# Mathematics Grade 12 Paper 1 NSC November 2017 Memo > Grade 12 Mathematics Paper 1 memorandum from the NSC November Exam 2017, 19 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/mathematics-p1-nov-2017-memo-afr-eng/a0a91548-6e8d-4475-a62d-081ae396a0b7 - Updated: 2026-09-27 - Subject: Mathematics - Grade: Grade 12 - Document type: Memorandum - Year: 2017 - Exam period: NSC November Exam - Paper: 1 - Curriculum: CAPS - Pages: 19 Breadcrumbs: [Home](https://www.hlayiso.com/) › [Library](https://www.hlayiso.com/library) › [Grade 12](https://www.hlayiso.com/library/grade-12) › [Mathematics](https://www.hlayiso.com/library/grade-12/mathematics) › [Mathematics Grade 12 Paper 1 NSC November 2017 Memo](https://www.hlayiso.com/library/mathematics-p1-nov-2017-memo-afr-eng/a0a91548-6e8d-4475-a62d-081ae396a0b7) ## About this document Mathematics · Grade 12 · NSC November Exam · 2017. Memorandum, 19 pages. Read online or download the PDF. Read it free online on the page above; download the PDF from there. ## Document text Text extracted from the PDF. Layout, diagrams and tables may not survive extraction. ### Page 1 Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE NASIONALE SENIOR SERTIFIKAAT GRADE 12/GRAAD 12 MATHEMATICS P1/WISKUNDE V1 NOVEMBER 2017 MARKING GUIDELINES/NASIENRIGLYNE MARKS/PUNTE: 150 This memorandum consists of 22 pages. Hierdie memorandum bestaan uit 22 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 2 Downloaded from hlayiso.com Mathematics/P1/Wiskunde/V1 2 DBE/November 2017 NSC/NSS – Marking Guidelines/Nasienriglyne NOTE: • If a candidate answers a question TWICE, only mark the FIRST attempt. • Consistent accuracy applies in ALL aspects of the marking guidelines. LET WEL: • Indien 'n kandidaat 'n vraag TWEE KEER beantwoord, merk slegs die EERSTE poging. • Volgehoue akkuraatheid is op ALLE aspekte van die nasienriglyne van toepassing. QUESTION/VRAAG 1 1.1.1 x 2 + 9 x + 14 = 0  factors (x + 7 )(x + 2) = 0  x = −7 x = −7 or x = −2  x = −2 (3) 1.1.2 2 4x + 9x − 3 = 0 − b ± b 2 − 4ac x= 2a − 9 ± 92 − 4(4 )(− 3)  substitution = 2(4 ) − 9 ± 129  simplification =  x = 0,29 8 x = 0,29 or x = − 2,54  x = − 2,54 OR/OF OR/OF 9 81 3 81 x2 + x+ = + 81 4 64 4 64 for adding on 2 64  9 129 both sides x +  =  8 64 9 129 x+ =±  simplification 8 8 − 9 ± 129 x= 8  x = 0,29 x = 0,29 or x = −2,54  x = − 2,54 (4) 1.1.3 2 x −5 = 2 x 2 x − 5 = 4x  x2 − 5 = 4x 2 x − 4x − 5 = 0  standard form (x − 5)(x + 1) = 0  both answers x = 5 or x = −1 x=5  select x = 5 (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 3 Downloaded from hlayiso.com Mathematics/P1/Wiskunde/V1 3 