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2d · E2.9–E2.11

Graphs of functions and practical graphs

Topic 2 Algebra and graphs · Core and Extended · Papers 1–4

🎯What you need to be able to do

  • Use and draw travel and conversion graphs, and interpret the gradient of a straight line as a rate of change.
  • Use distance–time and speed–time graphs: acceleration, deceleration, and distance as the area under a speed–time graph EXTENDED.
  • Estimate the gradient of a curve by drawing a tangent EXTENDED.
  • Complete tables of values and draw graphs of linear, quadratic and reciprocal functions; for Extended also cubics, sums of powers and exponentials \( ab^{x} + c \) EXTENDED.
  • Solve equations graphically, including by drawing a straight line.
  • Sketch linear and quadratic graphs; for Extended also cubic, reciprocal and exponential graphs with roots, turning points and asymptotes EXTENDED.

📚The mathematics

Practical graphs

On a distance–time graph the gradient is the speed, and a horizontal section means stopped. On a speed–time graph EXTENDED the gradient is the acceleration (negative gradient: deceleration), and the area under the graph is the distance travelled — split it into triangles, rectangles and trapezia. Watch the units: a graph in minutes and a speed in km/h need converting.

A speed-time graph: speed rises from 0 to 20 metres per second over the first 10 seconds, stays at 20 until 40 seconds, then falls to 0 at 55 seconds. The area underneath is split into a triangle of 100 metres, a rectangle of 600 metres and a triangle of 150 metres, total 850 metres.
Worked example (a): gradients give the acceleration (2 m/s²) and deceleration; the area gives the distance, 850 m.

EXTENDED To estimate the gradient of a curve at a point, draw the tangent there with a ruler, pick two points far apart on it, and compute rise over run.

Graphs of functions

Complete the table of values carefully (a calculator helps on Paper 3 and 4 — watch negative \(x\) squared), plot with small crosses, and join with a smooth curve, not straight segments. A quadratic has a rounded bottom (or top), never a point.

To solve an equation with a graph you already have, rearrange it so one side is the plotted function; the other side is the line to draw. With \( y = x^{2} - 2x - 3 \) drawn, the equation \( x^{2} - 3x - 4 = 0 \) is \( x^{2} - 2x - 3 = x + 1 \): draw \( y = x + 1 \) and read the \(x\)-coordinates of the intersections.

The parabola y = x squared minus 2x minus 3 plotted from x = minus 2 to 4 with its plotted points marked as crosses, and the straight line y = x + 1 crossing it at x = minus 1 and x = 4, where dashed lines drop to the x-axis.
Worked example (b): the line \( y = x + 1 \) meets the curve where \( x = -1 \) and \( x = 4 \).

Sketching curves

A sketch shows shape and key features, not plotted points. Label where it crosses the axes, and for a quadratic the turning point (found by completing the square) and line of symmetry.

Four small sketches. Quadratic: a U-shaped parabola with its turning point. Cubic: y = x cubed minus 4x, crossing the axis three times. Reciprocal: y = 2 over x plus 1, with dashed asymptotes x = 0 and y = 1. Exponential: y = 2 to the x minus 3, with dashed asymptote y = minus 3.
The shapes to recognise. Reciprocal and exponential graphs have asymptotes EXTENDED: lines the curve approaches but never reaches.

EXTENDED For \( y = \dfrac{a}{x} + b \) the asymptotes are \( x = 0 \) and \( y = b \); for \( y = ar^{x} + b \) the asymptote is \( y = b \) and the \(y\)-intercept is \( a + b \). A cubic \( y = ax^{3} + \ldots \) with \( a > 0 \) goes from bottom left to top right.

✏️Worked example

(a) (Non-calculator.) EXTENDED A car starts from rest and accelerates uniformly to 20 m/s in 10 s. It travels at 20 m/s for 30 s, then decelerates uniformly to rest in 15 s. Find the acceleration and the total distance travelled. [4] (b) (Non-calculator.) Complete a table of values for \( y = x^{2} - 2x - 3 \) for \( -2 \le x \le 4 \), and use the graph to solve \( x^{2} - 3x - 4 = 0 \). [5]

(a) Acceleration \( = \dfrac{20}{10} = 2 \) m/s². Distance = area \( = \tfrac{1}{2}(10)(20) + 30(20) + \tfrac{1}{2}(15)(20) = 100 + 600 + 150 = 850 \) m.

