2016 2014 CAPS Graphs P1 28 Sept memo

Western Cape Education Department Examination Preparation Learning Resource 2016 Functions and Graphs Memorandum MATHEMATICS Grade 12 Razzia Ebrahim Senior Curriculum Planner for Mathematics E-mail: Razzia.Ebrahim@wced.info / Razzia.Ebrahim@westerncape.gov.za Website: http://www.wcedcurriculum.westerncape.gov.za/index.php/component/jdownloads/category/1835- grade-12?Itemid=-1 Website: http://wcedeportal.co.za

Mathematics P1 Graphs/Wiskunde V1 Grafieke Memorandum 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 – 32 Copyright reserved/Kopiereg voorbehou 2 Please turn over/Blaai om asseblief

Mathematics P1 Graphs/Wiskunde V1 Grafieke Memorandum DBE/2016 QUESTION/VRAAG 4 X 4.1 (0; 3) ./ (0; 3) (1) 4.2 x=-- b 2 or -2x-2=0 ./ x=- (- ) or -2x-2=0 2(-1) 2a =--- (-2) .'. X =-1 2(-1) ./ simplification =-1 4ac-b2 ./ in the context of a turning y=-(-1) -2(-1)+3 2 or y= point 4a =4 = 4(-1X3)-(-2)2 -(-1)2 -2(-1)+3 4(-1) 4(-IX3)-(-2)2 4(-1) c(- 1;4) (3) 4.3 B(l; o) By symmetry/Deursimmetrie A(-3; o) � A(-3; 0) OR/OF (1) x 2 +2x-3 = 0 (x+3Xx-1)= 0 x=-3 or x=l � A(-3; 0) A(-3;0) (1) Copyright reserved/Kopiereg voorbehou 3 Please turn over/Blaai om asseblief

Mathematics P1 Graphs/Wiskunde V1 Grafieke Memorandum DBE/2016 Copyright reserved/Kopiereg voorbehou 4 Please turn over/Blaai om asseblief

Mathematics P1 Graphs/Wiskunde V1 Grafieke Memorandum DBE/2016 4.7 g(x)� g- 1 (x) 2x+6�-- x-6 x-6 ..1'2x+6�-- 2 4x+12�x-6 ,I' 4x+12 � x-6 3x�-18 x�--6 ..l'x�-6 (3 ) [211 QUESTION/nuAG 5 5.1 r=2 ./r=2 (1) x 5.2 g(x)=2 +2 ..I' g(O) = 2 + 2 ° g{0>::2 ° +2=3 B(O; 3) ..l'y=3 3=_3_+2 ..l'substitute B(O; 3) and q= 2 0-p ..l'p=-3 p=-3 (4) 5.3 atA: x=-3 ./ at A: x- -3 (p-value) y=2-3 +2 =2{ ..I' substitute x = -3 into exponential eqution A(-3; 2i) or ,{-3; 1;) or A{-3; 2,125) ..I' y-value (3) 5.4 -3 < x � 0 ORI OF (-3; o] ./-3<x ./x�O (2 ) 3 5.5 f(x)=-+2 x+3 3 f(x-2) = +2 ./substitution of x - 2 x-2+3 3 3 ./ h(x) =-+2 h(x)==- -+2 x+l x+l (2) r121 Copyright reserved/Kopiereg voorbehou 5 Please turn over/Blaai om asseblief

Mathematics P1 Graphs/Wiskunde V1 Grafieke Memorandum DBE/Feb.-Mrt. 2016 QUESTION/VRAAG 4 4.1 (0 ; 2) answer (1) 4.2 y shape  (0; 2 ) (0; 2) y =1 asymptote x (3) 4.3 f (− 2 ) = 5  f (− 2 ) = 5 3 3 f (1) = 2 −1 + 1 =  f (1) = 2 2 f (1) − f (− 2 ) Average gradient = 1 − (− 2 ) 3 −5 = 2 3 7  answer =− (3) 6 4.4 Since the asymptote of f is y = 1 , the asymptote of h( x ) = 3 f (x ) will be y = 3 . answer Omdat die asimptoot van f y = 1 is, (1) sal die asimptoot van h( x ) = 3 f (x ) y = 3 wees. [8] Copyright reserved/Kopiereg voorbehou 6 Please turn over/Blaai om asseblief

