2 Ͻ x Ͻ 4.

3. (a) p =1 + 5

——– 2

= 3

(b) y = ( x – 1)( x – 5)

= x 2 – 6 x + 5

4. y = a( x – 2)2 + 1

Substitute x = 0, y = 9 into the equation,

9 = a(–2)2 + 1

8 = 4a

a = 2

Therefore, the quadratic function is

f ( x ) = 2( x – 2)2 + 1.

5. x 2 + (1 + k ) x – k 2 + 1 = 0

For quadratic equation to have real roots,

b2 – 4ac у 0

(1 + k )2 – 4(1)(1 – k 2) у 0

1 + 2k + k 2 – 4 + 4k 2 у 0

5k 2 + 2k – 3 у 0

(5k – 3)(k + 1) у 0

0 –1 3 –

5

f (k )

k

The range of values of k is k р –1 or k у 3

— 5

.



13



6. y = x 2 + 7 x – 8 – 2k

For y to be positive for all real values of x , there is

no roots for y = 0.

Therefore, b2 – 4ac Ͻ 0

72 – 4(1)(–8 – 2k ) Ͻ 0

49 + 32 + 8k Ͻ 0

8k Ͻ –81

k Ͻ – 81

—– 8

Alternative

y = x 2 + 7 x – 8 – 2k

= x 2 + 7 x + 7 — 2

2

– 7 — 2

2

– 8 – 2k

= x + 7 — 2

2

– 49

—– 4

– 8 – 2k

For y to be positive for all real values of x , –

49 —–

4– 8 – 2k Ͼ 0

–2k Ͼ 49

—– 4

+ 8

–2k Ͼ 81

—– 4

k Ͻ – 81

—– 8

7. Substitute x = 6, y = 0 into y = px 2 + qx ,

0 = p(6)2 + q(6)

0 = 36 p + 6q

q + 6 p = 0 .........................ᕃ

y = px 2 + qx

= p x 2 +q — p x

= p ́x 2 +q — p x +

q —– 2 p

2

– q

—– 2 p

2

΅

= p΄ x +q

—– 2 p

2

– q2

—–– 4 p2 ΅

= p x +q

—– 2 p

2

– q2

—– 4 p

– q2

—–

4 p

= –12

q2 = 48 p

p =q2

—– 48

.......................................................... ᕄ

Substitute ᕄ into ᕃ,

q + 6 q2

—– 48

= 0

q + q2

— 8

= 0

8q + q2 = 0

q(8 + q) = 0

q = 0 or q = –8

When q = 0, p =02

—– 48

= 0

When q = –8, p =(–8)2

—––– 48

=64

—– 48

=4

— 3

Therefore, the values of p =4

— 3

and q = –8.

8. (2 – 3k ) x 2 + x +3

— 4k = 0

b2 – 4ac = 12 – 4(2 – 3k ) 3 — 4k

= 1 – 6k + 9k 2

= 9k 2 – 6k + 1

= (3k – 1)2

Since (3k – 1)2 у 0 for all values of k ,

therefore, (2 – 3k ) x 2 + x +3

— 4k = 0 has real roots for

all values of k .

9. f ( x ) = 3( x 2 + 2mx + m2 + n)

= 3[( x + m)2 + n]

= 3( x + m)2 + 3n

The minimum point is (– m, 3n)

Compare to A(t , 3t 2),

∴ – m = t and 3n = 3t 2

m = – t n = t 2

10. (a) y = px 2 + 8 x + 10 – p

When the graph does not intercept the x -axis,

there are no roots for px

2

+ 8 x + 10 – p = 0.Therefore, b2 – 4ac Ͻ 0

82 – 4 p(10 – p) Ͻ 0

64 – 40 p + 4 p2 Ͻ 0

p2 – 10 p + 16 Ͻ 0

( p – 2)( p – 8) Ͻ 0

2 8

Hence, r = 2 and t = 8



14



(b) When p = 2,

y = 2 x 2 + 8 x + 8

= 2( x 2 + 4 x ) + 8

= 2( x 2

+ 4 x + 22

– 22

) + 8= 2[( x + 2)2 – 4] + 8

= 2( x + 2)2 – 8 + 8

= 2( x + 2)2

Therefore, the minimum point is (–2, 0).

When x = 0, y = 8

When y = 0, 2( x + 2)2 = 0

x = –2

When p = 8,

y = 8 x 2 + 8 x + 2

= 8( x 2 + x ) + 2

= 8 ́x 2 + x + 1 — 2

2

– 1 — 2

2

΅ + 2

= 8΄ x + 1 — 2

2

– 1

— 4

΅ + 2

= 8 x + 1 — 2

2

– 2 + 2

= 8 x + 1 — 2

2

Therefore, the minimum point is (– 1

— 2

, 0).