DBE/November 2017 NSC/NSS – Marking Guidelines/Nasienriglyne 1.2 3x − y = 4 y = 3x − 4  y subject of formula x 2 + 2 xy − y 2 = −2 x 2 + 2 x(3 x − 4 ) − (3 x − 4 ) = −2 2  substitution ( ) x 2 + 6 x 2 − 8 x − 9 x 2 − 24 x + 16 = −2 7 x 2 − 8 x − 9 x 2 + 24 x − 16 = −2 − 2 x 2 + 16 x − 14 = 0 x 2 − 8x + 7 = 0  correct standard form (x − 7 )(x − 1) = 0  factors x =1 or x=7 x-values y = 3(1) − 4 y = 3(7 ) − 4 y = −1 or y = 17 y-values OR/OF 3x − y = 4 OR/OF y+4 x= x subject of formula 3 2 2 x + 2 xy − y = −2 x 2 + 2 xy − y 2 = −2 2  y + 4  y + 4 2  substitution   + 2  y − y = −2  3   3  y + 8 y + 16 + 6 y + 24 y − 9 y 2 = −18 2 2 − 2 y 2 + 32 y + 34 = 0 y 2 − 16 y − 17 = 0  correct standard form  factors ( y − 17 )( y + 1) = 0 y = −1 or y = 17 y-values −1+ 4 17 + 4 x= x= 3 3 x =1 or x=7 x-values (6) 1.3.1 x 2 + 8 x + 16 > 0 (x + 4)(x + 4) > 0  (x + 4 )(x + 4 )  any one of the x ∈ R, x ≠ − 4 or solutions x ∈ (− ∞ ; − 4) or x ∈ (− 4 ; ∞) or x < − 4 or x > − 4 OR/OF OR/OF x 2 + 8 x + 16 > 0 (x + 4)(x + 4) > 0 + +  (x + 4 )(x + 4 ) –4  any one of the The function values remain positive solutions x ∈ R, x ≠ −4 (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 4 Downloaded from hlayiso.com Mathematics/P1/Wiskunde/V1 4 DBE/November 2017 NSC/NSS – Marking Guidelines/Nasienriglyne y 1.3.2 (0;16) y=p  0  16 (-4 ; 0) For two negative unequal roots:   0 < p < 16 (4) 0 < p < 16 OR/OF OR/OF x 2 + 8 x + 16 = p x 2 + 8 x + 16 − p = 0 0 < 16 − p < 16  0 − 16 < − p < 0  16 0 < p < 16   0 < p < 16 (4) OR/OF x 2 + 8 x + 16 − p = 0 − 8 ± 64 − 4(16 − p ) x= 2  0 0 < 64 − 4(16 − p ) < 64  16 0 < 4 p < 64 0 < p < 16   0 < p < 16 (4) OR/OF x 2 + 8 x + 16 = p x 2 + 8 x + 16 − p = 0 Roots are real and unequal : 8 2 − 4(16 − p ) > 0 4p > 0 p>0 −8± 4p Roots are : 2 For both roots to be negative : 4p < 8  0 4 p < 64  16 p < 16 0 < p < 16   0 < p < 16 (4) [24] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 5 Downloaded from hlayiso.com Mathematics/P1/Wiskunde/V1 5 DBE/November 2017 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 2 2.1.1 5 –4 –19 –40 –9 –15 –21 –6 –6 first differences: –9; –15; –21  first differences second difference = – 6 –6 (2) 2.1.2 2 Tn = an + bn + c second difference a= = −3  a = −3 2 3a + b = −9 3(− 3) + b = −9  b=0 b=0 a+b+c =5 −3+0+c = 5  c=8 c=8  Tn = −3n 2 + 8 Tn = −3n 2 + 8 OR/OF OR/OF (n − 1)(n − 2)d 2 Tn = T1 + (n − 1)d 1 + 2 = 5 + (n − 1)(− 9 ) + (n − 1)(n − 2)(− 6)  a = −3 2  b=0 2 = 5 − 9n + 9 − 3n + 9n − 6 c = 8  Tn = −3n 2 + 8 Tn = −3n 2 + 8 (4) 2.1.3 2 − 3n + 8 = −25 939  Tn = −25 939 2 − 3n = −25947 n 2 = 8649  n 2 = 8649 n = −93 or n = 93  answer The 93rd term