Mathematics P1 Graphs/Wiskunde V1 Grafieke Memorandum DBE/Feb.-Mrt. 2016 QUESTION/VRAAG 5 5.1 y = a( x + p )2 + q Turning point (1 ; − 8) : y = a( x − 1)2 − 8  y = a(x − 1) − 8 2 Substitute (0 ; − 4) : − 4 = a(0 − 1)2 − 8 substitute (0;−4) − 4 = a −8 a = 4 p = −1 q = −8 a = 4  p and q values y = 4( x − 1) − 8 2 (4) 5.2 Asymptote is y = −2 ⇒ d = −2  d = −2 Substitute (1; − 8) : k −8 = −2 1+ r k = −6(1 + r )  k = −6 − 6r k = −6 − 6r ……………….line 1 Substitute (0 ; − 4 ) : k −4= −2 r k = −2 r  k = −2r k = −2r …………..………line 2 Equating lines 1 and 2:  − 6 − 6r = −2r − 6 − 6r = −2r − 4r = 6 3  value of r r=− 2 Substituting into line 2 or line 1:  3  3 k = (− 2 ) −  = 3 k = −6 − 6 −  = 3  value of k  2  2 (6) 5.3 g (x ) ≥ f (x )  0≤ x ∴0 ≤ x ≤ 1  x ≤1 (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  −8 < k kant van die x-as sny.  k < −4 − 8 < k < −4 (2) Copyright reserved/Kopiereg voorbehou 7 Please turn over/Blaai om asseblief

Mathematics P1 Graphs/Wiskunde V1 Grafieke Memorandum DBE/Feb.-Mrt. 2016 5.5 y = −x + c  y = −x + c 3  Substitute the intersection point of the asymptotes, i.e.  ; − 2  : 2  3 3    −2= − +c Vervang die snypunt van die asimptote, m.a.w.  ; − 2  : 2 2  3 −2=− +c 2 1 answer c=− 2 (3) 1 y = −x − 2 OR/OF 3 y = − x is translated units right and 2 units down/ 2  y = −x 3 y = − x transleer eenhede na regs en 2 eenhede na onder ⇒ 2 3 3  y = − x −  − 2   2 y = − x −  − 2  2 answer 1 (3) y = −x − 2 5.6 By symmetry, 3 3  Q =  + 8 − 2; − 2 + − 1 15 2 2  x= 2  15 3  =  ;−  3  2 2 y=− 2 (2) [19] Copyright reserved/Kopiereg voorbehou 8 Please turn over/Blaai om asseblief

Mathematics P1 Graphs/Wiskunde V1 Grafieke Memorandum DBE/Feb.-Mrt. 2016 QUESTION/VRAAG 6 6.1 1 2 f : y= x 4 1 f −1 : x = y 2 interchanging x and 4 y 2 y = 4x  y2 = 4x y = ± 4x f −1 ( x) = − 4 x OR/OF f −1 ( x) = −2 x answer (3) 6.2 y both graphs pass f through (0 ; 0) (-2; 1) x shape for both one additional (1; -2) point on both graphs f -1 (3) 6.3 Yes. No value of x in the domain of f -1 maps onto more than one yes y-value. reason Ja. Geen waarde van x in die definisieversameling van f -1 (2) assosieer met meer as een y-waarde nie. OR/OF Yes. One to one function./Ja. Een-tot-een-funksie. yes reason OR/OF (2) Yes. Vertical line test holds./Ja. Die vertikale lyntoets werk. yes reason (2) [8] Copyright reserved/Kopiereg voorbehou 9 Please turn over/Blaai om asseblief

Mathematics P1 Graphs/Wiskunde V1 Grafieke Memorandum DBE/November 2015 QUESTION/VRAAG 4 Given: f ( x) = 2 x +1 − 8 4.1 y=–8 answer (1) 4.2 y 8 6 4 x-intercept f y-intercept 2 shape x asymptote -8 -6 -4 -2 0 2 4 6 8 -2 -4 -6 y = -8 -8 (4) 4.3 g (x ) = 2 − x +1 − 8 answer (1) OR/OF answer x −1 1 g (x ) =   −8 (1) 2 [6] Copyright reserved/Kopiereg voorbehou 10 Please turn over/Blaai om asseblief

