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
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
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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)
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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
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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
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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)
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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]
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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]
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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
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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)
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( ) 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)
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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]
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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]
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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 ]
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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)
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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=
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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)
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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]
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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]
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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]
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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
equation/vergelyking 2
a = 2
substitution/substitusie
b = 2 and/en c = – 12
(5)
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( )
( )
122
284:)2()1(
)2(828:8;1
)1(06039:0;3
0212
21
2)(
−=∴=⇒=∴
=−
−=+∴−=++−
=+∴=+−−
=∴
=+
−=
−′
+=′
cb
aa
cacba
cacba
ba
baf
baxxf
equation/vergelyking 1
equation/vergelyking 2
a = 2
b = 2c = – 12
(5) 6.3
−−
−=
−−=
−=−=
−=
2112;
21TP
2112
12121
21
)2(22
2
y
y
x
abx
OR/OF
21
−=x
substitution/substitusie
y-value/waarde
(3)
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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]
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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)
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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]
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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]
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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
substitution/substitusie
answer/antwoord (6) [13]
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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)
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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]
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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]
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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]
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