(b) For \( x = -2, -1, 0, 1, 2, 3, 4 \): \( y = 5, 0, -3, -4, -3, 0, 5 \). Rearranging, \( x^{2} - 3x - 4 = 0 \) is the same as \( x^{2} - 2x - 3 = x + 1 \), so draw \( y = x + 1 \): it meets the curve at \( x = -1 \) and \( x = 4 \).

Check it. (b) \( (x + 1)(x - 4) = x^{2} - 3x - 4 \), so \( -1 \) and 4 are exact ✓. (a) The deceleration is \( 20 \div 15 = 1.33 \) m/s², gentler than the acceleration, as the longer slope on the graph shows ✓.
Reading \(y\) instead of \(x\). The solutions of an equation from a graph are the \(x\)-coordinates of the crossing points. Writing the \(y\)-values (0 and 5) loses both marks.

📝Practise

Questions in the style of the current papers. EXTENDED marks Extended-only content.

1. (Non-calculator.) A conversion graph is a straight line through \( (0, 0) \) and \( (5 \text{ miles}, 8 \text{ km}) \). Convert 45 miles to kilometres. [1]
45 miles is \( 9 \times 5 \) miles, so \( 9 \times 8 = 72 \) km.
2. (Non-calculator.) Ana walks 3 km from home in 45 minutes, rests for 15 minutes, then walks home in 30 minutes. Find her speed on the way home, in km/h. [2]
3 km in 0.5 h: 6 km/h. On a distance–time graph this is the steepest (downward) section.
3. (Non-calculator.) EXTENDED A cyclist decelerates uniformly from 24 m/s to rest in 8 s. Find the deceleration and the distance travelled while decelerating. [3]
Deceleration \( 24 \div 8 = 3 \) m/s². Distance = triangle area \( = \tfrac{1}{2} \times 8 \times 24 = 96 \) m.
4. (Calculator.) EXTENDED For \( y = x^{3} - 3x + 1 \), find \(y\) when \( x = -2, 0, 2 \). The graph crosses the \(x\)-axis three times; use it (or trial) to give the roots to 1 decimal place. [4]
\( y = -1, 1, 3 \). The roots are about \( x = -1.9 \), \( 0.3 \) and \( 1.5 \) (more precisely \(-1.88\), \(0.35\), \(1.53\)). Each sign change in the table shows a root between those \(x\)-values.
5. (Non-calculator.) EXTENDED Write down the equations of the asymptotes of \( y = \dfrac{2}{x} + 1 \), and the asymptote and \(y\)-intercept of \( y = 3 \times 2^{x} - 4 \). [3]
\( y = \tfrac{2}{x} + 1 \): \( x = 0 \) and \( y = 1 \). \( y = 3 \times 2^{x} - 4 \): asymptote \( y = -4 \), \(y\)-intercept \( 3 - 4 = -1 \).
6. (Non-calculator.) EXTENDED By completing the square, sketch \( y = -x^{2} + 4x + 5 \), labelling where it meets the axes and its turning point. [4]
\( -x^{2} + 4x + 5 = -(x - 2)^{2} + 9 \): a maximum at \( (2, 9) \). Roots: \( -(x - 5)(x + 1) = 0 \), so \( x = -1 \) and 5. \(y\)-intercept 5. An upside-down parabola symmetric about \( x = 2 \).
7. (Calculator.) EXTENDED The tangent to \( y = x^{2} \) at \( (3, 9) \) passes through \( (5, 21) \). Find the gradient of the curve at \( x = 3 \). [1]
\( \dfrac{21 - 9}{5 - 3} = 6 \). (This agrees with differentiation, \( 2x = 6 \) — see 2e.)

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • Desmos — plot a curve and a line together to check graphical solutions
  • PhET “The Moving Man” — distance–time and speed–time graphs drawn live