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Activity Sheet Lesson 1Introduction to Relations

________________________________________________________________Name:

Class: Date:

1. Given that set P = {1, 2, 3, 4} and set Q = {2, 4, 6, 8} and the relation is Q istwo times P. Represent the relation using;a) An arrow diagramb) Ordered pairsc) A graph

SOLUTION:

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________________________________________________________________

2. The following arrow diagram shows a relation.

q

rn

pm

F G

i) Determine the domain, codomain and the range of relation.

ii) State the object of q.iii) State the image of m.

SOLUTION:

3. The following set of ordered pairs shows a relation.{(1, 1), (2, 4), (3, 9), (4, 16)}

i) Determine the domain, codomain and range of this relation.ii) State the object of 9.iii) State the image of 4.

SOLUTION:

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________________________________________________________________

4. The following graph shows a relation.

32SetA

1

12

9

6

3

0

SetB

i) Determine the domain, codomain and range of this relation.ii) State the object of 12.iii) State the images of 1.

SOLUTION:

5. State the type of relation for the following arrow diagram.

SOLUTION:

3

6

9

b

c

a

A B

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________________________________________________________________

6. State the type of relation for the following ordered pairs.{(2, 3), (2, 4), (2, 5)}

SOLUTION:

7. State the type of relation for the following graph.

SOLUTION:

18

17

16

4321

15

0

8. The relation between setXand set Yis Yis the square root ofX. Draw anarrow diagram using any suitable objects and images. State the relation forthe two sets.

SOLUTION:

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________________________________________________________________

9. State the type of relation for the following cases. Draw the arrow diagram withany suitable objects and images.

P = {Age of students}Q = {Students in a class}

i) Set P is the domain and set Q is the range.ii) Set Q is the domain and set P is the range.

SOLUTION:

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________________________________________________________________

1. Given that set P = {1, 2, 3, 4} and set Q = {2, 4, 6, 8} and the relation is Q istwo times P. Represent the relation using;a) An arrow diagram

b) Ordered pairsc) A graph

SOLUTION:

a)

2

4

684

3

2

1

P Q

b) {(1, 2), (2, 4), (3, 6), (4, 8)}

c)

42 3

Set P

1

2

4

6

8

Set Q

0

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________________________________________________________________2. The following arrow diagram shows a relation.

q

rn

pm

F G

i) Determine the domain, codomain and the range of relation.ii) State the object of q.

iii) State the image of m.

SOLUTION:i) Domain = {m, n}

Codomain = {p, q, r}Range = {p, q, r}

ii) Object of qis n.iii) Image of mis p.

3. The following set of ordered pairs shows a relation.

{(1, 1), (2, 4), (3, 9), (4, 16)}

i) Determine the domain, codomain and range of this relation.ii) State the object of 9.iii) State the image of 4.

SOLUTION:i) Domain = {1, 2, 3, 4}

Codomain = {1, 4, 9, 16}Range = {1, 4, 9, 16}

ii) Object of 9 is 3.

iii) Image of 4 is 16.

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________________________________________________________________4. The following graph shows a relation.

32SetA

1

12

9

6

3

0

SetB

i) Determine the domain, codomain and range of this relation.

ii) State the object of 12.iii) State the images of 1.

SOLUTION:i) Domain = {1, 2, 3}

Codomain = {3, 6, 9, 12}Range = {3, 6, 9, 12}

ii) Object of 12 is 3.iii) Images of 1 are 3 and 6.

5. State the type of relation for the following arrow diagram.

SOLUTION:one-to-one.

3

6

9

b

c

a

A B

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________________________________________________________________6. State the type of relation for the following ordered pairs.

{(2, 3), (2, 4), (2, 5)}

SOLUTION:

one-to-many.

7. State the type of relation for the following graph.

SOLUTION:many-to-one.

18

17

16

4321

15

0

8. The relation between setXand set Yis Yis the square root ofX. Draw anarrow diagram using any suitable objects and images. State the relation for

the two sets.

SOLUTION:

The type of relation is one-to-many.

2

2

3

3

9

4

X Y

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________________________________________________________________9. State the type of relation for the following cases. Draw the arrow diagram with

any suitable objects and images.P = {Age of students}Q = {Students in a class}

i) Set P is the domain and set Q is the range.ii) Set Q is the domain and set P is the range.

SOLUTION:i) Set P is the domain and set Q is the range.

Ram

Chong

Diana15

John14

P Q

The type of relation is one-to-many.

ii) Set Q is the domain and set P is the range.

15

Chong

Diana

Ram

14John

P Q

The type of relation is many -to-one.

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Activity Sheet Lesson 2Concept of a Function

________________________________________________________________Name:

Class: Date:

1. Determine whether each of the following relations are functions. Give areason for your answer.

(a)

______________________________________________________

(b)

_____________________________________________________


(a) Ordered pairs of functionf= {(2,3), (1,4), (0,4), (1,5)}.

________________________________________________

(b) Ordered pairs of functionf= {(a,a), (b,b), (c,c)}.

________________________________________________

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________________________________________________________________

3. Determine whether each of the following relations are functions. Give areason for your answer.(e)

_____________________________________________________

(f)

__________________________________________________

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________________________________________________________________

4.

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________________________________________________________________1. Determine whether each of the following relations are functions. Give a

reason for your answer.

(a)

It is not a function because object b has two images, m and o.

(b)

It is a function because every object has only one image.


(a) Ordered pairs of functionf= {(2,3), (1,4), (0,4), (1,5)}.It is a function because every object has only one image.

(b) Ordered pairs of functionf= {(a,a), (b,b), (c,c)}.It is a function because every object has only one image.


(a)

It is not a function because object c has no image.

(b)

It is a function because every object has only one image.

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________________________________________________________________

4. Complete the following function notations based on the respective

representations.

(a)

(i) d (3) = 9 (ii) d( 2 ) = 4

(iii) d( 2 ) = 4 (iv) d(x) = x2

(b) Ordered pairs of function g = {(4,2),(3,1.5),(2,1)}

(i) g (3) = 1.5 (ii) g ( 4 ) = 2

(iii) g (x) =2

x

(c) (i) (i) f(8) = 2

(ii) (ii) f( 0 )=0

(iii) (iii) f(x) = 3 x

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Activity Sheet Lesson 3Range of function

________________________________________________________________Name:

Class: Date:

1. Determine the domain, objects, images and range for each of the followingfunctions.

(a) h:x 2cosx

x 2cosx

90 -1

120 0

240 1

270

(b) Ordered pairs of functions

f : x -2

x = {( -1,21 ), (2, -1), (4, -2)}

(c) p(x) = -x

-2 -1 0 1 2-1

2

- 3

- 4

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________________________________________________________________SOLUTION:

2. Sketch the graph of function d(x) = 2x 4in the domain -3 0x .

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________________________________________________________________Then determine the range of the function.

SOLUTION:

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________________________________________________________________1. Determine the domain, objects, images and range for each of the following

functions.

(a) h:x 2cosx

x 2cosx

90 -1

120 0

240 1

270

(b) Ordered pairs of functions

f : x -2

x = {( -1,21 ), (2, -1), (4, -2)}

(c) p(x) = -x

p(x)

-2 -1 0 1 2-1

2

- 3

- 4

x

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________________________________________________________________SOLUTION:a) Domain = {90, 120, 240, 270}.

Objects are 90, 120, 240 and 270Images are 1 and 0

Range = {1, 0}

b) Domain = {1, 2, 4}.Objects are 1, 2 and 4.

Images are 2, 1 and2

1

Range = {2, 1 and2

1}

c) Domain = {2, 1, 0, 1, 2}.Objects are 2, 1, 0, 1 and 2Images are 4, 1 and 0Range = {4, 1 and 0}

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________________________________________________________________2. Sketch the graph of function d(x) = 2x 4in the domain -3 0x .

Then determine the range of the function.

