Gradient of linear graphs — Paper 4 · IGCSE Mathematics 0580

3 y A NOT TO SCALE B C x 0 The diagram shows a sketch of y = x2 + 1 and y = 4 – x. (a) Write down the co-ordinates of (i) the point C, [1] (ii) the points of intersection of y = 4 – x with each axis. [2] (b) Write down the gradient of the line y = 4 – x. [1] (c) Write down the range of values of x for which the gradient of the graph of y = x2 + 1 is negative. [1] (d) The two graphs intersect at A and B. Show that the x co-ordinates of A and B satisfy the equation x2 + x – 3 = 0. [1] (e) Solve the equation x2 + x – 3 = 0, giving your answers correct to 2 decimal places. [4] (f) Find the co-ordinates of the mid-point of the straight line AB. [2]

12 marks

Mark scheme: 3 (a) (i) (0, 1) B1 Accept w/out brackets/ commas, condone (ii) (4, 0) and (0, 4) B1B1 vectors, or states x = , y = (b) -1 cao B1 (c) (x) < 0 (allow ≤) B1 Any other variable < 0 B0 (d) x 2 + 1 = 4 − x o.e. B1 must be these 4 terms (e) M1 p +(-)√q where p = −1 and r = 2×1 r and q = 1² − 4(1)(-3) o.e. M1 q Allow second mark if in form p± r -2.30 , 1.30 cao www4 A1A1 If ww ans.correct but wrong acc - SC3 After A0, A0, SC1 for -2.3027756 and 1.3027756 rounded or truncated (f) (-0.5, 4.5 or 4.49) B1ft f.t (their –2.30 + their 1.30) ÷2 B1 ft ft (4 – their x co-ord dep on attempt at mid value of x from values in e) [12]

This question in 0580/41 Oct/Nov 2007

x _ 25 (a) The table shows some values for the equation y = for – 4 Y x Y=–0.5 and 0.5 Y x Y 4. Examiner'sFor 2 x Use x –4 –3 –2 –1.5 –1 –0.5 0.5 1 1.5 2 3 4 y –1.5 –0.83 0 0.58 –3.75 –0.58 0 0.83 1.5 (i) Write the missing values of y in the empty spaces. [3] x _ 2 for – 4 Y x Y=–0.5 and 0.5 Y x Y 4. (ii) On the grid, draw the graph of y = 2 x y 4 3 2 1 x –4 –3 –2 –1 0 1 2 3 4 –1 –2 –3 –4 [5] x _ 2 For (b) Use your graph to solve the equation = 1 . Examiner's 2 x Use Answer(b) x = or x = [2] (c) (i) By drawing a tangent, work out the gradient of the graph where x = 2. Answer(c)(i) [3] (ii) Write down the gradient of the graph where x = –2. Answer(c)(ii) [1] (d) (i) On the grid, draw the line y = – x for – 4 Y x Y4. [1] x _ 2 _ (ii) Use your graphs to solve the equation = x . 2 x Answer(d)(ii) x = or x = [2] (e) Write down the equation of a straight line which passes through the origin and does not x _ 2 intersect the graph of y = . 2 x Answer(e) [2]

19 marks

Mark scheme: 5 (a) (i) 1.5, 3.75, –1.5 B1,B1,B1 (ii) 12 points plotted ft P3 ft P2 ft for 10 or 11 points, Curve through at least 10 points and correct P1 ft for 8 or 9 points shape over full domain C1 i.s.w. if two branches joined Two separate branches, one on each side of y-axis, neither in contact with y-axis B1 Independent (b) –1.4 ≤ x ≤ –1.1 and 3.1 ≤ x ≤ 3.4 B1,B1 i.s.w. 3rd answer if curve cuts y = 1 again (c) (i) Correct ruled tangent at x = 2 or x = –2 M1 Long enough to be able to find gradient Evidence of rise/run M1 Dependent – check their graph against gradient of 1 – must be correct side of 1 No tangent drawn M0M0 0.8 to 1.2 A1 (ii) 0.8 to 1.2 inc. or same answer as (i) ft B1 ft (d) (i) Correct ruled line to cut curve for all B1 Within ½ square of (–1, 1) and (1, –1) possible intersections (at least 2) (ii) –1.3 to –1.05, 1.05 to 1.3 inclusive B1, B1 i.s.w. any extra answers (e) y = kx with k ≥ 12 o.e. or x = 0 B2 If B0, allow SC1 for y = kx with k < 12 or for y-axis stated [19] IGCSE – May/June 2009 0580, 0581 04

