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Western Cape Education Department

Examination Preparation Learning Resource 2016

Functions and GraphsMemorandum

MATHEMATICS Grade 12

Razzia Ebrahim

Senior Curriculum Planner for Mathematics

E-mail: [email protected] / [email protected]

Website: http://www.wcedcurriculum.westerncape.gov.za/index.php/component/jdownloads/category/1835-grade-12?Itemid=-1

Website: http://wcedeportal.co.za

mailto:[email protected]

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Index Page

1. 2016 June Paper 1 3 – 5

2. 2016 Feb-March Paper 1 6 – 9

3. 2015 November Paper 1 10 – 14

4. 2015 June Paper 1 15 – 19

5. 2015 Feb-March Paper 1 20 – 24

6. 2014 November Paper 1 25 – 28

7. 2014 Exemplar Paper 1 29 – 23

Mathematics P1 Graphs/Wiskunde V1 Grafieke Memorandum

Copyright reserved/Kopiereg voorbehou 2 Please turn over/Blaai om asseblief

QUESTION/VRAAG 4

4.1 (0; 3) 4.2 b

x=--2a

(-2) =---

2(-1) =-1

or

y=-(-1)2 -2(-1)+3

=4

c(- 1;4)

-2x-2=0

.'. X =-1

or 4ac-b2

y= 4a

= 4(-1X3)-(-2)2 4(-1)

4.3 B(l; o) By symmetry/Deursimmetrie A(-3; o) OR/OF

x2 +2x-3 = 0

(x+3Xx-1)= 0 x=-3 or x=l A(-3;0)

X

./ (0; 3) (1)

./ x=- (-2) or -2x-2=02(-1)

./ simplification

./ in the context of a turningpoint-(-1)2 -2(-1)+3

4(-IX3)-(-2)2

4(-1)

� A(-3; 0)

� A(-3; 0)

(3)

(1)

(1)



DBE/2016



DBE/2016

4.7 g(x)� g-1 (x) x-62x+6�--

2

4x+12�x-6 3x�-18 x�--6

QUESTION/nuAG 5

5.1 r = 25.2 g(x)=2x

+2

g{0>::2° +2=3B(O; 3)

3=_3_+20-p

p=-3 5.3 atA: x=-3

y =2-3 +2 =2{

A(-3; 2i) or ,{-3; 1;) or A{-3; 2,125)

5.4 -3 < x � 0 ORI OF

5.5 3 f(x)=-+2

x+3

f(x-2) =

3 +2x-2+3

h(x)==-3-+2x+l

(-3; o]

x-6 ..1'2x+6�--

,I' 4x+12 � x-6

..l'x�-6

./r=2

..I' g(O) = 2° + 2

..l'y=3

(3)[211

(1)

..l'substitute B(O; 3) and q= 2..l'p=-3

./ at A: x- -3 (p-value)..I' substitute x = -3 intoexponential eqution ..I' y-value./-3<x

./x�O

./ substitution of x - 23 ./ h(x)=-+2x+l

(4)

(3)(2)

(2)r121



DBE/2016

QUESTION/VRAAG 4

4.1 ( )2;0 answer(1)

4.2

shape

( )2;0

asymptote

(3) 4.3 ( )

( )23121

52

1 =+=

=−

−f

f

Average gradient ( ) ( )( )21

21−−−−

=ff

673

523

−=

−=

( ) 52 =−f

( )231 =f

answer(3)

4.4 Since the asymptote of f is 1=y , the asymptote of ( ) ( )xfxh 3= will be 3=y .

Omdat die asimptoot van f 1=y is, sal die asimptoot van ( ) ( )xfxh 3= 3=y wees.

answer(1) [8]

x

y

( )2;0 1=y



DBE/Feb.-Mrt. 2016

QUESTION/VRAAG 5

5.1 ( )( ) ( )

( ) 8104:)4;0(Substitute

81:8;1pointTurning2

2

2

−−=−−

−−=−

++=

a

xay

qpxay

( ) 814

81484

2 −−=

−=−==−=−

xy

qpaa

( ) 81 2 −−= xaysubstitute )4;0( −

4=a p and q values

(4)

