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By-Hand Skills Sheet 3 — Graphs, Curves and Tangents

Cambridge IGCSE Mathematics 0580 Extended · E2.9–E2.11 · Name: ____________________   Date: __________
Drawing a curve, ruling a tangent, and reading solutions off a grid are marks that cannot be earned by typing into a box — the examiner marks what your pencil did on printed axes. Every table on this sheet is deliberately head arithmetic, because Paper 2 is non-calculator. Print the sheet and work on paper; the model answers are hidden in print and checked on screen afterwards.

In the print dialog choose A4 and Scale: 100% — not “Fit to page”. Then work through it with a sharp pencil, a 30 cm ruler and an eraser, exactly as you would in the exam. Check your answers on screen afterwards: the model answers below are hidden when you print, so the sheet comes out blank to work on.

What you need before you start. A sharp HB pencil, a 30 cm ruler and an eraser. No compasses and no protractor this time — and nothing on this sheet needs a calculator: every table value is a whole number or a simple fraction you can do in your head.
About reading answers off a graph. Where an answer is read from your own drawing, the model answer states an accepted range, not one exact value — that is exactly how Cambridge marks it. Examiner tolerance is about half a small square, so plot with a small sharp cross and read with a dashed line to the axis: a fat dot or a guessed read-off can put a correct method outside the range.

Section 1  ·  Straight lines from gradient and intercept (E2.9)

Every line on this sheet is ruled. A freehand straight line loses the accuracy mark even when it passes through the right points.
1.1 On the grid, draw the line y = 2x − 3 for values of x from −2 to 4. [2]
Start at the intercept (0, −3). Gradient 2 means: along 1, up 2. Step it out at least three times, then rule through.
-2 -1 1 2 3 4 -4 -3 -2 -1 1 2 3 4 5 6 x y
How the marks are awarded B1 for a ruled line with gradient 2 (a line through (0, −3) with the wrong steepness scores it away). B1 for passing through (0, −3) and covering the full x-range asked for. A short stub of line in the middle of the grid loses the second mark.
-2 -1 1 2 3 4 -4 -3 -2 -1 1 2 3 4 5 6 x y y = 2x - 3
  • Marking points
  • Plot (0, −3) first, then step the gradient: (1, −1), (2, 1), (3, 3). Small crosses, not dots.
  • One ruled line through all of them, drawn right across the stated x-range.
  • Check one point at the end by arithmetic: x = 3 gives y = 6 − 3 = 3. On the line ✓
1.2 The line through A(−2, 5) and B(4, −4) is drawn on the grid. Find the gradient of the line and write down its equation in the form y = mx + c. [3]
Draw the gradient triangle on the printed grid and count squares: rise over run. Then read c where the line crosses the y-axis.
-3 -2 -1 1 2 3 4 5 -5 -4 -3 -2 -1 1 2 3 4 5 6 x y A B
How the marks are awarded M1 for a correct rise-over-run using the two marked points (a drawn triangle is the safest evidence). A1 for gradient −3/2 (accept −1.5; the minus sign is where this mark dies). B1 for y = −3/2 x + 2 — the intercept can be read straight off the grid.
-3 -2 -1 1 2 3 4 5 -5 -4 -3 -2 -1 1 2 3 4 5 6 x y A B run = 6 rise = -9 c = 2
  • Marking points
  • Triangle from A across to (4, 5), down to B: run = 6, rise = −9. Gradient = −9/6 = −3/2.
  • Falling line ⇒ negative gradient. If your triangle gives +3/2, the sign was dropped, not the arithmetic.
  • The line cuts the y-axis at (0, 2), so y = −3/2 x + 2. Check with B: −3/2 × 4 + 2 = −4 ✓
1.3 On the same grid, draw the lines y = −x + 4 and y = 2x − 2. Write down the coordinates of the point where they cross. [4]
This is solving simultaneous equations by drawing. Two ruled lines, then read the crossing point exactly.
-2 -1 1 2 3 4 5 -4 -3 -2 -1 1 2 3 4 5 6 x y
How the marks are awarded B1 for each correct ruled line (gradient and intercept both right). B2 for the crossing point (2, 2) — both coordinates, as coordinates. Reading only the x-value, or writing 2,2 without brackets, is how full method turns into 3 of 4.
-2 -1 1 2 3 4 5 -4 -3 -2 -1 1 2 3 4 5 6 x y (2, 2) y = -x + 4 y = 2x - 2
  • Marking points
  • y = −x + 4 through (0, 4) and (4, 0); y = 2x − 2 through (0, −2) and (3, 4). Both ruled edge to edge.
  • They cross at (2, 2) — and 2 × 2 − 2 = 2 = −2 + 4, so the algebra agrees with the drawing.
  • Dashed lines to both axes show the examiner where you read the crossing from.