has a value of –25 939 (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 6 Downloaded from hlayiso.com Mathematics/P1/Wiskunde/V1 6 DBE/November 2017 NSC/NSS – Marking Guidelines/Nasienriglyne 2.2.1 2k − 7 ; k + 8 and 2k − 1 k + 8 − (2k − 7 ) = 2k − 1 − (k + 8)  k + 8 − (2k − 7 ) = 2k − 1 − (k + 8) − k + 15 = k − 9 2k = 24 k = 12 k = 12 2k − 7; k + 8 and 2k − 1  17 17 ; 20 ; 23.......  d =3 d =3 T15 = 17 + 14(3) = 59  T15 = 59 (5) 2.2.2 Sequence is 17 ; 20 ; 23 ; 26 ; 29 ; 32 ……. Every alternate term of the sequence will be even / Elke tweede term van die ry sal ewe wees 20 + 26 + 32 + .......  20 + 26 + 32 + ....... 30 S 30 = [2(20) + (29)(6)]  a = 20 d = 6 2 = 15[40 + 174] subst into correct formula = 3210  answer (4) OR/OF T 30 = 20 + 29(6)  a = 20 d = 6 = 94  T30 = 94 30 S 30 = (20 + 194) 30  S 30 = (20 + 194 ) 2 2 = 3210  answer (4) [18] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 7 Downloaded from hlayiso.com Mathematics/P1/Wiskunde/V1 7 DBE/November 2017 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 3 3.1 a + ar = 2  a + ar = 2 a(1 + r ) = 2 2 2 a=  a= 1+ r 1+ r OR/OF (2) a 1 −2= a 1 1− r 4  −2= 4a − 8(1 − r ) = 1 − r 1− r 4 4a − 8 + 8r = 1 − r 4a = 9 − 9r 9 − 9r  a= 9 − 9r 4 a= (2) 4 OR/OF OR/OF a (r n − 1) Sn = r −1 a (r 2 − 1)  2= ( a r 2 −1 ) 2= r −1 r −1 a(r − 1)(r + 1) 2= r −1 2 = a(r + 1) 2 2  a= a= 1+ r r +1 (2) OR/OF OR/OF ar 2 1 ar 2 1  = = 1− r 4 1− r 4 1− r 1− r a= 2  a= 4r 4r 2 (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 8 Downloaded from hlayiso.com Mathematics/P1/Wiskunde/V1 8 DBE/November 2017 NSC/NSS – Marking Guidelines/Nasienriglyne ∞ 3.2 S ∞ = T1 + T2 + ∑ Tn n =3 1 1 S∞ = 2 +  S∞ = 2 + 4 4 a 1 = 2+ 1− r 4 a 9 a 9 =  = 1− r 4 1− r 4  2   1  9  × =  substitution of a into the 1+ r  1− r  4 correct formula 2 9 = 1− r 2 4 8 = 9 − 9r 2 9r 2 = 1  9r 2 = 1 1 1 r=  r= 3 3 3 3 a= a = 2 2 (6) OR/OF OR/OF ∞ S ∞ = T1 + T2 + ∑ Tn n =3 1 1 S∞ = 2 +  S∞ = 2 + 4 4 a 1 = 2+ 1− r 4 a 9 a 9 =  = 1− r 4 1− r 4 4a = 9 − 9r 9 − 4a 9 − 4a r=  r= 9 9  9 − 4a  a + a =2  9  9a + 9a − 4a 2 = 18 2a 2 − 9a + 9 = 0  substitution of a into the (a − 3)(2a − 3) = 0 correct formula 3 a= or a = 3 2 1 1 r= or r = − 3 3 3  a= N/A 2 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 9 Downloaded from hlayiso.com Mathematics/P1/Wiskunde/V1 9 DBE/November 2017 NSC/NSS – Marking Guidelines/Nasienriglyne 1  r= OR/OF 3 2−a (6) r= a ar 2 1 OR/OF = 2−a 1− r 4 r = a 4ar 2 = 1 − r 2 ar 1 2−a 2 2−a  = 4a   = 1− 1− r 4  a  a 2 16 − 16a + 4a = 2a + 2  substitution of a 2a 2 − 9a + 9 = 0 (2a − 3)(a − 3) = 0 3 a= 2 a≠3  (2a − 3)(a − 3) = 0 1 1 3 r= r≠− a = 3 3 