5.2 Asymptote is 22 −=⇒−= dySubstitute ( ) :8;1 −

21

8 −+

=−r

k

( )rk +−= 16 rk 66 −−= ……………….line 1

Substitute ( ) :4;0 −

24 −=−rk

2−=rk

rk 2−= …………..………line 2 Equating lines 1 and 2:

23

64266

−=

=−−=−−

r

rrr

Substituting into line 2 or line 1:

( ) 3232 =

−−=k 3

2366 =

−−−=k

2−=d

rk 66 −−=

rk 2−=

rr 266 −=−−

value of r

value of k(6)

5.3 ( ) ( )10 ≤≤∴

≥x

xfxg x≤0 1≤x

(2) 5.4 The line y = k must pass through f twice on the positive side of

the x-axis./Die lyn y = k moet twee keer deur f aan die positiewe kant van die x-as sny.

48 −<<− k

k<−8 4−<k

(2)



DBE/Feb.-Mrt. 2016

5.5 cxy +−=

Substitute the intersection point of the asymptotes, i.e.

− 2;

23 :

Vervang die snypunt van die asimptote, m.a.w.

− 2;

23 :

21

21232

−−=

−=

+−=−

xy

c

c

OR/OF

xy −= is translated 23 units right and 2 units down/

xy −= transleer 23 eenhede na regs en 2 eenhede na onder ⇒

21

223

−−=

−

−−=

xy

xy

cxy +−=

c+−=−232

answer (3)

xy −=

223

−

−−= xy

answer (3)

5.6 By symmetry,

−=

−+−−+=

23;

215

1232;28

23Q

2

15=x

23

−=y

(2) [19]



DBE/Feb.-Mrt. 2016

QUESTION/VRAAG 6

6.1

21

2

41:

41:

yxf

xyf

=

=

−

xxfxxf

xy

xy

2)(OF4)(

4

4

11

2

−=−=

±=

=

−− OR/

interchanging x andy xy 42 =

answer(3)

6.2

x

y

f

f -1

(-2; 1)

(1; -2)

both graphs passthrough (0 ; 0)

shape for both

one additionalpoint on both graphs

(3)

6.3 Yes. No value of x in the domain of f -1 maps onto more than one y-value.Ja. Geen waarde van x in die definisieversameling van f -1assosieer met meer as een y-waarde nie.

OR/OF

Yes. One to one function./Ja. Een-tot-een-funksie.

OR/OF

Yes. Vertical line test holds./Ja. Die vertikale lyntoets werk.

yesreason

(2)

yesreason

(2)

yesreason

(2) [8]



DBE/Feb.-Mrt. 2016

QUESTION/VRAAG 4

Given: 82)( 1 −= +xxf4.1 y = – 8 answer

(1) 4.2

x-intercepty-interceptshapeasymptote

(4)

4.3 ( ) 82 1 −= +−xxg

OR/OF

( ) 821 1

−

=

−x

xg

answer(1)

answer

(1) [6]

-8 -6 -4 -2 2 4 6 8

-8

-6

-4

-2

2

4

6

8

x

y

f

y = -8

0



DBE/November 2015

QUESTION/VRAAG 5

Given ( ) 32 −= xxh for 42 ≤≤− x .

x

y

h

-3

OQ

-2 4P

5.1 For x-intercepts, y = 0

( )0;5,1Q5,1

032=

=−x

x 5,1=x y = 0

(2) 5.2

57: ofDomain 53)4(2:4

73)2(2:2:

1 ≤≤−

=−==−=−−=−=

− xhyxyx

h–75

57 ≤≤− x(3)

5.3

x

y

h-1

1,5

0-3

(-7; -2)

(5 ; 4)

interceptsshapeendpoints

(3)

5.4 ( ) 32 −= xxh

For the inverse of h,

2332

+=

−=xy

yx

( ) ( )

393

3642

332

1

==

+=−

+=−

= −

xx

xx

xx

xhxh

OR/OF

2

3+=

xy

2

332 +=−

xx

3=x(3)



DBE/November 2015

( ) 32 −= xxhh and 1−h intersect when xy =

( )