This question in 0580/41 May/June 2009

6 For y Examiner's Use 50 A 40 y = f(x) 30 20 y = g(x) 10 B x –5 –4 –3 –2 –1 0 1 2 –10 –20 –30 The graphs of y = f(x) and y = g(x) are shown above. (a) Find the value of (i) f(−2), Answer(a)(i) [1] (ii) g(0). Answer(a)(ii) [1] (b) Use the graphs to solve For Examiner's (i) the equation f(x) = 20, Use Answer(b)(i) x = or x = [2] (ii) the equation f(x) = g(x), Answer(b)(ii) x = or x = [2] (iii) the inequality f(x) < g(x). Answer(b)(iii) [1] (c) Use the points A and B to find the gradient of y = g(x) as an exact fraction. Answer(c) [2] (d) On the grid, draw the graph of y = g(x) − 10. [2] (e) (i) Draw the tangent to the graph of y = f(x) at ( −3, −27 ). [1] (ii) Write down the equation of this tangent. Answer(e)(ii) [1] (f) A region, R, contains points whose co-ordinates satisfy the inequalities −3 Y x Y −2, y Y 40 and y [ g(x). On the grid, draw suitable lines and label this region R. [2]

15 marks

Mark scheme: 6 (a) (i) –16 B1 (ii) 18 to 19 B1 (b) (i) –4.3 to –4.2, 1.5 to 1.6 B1,B1 (ii) –4.5 to –4.4 , 1.3 to 1.4 B1,B1 (iii) –4.5 to –4.4 < x < 1.3 to 1.4 ft B1ft Ft their (ii). Allow clear worded explanations and condone Y signs (c) − 30 oe isw conversion B2 Accept − 4 2 , 30/–7 7 7 M1 for 30/7 oe fracts, isw conversion or for –30/7 oe soi (d) Ruled line passing within 2 mm of B2 B1 for ruled line parallel to g(x). By eye (21° (–5, 30) and (2, 0) to 25° to horizontal if in doubt) allow broken line (e) (i) Ruled horizontal line through (–3, –27) B1 No daylight, not chord (allow broken) (ii) y = –27 B1 (f) Ruled lines x = –3, x = –2, y = 40 B1 Long enough to be boundary of region – allow broken or solid ruled lines Region enclosed by lines x = –3, x = –2, y = 40 and y = g(x) B1 Allow any clear indication [15] IGCSE – October/November 2009 0580 04

This question in 0580/41 Oct/Nov 2009

4 For 7 (a) Complete the table of values for the equation y = , x ≠ 0. Examiner's x 2 Use x O4 O3 O2 O1 O0.6 0.6 1 2 3 4 y 0.25 0.44 11.11 4.00 0.44 [3] 4 (b) On the grid, draw the graph of y = for O4 Y x Y O0.6 and 0.6 Y x Y 4 . x 2 y 12 11 10 9 8 7 6 5 4 3 2 1 x –4 –3 –2 –1 0 1 2 3 4 –1 –2 [5] 4 For (c) Use your graph to solve the equation = 6 . Examiner's x 2 Use Answer(c)x = or x = [2] (d) By drawing a suitable tangent, estimate the gradient of the graph where x = 1.5. Answer(d) [3] 4 (e) (i) The equation O x + 2 = 0 can be solved by finding the intersection of the graph x 2 4 of y = and a straight line. x 2 Write down the equation of this straight line. Answer(e)(i) [1] (ii) On the grid, draw the straight line from your answer to part (e)(i). [2] 4 (iii) Use your graphs to solve the equation O x + 2 = 0. x 2 Answer(e)(iii) x = [1]

