E2.10 Graphs of functions
- Syllabus
- 0580–2028–2029
- Topic
- E2.10
- Level
- Extended
Learning objectives
E2.10.1Draw and interpret function graphs• Construct tables of values and draw, recognise and interpret linear, quadratic, cubic, reciprocal, exponential and other specified power-function graphs.E2.10.2Solve equations graphically• Solve associated equations graphically, including finding and interpreting roots and intersections of lines and curves.E2.10.3Exponential growth and decay graphs• Draw and interpret graphs representing exponential growth and decay problems.
Construct and interpret function graphs
A function graph shows every plotted pair (x,f(x)). A reliable graph begins with an accurate table of values, then connects points according to the function’s continuous shape and domain.
Choose the stated x values; substitute each one carefully; keep enough decimal accuracy for plotting; label axes and use a uniform scale; plot coordinates; draw a smooth curve or straight line as appropriate; check intercepts, separate branches and overall shape.
| Form | Recognition feature |
|---|---|
| ax+c | straight line |
| ax2+bx+c | parabola with one turning point |
| cubic power combination | S-like or turning cubic shape |
| a/x+c or a/x2+c | separated reciprocal branches; excluded denominator values |
| abx+c | exponential curve approaching a horizontal level |
| specified axn combinations | shape depends on the allowed power n and coefficients |
For the specified power forms, n may be −2,−1,−21,0,21,1,2,3, and no more than three axn terms are combined. Respect real-domain restrictions for roots and negative powers.
Do not join reciprocal branches across an undefined value or force a curve through an uncalculated point. A sketch shows key structure; a drawn graph from a table must also preserve scale and plotted accuracy.
Solve equations using roots and intersections
A graphical solution is an x-coordinate where two required expressions have equal y values. Roots are intersections with the x-axis; simultaneous solutions are intersections of two graphs.
| Equation | Graphical target |
|---|---|
| f(x)=0 | where y=f(x) crosses or touches the x-axis |
| f(x)=k | intersections of y=f(x) with horizontal line y=k |
| f(x)=g(x) | intersections of y=f(x) and y=g(x) |
To solve x3+4x2−x−6=0 using an existing graph of y=x3+4x2−4, rearrange to x3+4x2−4=x+2. Draw y=x+2 and read the intersection x-coordinates.
Draw any required line accurately with a ruler, identify every intersection within the stated domain, project vertically to the x-axis and report values to the precision supported by the grid.
Zero, one, two or more visible intersections mean the equation has that many graphical solutions in the shown interval. A tangent contact counts as one repeated root.
Do not read the y-coordinate when the question asks for x, and do not invent accuracy beyond the graph scale. Every algebraic rearrangement must preserve the same equality.
Draw and interpret exponential growth and decay graphs
Exponential change multiplies by the same factor over equal time intervals. Growth curves rise increasingly quickly; decay curves fall quickly at first and then level towards zero.
| Situation | Model | Multiplier | Graph behaviour |
|---|---|---|---|
| growth at r% per interval | P=P0(1+r/100)t | greater than 1 | increasing |
| decay at r% per interval | P=P0(1−r/100)t | between 0 and 1 | decreasing |
Choose sensible time values including t=0; calculate the corresponding quantities; plot (t,P) with labelled units; join with a smooth curve; check that the vertical intercept is the initial value P0.
P=40000(1.15)t
The bacteria model starts at 40000 and multiplies by 1.15 each hour. At t=3, P=40000(1.15)3=60835. Equal vertical additions would indicate linear, not exponential, growth.
For M=20(0.9)t, the mass stays positive and approaches 0 as time increases. Read a threshold time from the first whole t for which the curve is below the stated level.
A percentage decrease uses 1−r/100, not r/100. Do not draw exponential decay crossing below zero when the model has a positive initial amount and positive multiplier.