Mathematics P1 Graphs/Wiskunde V1 Grafieke Memorandum DBE/November 2015 QUESTION/VRAAG 5 Given h( x ) = 2 x − 3 for − 2 ≤ x ≤ 4 . y h Q x -2 O 4 P -3 5.1 For x-intercepts, y = 0 2x − 3 = 0 x = 1,5  x = 1,5 y=0 Q(1,5 ; 0) (2) 5.2 h: x = −2 : y = 2(−2) − 3 = −7 –7 x = 4: y = 2(4) − 3 = 5 5 −1 Domain of h : −7 ≤ x ≤ 5 −7 ≤ x ≤ 5 (3) intercepts y 5.3 (5 ; 4) shape endpoints h-1 1,5 x -3 0 (-7; -2) (3) 5.4 h( x ) = 2 x − 3 For the inverse of h, x = 2y − 3 x+3 x+3 y= y= 2 2 h( x ) = h −1 ( x ) x+3 x+3 2x − 3 =  2x − 3 = 2 2 4x − 6 = x + 3 3x = 9 x=3 x=3 (3) OR/OF Copyright reserved/Kopiereg voorbehou 11 Please turn over/Blaai om asseblief

Mathematics P1 Graphs/Wiskunde V1 Grafieke Memorandum DBE/November 2015 h( x ) = 2 x − 3 h and h −1 intersect when y = x  h( x ) = x h( x ) = x 2x − 3 = x  2x − 3 = x x=3 x=3 (3) 5.5 OP = x + y 2 2 2  OP 2 = x 2 + y 2 = x 2 + (2 x − 3) 2 substitute = x 2 + 4 x 2 − 12 x + 9 y = 2x − 3 = 5 x 2 − 12 x + 9  5 x 2 − 12 x + 9 For OP to be at its minimum, OP 2 has to be a minimum Vir OP om minimum te wees, moet OP2 'n minimum wees d (OP2 ) b =0 or x=− dx 2a − 12 10 x − 12 = 0 =− 2(5) 6 x-value ∴x = 5 2 6 6 9 3 Minimum length of OP = 5  − 12  + 9 = or or 1,34 units 5 5 5 5 answer (5) OR/OF \ m h = 2 (given) −1 mOP = −1 2  mOP = −1 2 ∴ OP has equation y = x equation of 2 −1 OP x = 2x − 3 2 −1 − x = 4x − 6  x = 2x − 3 2 5x = 6 x P = 1,2 yP = − 1 (1,2) = −0,6  substitution 2 into distance OP = (1,2 − 0)2 + (− 0,6 − 0)2 formula answer = 1,34 or 1,8 units (5) OR/OF Copyright reserved/Kopiereg voorbehou 12 Please turn over/Blaai om asseblief

Mathematics P1 Graphs/Wiskunde V1 Grafieke Memorandum DBE/November 2015 Minimum distance between the line ax + by + c = o and the point ( xO ; yO ) Minimum afstand tussen die lyn ax + by + c = o en die punt ( xO ; yO ) axO + byO + c formula = a2 + b2  substitution 2(0 ) + (− 1)(0 ) − 3 = 2 2 + (− 1) 2 answer 3 (5) = 5 5.6.1 h( x) = f ′( x) < 0 for x ∈ (− 2 ; 1,5) ⇒ f is decreasing on the left of Q / decreasing f is dalend links van Q. left of Q h( x) = f ′( x) > 0 for x ∈ (1,5 ; 4 ) ⇒ f is increasing on the right of Q / increasing right of Q f is stygend regs van Q. (2) OR/OF 3 3 h  = f ′  = 0 3 2 2  f ′  = 0 2 Since h ′( x ) = 2 f ′′( x ) = 2 > 0  f ′′( x ) = 2 > 0 3 f ′′  > 0 2 3 f has a local minimum at x = by the second derivative test. 2 (2) 3 f het 'n lokale minimum by x = deur die tweede afgeleide toets. 2 OR/OF  f ( x) = x − 3 x + c 2 f ( x) = x 2 − 3 x + c f has a minimum value since a > 0/f het 'n minimum waarde omdat  explanation a>0 (2) 5.6.2 m = f ′(4) = h(4 ) = 5 answer (1) [19] Copyright reserved/Kopiereg voorbehou 13 Please turn over/Blaai om asseblief

Mathematics P1 Graphs/Wiskunde V1 Grafieke Memorandum DBE/November 2015 QUESTION/VRAAG 6 6.1.1 T (0 ;18) answer (1) 6.1.2 − 2 x + 18 = 0 2 equating to 0 (x − 3)(x + 3) = 0 factors Q(3 ; 0) coordinates of Q OR/OF (3) − 2 x 2 + 18 = 0 equating to 0 x2 = 9  x2 = 9 Q(3 ; 0) coordinates of Q (3) 6.1.3 x-coordinate of S is 4,5/x-koördinaat van S is 4,5  R(6 ; 0) By symmetry about the line x = 4,5/Deur simmetrie om die (2) lyn x = 4,5: R(6 ; 0) 6.1.4 For all x ∈ R answer (2) 6.2 If C ( x; y ) is the centre of the hyperbola/As C ( x; y ) die middelpunt is van die hiperbool asymptote y = x + 6 and x = −2 y=4 ∴ y = −2 + 6 = 4 y asymptote y=4 x = −2 shape x 0 (4) x = -2 [12] Copyright reserved/Kopiereg voorbehou 14 Please turn over/Blaai om asseblief