Section 2  ·  A quadratic curve, and reading answers off it (E2.10, E2.11)

Exercises 2.2, 2.3 and 2.4 are all done on the grid you draw in 2.1 — exactly as one long Paper 4 question would do it.
2.1 Complete the table of values for y = x² − 2x − 3, then plot all seven points on the grid and join them with a smooth curve. [2 + 3]
Squaring a negative gives a positive: (−2)² = 4, so y = 4 + 4 − 3 = 5. Every value here is head arithmetic.
x−2−101234
y5 −3−4 0 
-3 -2 -1 1 2 3 4 5 -5 -4 -3 -2 -1 1 2 3 4 5 6 x y
How the marks are awarded 2 marks for the table (0, −3 and 5 — one lost per error). P2 for all seven points plotted correctly, P1 if five or six are. C1 for one smooth curve through all seven: no straight-line segments, no flat bottom, no double line. The turning point is at (1, −4) — the curve turns at a point, it does not sit on a shelf.
-3 -2 -1 1 2 3 4 5 -5 -4 -3 -2 -1 1 2 3 4 5 6 x y y = x² - 2x - 3
  • Marking points
  • Missing values: x = −1 gives 1 + 2 − 3 = 0; x = 2 gives 4 − 4 − 3 = −3; x = 4 gives 16 − 8 − 3 = 5.
  • One freehand smooth curve through all seven crosses, drawn in a single movement — turn the page sideways if it helps your wrist.
  • Symmetry check: the table is symmetric about x = 1, so the curve must be too.
2.2 By drawing the line y = 3 on your grid, solve the equation x² − 2x − 3 = 3. [3]
The solutions are the x-coordinates where the line crosses the curve — there are two, and neither is a whole number.
How the marks are awarded B1 for the ruled horizontal line y = 3. A1 for each solution read from a crossing point, with dashed lines down to the x-axis as evidence. Accepted: x between −1.8 and −1.5, and x between 3.5 and 3.8 (true values 1 ± √7 ≈ −1.65 and 3.65). Solving the quadratic by formula instead of by graph scores 0 here — the question says “by drawing”.
-3 -2 -1 1 2 3 4 5 -5 -4 -3 -2 -1 1 2 3 4 5 6 x y x ≈ -1.65 x ≈ 3.65
  • Marking points
  • Rule y = 3 right across the grid, not just near the curve.
  • Read both crossings and drop dashed lines to the x-axis: x ≈ −1.65 and x ≈ 3.65. Anything inside the accepted ranges scores.
  • Sense check: the answers sit either side of x = 1, the line of symmetry, at equal distances.
2.3 The point P(2, −3) is marked on the curve. Draw a tangent to the curve at P and use it to estimate the gradient of the curve at P. [3]
Slide your ruler until it touches the curve at P without cutting across it, rule the line long, then draw a big gradient triangle on the grid.
How the marks are awarded B1 for a ruled tangent touching at P (a chord that crosses the curve loses this mark). M1 for a gradient triangle read from the printed grid, the wider the better. A1 for a gradient in the accepted range 1.6 to 2.4 (the true value is 2). A tangent is an estimate — the examiner accepts a range, never one exact value.
-3 -2 -1 1 2 3 4 5 -5 -4 -3 -2 -1 1 2 3 4 5 6 x y P run 2 rise 4
  • Marking points
  • The tangent touches at P only — the curve stays on one side of it near P.
  • Triangle: run 2, rise 4, gradient ≈ 4/2 = 2. Accepted range: 1.6 to 2.4.
  • State the answer as an estimate from your triangle, and leave the triangle on the page — it is the method mark.
2.4 By drawing the line y = x − 1 on your grid, solve the equation x² − 2x − 3 = x − 1. [3]
Same idea as 2.2, but the line is sloping: the x-values where line meets curve are the solutions.
How the marks are awarded B1 for the ruled line y = x − 1 (through (0, −1) with gradient 1). A1 per solution with evidence. Accepted: x between −0.8 and −0.4, and x between 3.3 and 3.8 (true values (3 ± √17)/2 ≈ −0.56 and 3.56).
-3 -2 -1 1 2 3 4 5 -5 -4 -3 -2 -1 1 2 3 4 5 6 x y x ≈ -0.56 x ≈ 3.56
  • Marking points
  • Rearranged, this is x² − 3x − 2 = 0 — but the graph does the solving, not the formula.
  • Read the two crossings: x ≈ −0.56 and x ≈ 3.56, dashed lines down to the axis.
  • Note the exam pattern: the line you are told to draw is always (curve) − (equation to solve).