332==−=

xxxxxh

( ) xxh =

xx =−32

3=x(3)

5.5

( )

91259124

32

OP

2

22

22

222

+−=

+−+=

−+=

+=

xxxxx

xx

yx

minimum a be tohasOP minimum, itsat be toOPFor 2

Vir OP om minimum te wees, moet OP2

( )

( )

units1,34or 5

3or 599

5612

565 OP oflength Minimum

56

521201210

20

2

=+

−

=

=∴

−−==−

−==

x

x

abx

dxd orOP2

'n minimum wees

OR/OF

( )

( ) ( )units8,1or34,1

06,002,1OP

6,02,1212,1

6564

3221

21equation has OP

21(given)2

22

OP

=

−−+−=

−=−=

==

−=−

−=−

−=∴

−=

=

P

P

h

y

xx

xx

xx

xy

m

m

OR/OF

222OP yx +=

substitute32 −= xy

9125 2 +− xx

x-value

answer(5)

\

21

OP−

=m

equation ofOP

3221

−=− xx

substitutioninto distanceformulaanswer

(5)



DBE/November 2015

Minimum distance between the line ocbyax =++ and the point ( )OO yx ;Minimum afstand tussen die lyn ocbyax =++ en die punt ( )OO yx ;

( ) ( )( )( )

53

12

3010222

22

=

−+

−−+=

+

++=

ba

cbyax OO formula

substitution

answer(5)

5.6.1 ( )⇒−∈<′= 5,1;2for 0)()( xxfxh f is decreasing on the left of Q / f is dalend links van Q.

( )⇒∈>′= 4;5,1for 0)()( xxfxh f is increasing on the right of Q / f is stygend regs van Q.

OR/OF

( )( )

023

022 Since

023

23

>

′′

>=′′=′

=

′=

f

xfxh

fh

f has a local minimum at 23

=x by the second derivative test.

f het 'n lokale minimum by 23

=x deur die tweede afgeleide toets.

OR/OF

cxxxf +−= 3)( 2 f has a minimum value since a > 0/f het 'n minimum waarde omdat

a > 0

decreasingleft of Qincreasingright of Q

(2)

023

=

′f

( ) 02 >=′′ xf

(2)

cxxxf +−= 3)( 2

explanation(2)

5.6.2 ( ) 54)4( ==′= hfm answer(1)

[19]



DBE/November 2015

QUESTION/VRAAG 6

6.1.1 )18;0(T answer(1)

6.1.2

( )( ))0;3(Q

0330182 2

=+−=+−

xxx

OR/OF

)0;3(Q9

01822

2

=

=+−

xx

equating to 0factors

coordinatesof Q

(3)

equating to 0

92 =x

coordinatesof Q

(3)

6.1.3 x-coordinate of S is 4,5/x-koördinaat van S is 4,5By symmetry about the line x = 4,5/Deur simmetrie om dielyn x = 4,5: R(6 ; 0)

0);R(6

(2)

6.1.4 For all R∈x answer(2)

6.2 If C ( )yx; is the centre of the hyperbola/As C ( )yx; die middelpunt is vandie hiperbool

6+= xy and 2−=x 462 =+−=∴ y

x

y

0

y = 4

x = -2

asymptote4=y

asymptote2−=x

shape

(4) [12]



DBE/November 2015

QUESTION/VRAAG 4

4.1 2, −≠yR

OR/OF

( ) ( )∞−∪−∞− ;22;

2−≠y(2)

( )2;−∞− ( )∞− ;2

(2)

4.2

325

21

5

21

)(

=+=

−−

=−

−−

=

aa

ax

axgsubstitution of

the point (0; –5)in to g(x)

answer(2)

4.3 For g, asymptotes intersect at/Vir g, asimptote sny by ( )2;1 −∴ For/Vir ( ) 73 +−= xgy , asymptotes will intersect at/

asimptote sal sny by ( )72;31 +−+i.e./d.i. at/by ( )5;4

OR/OF

( )

( )5;4

54

3

7213

373

21

)(

+−

=

+−−−

=

+−=

−−

=

x

x

xgyx

axg

( )2;1 − for gx = 4y = 5

(3)

subs

x = 4y = 5

(3) [ 7 ]



DBE/2015

QUESTION/VRAAG 5

5.1

164

41

2

2

==

=

−

y substitution

answer(2)

5.2

xxy

xf

y

y

x

441

1

logyorlog41:

41

−==

=

=

−interchange

x and y

answer(2)

5.3 f and f –1 are symmetrical about the line y = x, to obtain f –1, reflect f in the line y = x. f en f –1 is simmetries om die lyn y = x, om dus f –1 te kry reflekteer f in die lyn y = x.