Mathematics P1 Graphs/Wiskunde V1 Grafieke Memorandum DBE/2015 QUESTION/VRAAG 4 4.1 R , y ≠ −2  y ≠ −2 (2) OR/OF  (− ∞;−2) (− ∞;−2) ∪ (− 2; ∞ )  (− 2; ∞ ) (2) 4.2 a g ( x) = −2 x −1 a substitution of −5 = −2 the point (0; –5) −1 in to g(x) 5=a+2 a=3 answer (2) 4.3 For g, asymptotes intersect at/Vir g, asimptote sny by (1; − 2 ) ∴ For/Vir y = g ( x − 3) + 7 , asymptotes will intersect at/  (1; − 2 ) for g asimptote sal sny by (1 + 3 ; − 2 + 7 ) x = 4 y = 5 i.e./d.i. at/by (4 ; 5) (3) OR/OF a g ( x) = −2 x −1 y = g ( x − 3) + 7 subs 3 = −2+7 x − 3 −1 3 x = 4 = +5 x−4 y = 5 (4 ; 5) (3) [7] Copyright reserved/Kopiereg voorbehou 15 Please turn over/Blaai om asseblief

Mathematics P1 Graphs/Wiskunde V1 Grafieke Memorandum DBE/2015 QUESTION/VRAAG 5 −2 5.1 1 y=  substitution 4 = 42 = 16  answer (2) x 5.2 1 y=  4 y 1 f −1 : x =   interchange 4 x and y y = log 1 x or y = − log 4 x 4  answer (2) –1 –1 5.3 f and f are symmetrical about the line y = x, to obtain f , reflect f in the line y = x. reflect in 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. (1) 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 swop x and y −1 will lie on 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 (1) is (0 ; 1) en (–2 ; 16). Die ooreenstemmende punte op f −1 sal dus (1 ; 0) and/en (16 ; –2) wees. y 5.4  shape of f –1 (1 ; 0) x x-int of f –1at 1 f -1 (2) Copyright reserved/Kopiereg voorbehou 16 Please turn over/Blaai om asseblief

Mathematics P1 Graphs/Wiskunde V1 Grafieke Memorandum DBE/2015 5.5 x>0 x>0 (1) OR/OF  (0 ; ∞ ) (0 ; ∞ ) (1) 5.6 f −1 ( x ) ≥ −2 From 5.1, f −1 (16 ) = −2 x>0 0 < x ≤ 16 or x ∈ (0 ; 16 ]  x ≤ 16 (2) 5.7.1 1 1 q= (using a calculator/gebruik 'n sakrekenaar) q = 2 2 (1) OR/OF Without a calculator (not necessary)/Sonder sakrekenaar (nie nodig) 1 1 q = log 1 q = log 1 2 4 2 4 1 q 1 1 = log 4 2 q= 2 1 2 −2 q = 2 −1 OR/OF log 4 2q = 1 1 − log 2 q = 1 q= q= − 2 log 2 2 2 (1) 1 q= 2 5.7.2 At the intersection point of f and f -1, y = x (by symmetry). Thus need only solve f −1 ( x ) = x (instead of f ( x ) = f −1 (x ) ) By die snypunt van f en f -1 , y = x (deur simmetrie). Slegs nodig om f −1 ( x ) = x op te los (in plaas van f ( x ) = f −1 (x ) log 1 x = x 4 1 1  = log 1 1 1 2 2 log 1 = from 5.7.1 4 2 2 1 4 x= 1 2 x= 2 1 1 y= y= 2 2 (3) 1 1  ;  2 2 OR/OF 1 1 By/Van 5.7.1, = log 1 2 2 1 1 4  = log 1 2 2 1 1 4 Which means that  ;  lies on the graph of f −1 ./ 2 2 Copyright reserved/Kopiereg voorbehou 17 Please turn over/Blaai om asseblief