Section 3  ·  A reciprocal curve (E2.10, E2.11)

Two separate branches, and the curve never touches either axis. Joining the branches across x = 0 is the classic lost mark.
3.1 Complete the table of values for y = 6/x, plot all eight points, and draw the curve. It has two branches. [2 + 3]
All eight values are whole numbers — 6 divided by things that divide 6.
x−6−3−2−11236
y−1 −3−66 2 
-6 -5 -4 -3 -2 -1 1 2 3 4 5 6 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 x y
How the marks are awarded 2 marks for the table (−2, 3 and 1). P2 for all eight points, P1 for six or seven. C1 for two smooth branches that approach the axes without touching them — one branch bottom-left, one top-right. Any ink crossing the y-axis, and the curve mark is gone.
-6 -5 -4 -3 -2 -1 1 2 3 4 5 6 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 x y y = 6/x
  • Marking points
  • Missing values: 6/(−3) = −2, 6/2 = 3, 6/6 = 1.
  • Two separate freehand curves, each through its four crosses, hugging the axes ever closer — never touching, never joined.
  • Symmetry check: the curve maps to itself if you swap x and y; (2, 3) and (3, 2) must both sit on it.
3.2 By drawing the line y = x on your grid, solve the equation 6/x = x. [3]
The line passes through the origin at 45° to the grid. It meets each branch once.
How the marks are awarded B1 for the ruled line y = x, corner to corner. A1 per solution. Accepted: x between 2.3 and 2.6, and x between −2.6 and −2.3 (true values ±√6 ≈ ±2.45). The algebra says x² = 6 — so this graph has just found √6 with a pencil.
-6 -5 -4 -3 -2 -1 1 2 3 4 5 6 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 x y x ≈ 2.45 x ≈ -2.45
  • Marking points
  • y = x through (−6, −6) and (6, 6), ruled.
  • Crossings at x ≈ 2.45 and x ≈ −2.45, dashed lines to the axis. Anything in the accepted ranges scores.
  • Both solutions have the same size — the two branches are point-symmetric through the origin.
3.3 The point Q(2, 3) is marked on the curve. Draw a tangent to the curve at Q and estimate the gradient of the curve at Q. [3]
The curve is falling at Q, so your answer must be negative before you even measure.
How the marks are awarded B1 for a ruled tangent touching at Q. M1 for a gradient triangle on the grid. A1 for a gradient in the accepted range −1.9 to −1.1 (true value −3/2). A positive answer scores 0 of 3 — the sign is part of the reading, not a decoration.
-6 -5 -4 -3 -2 -1 1 2 3 4 5 6 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 x y Q run 2 rise -3
  • Marking points
  • The tangent touches the falling branch at Q and the curve bends away above it on both sides.
  • Triangle: run 2, rise −3, gradient ≈ −3/2. Accepted: −1.9 to −1.1.
  • For y = 6/x the exact gradient at x = 2 is −6/2² = −1.5 — your triangle should land near it.