17 marks

Mark scheme: 7 (a) 1(.00) 4(.00) 11.1(1) 1(.00) 0.25 3 B2 for 4 correct, B1 for 3 correct (b) 10 points plotted P3 ft B2 for 8 or 9 points correct ft B1 for 6 or 7 points correct ft Correct shaped curve through 10 points C1 ft ft their points if shape correct – ignore anything (condone 2 points slightly missed) between – 0.6 and 0.6 2 separate curves not crossing x-axis and B1 Independent not touching or crossing y-axis (c) −0.85 to – 0.75 cao 1 0.75 to 0.85 cao 1 (d) Tangent drawn (ruled) at x = 1.5 T1 Allow slight daylight – 3 to −2 2 Dep on T1 M1 evidence rise/run dependent on tangent SC1 for answer in range 2 to 3 Answer implies M but not the T mark (e) (i) y = x − 2 oe 1 (ii) line ruled to cross curve 2 ft Dependent on (i) in form y = mx + c, m ≠ 0, c ≠ 0 B1 for gradient ft or y intercept ft but again to cross curve at all possible points (iii) 2.5 to 2.7 cao 1 Dependent on (e)(i) correct

This question in 0580/41 May/June 2011

7 (a) The co-ordinates of P are (–4, –4) and the co-ordinates of Q are (8, 14). For Examiner′s Use (i) Find the gradient of the line PQ. Answer(a)(i) … [2] (ii) Find the equation of the line PQ. Answer(a)(ii) … [2] (iii) Write as a column vector. Answer(a)(iii) = [1] f p (iv) Find the magnitude of . Answer(a)(iv) … [2] (b) For Examiner′s T Use A NOT TO SCALE R 4a O 3b B In the diagram, = 4a and = 3b. 1 R lies on AB such that = 5 (12a + 6b). 3 T is the point such that = 2 . (i) Find the following in terms of a and b, giving each answer in its simplest form. (a) Answer(b)(i)(a) = … [1] (b) Answer(b)(i)(b) = … [2] (c) Answer(b)(i)(c) = … [1] (ii) Complete the following statement. The points O, R and T are in a straight line because … … [1] (iii) Triangle OAR and triangle TBR are similar. area of triangle TBR Find the value of . area of triangle OAR Answer(b)(iii) … [2] _____________________________________________________________________________________

14 marks

Mark scheme: 3 14 ( 4 ) 7 (a) (i) or 1.5 2 M1 for oe 2 8 − ( −4 ) 3 3 (ii) y = x + 2 oe 2 B1 for y = their 2 x + c o.e. 2 or y = mx + 2, m ≠ 0 3 SC1 for x + 2 2 (iii)  12  1   18   (iv) 21.6 or 21.63[…] 2 M1 FT for their 122 + their 182 oe (b) (i) (a) 3b – 4a 1 1 1 (b) (6b – 8a) oe simplified 2 M1 for (12a + 6b) – 4a or AR = AO + OR 5 5 (c) 6a + 3b oe simplified 1 (ii) OR is parallel to OT 1 Dep on OT correct 9 2  3  (iii) or 2.25 2 M1 for   4  2  IGCSE – October/November 2013 0580 43 2 (s – ut )