Mathematics P1 Graphs/Wiskunde V1 Grafieke Memorandum DBE/2015 1 1 Wat beteken  ;  lê op die grafiek van f −1 . 1 2 2 x= 2 1 1 1 1 But clearly,  ;  lies on y = x /Maar,  ;  lê op y = x 1 2 2 2 2 y= 2 1 1 1 1 Hence  ;  is the intersection point of f and f -1 /Dus is  ;  2 2 2 2 (3) -1 die snypunt van f en f [14] QUESTION/VRAAG 6 6.1 − 3x 2 − 9 x + 30 = 0  − 3x 2 − 9 x + 30 = 0 x 2 + 3x − 10 = 0 factors (x + 5)(x − 2) = 0 answers x = −5 or x = 2 AB = 7 AB = 7 units (4) 6.2 2 − 3 x − 9 x + 30 = −12 x + 12 equating of equations − 3 x 2 + 3 x + 18 = 0 x2 − x − 6 = 0  x2 − x − 6 = 0 (x − 3)(x + 2) = 0 factors x = −2 or x = 3 x =3 At K, x > 0 , hence y = −12(3) + 12 = −24 y= –24 K (3 ; − 24 ) (5) 6.3 f (x ) ≤ g (x )  x ≤ −2 x ≤ −2 or x ≥ 3  x≥3  or OR/OF (3) f (x ) ≤ g (x )  (−∞ ; − 2] x ∈ (−∞ ; − 2] or [ 3; ∞)  [ 3; ∞)  or OR/OF (3) − 3x 2 − 9 x + 30 − (− 12 x + 12) ≤ 0 − 3x 2 + 3x + 18 ≤ 0 x2 − x − 6 ≥ 0  x ≤ −2 (x − 3)(x + 2) ≥ 0  x≥3 x ≤ −2 or x ≥ 3  or (3) Copyright reserved/Kopiereg voorbehou 18 Please turn over/Blaai om asseblief

Mathematics P1 Graphs/Wiskunde V1 Grafieke Memorandum DBE/2015 6.4 CD = −3 x 2 − 9 x + 30 − (−12 x + 12)  CD = y f − y g = −3 x 2 + 3 x + 18  − 3 x 2 + 3 x + 18 b x=− 2a OR/OF f ′( x ) = 0 ( ) OR/OF CD = -3 x 2 − x + 18  1 1  method 2 2 −3 = − 6x + 3 = 0 = −3 x −  −    + 18 2(− 3)  2   2   2 1 1  1 3 1 = x= = −3 x −  + + 18 x = 2 2  2 4 2 2  1 3 = −3 x −  + 18  2 4 Max length /Maks lengte CD OR/OF Max length /Maks lengte CD 2 1  1  3 = −3  + 3  + 18 = 18 2  2  4 75 = 4 max length 75 3 3 CD = or 18 = 18 4 4 4 (5) [17] Copyright reserved/Kopiereg voorbehou 19 Please turn over/Blaai om asseblief

Mathematics P1 Graphs/Wiskunde V1 Grafieke Memorandum DBE/Feb.-Mrt 2015 QUESTION/VRAAG 4 4.1 x = −2  x = −2 y = −1  y = −1 (2) 4.2.1 6 g (0) = −1 0+2 =2 y-intercept/afsnit (0 ; 2)  answer/antwoord (1) 4.2.2 6  equating to/stel gelyk aan 0 0= −1 x+2 6 1= x+2 x+2=6 x=4  answer/antwoord x-intercept/afsnit (4 ; 0) (2) 4.3  asymptotes/asimptote  intercepts/afsnitte  shape/vorm (3) 4.4 y + 1 = −( x + 2 ) m = – 1 y = −x − 3 substitution of (–2 ; –1) answer OR/OF (3) Using general formula/Gebruik algemene formule: y = −( x + p ) + q formula/formule substitution of p and q y = −( x + 2) − 1 values/substitusie van p- en y = −x − 3 q-waardes answer/antwoord (3) 4.5 x > −2  answer (2) [13] Copyright reserved/Kopiereg voorbehou 20 Please turn over/Blaai om asseblief

Mathematics P1 Graphs/Wiskunde V1 Grafieke Memorandum DBE/Feb.-Mrt 2015 QUESTION/VRAAG 5 5.1 9 = a2  9 = a2 a=3 a =3 (2) OR/OF f −1 ( x) = log a x 2 = log a 9  9 = a2 a 2 = 9 = 32 a =3 (2) ∴a = 3 5.2 g ( x) = 3 − x  answer/antwoord (1) OR/OF 1 x g ( x) =    answer/antwoord 3 (1) 5.3 x≥9  answer/antwoord (2) OR/OF f −1 ( x) = log 3 x log 3 x = 2 x = 32 = 9 ∴x ≥ 9  answer/antwoord (2) OR/OF log 3 x ≥ 2 x ≥ 32 ∴x ≥ 9  answer/antwoord (2) 5.4 Yes/Ja. For every y-value there is only one x such that/Vir  Yes/Ja elke y-waarde is daar slegs een x sodanig dat y = f ( x ) .  Reason/Rede (2) OR/OF  Yes/Ja Yes/Ja. f is a one-to-one relation/is 'n een-tot-een-relasie.  Reason/Rede (2) [7] Copyright reserved/Kopiereg voorbehou 21 Please turn over/Blaai om asseblief