OR/OF

The x and y-coordinates of points on f may be swopped around to obtain the coordinates of the points on f –1. Two points that lie on the graph of f are (0 ; 1) and (–2 ; 16). The corresponding points that will lie on 1−f will therefore be (1 ; 0) and (16 ; –2). Die x- en y-koördinate van punte op f mag omgeruil word om die koördinate van punte op f –1 te kry. Twee punte op die grafiek van f is (0 ; 1) en (–2 ; 16). Die ooreenstemmende punte op 1−f sal dus (1 ; 0) and/en (16 ; –2) wees.

reflect in y = x

(1)

swop x and y

(1)

5.4

x

y

f -1

(1 ; 0)

shape of f –1

x-int of f –1at 1

(2)



DBE/2015

5.5 0>x

OR/OF

( )∞;0

0>x (1)

( )∞;0(1)

5.6 ( )( )

]16;0(or160216 5.1, From

21

1

∈≤<−=

−≥−

−

xxf

xf

0>x 16≤x

(2) 5.7.1

21

=q (using a calculator/gebruik 'n sakrekenaar)

OR/OF

Without a calculator (not necessary)/Sonder sakrekenaar (nie nodig)

2112

2221

41

21log

12

41

=

==

=

=

−−

q

q

q

q

q

OFR/O

21

2log22log

41log

21log

21log

41

=

−−

=

=

=

q

q

q

q

21

=q

(1)

21

=q

(1)

5.7.2 At the intersection point of f and f -1, xy = (by symmetry). Thus need only solve ( ) xxf =−1 (instead of ( ) ( )xfxf 1−= )By die snypunt van f en f -1 xy =, (deur simmetrie). Slegs nodig om ( ) xxf =−1 op te los (in plaas van ( ) ( )xfxf 1−=

=

=

=

=

21;

212121

5.7.1 from21

21log

log

41

41

y

x

xx

OR/OF

By/Van 5.7.1, 21log

21

41=

Which means that

21;

21 lies on the graph of 1−f ./

21log

21

41=

21

=x

21

=y

(3)

21log

21

41=



DBE/2015

Wat beteken

21;

21 lê op die grafiek van 1−f .

But clearly,

21;

21 lies on xy = /Maar,

21;

21 lê op xy =

Hence

21;

21 is the intersection point of f and f -1 /Dus is

21;

21

die snypunt van f en f -1

21

=x

21

=y

(3) [14]

QUESTION/VRAAG 6

6.1

( )( )

units7AB2or5

0250103

030932

2

==−==−+=−+

=+−−

xxxx

xxxx 03093 2 =+−− xx

factorsanswers

AB = 7(4)

6.2

( )( )3or2

0230601833

12123093

2

2

2

=−==+−=−−

=++−

+−=+−−

xxxxxx

xxxxx

At K, 0>x , hence ( ) 2412312 −=+−=yK ( )24;3 −

equating ofequations

062 =−− xxfactors

x =3y= –24

(5)6.3 ( ) ( )

3or2 ≥−≤≤

xxxgxf

OR/OF

( ) ( ));3[or]2;( ∞−−∞∈

≤x

xgxf

OR/OF

( )

( )( ) 0230601833012123093

2

2

2

≥+−≥−−

≤++−

≤+−−+−−

xxxx

xxxxx

3or2 ≥−≤ xx

2−≤x 3≥x or

(3)

]2;( −−∞ );3[ ∞ or

(3)

2−≤x 3≥x or

(3)



DBE/2015

6.4

( ) ( )

( )