SOLUTION:

x 3 2 1 0d(x) 10 8 6 4

d x

10

8

6

4

0123x

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Activity Sheet Lesson 4Determine the image of a function given the object and vice

versa________________________________________________________________Name:

Class: Date:

1. Given h (x) =x

12 2 forx 0. Find the images whenx = 1, 2 and 4

SOLUTION:

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versa________________________________________________________________2. Givenf(x) = 4 x. Find the object for the images 3 and 0

SOLUTION :

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versa________________________________________________________________3. Given g (x) = 2 cosx for 0 x 90. Find the object for the image 2

SOLUTION:

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versa________________________________________________________________4. Givenf(x) = 2x + 3. Find

a) the images of

i) x = 1ii) x = 3

b) the objects of 5

SOLUTION:

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versa________________________________________________________________

SOLUTION 3

g (x) = 2

2 = 2 cosx

cosx =2

2

cosx = 1x = cos (1)x = 0

4. Givenf(x) = -2x + 3. Find

a) the images of

i) x = 1ii) x = 3

b) the objects of 5

SOLUTION:

(a) i) x =1 ii) x =1

f(1) = 2(1) +3 f(3) = 2(3) +3= 2 + 3 = 6 + 3= 1 = 3

= 3

(b) f(x) = 2x +3 and 2x + 3 = 5 2x +3= 5 2x = 5 3

( 2x + 3) = 5 2x = 22x = 8 x = 1x = 4

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Activity Sheet Lesson 5Composite Functions.

________________________________________________________________

Name:

Class: Date:

1. Givenf(x) = 3x 6 and g(x) = 6x + 1. Find

a) fgb) gfc) ffd) g2

e) fg(0)

f) fg(1)

g) gf(2

1)

h) g2(2)

i) x given gf(x) = 3

j) x given g2(x) = 2

Solution:

1

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________________________________________________________________

2

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________________________________________________________________

2. Givenf(x) =9

42 +xfor2 x 2 and g(x) =x + 2.

a) Find(i)fg(ii)gf

b) Find the range of

(i)fg(ii)gf

Solution:

3

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________________________________________________________________

4

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________________________________________________________________

1. Givenf(x) = 3x 6 and g(x) = 6x + 1. Find

k) fgl) gfm)ffn) g2

o) fg(0)

p) fg(1)

q) gf(2

1)

r) g2(2)

s) x given gf(x) = 3

t) x given g2(x) = 2

Solution:

a) fg(x) =f[g(x)]

=f(6x +1)

= 3(6x + 1) 6

= 18x + 3 6

= 18x 3

fg(x) = 18x 3

b) gf(x) = g[f(x)]

= g(3x 6)

= 6(3x 6) + 1

= 18x 36 + 1

= 18x 35

gf(x) = 18x 35

5

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________________________________________________________________

c) ff(x) =f[f(x)]

=f(3x 6)

= 3(3x 6) 6= 9x 18 6

= 9x 24

ff(x) = 9x 24

d) g2(x) = gg(x)

= g[g(x)]

= g(6x + 1)

= 6(6x+1) +1

= 36x + 6 +1

= 36x + 7

g2(x) = 36x + 7

e) fg(0) = 18x 3

= 18(0) 3

= 3

f) fg(1) = 18(x) 3

= 18(1) 3

= 18 3

= 21

g) gf(2

1) = 18x 35

= 18(2

1) 35

= 9 35

= 26

6

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________________________________________________________________

h) g2(2) = 36x + 7

= 36(2) + 7

= 72 + 7= 65

i) gf(x) = 3

18x 35 = 3

18x = 3 + 35

18x = 38

x =18

38

=9

12

j) g2(x) = 2

36x + 7 = 2

36x = 2 7

36x = 9

x =36

9

=4

1

7

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________________________________________________________________

2. Givenf(x) =9

42 +xfor2 x 2 and g(x) =x + 2.

b) Find(iii)fg

(iv)gfc) Find the range of

(iii)fg

(iv)gf

Solution:

a)(i) fg(x) =f(x + 2)

=9

4)2(2 ++x

=9

442 ++x

fg(x) =9

82 +x

(ii) gf(x) =f(9

42 +x)

=9

42 +x+ 2

=9

1842 ++x

gf(x) =9

222 +x

b)

(i) To find the range offg, substitutex = 2 andx = 2 intofg(x)

fg(2) =9

8)2(2 +

=9

84+

8

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________________________________________________________________

=9

12

=34

fg(2) =9

8)2(2 +

=9

84 +

=9

4

= 9

4

The range offg is9

4fg(x)

3

4

(ii) To find the range ofgf, substitutex = 2 andx = 2 into gf(x)

gf(2) =9

22)2(2 +

=9

82

gf(2) =9

22)2(2 +

= 2

The range ofgfis 2 gf(x) 9

82

9

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Activity Sheet Lesson 6Image of a Composite Function

________________________________________________________________Name:

Class: Date:

1. Given gf:x2x2 + 1 and g :xx +1, determine the function off(x).

Solution:

2. Given gf:x8x 9 andf:xx + 4, determine the function ofg(x).

Solution:

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________________________________________________________________

3. Iffg :x 18

7+x andf:x 1

8

1x , what is the function of g(x)?

Solution:

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________________________________________________________________

1. Given gf:x2x2 + 1 and g :xx +1, determine the function off(x).

Solution:Given gf:x2x2 + 1 g :xx + 1

gf(x) = 2x2 + 1 g(x) =x + 1When we write the composite function in the form ofg, we will get

g[f(x)] = 2x2 + 1When we apply the rule of g

f(x) + 1 = 2x2 + 1We will get,

f(x) = 2x2

2. Given gf:x8x 9 andf:xx + 4, determine the function ofg(x).

Solution:Given gf:x8x 9 g :xx + 4

gf(x) = 8x 9 g(x) =x + 4When we write the composite function in the form ofg, we will get

g[f(x)] = 8x 9When we apply the rule off

g(x + 4) = 8x 9Lety =x + 4, thereforex =y 4

g(y) = 8x 9= 8(y 4) 9= 8y 32 9= 8y 41

So g(y) = 8y 41Replacingy withx,

g(x) = 8x 41

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________________________________________________________________

3. Iffg :x 18

7+x andf:x 1

8

1x , what is the function of g(x)?

Solution:

Given fg :x 18

7+x f:x 1

8

1x

fg(x) = 18

7+x f(x) = 1

8

1x

When we write the composite function in the form off, we will get

f[g(x)] = 18

1x

When we apply the rule off

1)(8

1xg = 1

8

7+x

And make the necessary calculation,

1)(8

1xg = 1

8

7+x

11)(8

1+xg = 11

8

7++x

)(8

18 xg = 8)2

8

7( +x

We will get,

g(x) = 7x + 16

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Activity Sheet Lesson 7Finding the object by Inverse mapping given its image and

function________________________________________________________________

Name:

Class: Date:

1. Use the method of inverse mapping to find the corresponding objects ofimages, for each of the given functions belowa) f:x5x + 2

Images 6 and 4.

b) h :x 2x + 9Images 0 and 5.

SOLUTION:

1

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function________________________________________________________________

2. Given d(x) =

x

x 2wherex 0. Find (i) d1(2) (ii) d1(3).

SOLUTION:

2

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function________________________________________________________________

3. Given g(x) =1

12

+

x

xwherex1. Find (i) g1(1) (ii) g1(5)

SOLUTION:

3

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function________________________________________________________________

1. Use the method of inverse mapping to find the corresponding objects ofimages, for each of the given functions belowa) f:x5x + 2

Images 6 and 4.

b) h :x 2x + 9Images 0 and 5.

SOLUTION:a) Image is 6

5x + 2 = 65x = 6 2 = 8

x = 5

8

x =5

8or

5

31

Therefore the object for6 is5

8or

5

31 .

Image is 45x + 2 = 45x = 4 2 = 2

x =5

2

Therefore the object for 4 is5

2 .

b) Image is 02x + 9 = 02x = 9

x =2

9

x =2

9 or

2

14

Therefore the object for 0 is 2

9

or 2

14

.Image is 52x + 9 = 52x = 5 9 = 4

x =2

4 = 2

Therefore the object for 5 is 2.

4

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function________________________________________________________________

2. Given d(x) =

x

x 2wherex 0. Find (i) d1(2) (ii) d1(3).