Mathematics P1 Graphs/Wiskunde V1 Grafieke Memorandum DBE/Feb.-Mrt 2015 QUESTION/VRAAG 6 6.1 −3≤ x ≤ 2  critical values/ kritiese waardes  notation/notasie (2) 6.2 f : y = a( x − x1 )( x − x 2 ) y = a( x + 3)( x − 2)  y = a ( x + 3)( x − 2) − 8 = a(1 + 3)(1 − 2)  substitute/vervang (1 ; – 8) − 8 = −4a 2=a a = 2 y = 2( x + 3)( x − 2 ) y = 2 x 2 + 2 x − 12 b = 2 and/en c = – 12 b = 2 and/en OR/OF  c = – 12 (5) 2  1 y = a x +  + q  2 2  1 25 0 = a 2 +  + q → 0= a+q ....(1) equation/vergelyking 1  2 4 2  1 9 equation/vergelyking 2 − 8 = a 1 +  + q → −8= a+q .....(2)  2 4 (1) − (2) : 8 = 4a a = 2 a=2 25 25 q=0− (2) = − = −12,5 4 2 2  1 1  substitution/substitusie y = 2 x +  − 12  2 2  1 1 y = 2 x 2 + x +  − 12  4 2 1 1 y = 2 x 2 + 2 x + − 12 2 2 2 y = 2 x + 2 x − 12 b = 2 and/en ∴ b = 2 and c = −12  c = – 12 (5) OR/OF Copyright reserved/Kopiereg voorbehou 22 Please turn over/Blaai om asseblief

Mathematics P1 Graphs/Wiskunde V1 Grafieke Memorandum DBE/Feb.-Mrt 2015 f ′( x) = 2ax + b  1  1 f ′ −  = 2a −  + b = 0  2  2 ∴a = b (− 3;0) : 9a − 3b + c = 0 ∴ 6a + c = 0(1) equation/vergelyking 1 (1;− 8) : a + b + c = −8 ∴ 2a + c = −8(2) equation/vergelyking 2 (1) − (2) : 4a = 8 ∴a = 2 a = 2 ⇒b=2 ∴ c = −12 b=2 c = – 12 (5) 6.3 b x=− 2a 2 1 1 x=− =− x=− 2 2(2) 2 1 substitution/substitusie y = − 1 − 12 2 1 y = −12 2  1 1 TP − ; − 12  y-value/waarde  2 2 OR/OF (3) Copyright reserved/Kopiereg voorbehou 23 Please turn over/Blaai om asseblief

Mathematics P1 Graphs/Wiskunde V1 Grafieke Memorandum DBE/Feb.-Mrt 2015 y = 2[ x 2 + x − 6] 2 2  1  1   y = 2  x 2 + x +  .1 − 6 −  .1   2   2    method/metode  1 2  = 2  x +  − 6,25  2  2  1 = 2 x +  − 12,5  2 x-value/waarde  1  TP − ; − 12,5  y-value/waarde  2  (3) OR/OF −3+ 2 1 x= =− 2 2  1  1  method/metode y = 2 −  + 2 −  − 12 x-value/waarde  2  2 1 y = −12 2  1  TP − ; − 12,5  y-value/waarde  2  (3) OR/OF f ( x) = y = 2 x 2 + 2 x − 12 f / ( x) = 4 x + 2 4x + 2 = 0  method/metode 4 x = −2 1 x=− 2 2 x-value/waarde  1  1 25 ∴ y = 2 −  + 2 −  − 12 = −  2  2 2  1 25  y-value/waarde (3) TP − ; −   2 2  6.4 13 x=  answer/i 2 (2) 6.5 / f ( x) = 4 x + 2 /  y = 4x + 2 m = f ′(1) = 4(1) + 2  subst. x = 1  answer/antwoord m=6 (3) [15] Copyright reserved/Kopiereg voorbehou 24 Please turn over/Blaai om asseblief