4318

475

431818

213

213

CD/lengthMax CD/lengthMax

4318

213

1843

213

21

21

1821

213036

323

183CD02

1833)1212(3093CD

2

2

2

22

2

2

2

=

=

=+

+

−=

+

−−=

++

−−===

+

−

−−==+−

−−

=

+−==′−=

++−=

+−−+−−=

lengte Maks lengte Maks

x

xx

xx

xx-xfa

bx

xxxxx

OF

OFOF

OR/

OR/OR/

gf yy −= CD

1833 2 ++− xx

method

21

=x

max length

4318or

475CD =

(5) [17]



DBE/2015

QUESTION/VRAAG 4

4.1 12−=−=

yx 2−=x

1−=y (2)

4.2.1

2

120

6)0(

=

−+

=g

y-intercept/afsnit (0 ; 2) answer/antwoord (1)4.2.2

462

261

12

60

==+

+=

−+

=

xx

x

x

x-intercept/afsnit (4 ; 0)

equating to/stel gelyk aan 0

answer/antwoord(2)

4.3

asymptotes/asimptote intercepts/afsnitte shape/vorm

(3)

4.4 ( )21 +−=+ xy3−−= xy

OR/OF

Using general formula/Gebruik algemene formule: ( )

31)2(

−−=−+−=++−=

xyxy

qpxy

m = – 1substitution of

(–2 ; –1)answer

(3)

formula/formulesubstitution of p and qvalues/substitusie van p- enq-waardesanswer/antwoord (3)

4.5 2−>x answer (2)[13]



DBE/Feb.-Mrt 2015

QUESTION/VRAAG 5

5.1 3

9 2

==

aa

OR/OF

3399log2log)(

22

1

=∴==

==−

aa

xxf

a

a

29 a= 3=a (2)

29 a= 3=a (2)

5.2 xxg −= 3)(

OR/OF x

xg

=

31)(

answer/antwoord (1)

answer/antwoord(1)

5.3 9≥x

OR/OF

9932loglog)(

23

31

≥∴==

==−

xx

xxxf

OR/OF

93

2log2

3

≥∴≥

≥

xx

x

answer/antwoord(2)

answer/antwoord(2)

answer/antwoord(2)

5.4 Yes/Ja. For every y-value there is only one x such that/Vir elke y-waarde is daar slegs een x sodanig dat ( )xfy = .

OR/OF

Yes/Ja. f is a one-to-one relation/is 'n een-tot-een-relasie.

Yes/Ja Reason/Rede (2)

Yes/Ja Reason/Rede (2)

[7]



DBE/Feb.-Mrt 2015

QUESTION/VRAAG 6

6.1 23 ≤≤− x critical values/kritiese waardes

notation/notasie(2)

6.2 ( )( )

( )( )1222232

248

)21)(31(8)2)(3(

:

2

21

−+=

−+==

−=−−+=−−+=

−−=

xxyxxy

aa

axxay

xxxxayf

b = 2 and/en c = – 12

OR/OF

12 and2

12222112

2122

2112

412

2112

212

5,12225)2(

4250

248:)2()1(

)2.....(498

2118

)1....(4250

2120

21

2

2

2

2

2

2

2

−==∴

−+=

−++=

−

++=

−

+=

−=−=−=

==−

+=−→+

+=−

+=→+

+=

+

+=

cb

xxy

xxy

xxy

xy

q

aa

qaqa

qaqa

qxay

OR/OF

)2)(3( −+= xxay substitute/vervang (1 ; – 8)

a = 2

b = 2 and/en c = – 12

(5)

equation/vergelyking 1


a = 2

substitution/substitusie

b = 2 and/en c = – 12

(5)



DBE/Feb.-Mrt 2015

( )

( )

122

284:)2()1(

)2(828:8;1

)1(06039:0;3

0212

21

2)(

−=∴=⇒=∴

=−

−=+∴−=++−

=+∴=+−−

=∴

=+

−=

−′

+=′

cb

aa

cacba

cacba

ba

baf

baxxf



a = 2

b = 2c = – 12

(5) 6.3

−−

−=

−−=

−=−=

−=

2112;

21TP

2112

12121

21

)2(22

2

y

y

x

abx

OR/OF

21

−=x


y-value/waarde

(3)



DBE/Feb.-Mrt 2015

OR/OF

−−

−

+=

−

+=

−−

++=

−+=

5,12;21TP

5,12212

25,6212

1.2161.