SOLUTION:(i) Find d1(2)

Let d1(2) =xd(x) = 2

x

x 2= 2

x 2 = 2xx + 2x = 2

3x = 2

x =3

2

d1(2) =

3

2

(ii) Find d1(3)

Let d1(3) =xd(x) = 3

x

x 2

= 3

x2 = 3x3xx = 2

2x = 2

x =2

2 = 1

d1(3) = 1

5

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function________________________________________________________________

3. Given g(x) =1

12

+

x

xwherex1. Find (i) g1(1) (ii) g1(5)

SOLUTION:(i) Find g1(1)

Let g1(1) =xg(x) = 1

1

12

+

x

x= 1

2x 1 =x + 12xx = 1 + 1

x = 2g1(1) = 2

(ii) Find g1(5)Let g1(5) =x

g(x) = 5

1

12

+

x

x= 5

2x 1 = 5x + 52x 5x = 5 + 1

3x = 6

x =3

6 = 2

g1(5) = 2

6

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Activity Sheet Lesson 8Determining the Inverse Function Using Algebra

________________________________________________________________

Name:

Class: Date:

1. Find the inverse of the following functions.a) f(x) = 3xb) h(x) = 2x 5c) g(x) = 7x + 14

d) p(x) =3

102 +x

e) q(x) =2

23 x

f) f(x) = 43

7

+x wherex 3

4

g) h(x) =x2

5wherex 0

h) f(x) =14

3

xwherex

4

1

SOLUTION:

1

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________________________________________________________________

f(x) =2.5

3 x

f1(x) = 3 ______

OLUTION:

. f(x) = 2x 6

S

3

f1(x) =

_____

__ x __

OLUTION:S

2

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________________________________________________________________

4. f(x) =1

13

x

x

f1

(x) = _____

____ x

SOLUTION:

5. f(x) = 3x 11

f1(x) =

_____

____+x

SOLUTION:

3

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________________________________________________________________

1. Find the inverse of the following functions.i) f(x) = 3x

j) h(x) = 2x 5

k) g(x) = 7x + 14

l) p(x) =3

102 +x

m) q(x) =2

23 x

n) f(x) =43

7

+xwherex

3

4

o) h(x) =x2

5wherex 0

p) f(x) =14

3

xwherex

4

1

SOLUTION:a) letf(x) =y

y = 3x

x =3

y

f1(y) =

3

y

f1(x) =

3

x

b) let h(x) =yy = 2x 52x =y + 5

x =2

5+y

h1(y) =

2

5+y

h1(x) =

2

5+x

4

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________________________________________________________________

c) let g(x) =yy = 7x + 14

7x = 14 y

x =7

14 y

g1(y) =

7

14 y

g1(x) =

7

14 x

d) letp(x) =y

y =

3

102 +x

3y = 2x + 102x = 3y 10

x =2

103 y

p1(y) =

2

103 y

p1(x) =

2

103 x

e) let q(x) =y

y =2

23 x

2y = 3x 23x = 2y 2

x =3

22 y

q1(y) =

3

22 y

q1(x) =

3

22 x

5

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________________________________________________________________

4. f(x) =1

13

x

x

f1

(x) = _____

____ x

SOLUTION:

f(x) =1

13

x

x

y =1

13

x

x

xyy = 3x1xy 3x =y1

x(y 3) =y1

x =3

1

y

y

f1(x) =

3

1

x

x

5. f(x) = 3x 11

f1(x) =

_____

____+x

SOLUTION:

f(x) = 3x 11y = 3x 113x =y + 11

x =3

11+y

f1(x) =

3

11+x

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Activity Sheet Lesson 9Determining the condition of an inverse function

________________________________________________________________Name:

Class: Date:

1. Determine whether the following functions have an inverse. Give a reasonfor your answer.(a)

0

f(x)

x

______________________________________________________

g(x)(b)

0

x

_____________________________________________________

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________________________________________________________________

(c)

a

b

c

u

v

BA

____________________________________________________________

(d)

w

x

y

p

r

q

QP

____________________________________________________________

(e) {(1, 3), (2, 3), (3, 7)}

____________________________________________________________

(f) {(6, 2), (7, 32), (8, 42)}

____________________________________________________________

(g)f:xtx2 + 3

____________________________________________________________

(h)p(x) =5

32 +x

____________________________________________________________

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________________________________________________________________Determine whether the following functions have an inverse. Give a reason for

your answer.(a)

0

f(x)

x

Has an inverse because it is a one-to-one function.

g(x)(b)

0

x

No inverse because it is a many-to-one function.

(c)

a

b

c

u

v

BA


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________________________________________________________________(d)

w

x y

p

r q

QP


(e) {(1, 3), (2, 3), (3, 7)}


(f) {(6, 2), (7, 32), (8, 42)}


(g)f:xtx2 + 3


(h)p(x) = 5 32 +

x


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Activity Sheet Lesson 10Recognising a Quadratic Equation

________________________________________________________________

Name:

Class: Date:

1. Which of the following is the general form of a quadratic equation?A) ax2 + bxy + c = 0, (x is a variable, a, b and c are real numbers and a 0)B) ax

2+ bx + c = 0, (x is a variable, a, b and c are real numbers and a 0)

C) ax3

+ bx2

+ cx + d= 0, (xis a variable, a, b, c and dare real numbers anda 0)

D) ax + b = 0, (xis a variable, a and b are real numbers and a 0)

SOLUTION:

2. Is 7

1x 10x2 = 4 a quadratic equation?

SOLUTION:

3. Is the following expression a quadratic equation?

x2 +

11

7x 18 = 5y

SOLUTION:

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________________________________________________________________

4. Is the following expression a quadratic equation?3x2 x 18 < 0

SOLUTION:

1. Which of the following is the general form of a quadratic equation?

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________________________________________________________________A) ax2 + bxy + c = 0, (x is a variable, a, b and c are real numbers and a 0)B) ax


C) ax3

+ bx2

+ cx + d= 0, (xis a variable, a, b, c and dare real numbers anda 0)

D) ax + b = 0, (xis a variable, a and b are real numbers and a 0)

SOLUTION:B) ax


2. Is 7

1x 10x2 = 4 a quadratic equation?

SOLUTION:First, we rewrite the expression as follows:

10x2 +

7

1x 4 = 0

We know the expression is an equation.It involves only one variable, which isx.Powers of the variable are 1 and 2, which are positive integers and thehighest power is 2.

Therefore,7

1x 10x2 = 4 is a quadratic equation.


x2 +117 x 18 = 5y

SOLUTION:First, we rewrite the expression as follows:

x2 +

11

7x 18 5y = 0

We know the expression is an equation.It involves two variables,x andy.Therefore,x2 +

11

7x 18 = 5y is not a quadratic equation.


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________________________________________________________________3x2 x 18 < 0

SOLUTION:

Note that this is an inequality, therefore 3x2

x 18 < 0 is not a quadraticequation.

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Activity Sheet Lesson 11Determining the roots of a quadratic equation by substituting

________________________________________________________________

Name:

Class: Date:

1. Determine ifx= 2 is the root of this quadratic equation,2x2 + 16x+ 10 = 106.

SOLUTION:

2. Determine ifx

= 5 is the root of this quadratic equation,x

2

10x

+ 35 = 10.

SOLUTION:

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Activity Sheet Lesson 11Determining the roots of a quadratic equation by substituting

________________________________________________________________

1. Determine ifx= 2 is the root of this quadratic equation,2x2 + 16x+ 10 = 106.

SOLUTION:To determine:

1. Substitutex= 2 into the equation.

2. Determine whether the value on the LHS = value on the RHS.

Left-hand side = 2x2 + 16x+ 10= 2(2)2 + 16(2) + 10

= 2(4) + 32 + 10

= 8 + 42= 50

Right-hand side

Since LHS RHS, hence x= 2 is not a root of the equation.

2. Determine ifx= 5 is the root of this quadratic equation,x2 10x+ 35 = 10.SOLUTION:To determine:

1. Substitutex= 5 into the equation.

2. Determine whether the value on the LHS = value on the RHS.

Left-hand side =x2 10x+ 35

= (5)2 10(5) + 35

= 25 50 + 35

= 10

= Right-hand side

Since LHS = RHS, hencex= 5 is a root of the equation.

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Activity Sheet Lesson 12Determining roots of quadratic equations by inspection

________________________________________________________________

Name:

Class: Date:

1. From the list below choose the correct roots of the equation (x + 5)(x 2) = 0.

SOLUTION

a) For ,5=x 2

c) For =x

5=x

,

b) For ,d) For

2

5=x ,

2. Determine the roots of 0)3)(7( =++ xx by inspection.

SOLUTION:For ,7=x

For 3=x ,

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Activity Sheet Lesson 12Determining roots of quadratic equations by inspect

__________________________________________ ion

______________________

2

=x

00

0)7)(0(

0)25)(55(

0)2)(5(

=

=

=+

1. From the list below choose the correct roots of the equation (x + 5)(x 2) = 0.

SOLUTIONA) For ,5

=+ xx

So it is a root.