Section 4  ·  Travel graphs (E2.9)

Distance–time first, then speed–time. They look alike and mean different things: the gradient of the first is speed; the area under the second is distance.
4.1 Draw the distance–time graph of this journey. A student leaves home at time 0 and walks 4 km in 20 minutes to a friend’s house, and stays 10 minutes. A bus then takes them 6 km further in 10 minutes to the pool, where they swim for 10 minutes. A car brings them straight home in the final 10 minutes. Then: (a) which part of the journey was fastest? (b) find the speed of the bus in km/h. [4 + 2]
Six ruled segments joined end to end. Flat means not moving. Coming home means the line comes back DOWN to zero.
0 10 20 30 40 50 60 2 4 6 8 10 12 Time (minutes) Distance from home (km)
How the marks are awarded B4 for the five ruled segments, one lost per wrong vertex — the two flat rests and the return to distance 0 are what examiners look at first. (a) B1 for the car home: steepest line = fastest. (b) M1+A1: 6 km in 10 minutes = 6 km in ⅙ hour = 36 km/h — convert the minutes, do not just write 0.6.
0 10 20 30 40 50 60 2 4 6 8 10 12 Time (minutes) Distance from home (km) walk rest bus swim car
  • Marking points
  • Vertices: (0,0) → (20,4) → (30,4) → (40,10) → (50,10) → (60,0). All ruled.
  • (a) The car home — 10 km in 10 minutes (60 km/h), the steepest segment on the page.
  • (b) Bus: 6 km in 10 min. In an hour it would do 6 × 6 = 36 km/h.
4.2 A tram accelerates uniformly from rest to 10 m/s in 20 seconds, holds 10 m/s for 20 seconds, then brakes uniformly to rest in 10 seconds. Draw the speed–time graph. Then find the total distance travelled, by counting grid squares under your graph. [3 + 2]
One large grid square is 10 s wide and 2 m/s tall, so each large square under the graph is 10 × 2 = 20 metres.
0 10 20 30 40 50 2 4 6 8 10 12 Time (seconds) Speed (m/s)
How the marks are awarded B3 for the three ruled segments (rising line to (20,10), flat to (40,10), falling to (50,0)). M1 for area = distance with the square-counting shown. A1 for 350 m. Counting squares only works because the numbers were chosen to make whole and half squares — in the exam, do it as triangles and rectangles if the squares are ugly.
0 10 20 30 40 50 2 4 6 8 10 12 Time (seconds) Speed (m/s) 5 10 2.5
  • Marking points
  • Vertices: (0,0) → (20,10) → (40,10) → (50,0), all ruled — a trapezium sitting on the time axis.
  • Squares under the graph: triangle 5 + rectangle 10 + triangle 2½ = 17½ large squares.
  • 17½ × 20 m = 350 m. (Check by areas: ½·20·10 + 20·10 + ½·10·10 = 100 + 200 + 50 = 350 ✓)

How examiners award the marks on this sheet

Read this before you start, and again before you hand the paper in.
  1. Rule every straight line. Lines from equations, tangents, lines drawn to solve f(x) = k, travel-graph segments. A freehand straight line loses the accuracy mark even when the numbers read off it are right.
  2. Curves are smooth, drawn freehand in one movement, through every plotted point. Never join plotted points with straight segments, never flatten the bottom of a parabola, never let a reciprocal branch touch an axis.
  3. Plot with a small cross, not a blob. The tolerance is half a small square; a 2 mm dot spends it all.
  4. A tangent touches, it does not cross. Slide the ruler until the curve stays on one side of it at the marked point. Then make the gradient triangle wide — a bigger triangle divides away your drawing error.
  5. Show where you read a value from. Dashed lines from the crossing point to the axis. They carry marks of their own and cost ten seconds.
  6. Graph answers are ranges, not bull’s-eyes. If your read-off lands in the accepted range, it is full marks; chasing a third decimal place from a pencil line is wasted time.
  7. Gradient needs its sign. Falling line, negative gradient — decide the sign from the picture before you measure the triangle.
  8. On a distance–time graph the gradient is the speed; on a speed–time graph the area is the distance. Mixing these two up is the most expensive confusion on the paper.
  9. Check one value at the end by arithmetic. Substitute one read-off back into the equation in your head. It catches a mis-drawn line faster than redrawing ever will.
Cambridge IGCSE Mathematics 0580 Extended · by-hand skills sheet 3 · 12 exercises · E2.9 straight lines and travel graphs, E2.10 curves, E2.11 tangents and graphical solution of equations · Print at 100% on A4 and complete in pencil.