When x = 0, y = 2

When y = 0, 0 = 8 x + 1 — 2

2

x = – 1

—

2

0

2

8

1 – –

2 –2

y

p = 2 p = 8

x

11. (a) f ( x ) = 24 x – 4 x 2 + r

= – 4 x 2 + 24 x + r

= – 4( x 2 – 6 x ) + r

= – 4( x 2 – 6 x + 32 – 32) + r = – 4[( x – 3)2 – 9] + r

= – 4( x – 3)2 + 36 + r

Compare to f ( x ) = p( x – q)2 + 16

Therefore, p = – 4, q = 3 and 36 + r = 16

r = –20

(b) The turning point is (3, 16).

(c) f ( x ) = 24 x – 4 x 2 – 20

When x = 0, f ( x ) = –20

When f ( x ) = 0, – 4( x – 3)2 + 16 = 0

4( x – 3)2 = 16( x – 3)2 = 4

x – 3 = ±2

x = ±2 + 3

= –2 + 3 or 2 + 3

= 1 or 5

0 1

(3, 16)

5

–20

y

x

12. (a) y = – | p( x – 3)2 + q|

Substitute x = 3, y = –5 into the equation,

–5 = – | p(3 – 3)2 + q|5 = |q|

q = ±5

Substitute x = 4, y = 0 into the equation,

0 = – | p(4 – 3)2 ± 5| p ± 5 = 0

p = 5

Therefore, p = 5, q = –5 or p = –5, q = 5.

(b) When x = 3, y = –5

For p = 5, q = –5,

When x = 6, y = – |5(6 – 3)2 – 5|= – |40|= –40

Based on the graph, the range of values of y is

– 40 р y р 0.

For p = –5, q = 5,

When x = 6, y = – | –5(6 – 3)2 + 5|= – | – 40|= – 40

Therefore, the range of values of y is

– 40 р y р 0.

13. (a) y = –2( x – 3)2 + 2k

= – x 2 + 2 x + px – 8

= – x 2 + (2 + p) x – 8

= –[ x 2 – (2 + p) x ] – 8

= – ́ x 2 – (2 + p) x + 2 + p ——–

2

2

– 2 + p ——–

2

2

΅ – 8

= – ́ x – 2 + p

——– 2

2

– 2 + p ——–

2

2

΅ – 8

= – x – 2 + p

——– 2

2

+(2 + p)2

———– 4

– 8



15



Since the x -coordinate of the maximum point for

both the graphs are same,

therefore,2 + p

——– 2

= 3

p = 4

y = – x 2 + 2 x + px – 8 becomes

y = – x 2 + 2 x + 4 x – 8

y = – x 2 + 6 x – 8

When y = 0,

– x 2 + 6 x – 8 = 0

x 2 – 6 x + 8 = 0

( x – 2)( x – 4) = 0

x = 2 or 4

Hence, A(2, 0) and B(4, 0).

Substitute x = 2, y = 0 into y = –2( x – 3)2 + 2k ,

0 = –2(2 – 3)2 + 2k 2k = 2

k = 1

Hence, k = 1 and p = 4.

(b) For y = –2( x – 3)2 + 2k

= –2( x – 3)2 + 2(1)

= –2( x – 3)2 + 2

Maximum value of the curve is 2.

For y = – x 2 + 2 x + px – 8

= – x 2 + 2 x + 4 x – 8

= – x 2 + 6 x – 8

When x = 3,

y = –9 + 18 – 8

= 1

Maximum value of the curve is 1.

14. Since 3 x 2 у 0 for all values of x ,

therefore3 x 2

——————– (2 x – 1)( x + 4)

р 0

(2 x – 1)( x + 4) р 0

0 –4 1 –

2

f (x )

x

Hence, – 4 р x р 1

— 2

.

15. Since x 2 + 1 Ͼ 0,

therefore x 2 + 3 x + 2 —————

x 2 + 1 Ͼ 0

x 2 + 3 x + 2 Ͼ 0( x + 1)( x + 2) Ͼ 0

0 –2 –1

f (x )

x

Hence, x Ͻ –2, x > –1.

16. – 4

——— 1 – 3 x

р x

0 р x +4

——— 1 – 3 x

0 р

x (1 – 3 x ) + 4

—————— 1 – 3 x

0 р x – 3 x 2 + 4

————— 1 – 3 x

0 р –3 x 2 + x + 4 —————–

1 – 3 x

0 р (–3 x + 4)( x + 1)

——————— 1 – 3 x

For –3 x + 4 у 0,

4 у 3 x

x р 4

— 3

For x + 1 у 0,

x у –1

For 1 – 3 x Ͼ 0,

–3 x Ͼ –1

x Ͻ 1

— 3

–1 – + – +

1 4 – –

3

4x р –3

1x < –

3

x у –1

3

x

Therefore, the range is –1 р x Ͻ 1

— 3

, x у 4

— 3

.



16



17. y2 – 9 = x

x = y2 – 9

When x = 0,

y2 = 9 y = ±3

When y = 0,

x = –9

When x = 7,

7 = y2 – 9

y2 = 16

y = ±4

x –9 0 7

y 0 ±3 ±4

0 –9 7

–4 –3

34

y

x

The range of values of y is – 4 р y р 4.

3.quadratic function

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