2 1  r= OR/OF 3 ∞ (6) S ∞ = T1 + T2 + ∑ Tn OR/OF n =3 1 S∞ = 2 + 4 a 1 = 2+ 1 1− r 4  S∞ = 2 + a 9 4 = 1− r 4 1− r   1  9 a 9  2 ×  =  =  4r   1 − r  4 1− r 4 1 9 = 4r 2 4  substitution of a 4 = 36r 2 9r 2 = 1 1 r= 3  9r 2 = 1 3 1 a=  r= 2 3 3 a = 2 (6) [8] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 10 Downloaded from hlayiso.com Mathematics/P1/Wiskunde/V1 10 DBE/November 2017 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 4 4.1 f (x ) = −ax 2 + bx + 6 f / (x ) = −2ax + b – 2ax + b − 2ax + b = 3 at x = −1  2a + b = 3 2a + b = 3 [1] 7 f (− 1) = 2 7 7 −a−b+6 = − a −b + 6= 2 2 − 2a − 2b + 12 = 7 2a + 2b = 5 [2] solve simultaneously [2] − [1] (5) b=2 2a + 2 = 3 1 a= 2 OR/OF f / (x ) = −2ax + b 3 = 2a + b b = 3 − 2a – 2ax + b  2a + b = 3 7 = −a(− 1) + (3 − 2a )(− 1) + 6 2 2 7  a+3= 7 2 = −a(− 1) + (3 − 2a )(− 1) + 6 2 2 1 a= 2 solve simultaneously b=2 (5) 4.2 1 2 f (x ) = − x + 2x + 6 2 x − intercepts : 1 2 1  − x + 2x + 6 = 0 − x 2 + 2x + 6 = 0 2 2 -x 2 + 4 x + 12 = 0 x 2 − 4 x − 12 = 0  (− 2 ; 0 ) (x − 6)(x + 2) = 0  (6 ; 0) (− 2 ; 0) (6 ; 0) (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 11 Downloaded from hlayiso.com Mathematics/P1/Wiskunde/V1 11 DBE/November 2017 NSC/NSS – Marking Guidelines/Nasienriglyne 1 4.3 f (x ) = − x 2 + 2 x + 6 2 2 − x+2 / − / b −2+6  1 f / (x ) = 0 or x=− or x= 2. −  2a 2  2 2 −2+6 −x+2=0 x=− x=2 2  1 2. −   2 x=2 x=2 x =2 1 2 y=− (2) + 2(2) + 6 2  y=8 = −2 + 4 + 6 =8 TP (2 ; 8) OR/OF OR/OF `1 − (x − 2)2 + 8 2 1 2 y=− 2 ( x − 4 x − 12 ) 1 [ = − ( x − 2 ) − 4 − 12 2 2 ] x =2  y=8 `1 (3) = − (x − 2) + 8 2 2 TP (2 ; 8) 4.4 y (2 ; 8) 4.6 4.4: (0 ; 6) f: shape f x- intercepts (- 1 ; 0) x y- intercept (- 2 ; 0) (2 ; 8) (6 ; 0) (4) (0 ; - 1) g 4.6: g:  x- intercept  y- intercept (2) 4.5 0 < x < 4 or (0 ; 4) 4  0 < x < 4 (3) 4.7 x ≤ −2 or − 1 ≤ x ≤ 6  x ≤ −2  − 1 ≤ x ≤ 6 OR/OF (3) (−∞ ; − 2] or [-1 ; 6] [23] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 12 Downloaded from hlayiso.com Mathematics/P1/Wiskunde/V1 12 DBE/November 2017 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 5 5.1 y ∈ R ; y ≠ −1 OR/OF y < −1 or y > −1  answer OR/OF y ∈ (− ∞ ; − 1) or y ∈ (−1; ∞) OR/OF (2) R − {−1} 5.2 D(2 ; − 1)  D(2 ; − 1) 2 2 g (x ) = −1  −1 (2) x−2 x−2 5.3 f (x ) = log 3 x. 2 correct substitution of A log 3 t = 1 OR/OF g ( x) = −1 x−2  t = 3 t =3 (3) 2 1= −1 t−2 2 2= t−2 2t − 4 = 2 t =3 5.4 x = log 3 y interchange x and y y = 3x  y = 3x (2) 5.5 3 x < 31  3 x < 31 x <1  x <1 (2) OR/OF 3 x < 31  3 x < 31 x ∈ (− ∞ ; 1)  x ∈ (− ∞ ; 1) (2) 5.6 Equation of the axis of symmetry: y = − x + 1  equation of axis of x-intercept of the axis of symmetry is at x = 1 symmetry f has an x-intercept at B(1 ; 0) which is the same as the x-intercept of the axis of symmetry Point of intersection: B (1 ; 0 ) B or (1 ; 0 ) OR/OF OR/OF Since BE = ED = 1 and D lies on the axis of symmetry and the gradient of the axis of symmetry is –1, B will also lie on the axis of symmetry. But B also lies on f.  