212

]6[2

2

2

222

2

x

x

xxy

xxy

−−

−=

−

−+

−=

−=+−

=

5,12;21TP

2112

12212

212

21

223

y

y

x

OR/OF

−−

−=−

−+

−=∴

−=

−==+

+=

−+==

225;

21TP

22512

212

212

2124

02424)(

1222)(

2

/

2

y

x

xx

xxfxxyxf

method/metode

x-value/waardey-value/waarde

(3)

method/metodex-value/waarde

y-value/waarde(3)

method/metode

x-value/waarde

y-value/waarde (3)

6.4 2

13=x answer/i

(2) 6.5

62)1(4)1(

24)(/

=+=′=

+=

mfm

xxf 24/ += xy subst. x = 1 answer/antwoord

(3) [15]



DBE/Feb.-Mrt 2015

QUESTION/VRAAG 4

4.1 11

==

qp p value /waarde

q value /waarde (2)

4.2

321

11

20

−==−−

++

=

xx

x

OR/OF

Reflect (0 ; 3) across y = – x to get T( –3 ; 0) 3−=x

Reflekteer (0 ; 3) om y = – 1 om T( –3 ; 0) te kry 3−=x

11

20 ++

=x

3−=x ( 2)

reflect across/reflekteer omy = – x

3−=x(2)

4.3 Shifting g five units to the left shifts (– 1 ; 0) five units to the left. x = – 6 answer/antwoord (1)

4.4

eenhede

yxy

xx

xxxx

xx

units/45,2 6OS

633OS...73,13

0 S,at since 3

312

11

2

222

2

2

==∴

=+=+=

==

>=∴

=

+=++

=++

OR/OF

equating both graphs/stelgrafieke gelyk

32 =x 3and3 == yx

OS 2 = 6

answer/antwoord (5)



DBE/November 2014

Translate g one unit down and one unit to the right/Transleer g een eenheid af en een eenheid na regs

The new equation/Die nuwe vergelyking :x

xp 2)( =

Therefore the image of S is ( )2;2S/ /Daarom is die beeld van S nou ( )2;2S/

Now translate p back to g/Transleer p terug na g: ( )

( ) ( )eenhedeunits/45,26OS

122212221212OS

12;12S222

==∴

++++−=++−=

+−

x

xp 2)( =

coord. of/koörd. van /S

coord. of/koörd. van S

answer/antwoord (5) 4.5 k < 3 will give roots with opposite signs/

k < 3 sal wortels met teenoorgestelde tekens gee k < 3 (1)

[11]



DBE/November 2014

QUESTION 5

5.1

33131

31log1

log

1

1

=∴

=

=

=−

=

−

−

a

a

a

xy

a

a

subt.

−1;

31

311 =−a or

1

31 −

=a

(2)

5.2 xy

yxh3

log: 3

=∴

= swop x and y/ruil x en y

answer/antwoord(2)

5.3

xxg

xxg

xxg

31

3

3

log)(

1log)(

log)(

=

=

−=

OF

OF

OR/

OR/

OR/OF

yx −= 3

OR/OF y

x

=

31

answer/antwoord(1)

answer/antwoord (1)

answer/antwoord (1)

answer/antwoord (1)

answer/antwoord (1) 5.4 0>x

OR/OF

( )∞;0

answer/antwoord(1)

answer/antwoord (1)

5.5

271271

33log

33

≥

=

=

−=−

x

x

xx

exponential form/eksponensiële vormsimplification/vereenvoudiging

answer/antwoord (3) [9]