Mathematics P1 Graphs/Wiskunde V1 Grafieke Memorandum DBE/November 2014 QUESTION/VRAAG 4 4.1 p =1 p value /waarde q =1 q value /waarde (2) 4.2 2 2 0= +1 0 = +1 x +1 x +1 − x −1 = 2  x = −3 x = −3 ( 2) OR/OF  reflect across/reflekteer om Reflect (0 ; 3) across y = – x to get T( –3 ; 0) y=–x x = −3 Reflekteer (0 ; 3) om y = – 1 om T( –3 ; 0) te kry x = −3  x = −3 (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 2  equating both graphs/stel +1 = x x +1 grafieke gelyk 2 + x + 1 = x2 + x x2 = 3  x2 = 3  x = 3 and y = 3 ∴ x = 3 since at S, x > 0 y = 3 = 1,73... OS2 = x 2 + y 2 = 3 + 3 = 6  OS 2 = 6 ∴ OS = 6 = 2,45 units/eenhede  answer/antwoord (5) OR/OF Copyright reserved/Kopiereg voorbehou 25 Please turn over/Blaai om asseblief

Mathematics P1 Graphs/Wiskunde V1 Grafieke Memorandum DBE/November 2014 Translate g one unit down and one unit to the right/Transleer g een eenheid af en een eenheid na regs 2 2 The new equation/Die nuwe vergelyking : p ( x) =  p( x) = x x ( Therefore the image of S is S 2 ; 2 / / ) ( Daarom is die beeld van S nou S / 2 ; 2 ) coord. of/koörd. van S / Now translate p back to g/Transleer p terug na g: coord. of/koörd. van S ( S 2 − 1; 2 + 1 ) OS2 = ( 2 − 1) + ( 2 + 1) = 2 − 2 2 + 1 + 2 + 2 2 + 1 2 2 ∴ OS = 6 = 2,45 units/eenhede answer/antwoord (5) 4.5 k < 3 will give roots with opposite signs/ k < 3 (1) k < 3 sal wortels met teenoorgestelde tekens gee [11] Copyright reserved/Kopiereg voorbehou 26 Please turn over/Blaai om asseblief

Mathematics P1 Graphs/Wiskunde V1 Grafieke Memorandum DBE/November 2014 QUESTION 5 5.1 y = log a x 1 1  − 1 = log a subt.  ; − 1 3 3  −1 −11 −11 1 a =  a = or a =   3 3 3 1 −1 (2) a=  3 ∴a = 3 5.2 h : x = log 3 y swop x and y/ruil x en y ∴ y=3 x  answer/antwoord (2) 5.3 g ( x) = − log 3 x answer/antwoord (1) OR/OF 1 answer/antwoord g ( x) = log 3 x (1) OR/OF g ( x) = log 1 x 3 answer/antwoord (1) OR/OF x = 3− y answer/antwoord (1) OR/OF 1 y x=  3 answer/antwoord (1) 5.4 x>0  answer/antwoord (1) OR/OF (0 ; ∞ )  answer/antwoord (1) 5.5 log 3 x = −3 x = 3 −3  exponential form/ 1 eksponensiële vorm x= 27 simplification/vereenvoudiging 1 x≥ answer/antwoord (3) 27 [9] Copyright reserved/Kopiereg voorbehou 27 Please turn over/Blaai om asseblief

Mathematics P1 Graphs/Wiskunde V1 Grafieke Memorandum DBE/November 2014 QUESTION/VRAAG 6 6.1 4x2 − 6 = 0 y = 0 3 x2 = 2 x = 1,22 ( x - coordinate of S is positive) 1,22 (2) 6.2 ( 0 ; – 6) 0 – 6 (2) 6.3.1 QT = f ( x) − g ( x)  correct formula/ korrekte formule = 2 x − (4 x 2 −6) or = 2 x − 4 x 2 +6 substitution/substitusie (3) 6.3.2 1 QT = 2 x − 4 x 2 + 6 2 −1 derivative/afgeleide Deravitive of QT = x 2 − 8 x = 0 derivative equal to 0/ 1 afgeleide gelyk aan 0 = 8x x 3 3 1 1 1 x2 = or = 64 x 2 x =2 8 x 8 2 13 x=  8 2 1 1 x =   or x 3 = 2 64 1 x-value/x-waarde x = = 0,25 4 1 2  12 1  substitution/substitusie Max/Maks QT = 2  − 4  + 6 4 4 3 answer/antwoord (6) = 6 = 6,75units/eenhede 4 [13] Copyright reserved/Kopiereg voorbehou 28 Please turn over/Blaai om asseblief