C) For 2=x ,

012

0)4)(3(

0)22)(52(

0)2)(5(

=

=+

=+ xx

So it is not a root.

B) For ,5=x

030

0)3)(10(0)25)(55(

0)2)(5(

=

=+

=+ xx


D) For 2

5=x ,

04

15

0)2

1)(

2

15(

0)22

5)(5

2

5(

0)2)(5(

=

=+

=+ xx


2. Determine the roots of 0)3)(7( =++ xx by inspection.

SOLUTION:For ,7=x

0)4)(0(

0)37)(77(

0)3)(7(

=

=++

=++ xx

So is a root.7=x

For 3=x ,

0)0)(4(

0)33)(73(

0)3)(7(

=

=++

=++ xx

So 3=x is a root.

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Activity Sheet Lesson 13Determining roots of quadratic equations by trial and

improvement________________________________________________________________

Name:

Class: Date:

1. Find the roots of the following equations.a) x2 +x 6 = 0b) x2 16 = 0c) 6x2 + 42x 60 = 0

Solution:

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Activity Sheet Lesson 13Determining roots of quadratic equations by trial and

improvement________________________________________________________________

1. Find the roots of the following equations.a) x2 +x 6 = 0b) x2

16 = 0

c) 6x2 + 42x 60 = 0

Solution:a) x= 3

x2 +x 6 = (3)2 + (3) 6

= 9 3 6= 9 9= 0

x= 2x2 + x

6 = (2)2 + (2)

6

= 4 + 2 6= 6 6= 0

The roots ofx2 +x 6 are 3 and 2.

b) x= 4x

2 16 = (4)2 16

= 16 16= 0

x= 4x2

16 = (

4)2

16

= 16 16= 0

The roots ofx2 16 are 4 and 4.

c) x= 56x2 + 42x 60 = 6(5)2 + 42(5) 60

= 6(25) + 210 60= 150 + 150= 0

x= 2

6x2 + 42x

60 =

6(2)2 + 42(2)

60

= 6(4) + 84 60= 24 + 24= 0

The roots of6x2 + 42x 60 are 5 and 2.

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Activity Sheet Lesson 14Determining the roots of quadratic equations by factorization

________________________________________________________________Name:

Class: Date:

1. Find the roots of the quadratic equation.x(6x+ 5) = 4?

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Activity Sheet Lesson 14Determining the roots of quadratic equations by factorization

________________________________________________________________

1. Find the roots of the quadratic equation.x(6x+ 5) = 4?

SOLUTION:x(6x+ 5) = 46x2 + 5x = 4

6x2 + 5x 4 = 0(2x 1)(3x+ 4) = 0

2x 1 = 0 or 3x+ 4 = 0

x=21 or x =

34

Hence,x=21andx=

34

are the roots of the equation.

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Activity Sheet Lesson 15Determining roots of quadratic equation by completing the

square________________________________________________________________

Name:

Class: Date:

QUESTION 1(a)x2 2x 1 = 0

SOLUTION:

(b)x2 10x + 5 = 0

SOLUTION:

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square________________________________________________________________

(c)x2 = 3(x + 2)

SOLUTION:

(d) 2p2 + 5p 6 = 0

SOLUTION:

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square________________________________________________________________

(e) 6x2 + 7x = 3

SOLUTION:

(f) 8x = 12x2 5

SOLUTION:

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square________________________________________________________________

QUESTION 1(a)x2 2x 1 = 0

SOLUTION:x

2 2x = 1

x2 2x + (

22 )

2 = 1 + (

22 )

2

(x 1)2 = 1 + 1(x 1)2 = 2

x 1 = 2

x 1 = 1.4142x = 1 + 1.4142 or x = 1 1.4142

= 2.4142 = 0.4142

The solutions are 2.4142 and 0.4142.

(f) x2 10x + 5 = 0

SOLUTION:x

2 10x + 5 = 0x

2 10x =5

x2 10x + (

210

)2 = 5 + (2

10 )2

(x 5)2 = 5 + (5)2

(x 5)2 = 20

x 5 =

20x 5 = 4.4721

x = 5 + 4.4721 or x = 5 4.4721= 9.4721 = 0.5279


(g)x2 = 3(x + 2)

SOLUTION:x

2 = 3(x + 2)x

2 = 3x + 6

x

2

3x 6 = 0x

2 3x = 6

x2 3x + (

23 )2 = 6 + (

23 )2

(x 23 )2 = 6 + 4

9

(x23 )2 =

433

x23

= 433

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square________________________________________________________________

x +127 =

1211

x +127 =

1211

or x +127 =

1211

x =127

+1211

x =127

1211

= 0.3333 = 1218

= 1.5The solutions are 0.3333 and 1.5.

(f) 8x = 12x2 5

SOLUTION:

8x = 12x2 512x

2 8x 5 = 0

x2

128 x

125 = 0

x2

32x

125 = 0

x2

32x =

125

x2

32x+ ( 2

32

)2 =125 + ( 2

32

)2

(x 31 )2 =

3619

x 31 = 36

19

x 31 = 0.7265

x = 0.3333 + 0.7265 or x = 0.3333 0.7265= 1.0598 = 0.3932


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Activity Sheet Lesson 16Determining roots of quadratic equations by using a formula

________________________________________________________________

Name:

Class: Date:

1. Solve the following equation by using the formula.(x 2)(x + 1) = 3x 4

Solution:

2. Ifx = 2 is a root of the quadratic equation below, find the value oftand theother root using the formula.x

2 + tx 6t= 0

Solution:

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________________________________________________________________

1. Solve the following equation by using the formula.(x 2)(x + 1) = 3x 4

Solution:(x 2)(x + 1) = 3x 4

x2x 2 = 3x 4

x2 4x + 2 = 0

a

acbbx

2

42

=

)1(2

)2)(1(4)4()4(2

=x

2

8164 =x

2

84 =x

2

224=x

22=x

x = 2 + 1.414 orx = 2 1.414x = 3.414 or x = 0.586

2. Ifx = 2 is a root of the quadratic equation below, find the value oftandthe other root using the formula.

x2 + tx 6t= 0

Solution:Sincex = 2 is a root, then it satisfies the equation, substitutex = 2 into theequation.

x2 + tx 6t= 0

(2)2 + t(2) 6t= 0

4

2t

6t= 04 8t = 08t= 4

t=2

1

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________________________________________________________________

Therefore,x2 +2

1x 6(

2

1) = 0

2x2 +x 6 = 0

To find the other root,a = 2, b = 1 and c = 6.

a

acbbx

2

42

=

)2(2

)6)(2(4)1()1(2

=x

4

)48(11 =x

4

4811 +

=x

4

491=x

4

71=x

4

71+=x or

4

71=x

4

6=x or

4

8=x

211=x or 2=x

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Activity Sheet Lesson 17Determine roots of a quadratic equation

________________________________________________________________

Name:

Class: Date:

1. If 2 and 31 are the roots of a quadratic equation, what is the quadratic

equation?

SOLUTION:

2. Letp and q be the roots of the quadratic equation 3x2 + 12x 6 = 0. Form aquadratic equation with roots (p + 1) and (q + 1).

SOLUTION:

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________________________________________________________________

1. If 2 and 31 are the roots of a quadratic equation, what is the quadratic

equation?

SOLUTION:1st way:First, we form the factorised form of the quadratic equation by using the roots.

x = 2 andx = 31 are the roots.

(x 2)(x (31 )) = (x 2)(x +

31 )

Then, we multiply the factors.

(x 2)(x +31 ) =x2 +

31x 2x

32

=x2 35x 3

2

Quadratic equation: x2

35x 32 = 03x2 5x 2 = 0

2nd way:x

2 (sum of roots)x + (product of roots) = 0sum of roots =p + q

= 2 31

=35

product of roots =pq

= 2 (31 )

= 32

x2 (sum of roots)x + (product of roots) = 0

x2 ( 3

5 )x + (32 ) = 0

x2

35x

32 = 0

3x2 5x 2 = 0

2. Letp and q be the roots of the quadratic equation 3x2

+ 12x 6 = 0. Form aquadratic equation with roots (p + 1) and (q + 1).