DBE/November 2014

QUESTION/VRAAG 6

6.1

positive) is S of coordinate-( 22,123

064

2

2

xx

x

x

=

=

=− y = 0

1,22 (2)6.2 ( 0 ; – 6) 0

– 6 (2) 6.3.1

)64(2

)()(2−−=

−=

xx

xgxfQT

or 642 2+−= xx

correct formula/korrekte formulesubstitution/substitusie

(3) 6.3.2

QT 642 221

+−= xx

08QT of Deravitive 21

=−=−

xx x

x81

=

812

3

=x or 2641 xx=

32

81

=x

2

21

=x or

6413 =x

41

=x = 0,25

6414

412QTMax/

221

+

−

=Maks

eenhedeunits/75,6436 ==

derivative/afgeleidederivative equal to 0/afgeleide gelyk aan 0

812

3

=x

x-value/x-waarde


answer/antwoord (6) [13]



DBE/November 2014

QUESTION/VRAAG 4

4.1.1 3

12)( −+

=x

xf

1

310

2)0(

−=

−+

=

= fy

( )1;0 −

subst 0=x

( )1;0 − (2)

4.1.2

31

2331

23

31

20

−=

=++

=

−+

=

x

xx

x

− 0;

31

subs y = 0

− 0;

31

(2) 4.1.3

x

y

f

A(-1/3 ; 0)

B(0;-1)x = -1

y = -3

0 shape

both interceptscorrect horizontal and

vertical asymptote

(3)

4.1.4 4

3)1(−−=

−+−=xyxy

OR

( )

44

13

−−=−=

+−−=−+−=

xyk

kkxy

3)1( −+−= xy 4−−= xy

(2)

( ) k+−−=− 13

4−−= xy(2)



DBE/2014

4.2.1

( )

( )

( ) 322

31]1;1subs[3.1

1]2;0subs[3.2

3..

1

0

−=

=−=−

−−=

=−−=−

−=

+=

x

x

x

x

xfb

bby

ababay

qbay

subs q = – 3

1=a

2=b

( ) 32 −= xxf (4)

4.2.2 A translation of 4 units up and 1 unit to the left. 'n Translasie van 4 eenhede na bo en 1 eenheid na links.

OR

Dilation by a factor of 2 and 7 units up. Verkleining deur faktor van 2 en 7 eenhede na bo.

4 units up 1 unit to the left

(2)

dilation by factor 2 7 units up

(2) [15]



DBE/2014

QUESTION/VRAAG 5

5. 1

( )

125,6/849

3455

452

45

45

054)2(2

5

02

352)(

2

2

=

+

−−

−−=

−=−=

=−−

−−

−=

=′−=

+−−=

y

xx

xx

xfa

bx

xxxf

or

TP (45

− ;849 )

OR

849)

45(2

]1649)

45[(2

]23

1625)

45[(2

)23

252(-

2

2

2

2

++−=

−+−=

−−+−=

−+=

x

x

x

xxy

TP (45

− ;849 )

/2abx −= ( ) 0=′ xf

45

−=x

y = 125,6/849

(3)

]23

1625)

45[(2 2 −−+− x

45

−=x

y = 125,6/849

(3) 5. 2 mtangent = tan 135

= – 1 –4x –5 = –1

– 4x = 4x = –1

y = – 2(–1) 2– 5(–1) + 3 = 6

Point of contact: P(–1; 6)

tan 135 = –1 –4x –5 = –1

x = –1

y = 6(4)

5. 3 Eq of g: y – y1 = m(x – x1) y – 6 = – 1(x + 1)

y = – x +5

substitute inequation

answer(2)

5. 4 d > 5 answer(1)

[10]



DBE/2014

QUESTION/VRAAG 6

6.1

2168

)8(4

)(

==

=

=

aa

a

axxg

subst (8 ; 4)

2=a(2)

6.2 0≥x answer(1)

6.3 0≥y answer(1)

6.4

0;2

20;2

2

2

≥=

=

≥=

yxy

yxxxy

interchange x and y

answer (2)

6.5

( )( )2 or 8

28016100

168242

2

2

==−−=+−=

+−=

−=

xxxxxx

xxxxx

when x = 2, LHS = 2 but RHS = –2 Hence 8=x only

1682 2 +−= xxx

(squaring both sides)

factors

2 or 8 == xx

selects 8=x

(4)

6.6 80 << x 8<x x<0

(2) [12]



DBE/2014

mathematics grade 12 - western cape · western cape education department . examination preparation...

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