Mathematics P1 Graphs/Wiskunde V1 Grafieke Memorandum DBE/2014 QUESTION/VRAAG 4 4.1.1 2 f ( x) = −3 x +1 y = f (0) 2 = −3 subst x = 0 0 +1 = −1  (0 ; −1) (0 ; −1) (2) 4.1.2 2 0= −3  subs y = 0 x +1 2 3= x +1 3x + 3 = 2 1  1  x=−   − ; 0 3  3   1   − ; 0  3  (2) 4.1.3 y A(-1/3 ; 0) 0 x  shape x = -1 B(0;-1)  both intercepts correct y = -3  horizontal and f vertical asymptote (3) 4.1.4 y = −( x + 1) − 3  y = −( x + 1) − 3 y = −x − 4  y = −x − 4 (2) OR y = −x + k − 3 = −(− 1) + k  − 3 = −(− 1) + k k = −4  y = −x − 4 y = −x − 4 (2) Copyright reserved/Kopiereg voorbehou 29 Please turn over/Blaai om asseblief

Mathematics P1 Graphs/Wiskunde V1 Grafieke Memorandum DBE/2014 4.2.1 y = a.b x + q y = a.b x − 3 subs q = – 3 0 − 2 = a.b − 3 [subs (0 ; − 2 )] a =1  a =1 y = 1.b x − 3 [subs (1;−1)] − 1 = b1 − 3 b=2 b = 2 f (x ) = 2 x − 3  f (x ) = 2 x − 3 (4) 4.2.2 A translation of 4 units up and 1 unit to the left. 4 units up 'n Translasie van 4 eenhede na bo en 1 eenheid na links.  1 unit to the left (2) OR Dilation by a factor of 2 and 7 units up.  dilation by factor 2 Verkleining deur faktor van 2 en 7 eenhede na bo.  7 units up (2) [15] Copyright reserved/Kopiereg voorbehou 30 Please turn over/Blaai om asseblief

Mathematics P1 Graphs/Wiskunde V1 Grafieke Memorandum DBE/2014 QUESTION/VRAAG 5 5. 1 f ( x ) = −2 x 2 − 5 x + 3 f ′(x ) = 0 b x=− or / f ′( x ) = 0 b 2a x =− 2a  −5  x = −   − 4x − 5 = 0  2(−2)  5 5 5 x =− x=−  x =− 4 4 4 2  5  5 y = −2 −  − 5 −  + 3  4  4 49 49 = / 6,125 y= / 6,125 8 8 5 49 (3) TP ( − ; ) 4 8 OR 5 3 y = - 2( x 2 + x − ) 2 2 5 25 3 5 25 3 = −2[( x + ) 2 − − ]  − 2[( x + ) 2 − − ] 4 16 2 5 49 4 16 2 TP ( − ; ) 5 49 4 8 5 = −2[( x + ) 2 − ]  x =− 4 16 4 5 49 49 = −2( x + ) 2 + y= / 6,125 4 8 8 (3) 5. 2 m tangent = tan 135  =–1  tan 135  = –1 –4x –5 = –1  –4x –5 = –1 – 4x = 4 x = –1  x = –1 y = – 2(–1) 2– 5(–1) + 3 =6 y=6 Point of contact: P(–1; 6) (4) 5. 3 Eq of g: y – y 1 = m(x – x 1 )  substitute in y – 6 = – 1(x + 1) equation y = – x +5  answer (2) 5. 4 d> 5  answer (1) [10] Copyright reserved/Kopiereg voorbehou 31 Please turn over/Blaai om asseblief

Mathematics P1 Graphs/Wiskunde V1 Grafieke Memorandum DBE/2014 QUESTION/VRAAG 6 6.1 g ( x) = ax 4 = a (8) subst (8 ; 4) 8a = 16 a=2 a = 2 (2) 6.2 x≥0 answer (1) 6.3 y≥0 answer (1) 6.4 y = 2x ; x ≥ 0 interchange x and y x2 = 2y x2 answer y= ; y≥0 (2) 2 6.5 2x = x − 4 2 x = x 2 − 8 x + 16  2 x = x 2 − 8 x + 16 0 = x 2 − 10 x + 16 (squaring both sides) 0 = ( x − 8)( x − 2 ) factors x = 8 or x = 2  x = 8 or x = 2 when x = 2, LHS = 2 but RHS = –2 Hence x = 8 only  selects x = 8 (4) 6.6 0< x<8 x<8 0 < x (2) [12] Copyright reserved/Kopiereg voorbehou 32 Please turn over/Blaai om asseblief