SOLUTION:First, rewrite the quadratic equation in the following format.x

2 (sum of roots)x + (product of roots) = 0

3x2 + 12x 6 = 0 x2 (4)x 2 = 0Then, sum of roots =p + q

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________________________________________________________________

= 4

product of roots =pq

= 2

ets form a quadratic equation with roots (p + 1)(q + 1):

(sum of roots)x + (product of roots) = 0

x [(p + 1)(q + 1)]x + [(p + 1)(q + 1)] = 0

x (p + q + 2)x + (pq +p + q + 1) = 0

ubstitutep + q = 4 andpq = 2 in the above equation.

hen, we get:

x2 (4 + 2)x + (2 + 4 + 1) = 0

x + 2x 5 = 0

L

2x

2

2

ST

2

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Activity Sheet Lesson 18Discriminants of a quadratic equation:

________________________________________________________________

3. Determine the type of roots in this quadratic equation.4x

2 20x + 25 = 0

SOLUTION:

4. Determine the type of roots in this quadratic equation.

32x

2 +65x +

71 = 0

SOLUTION:

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________________________________________________________________


3x2+ 5x + 15 = 0

SOLUTION:

6. Determine the type of roots in this quadratic equation.ax

2 + (2a + 3)x + a + 3 = 0

SOLUTION:

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________________________________________________________________

1. Determine the type of roots in this quadratic equation.x

2 + 3x + 7 = 0

SOLUTION:x

2 + 3x + 7 = 0b

2 4ac = (3)

2 4(1)(7)

= 9 28

= 19

b2

4ac < 0no roots.


2 + 5x + 4 = 0

SOLUTION:

2x2 + 5x + 4 = 0b

2 4ac = (5)

2 4(2)(4)

= 25 + 32

= 57

b2

4ac > 0two different roots.


2 20x + 25 = 0

SOLUTION:

4x2 20x + 25 = 0b

2 4ac = (20)2 4(4)(25)

= 400 400

= 0

b2

4ac = 0two equal roots.


32x

2 +65x +

71 = 0

SOLUTION:

32x

2 +65x +

71 = 0

b2

4ac = (65 )

2 4(

32 )(

71 )

=3625 +

218

=25279

b2



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________________________________________________________________

3x2+ 5x + 15 = 0

SOLUTION:

3x2+ 5x + 15 = 0

b2

4ac = ( 5 )2

4( 3 )( 15 )

= 5 12 5

= 5 26.833

= 21.833

b2

4ac < 0no real roots.

6. Determine the type of roots in this quadratic equation.ax

2 + (2a + 3)x + a + 3 = 0

SOLUTION:ax

2 + (2a + 3)x + a + 3 = 0b

2 4ac = (2a + 3)2 4(a)( a + 3)

= 4a2 + 12a + 9 4a2 12a

= 9

b2


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Activity Sheet Lesson 19Solving problems involving the use of the discriminants

________________________________________________________________

Name:

Class: Date:

1. (a) Given that the quadratic equationx2 4qx +p2 = 0 has two different roots.Show that 4q2 >p2.

SOLUTION:

(b) Given that the equation m2x2 + 3nx + 2 = 0 has no roots. Find the relation

between m and n.

SOLUTION:

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________________________________________________________________

(c) If the equation 2kx2 (4k+ 6)x + 2k= 0 has no roots, what is the range ofvalues of k?

SOLUTION:

(d) If the equation rx2 6x + 2 = 0 has no roots, what is the range for r?

SOLUTION:

(e) If the following equation has two different roots, what is the range for h?x

2 x h = 0

SOLUTION:

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________________________________________________________________

(f) If the following equation has two different roots, what is the range for s?

0322

2= sxx

SOLUTION:

(g) Find the values of tif the following equation has two equal roots.x

2 8x = t2SOLUTION:

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________________________________________________________________

1. (a) Given that the quadratic equationx2 4qx +p2 = 0 has two different roots.Show that 4q2 >p2.

SOLUTION:a = 1, b = 4q, c =p2

b2 4ac > 0

(4q)2 4(1)(p2) > 016q2 4p2 > 0

16q2 > 4p2

4q2 >p2(b) Given that the equation m2x2 + 3nx + 2 = 0 has no roots. Find the relation

between m and n.

SOLUTION:a = m2, b = 3n, c = 2

b2 4ac < 0

(3n)2 4(m2)(2) < 09n2 8m2 < 0

9n2 < 8m2

(c) If the equation 2kx2 (4k+ 6)x + 2k= 0 has no roots, what is the range ofvalues of k?

SOLUTION:

a = 2k, b = (4k+ 6), c = 2kSince this quadratic equation has no roots

b2 4ac < 0

((4k+ 6))2 4(2k)(2k) < 016k2 + 48k+ 36 16k2 < 0

48k+ 36 < 048k< 36

k< 4836

k< 43

(d) If the equation rx2

6x + 2 = 0 has no roots, what is the range for r?

SOLUTION:a = r, b = 6, c = 2Since this quadratic equation has no roots

b2 4ac < 0

(6)2 4(r)(2) < 036 8r < 0

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________________________________________________________________

8r< 368r > 36

r >8

36

r > 29

r > 4 21

(e) If the following equation has two different roots, what is the range for h?x

2 x h = 0

SOLUTION:

x2 x h = 0

a = 1, b = 1, c = hSince this quadratic equation has two different roots

b2 4ac > 0(1)2 4(1)(h) > 0

1 + 4h > 04h > 1

h > 41

(f) If the following equation has two different roots, what is the range for s?

0322

2= s

xx

SOLUTION:

A =21 , b =

21

, c = -3s

Since this quadratic equation has two different rootsb

2 4ac > 0

(21 )2 4(

21 )(3s) > 0

41 + 6s > 0

6s > 41

s > 241

(g) Find the values of tif the following equation has two equal roots.x

2 8x = t2

SOLUTION:

x2 8x + t2 = 0

a = 1, b = 8, c = t2

b2 4ac = 0

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________________________________________________________________

(8)2 4(1)(t2) = 064 4(1)(t2) = 0

64 = 4t2

t2

= 16t = 16

t = 4 or t= 4

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Activity Sheet Lesson 20Recognising Quadratic functions

________________________________________________________________

Name:

Class: Date:

1. (a) Is f(x) = 4x2 + x + x5 a quadratic function?

SOLUTION:

(b) Is the following expression a quadratic function?x2 +

85 x 13 = f(x)

SOLUTION:

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________________________________________________________________

(c) Is the following expression a quadratic function?

7x 21

xf(x) = 0

SOLUTION:

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________________________________________________________________

1. (a) Is f(x) = 4x2 + x + x5 a quadratic function?

SOLUTION:

The function involves only one variable, which is x. The powers of the variable in the terms are 1, 2 and 5 which are

positive integers but the highest power is 5.Therefore, f(x) = 4x2 + x + x5 is not a quadratic function.


x2 + 85x 13 =f(x)

SOLUTION:

First, we rewrite the function as follows:

f(x) = x2 + 85x 13

The function involves only one variable, which is x. The powers of thevariable in the terms are 1 and 2 which are positive integers and thehighest power is 2.

Therefore, x2 + 85x 13 =f(x) is a quadratic function.


7x 21

xf(x) = 0

SOLUTION: First, we rewrite the function as follows:

f(x) = 7x +x2

The function involves only one variable, which is x.

The powers of the variable in the terms are 1 and 2. Note that 2 isnot a positive number.

Therefore, 7x 21

xf(x) = 0 is not a quadratic function.

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________________________________________________________________

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Activity Sheet Lesson 21Quadratic Functions

________________________________________________________________

2. Plot the graph of quadratic function based on the given tabulated values?

x 2 1 0 1 2 3

f(x) 16 6 0 2 0 6

SOLUTION

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________________________________________________________________

3. By using a suitable scale, plot the graph of the functiony =x2 + 2x + 2 for4 x 1.

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________________________________________________________________

5. By using a suitable scale, plot the graph of the functiony =x2 + 2x + 2 for4 x 1.To draw the graph for this function, we should follow the following steps.

i. Construct the table of values fory =x2

+ 2 x + 2.Whenx = 4, y = (4)2 + 2(4) + 2.

= 16 8 + 2= 10

Whenx = 3, y = (3)2 + 2(3) + 2.= 9 6 + 2= 5

Whenx = 2, y = (2)2 + 2(2) + 2.= 4 4 + 2= 2

Whenx = 1, y = (1)2 + 2(1) + 2.