This question in 0580/43 Oct/Nov 2013

8 (a) Complete the table of values for y = x3 – 3x + 1 . x –2.5 –2 –1.5 –1 –0.5 0 0.5 1 1.5 2 2.5 y –7.125 –1 3 1 –0.375 –1 –0.125 3 9.125 [2] (b) Draw the graph of y = x3 – 3x + 1 for –2.5 Ğ x Ğ 2.5 . y 10 9 8 7 6 5 4 3 2 1 x –3 –2 –1 0 1 2 3 –1 –2 –3 –4 –5 –6 –7 –8 [4] (c) By drawing a suitable tangent, estimate the gradient of the curve at the point where x = 2. Answer(c) … [3] (d) Use your graph to solve the equation x3 – 3x + 1 = 1 . Answer(d) x = … or x = … or x = … [2] (e) Use your graph to complete the inequality in k for which the equation x3 – 3x + 1 = k has three different solutions. Answer(e) … < k < … [2] __________________________________________________________________________________________

13 marks

Mark scheme: 8 (a) 2.125 and 2.375 2 B1 for one correct value (b) Correct curve B4 B3FT for 11 correct plots or B2FT for 9 or 10 correct plots or B1FT for 7 or 8 correct plots (c) Ruled tangent at x = 2 B1 No daylight at x = 2. Consider point of contact as midpoint between two vertices of daylight, this must be between x = 1.8 and 2.2 Gradient from 7.8 to 10.2 2 Dep on B1 awarded Allow integer/integer or a mixed number if within range or M1 dep for (change in y) ÷ (change in x) Dependent on any tangent drawn or close attempt at a tangent at any point Must see correct or implied calculation from a drawn tangent (d) 0 and –1.75 to –1.65 and 1.65 to 1.75 2 B1 for two correct values (e) –1.2 to –0.8 < k < 2.8 to 3.2 2 B1 for each correct or SC1 for reversed answers IGCSE – May/June 2014 0580 41 Qu Answers Mark Part Marks

This question in 0580/41 May/June 2014

12 5 y = x2 – 2x + , x ! 0 x (a) Complete the table of values. x –4 –3 –2 –1 –0.5 0.5 1 2 3 4 y 21 11 –9 –22.75 23.25 11 6 11 [2] 12 (b) On the grid, draw the graph of y = x2 – 2x + for –4  x  –0.5 and 0.5  x  4. x y 25 20 15 10 5 x –4 –3 –2 –1 0 1 2 3 4 –5 –10 –15 –20 –25 [5] (c) By drawing a suitable tangent, find an estimate of the gradient of the graph at the point (1, 11). Answer(c) … [3] 12 (d) The equation x2 – 2x + = k has exactly two distinct solutions. x Use the graph to find (i) the value of k, Answer(d)(i) k = … [1] 12 (ii) the solutions of x2 – 2x + = k. x Answer(d)(ii) x = … or x = … [2] (e) The equation x3 + ax2 + bx + c = 0 can be solved by drawing the line y = 3x + 1 on the grid. Find the value of a, the value of b and the value of c. Answer(e) a = … b = … c = … [3] __________________________________________________________________________________________

16 marks

Mark scheme: 5 (a) 2 and 7 2 B1 for each value (b) Complete correct curve 5 B3 FT for their 9 or 10 points or B2 FT for their 7 or 8 points or B1 FT for their 5 or 6 points and B1 independent for one branch on each side of the y-axis and not touching the y-axis SC4 for correct curve with branches joined (c) Correct tangent and 3 B2 for close attempt at tangent at x = 1 and –13 Y grad Y –8 answer in range OR B1 for ruled tangent at x = 1, no daylight at x = 1 Consider point of contact as midpoint between two vertices of daylight, the midpoint must be between x = 0.8 and 1.2 and M1 (dep on B1 or close attempt at tangent rise [at any point ] for run (d) (i) 5 to 6 1 (ii) 2 to 2.35 and –2.55 to –2.35 2FT FT their k B1FT for each correct solution (e) [a =] –5 3 B2 for two correct values [b =] –1 or for x3 – 5x2 – x + 12 [= 0] oe [c =] 12 or 12 M1 for x2 – 2x + = 3x + 1 x 2 2 955. 2 + 831.2 − AB 2 f