BE = ED = 1 Therefore B(1 ; 0) is the point of intersection between f and the axis of symmetry with a negative gradient./  B or (1 ; 0 ) Omdat BE = ED = 1 en D op die simmetrie-as lê en die simmetrie-as se gradiënt –1 is, sal B ook op die simmetrie-as lê. Maar B lê ook op f. Dus is B(1 ; 0) die snypunt van f en die simmetrie-as met negatiewe (3) gradiënt. [14] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 13 Downloaded from hlayiso.com Mathematics/P1/Wiskunde/V1 13 DBE/November 2017 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 6 6.1 A = P(1 + i ) n 36 r  r   12 146,72 = 10 0001 +  12  12   n = 36 36  r  1 +  = 1,214672  correct substitution into  12  formula r 36 1+ = 1,214672 12 r 36  1+ = 1,214672 = 1,005416 12 r = 0,005416 12 r = 0,06500 r = 6,5%  6,5% (5) 6.2.1 P= [ x 1 − (1 + i ) −n ] i   0,11  −54  0,11 x 1 − 1 +   i=   12   12 235 000 =  n = 54 0,11  correct substitution 12 in P 0,11 235 000 × x= 12   0,11  −54  1 − 1 +     12   = R 5 536,95  answer His monthly instalment is R 5 536,95 (4) 6.2.2 Amount paid for the year : (5 536,95 ×12 ) = R 66 443,40  R 66 443,40 12 0,11   235 0001 +   0,11 12   12  5 536,95 1 +  − 1  0,11  12  12    0,11 12  Balance = 235 0001 +  − 5 536,951 +  − 1 12  0,11   12    0,11 12 12 = 192 296,17  R192 296,17 Interest = (5 536,95 × 12) − (235 000 − 192 296,17 )  R42 703,83 = 66 443,40 − 42 703,83  R23 739,57 = 23 739,57 OR/OF OR/OF Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 14 Downloaded from hlayiso.com Mathematics/P1/Wiskunde/V1 14 DBE/November 2017 NSC/NSS – Marking Guidelines/Nasienriglyne Total amount paid in first year = R 5 536.95 × 12 = R 66 443,40 Balance on loan after 1 year = P of remaining installments  R 66 443,40 P= [ ] x 1 − (1 + i ) −n i   0,11  −42   n = – 42 5 536,951 − 1 +     12   =  substitution into 0,11 correct formula 12 = R192 296,20 Amount paid off in the first year:  R192 296,20 R235 000 – R192 296,20 = R42 703,80 Amount of interest = R 66 443,40 – R42 703,80  R42 703,80 = R23 739,60  R23 739,60 OR/OF (6) OR/OF   0,11  −12  5536,951 − 1 +     12   P= 0,11 12 = R 62 648,18  R62 648,18 235 000 – 62 648,18 = R172 351,82  R172 351,82 After 12 months, money owed on house is 12  0,11  172 351,821 +   12  = 192 296,17  R192 296,17 Amount paid after 12 