= 1 2 + 2= 1

Whenx = 0, y = (0)2 + 2(0) + 2.= 0 + 2= 2

Whenx = 1,y = (1)2 + 2(1) + 2.= 1 + 2 + 2= 5

The table of values is

x 4 3 2 1 0 1

f(x) 10 5 2 1 2 5

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________________________________________________________________

8

6. By using a suitable scale, plot the graph of the functiony = 2x2 + 8 for

3 x 3.

Solution

To draw the graph for this function, we should follow the following steps.i. Construct the table of values fory = 2x2 + 8.Whenx = 3, y = 2(3)2 + 8

= 18 + 8= 10

Whenx = 2, y = 2(2)2 + 8= 8 + 8= 0

Whenx = 1, y = 2(1)2

+ 8= 2 + 8= 6

Whenx = 0, y = 2(0)2 + 8= 0 + 8= 8

Whenx = 1,y = 2(1)2 + 8= 2 + 8= 6

Whenx = 2,y = 2(2)2 + 8= 8 + 8

= 0Whenx = 3,y = 2(3)2 + 8

= 18 + 8= 10

The table of values is

x 3 2 1 0 1 2 3

f(x) 10 0 6 8 6 0 10

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________________________________________________________________

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Activity Sheet Lesson 22Recognising shapes of graphs of quadratic functions

________________________________________________________________2. If the coefficient ofx in a quadratic functionf(x) = ax+ bx + c is positive, a >

0 the parabola opens .

SOLUTION:

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________________________________________________________________3. If the coefficient ofx in a quadratic functionf(x) = ax+ bx + c is negative, a 0 the parabola opens .

SOLUTION:

If the coefficient ofx in a quadratic functionf(x) = ax+ bx + c is positive, a > 0

the parabola opens upward

f(x)

f(x) = ax

+ bx + ca > 0

x

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________________________________________________________________

3. If the coefficient ofx in a quadratic functionf(x) = ax+ bx + c is negative,a < 0 the parabola opens .

SOLUTION:If the coefficient ofx in a quadratic functionf(x) = ax+ bx + c is negative, a < 0

the parabola opens downward

f(x)

f(x) = ax+ bx + ca < 0

x

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________________________________________________________________

4. If the absolute value of the coefficient ofx in a quadratic functionf(x) = ax+ bx + c increases, a , the parabola becomes .

SOLUTION:If the absolute value of the coefficient ofx in a quadratic function

f(x) = ax+ bx + c increases, a , the parabola becomes narrower

f(x)

f(x) = ax+ bx + cnarrow

a

x

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________________________________________________________________

5. If the absolute value of the coefficient ofx in a quadratic functionf(x) = ax+ bx + c decreases, a , the parabola becomes .

SOLUTION:If the absolute value of the coefficient ofx in a quadratic function

f(x) = ax+ bx + c increases, a , the parabola becomes wider

f(x) = ax+ bx + cwider

a

f(x)

x

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________________________________________________________________

6. Which function does the graph below represent?

f(x)

x

f(x) =x+ bx + c f(x) =x+ bx + c

f(x) = x+ bx + f(x) = bx + cc

SOLUTION:

The graph is a parabola, therefore, it is a quadratic function:f(x) = ax+ bx + c

Since the graph opens downward, therefore a is negative.

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Activity Sheet Lesson 23Relating the position of the graph of a quadratic function

________________________________________________________________Name:

Class: Date:

1. Determine the shape and position of the graphs of the following functions withrespect to thexaxis by considering the value ofa and b 4ac.

(a) f(x) = x+ 5x + 3(b) f(x) = x 4x + 4(c) f(x) = x+ 4(d) f(x) = 3xx(e) f(x) = 1 +2xx

(a)

(b)

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________________________________________________________________(c)

(d)

(e)

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________________________________________________________________QUESTION 1

Determine the shape and position of the graphs of the following functions withrespect to thexaxis by considering the value ofa and b 4ac.

(f) f(x) = x+ 5x + 3(g) f(x) = x 4x + 4(h) f(x) = x+ 4(i) f(x) = 3xx(j) f(x) = 1 +2xx

SOLUTION 1

(a) For the functionf(x) = x + 5x + 3, a = 1, b = 5, c = 3b 4ac = (5) 5(1)(3) = 10

Since a > 0 and b 4ac > 0, the graph off(x) has the shape of and

intersects thexaxis at two points as shown below

x

b) For the functionf(x) =x 4x + 4, a = 1, b = 4, c = 4b 4ac = ( 4) 4(1)(4) = 0

Since a > 0 and b 4ac = 0, the graph off(x) has the shape of and

touches thexaxis at one point as shown below

x

(c) For the functionf(x) =x + 4, a = 1, b = 0, c = 4b 4ac = (0) 4(1)(4) = 16

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________________________________________________________________Since a > 0 and b 4ac < 0, the graph off(x) has the shape of and

does not intersect thexaxis as shown below

x

(d) For the functionf(x) = 3xx,= x + 3x

a = 1, b = 3, c = 0b 4ac = (3) 4(1)(0) = 9

Since a < 0 and b 4ac > 0, the graph off(x) has the shape of and

intersects thexaxis at two points, as shown below

x

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________________________________________________________________

(e) For the functionf(x) = 1 +2xx= x + 2x1

a =

1, b = 2, c =

1b 4ac = (2) 4(1)(1) = 0

Since a < 0 and b 4ac = 0, the graph off(x) has the shape of and

touches thexaxis at only point, as shown below.

x

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________________________________________________________________

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Activity Sheet Lesson 24Maximum and minimum values of Quadratic Functions.

________________________________________________________________Name:

Class: Date:

1. Find the minimum value of the following quadratic function.f(x) = 4x 24x 13

2. Find the maximum value of the following quadratic function.f(x) = 2x+ 10x 1

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Activity Sheet Lesson 24Maximum and minimum values of Quadratic Functions.

________________________________________________________________

1.Find the minimum value of the following quadratic function.

f(x) = 4x 24x 13

SOLUTION:f(x) = 4x 24x 13

= 4(x 6x) 13= 4(x 6x + 9 9) 13= 4(x 6x + 9) 36 13= 4(x 3) 46

49 is the minimum value of functionf.

2. Find the maximum value of the following quadratic function.

f(x) =

2x+ 10x

1

SOLUTION:

f(x) = 2x+ 10x 1= 2(x 5x) 1

= 2(x 5x +4

25

4

25) 1

= 2(x 5x +4

25) +

2

23

=

2(x

25 ) +

223

2

23is the maximum value of functionf.

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Activity Sheet Lesson 25Sketching graphs of quadratic functions

________________________________________________________________Name:

Class: Date:

1. Sketch the graph of the quadratic function.f(x) = (x 1) 4

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________________________________________________________________f(x) =x+ 2x 8

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________________________________________________________________1. Sketch the graph of the quadratic function.

f(x) = (x 1) 4

SOLUTION:

f(x) = (x

1)

4= x 2x + 1 4= x 2x 3

a =1

a >0, therefore

f(x) = (x 1) 4

Turning point = (1, 4)

The dicriminant,

b 4ac = (2) 4(1)(3)= 4 + 12= 16

b 4ac > 0

The graph intersects thexaxis at two points.

x 2x 3 = 0(x 3) (x + 1) = 0

f(x)

x-1

-3

(1, -4)

3

x 3 = 0 or x + 1 = 0x = 3 x = 1

thexintercepts are (3, 0) and (1, 0).

When x = 0,f(0) = 0 2(0) 3

= 3Theyintercept is (0, 3)

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________________________________________________________________2. Sketch the graph of the quadratic function:.

f(x) = (x 2)+ 1

SOLUTION:

f(x) = (x 2)+ 1= (x 4x + 4) + 1= x+ 4x 4 + 1= x+ 4x 3

a =1

a 0


(x

2)

+ 1= 0(x + 3) (x 1) = 0

x + 3 = 0 or x = 1x = 3

The graph intersects thexaxis at (3, 0)

f(x)

x-3

(2, 1)

3

and (1, 0).

When x = 0,f(0) = (0 2)+ 1

=

( 2)

+ 1= 4+ 1= 3

Theyintercept is (0, 3)

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________________________________________________________________

3. Sketch the graph of the quadratic function:.f(x) =x+ 2x 8

SOLUTION:

f(x) = x+ 2x 8

a = 1

a >0, therefore

f(x) = x+ 2x - 8

= x

+ 2x +

2

2

2

2

- 8

= x+ 2x + 1 1 8= (x + 1) 9

Turning point = ( 1, 9)The dicriminant,

b- 4ac = (2) 4(1)(8)= 4 + 32= 36

b- 4ac > 0


x+ 2x 8 = 0(x 2) (x + 4) = 0

x 2 = 0 or x + 4 = 0x = 2 x = 4

f(x)

x4

8

(1, 9)

2

The graph intersects thexaxis at (2, 0)and (

4, 0).