This question in 0580/42 May/June 2015

17 The table shows some values of y = x + , x ! 0 . x 2 x –2 –1.5 –1 –0.75 –0.5 0.5 0.75 1 1.5 2 3 y –1.75 –1.06 0 1.03 4.50 2.53 2 2.25 (a) Complete the table of values. [3] 1 (b) On the grid, draw the graph of y = x + for – 2  x  – 0.5 and 0.5  x  3. x 2 y 5 4 3 2 1 x –2 –1 0 1 2 3 –1 –2 [5] 1(c) Use your graph to solve the equation x + = 1.5 . x 2 x = … [1] 1(d) The line y = ax + b can be drawn on the grid to solve the equation = 2.5 - 2 x . x 2 (i) Find the value of a and the value of b. a = … b = … [2] 1 (ii) Draw the line y = ax + b to solve the equation = 2.5 - 2 x . x 2 x = … [3] (e) By drawing a suitable tangent, find an estimate of the gradient of the curve at the point where x = 2. … [3]

17 marks

Mark scheme: 7 (a) 3.5[0] 1.94 3.11 3 B1 for each (b) Fully correct curve 5 B3 FT for 10 or 11 points or B2 FT for 8 or 9 points or B1 FT for 6 or 7 points B1 indep two separate branches not touching or cutting y-axis SC4 for correct curve, but branches joined (c) – 0.7 to – 0.6 1 Qu. Answers Mark Part Marks (d) (i) – 1 1 2.5 1 If 0,0, M1 for y = 2.5 – x oe seen in working (ii) – 0.6 to – 0.5 with correct ruled line 3 B2FT for drawing their ruled line from (d)(i) or M1 for ruled line through (0, 2.5)FT or gradient −1 FT (e) Correct tangent and 3 B2 for close attempt at tangent at x = 2 and 0.5 ⩽ grad ⩽ 0.85 answer in range OR B1 for ruled tangent at x = 2, no daylight at x = 2 Consider point of contact as midpoint between two vertices of daylight, the midpoint must be between x = 1.8 and 2.2 and M1 (dep on B1 or close attempt at tangent rise [at any point ] for run ( )

This question in 0580/42 Feb/March 2016

8 Line A has equation y = 5x - 4 . Line B has equation 3x + 2y = 18 . (a) Find the gradient of (i) line A, … [1] (ii) line B. … [1] (b) Write down the co-ordinates of the point where line A crosses the x-axis. ( … , … ) [2] (c) Find the equation of the line perpendicular to line A which passes through the point (10, 9). Give your answer in the form y = mx + c . y = … [4] (d) Work out the co-ordinates of the point of intersection of line A and line B. ( … , … ) [3] (e) Work out the area enclosed by line A, line B and the y-axis. … [3]

14 marks

Mark scheme: 8(a)(i) 5 1 8(a)(ii) 3 1 − oe 2 8(b)  4  2 M1 for 5x – 4 = 0 soi  , 0  oe  5  8(c) y = –0.2x + 11 final answer 4 M2 for y = –0.2x + c oe (any form) FT their (a) or −1 B1FT for grad = soi their (a)(i) and M1 for substitution of (10, 9) into their equation 8(d) (2, 6) 3 M1 for elimination of one variable A1 for x = 2 or y = 6 8(e) 13 3 M2 for (4 + 9) × their 2 ÷ 2 oe or B1 for 9 oe or 4 or –4 seen

This question in 0580/43 Oct/Nov 2017

4 (a) The equation of a straight line is 2y = 3x + 4 . (i) Find the gradient of this line. … [1] (ii) Find the co-ordinates of the point where the line crosses the y-axis. ( … , … ) [1] (b) The diagram shows a straight line L. y 6 4 2 –2–2 0 22 4 6 x L –2 (i) Find the equation of line L. … [3] (ii) Find the equation of the line perpendicular to line L that passes through (9, 3). … [3] (c) A is the point (8, 5) and B is the point (- 4, 1). (i) Calculate the length of AB. … [3] (ii) Find the co-ordinates of the midpoint of AB. ( … , … ) [2]