months is 5 536,95 × 12 = R 66 443, 40  R66 443,40 Amount of interest paid: R 66 443, 40 – (235 000 – 192 296,17)  235 000 – 192 296,17 = R 23 739, 57  R23 739,57 (6) [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 15 Downloaded from hlayiso.com Mathematics/P1/Wiskunde/V1 15 DBE/November 2017 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 7 7.1 f ( x + h) = 2( x + h) 2 − ( x + h ) = 2( x 2 + 2 xh + h 2 ) − x − h 2 2  2 x 2 + 4 xh + 2h 2 − x − h = 2 x + 4 xh + 2h − x − h f ( x + h) − f ( x) = 2 x 2 + 4 xh + 2h 2 − x − h − 2 x 2 + x  4 xh + 2h 2 − h = 4 xh + 2h 2 − h f ( x + h) − f ( x ) f ( x + h) − f ( x )  f / ( x) = lim f / ( x) = lim h →0 h h →0 h 4 xh + 2h 2 − h subst. into formula = lim h →0 h h(4 x + 2h − 1) = lim  lim(4 x + 2h − 1) h →0 h h →0 = lim (4 x + 2h − 1) h →0 4x – 1 = 4x −1 OR/OF OR/OF f ( x + h) − f ( x ) f ( x + h) − f ( x )  f / ( x) = lim / f ( x) = lim h →0 h h →0 h 2( x + h) 2 − ( x + h) − (2 x 2 − x) subst. into formula = lim h →0 h 2 x + 4 xh + 2h 2 − x − h − 2 x 2 + x 2  2 x 2 + 4 xh + 2h 2 − x − h = lim h →0 h  4 xh + 2h 2 − h 2 4 xh + 2h − h = lim h →0 h  lim(4 x + 2h − 1) h →0 h(4 x + 2h − 1) = lim h →0 h = lim (4 x + 2h − 1) h →0  4x – 1 = 4x −1 (6) 7.2.1 Dx [( x + 1)(3 x − 7 )] ( = Dx 3 x 2 − 4 x − 7 )  3x 2 − 4 x − 7  6x − 4 (2) = 6x − 4 5 1 7.2.2 y = x3 − + π 3 x 2  x 2 − 5 x −1 3 1 1 y = x 2 − 5 x −1 + π 3 2  x2 1 2 dy 3 2  + 5 x −2 = x + 5 x −2 dx 2 1  derivative of π is 0 2 (4) [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 16 Downloaded from hlayiso.com Mathematics/P1/Wiskunde/V1 16 DBE/November 2017 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 8 8.1 f (x ) = x 3 − 6 x 2 + 9 x  x3 − 6 x 2 + 9 x f / (x ) = 3 x 2 − 12 x + 9  3 x 2 − 12 x + 9  6 x − 12 f // (x ) = 6 x − 12 = 0  6 x − 12 = 0 x=2 f // (0) = 6(0) − 12 = −12 < >0 f // (3) = 6(3) − 12 2  explanation =6 Point of inflection at x = 2 (5) 8.2 y (1 ; 4) • x O(0 ; 0) (3 ; 0) shape  (0 ; 0)  (3 ; 0) as TP (1 ; 4) (4) 8.3 f concave up for x > 2 y = − f (x ) will be concave down for x > 2  x > 2 (2) 8.4.1 (3; 7 ) 3 7 (2) 8.4.2 Do not agree with Claire as her statement is incorrect. no Between x = 1 and x = 3 the graph of f is decreasing. Therefore at x = 2 the gradient will have a negative value. justification Stem nie saam met Claire nie, want haar stelling in verkeerd. Die grafiek van f is dalend/afnemend tussen x = 1 en x = 3. By x = 2 moet die gradiënt dus 'n negatiewe waarde hê. OR/OF f / (2) = 3(2) 2 − 12(2) + 9 = −3 ≠1 (2) [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Text truncated here. 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