When x = 0,f(0) = (0)+ 2(0) 8

= 8

Theyintercept is (0, 8)

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Activity Sheet Lesson 26Determining the range of values of x that satisfies a quadratic inequality

__________________________________________________________

1. Find the range of values of x for which x(2 x) >21

x2.

SOLUTION:

x(2 x) > 21x2

0 < x 0

34

0

2x x2 >21x

2

4x 2x2 >x2

4x 3x2 > 0Letf(x) = 4x 3x2

Whenf(x) = 04x 3x2 = 0

Forf(x) > 0, the range of

values ofx is 0 < x 0, the range of

values of m is m 4 and m 4.

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Lesson 27

Activity Sheet

Solving Simultaneous Equations

Name:

Class Date:

1. Solve the following simultaneous equations.

i) 2 3 4 0x y + =

2

5 1 0x xy =

ii) 8 3 7y x= +

3 1xy =

iii) 3 2 5x y =

2 2 3 0x y y =

iv)2

5 2 2 9y x x xy y+ = + + =2

2. The line 0y x = intersects the curve 1xy = at the points p and q . Find the

length ofpq .

3. Find the coordinates of the points of intersection between the curve2 2

2 3 3 0x y x y+ + = and the line 1 0x = .

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Answer

1. i) 2.698, 0.4653x y= = ;

0.1588, 1.227x y= =

ii)8 1

,3 8

x y= = ;

1, 1

3x y= =

iii)4 1

1 ,5 5

x y= = ;

3, 2x y= =

iv) 8, 5;x y= =

2, 1x y= =

2. 2.828 units

3. ( ) ( )1,1 , 1, 4

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Lesson 28

Activity Sheet

Simultaneous Equations Involving Real-Life Situations

Name:

Class: Date:

Solve the following questions.

1. A wire of length 120 cm is cut into two pieces. Each piece of the wire is then

bent to form a square. If the total area of the two squares formed is 650cm 2 , find

the length of each piece of the wire.

2. Given that the sum of the circumferences of two circles is 30 cm and the areas

of the circles differ by 117 cm 2 , find the diameter of each circle.

3. Given that the sum of two positive numbers is 18 and the sum of the square ofthe numbers is 194. Find the two numbers.

4. Rosma bought x birds and y rabbits for RM208. Given that she bought a total

of 20 birds and rabbits altogether. A bird costs RMx while a rabbit costs RMy

where y x> . Find the values ofx and y .

5. Given that triangle ABC is right angled at B, where ( 3)AB x= cm,

( 9)AC y= + cm and 15BC = cm. If the perimeter of the triangle is 36 cm, find

the values ofx and y .

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Lesson 29

Activity Sheet

Finding the Value of Numbers Given In Index Form

Name:

Class: Date:

Find the values of

1.3

( 6)

2.

5

23

3.

3

215

4. ( )3

1.3

5. ( )0

7

6.

0

1

16

7.

3

1

4

8.

2

223

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Answer

1. 216

2.32

243

3.93

2

125

4. 2.197

5. 1

6. 1

7. 64

8.

9

64

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Lesson 30

Activity Sheet

Finding the Value of Numbers Given in Index FormThat are Multiplied Divided or Raised to a Power.

Name:

Class: Date:

Simplify each of the following.

(a)4 3

5 5

(b)5 2

3 3

(c) ( ) ( )5 4

0.3 0.3

(d)3 5

9 3

(e) ( ) ( )2 44

2 8

(f)

3 49 3

27

(g) ( )3

64

4 16

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Answer

(a) 5

(b) 2187

(c) 0.3

(d) 177 147

(e) 1 048 576

(f) 2187

(g)1

16

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Lesson 31

Activity Sheet

Use Laws of Indices to Simplify Algebraic Expressions

Name:

Class: Date:

1. Simplify each of the following.

i)

1

2 1 1

16

2 4

x

x x

+

+

ii)

3 2

4 2

n n

n

y y

y y

+

+

iii)1

3 3n n+

2. Solve each of the following

i)2 4 3 2 1

5 5125

x x+ =

ii)3 2

32 4x x

=

iii)

2

4 2739

y

y

y

+

=

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Answer

1. i) 8

ii)

1

y

iii) (2)3n

2. i) 1x =

ii) 5x =

iii) 2y =

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1.

1. 8

2.1

y

3. (2)3n

b.

1. 1x =

2. 5x =

3. 2y =

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Lesson 32

Activity Sheet

Converting Equations in Index Forms to Logarithmic Forms and Vice Versa

Name:

Class: Date:

1. Convert each of the following index form to the logarithmic form.

i)4

3 81=

ii)

2 1

5 25

=

iii)

1

21

366

=

iv)

0

11

10

=

v)

3

2

81 729=

2. Convert each of the following logarithmic form to the index form.

i) 161

log 42

=

ii)5

1log 2

25=

iii) 101

log 102

=

iv)4

3 log 64=

v)2

log 3x=

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Answer

1. i)3

log 81 4=

ii) 51

log 225 =

iii)36

1 1log

6 2=

iv) 110

log 1 0=

v) 813

log 7292

=

2. i)

1

216 4=

ii)2 15

25

=

iii)

1

210 10=

iv) 34 64=

v)32 x=

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Lesson 33

Activity Sheet

Finding Logarithm of a Number

Name:

Class: Date:

Evaluate each of the following.

1.10

log 1000

2.10

log 0.001

3.5

10log 10

4.

1

3

10log 10

510

log 5.22

6.5

10log 1.2

7.3

10log 8

8.3

10log 1.55

9.3

10log 0.123

10 .

2

10

2log

9

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Answer

1. 3

2. 3

3. 5

4.1

3

5. 0.7177

6. 0.3959

7. 2.709

8. 0.5710

9. 2.730

10. 1.306

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Lesson 34

Activity Sheet

Finding Logarithm of Numbers by Using Laws of Logarithms

Name:

Class: Date:

1. Evaluate each of the following.

i)6 6

log 4 log 9+

ii)5 5

log 75 log 3

2. Given that log 2 0.631x

= and log 3 1.465x

= , evaluate

i) log 18x

ii) log 36x

iii) log 0.75x

iv) log 6x


i) 2log5 log4 log10+

ii)8 8 8

1log 24 log log 3

8

iii)10 10

2 1log 125 log 160

3 2+

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Answer

1. i) 2

ii) 2

2. i) 3.561

ii) 4.192

iii) 1.161

iv) 1.048

3. i) 1

ii) 2

iii) 3

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Lesson 35

Activity Sheet

Simplifying Logarithmic Expressions

Name:

Class: Date:

1. Simplify each of the following.i)

3 3log 6 log 3+

ii) 3 3log 18 log 2 iii)

5 5 5log 20 log 5 log 4x+


i)10 10

log 20 log 50+

ii)3 3

log 1 log 27

iii)2 2 2

1log 4 log 12 log

3+ +

iv)2

8log logx x

x x

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Answer

1. i)3

log 18ii) 2iii)

52 log x+

2. i) 3

ii) 3

iii) 4

iv) 4

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Lesson 36

Activity Sheet

Changing the Base of Logarithms

Name:

Class: Date:


i) 3log 7

ii) 5log 6

2. Given that log 3a x= and log 5a y= , express the following in terms ofxand/ory

i) 3log 5

ii) 3log 15

3. Given log 2p k= and log 5p m= , express each of following in terms ofk

and/ormi) 2log 20

ii) 5log 50

4. Given that 2 0.631logx

= and log 3 1.465x = , evaluate

i) 4log 12

ii) 6log 18

5. Without using a calculator, find the value of

2 9 6 7 18log 18 log 6 log 2 log 9 log 7

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Answer

1. i) 1.771

ii) 1.113

2. i)y

x

ii)y x

x

+

3. i)2k m

k

+

ii) 2m km

+

4. i) 2.161

ii) 1.699

5. 1

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Lesson 37

Activity Sheet

Solving Problems Involving Logarithms

Name:

Class: Date:

1. Given that log 3p

k= and log 5p

m= , express the following in terms of

kand m.i) log 45

p

ii)25

log 3p

2. Given that a = log 3 xand b= log 3 y, express the following in terms ofa and b.i) 3log 27

yx

ii) 3

3

1

log log 9yx

3. Given 2hA = and 4kB = , express the following in terms ofhand k.

i)

2

4log

A

B

ii)16

log AB

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Answer

1. i) 2k m+

ii) 12

k

m

+

2. i)3 3a

b

+

ii)6a

b

3. i) h k

ii)8 4

h k+

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Lesson 38

Activity Sheet

Equations involving indices

Name:

Class: Date:

1. Solve the following equations:

i) 4 16x =

ii) 19 3 243x x =

iii) 32 72 2 0x x+ + =

2. Solve the following equations:

i) 3 2 52 8x x+=

ii) 164 4x x+=

iii) 3 227 9x x+ =

iv) 1 3 13(9 ) 81x x+ +=

v) 4 127 1x+ =

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Answer

1.

i) 2x =

ii) 2x =

iii) 3x =

2.

i) 5x =

ii)1

2x =

iii) 9x =

iv)1

5x =

v) 3x =

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Lesson 39

Activity Sheet

Equations Involving Logarithms

Name:

Class: Date:

Answer

1. Solve the following equations.

i) 4 15x =

ii) 3(4 ) 5

x

=

iii)1

4 8x+

=

2. Solve each of the following equations.

i)2 3

4 10 0x

=

ii)

1

0.3 0.4

x x+

=

iii)1

5 2x x+

=

iv)1

2 (2 ) 7x x+

=

v)1

7 (2 ) 10x x+

=

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Answer

1. i) 1.953x =

ii) 0.3685x =

iii) 0.5x =

2. i) 2.330x =

ii) 3.185x =

iii) 1.756x =

iv) 0.9037x =

v) 0.1352x =

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Lesson 40

Activity Sheet

Equations Involving logarithms

Name:

Class: Date:

1. Find the distance between each of the following pairs of points.

i) ( 1, 4) ( 4,0)A B and

ii) (0,3) ( 4,1)A B and

iii) (8,3) ( 8, 6)A B and

iv) (6, 2) (7,3)A B and

2. Given that the distance between ( 3, )P m and (2,3)Q is 10 units, find the

value of m .

3. Show that A(1, 7), B( 3, 5) and C( 1, 1) are vertices of an isosceles

triangle.

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Answer

1.

i) 5 units

ii) 4.472 units

iii) 18.36 units

iv) 5.099 units

2. 5.660m = , 11.66m =

3. AB = 20 units

BC= 20 units

AC= 20 units

SinceAB=BC, therefore ABC is an isosceles triangle.

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Lesson 41

Activity Sheet

Midpoint of Two Given Points

Name:

Class: Date:

1. Find the coordinates of the midpoint of each of the following pairs of points.

i) ( 1, 4) and ( 4,0)A B

ii) (0,3) and ( 4,1)C D

iii) (8,3) and ( 8, 6)P Q

iv) (6, 2) and (7,3)R S

2. If ( 3, )P m is the midpoint of the straight line joining (2,3)Q andR( n,10),

find the values of m and n.

3. The coordinates of pointsA, B and Care (p, -7), (-3, q) and (-1, -1) respectively.IfCis the midpoint of the lineAB, find the values ofpand q.

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Answer

1. i)1

( 2 , 2)2

ii) ( 2,2)

iii)1

(0, 1 )2

iv)1 1

(6 , )

2 3

2.1

62

m = and 8n =

3. p= 2and q= 6

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Lesson 42

Activity Sheet

Finding the point that divides a line segment according to a given ratio

Name:

Class: Date:

1. For each of the following, find the coordinates ofQ for which Q divides PQ

according to the ratio as stated.

i) ( 1, 3), ( 4,0); : 1 :2P R PQ QR =

ii) (0,3) , ( 4,1); : 1 :3P Q PQ QR =

iii) (8,3) , ( 8, 6); : 2 :3P Q PQ QR =

iv) (6, 4) , Q(6,3); : 2 : 5P PQ QR =

2. Point B is a point lying on the straight line joining (2,1)A and ( 2,5)C so that

3AB BC= . Find the coordinates ofB .

3. Given thatA( 2, 7),B( 3, q) and C( 8, 1) lie on the straight line such that

: :AB BC m n= , find

i) the ratio :m n

ii) the value of q

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Answer

1.

i) ( 2, 2)Q

ii)1

( 1,2 )2

Q

iii)3 3

(1 , )5 5

Q

iv) (6, 2)Q

2. ( 1,4)B

3.

i) : 1:5m n =

ii) 6q =

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Lesson 43

Activity Sheet

Finding Areas of Triangles

Name:

Class: Date:

By determining the areas of specific geometrical shapes, find the area of each of the

following trianglesABC.

(i) A(2,0) ,B(2,10) , C(5,5)

(ii) A(1,3) ,B(5,4) , C(4,3)

(iii) A(-2,3) ,B(6,5) , C(2,12)

(iv) A(2,-3) ,B(-5,10) , C(-5,-5)

(v) A(-1,1) ,B(-5,0) , C(4,-3)

(vi)A(2,3) ,B(1,8) , C(-2,-5)

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Answer

(i)2

15 unit

(ii)2

1.5 unit

(iii)2

32 unit

(iv)2

52.5 unit

(v)2

8.5 unit

(vi)2

14 unit

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Lesson 44

Activity Sheet

Finding Areas of Triangles by Using the Formula

Name:

Class: Date:

1. For each of the following, find the area of the triangleABCwith the given vertices.

i) A(2, 0), B(2,0) , C(5, 5)

ii) A(

1,

3), B(5,

4), C(4, 3)

iii) A(2,3) ,B(6,15), C(4, 12)

iv) A(6,5) ,B(5,10), C(4, 9)

v) A(0, 1) ,B(5, 0), C(4 ,3)

vi) A(2, 3),B(1, 8), C(0, 0)

2. Show that pointsA(2, 3), B(1, 1) and C(2,5) are collinear.

3. Find the value ofm if the pointsA(

2, 3) ,B(6, 5) and C(2, m) are collinear.

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Answer

1.

i) 210 unit

ii)2

20.5 unit

iii) 212 unit

iv) 2102 unit

v) 212 unit

vi) 26.5 unit

2. The area ofABC= 0

3. m= 4

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Lesson 45

Activity Sheet

Finding Areas of Quadrilaterals

Name:

Class: Date:

1. For each of the following, find the area of quadrilateralABCDwith the given vertices.

i) (2,0), ( 2,0), ( 5,5), (8,8)A B C D

ii) ( 1, 3), (5, 4), (4,3), ( 7,3)A B C D

iii) (2, 3), (6,15), (4,12), (0,0)A B C D

iv) (6, 5), ( 5, 10), ( 4,9), (1,1)A B C D

v) (0,1), ( 5,0), (4, 3), (8,0)A B C D

2. Show that the pointsA(4,7) ,B(2,3) , C(1,1) and ( 2, 5)D are collinear.

3. Find the value ofm if the points ( 5, )A m , ( 1,3)B , C(3,2m) andD(7,5) are collinear.

4. Find the possible values ofpif the area of the quadrilateral with vertices

(3, 2)A ,B(6,p), C(4,5) and ( 2,4)D is1

29

2

unit 2 .

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Answer

1. i)2

53 unit

ii)2

53.5 unit

iii)2

30 unit

iv)2

97 unit

v)2

26 unit

2. Hint: The area ofABCD= 0

3. m = 2

4. p = 1 or p=119

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Lesson 46

Activity Sheet

Gradients of Straight Lines

Name:

Class: Date:

1. Find the gradient of the straight line that passes through each of the followingpairs of points.

i) A(2, 0) andB(2, 8)

ii) B(5, 4) and C(4, 3)

iii) A(2, 3) and D(0, 0)

iv)A( 6, 5) and C( 4, 9)

v) C(4, 3) andD(8, 0)

2. Given that the gradient of the straight line passing through P(1,m2) and

Q(4, 9) is1

2, find the value ofm.

3. Given that the gradient of the straight line passing throughA(1,m) and

B(4m, 9) is 2, find the value ofm.

4. Given that pointsA(1,t) andB(12t, 8). If the gradient of the lineAB is2

3 ,

find the value oft.

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Lesson 47

Activity Sheet

Intercepts and Finding the Gradient of a Straight Line Using the Intercepts

Name:

Class: Date:

Answer1. State thex-intercept andy-intercept of the stra

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