13 marks

Mark scheme: 4(a)(i) 1.5 oe 1 4(a)(ii) (0, 2) 1 4(b)(i) y = −2 x + 6 oe final answer 3 B2 for y = − 2 x + c oe or y = mx + 6 oe m ≠ 0 or for answer −x2 + 6 6 or B1 for [gradient =] − oe or c = + 6 soi 3 4(b)(ii) y = 5.0 x − 5.1 oe final answer 3 B1 for [gradient = ] – 1 divided by their gradient from (b)(i) evaluated soi M1 for substitution of (9, 3) into y = (their m)x+ c seen in working 4(c)(i) 12.6 or 12.64 to 12.65 3 2 2 M2 for (8 − − 4) + ( 5 − )1 oe or M1 for (8 − −4 ) 2 + (5 − 1)2 oe 4(c)(ii) (2, 3) 2 B1 for each

This question in 0580/42 May/June 2019

2 (a) The table shows some values for y = 2x 3 - 4x 2 + 3 . x -1 - 0.5 0 0.5 1 1.5 2 y -3 1.75 0.75 3 (i) Complete the table. [3] (ii) On the grid, draw the graph of y = 2x 3 - 4x 2 + 3 for - 1 G x G 2 . y 3 2 1 – 1 0 1 2 x – 1 – 2 – 3 [4] (iii) Use your graph to solve the equation 2x 3 - 4x 2 + 3 = 1 .5 . x = … or x = … or x = … [3] (iv) The equation 2x 3 - 4x 2 + 3 = k has only one solution for - 1 G x G 2 . Write down a possible integer value of k. … [1] (b) y 6 5 4 3 2 1 – 1 0 1 x – 1 – 2 – 3 – 4 (i) On the grid, draw the tangent to the curve at x = 1. [1] (ii) Use your tangent to estimate the gradient of the curve at x = 1. … [2] (iii) Write down the equation of your tangent in the form y = mx + c . y = … [2]

16 marks

Mark scheme: 2(a)(i) 3 2.25 1 3 B1 for each 2(a)(ii) Fully correct smooth curve 4 B3FT for 7 or 6 correct plots B2FT for 5 or 4 correct plots B1FT for 3 correct plots 2(a)(iii) −0.6 to −0.51, 0.75 to 0.85, 3 B1 for each 1.7 to 1.85 If 0 scored, SC1 for y = 1.5 drawn 2(a)(iv) −3 or −2 or −1 or 0 1 2(b)(i) Tangent ruled at x = 1 1 2(b)(ii) 4.4 to 5.6 2 Dep on tangent at x = 1 or close attempt M1 for rise/run for their line 2(b)(iii) y = (4.4 to 5.6)x – (1.8 to 2.2) 2 FT for any line but not horizontal or vertical or line for 2 marks or B1 [y =] their (b)(ii)x + their(y-intercept) B1FT for [m =] their 5 or for their y-intercept

This question in 0580/42 Feb/March 2020

11 M has coordinates (4, 1) and N has coordinates ( -2, -7). (a) Find the length of MN. … [3] (b) Find the gradient of MN. … [2] (c) Find the equation of the perpendicular bisector of MN. … [4] Question 12 is printed on the next page.

9 marks

Mark scheme: 11(a) 10 3 M2 for (1 – –7)2 + (4 – –2)2 oe or M1 for (1 – –7) or (4 – –2) oe 11(b) 4 8 2 1 7 or M1 for oe 3 6 4 2 11(c) 3 9 4 3 9 y  x  B3 for  x  4 4 4 4 or 4 y  3 x  9  0 oe OR final answers B1 for midpoint (1, − 3) 3 1 M1 for gradient  or  4 their (b) M1 for substituting their (1, −3) into y = (their m)x + c or for y 3 their m = oe x  1

This question in 0580/43 May/June 2023