4 Graphs

Syllabus
2024
Topic
4
Level

Translate between graphs and numerical values

A graph and a table can represent the same relationship: each plotted point pairs one value on the horizontal axis with one value on the vertical axis.

Task Construction Numerical result
find yy for a given xx move vertically from xx to the graph, then horizontally to the yy-axis read yy with its unit
find xx for a given yy move horizontally from yy to the graph, then vertically to the xx-axis read xx with its unit
make a table record the two coordinates of selected points one row is one (x,y)(x,y) pair

Read each axis label, unit and scale before estimating. Interpolation estimates between plotted values. An intersection gives the pair of values shared by two graphs; a plateau shows that the vertical value is no longer changing as the horizontal value increases.

Keep construction lines parallel to the axes and report a precision justified by the scale. Extrapolation beyond the measured range is less secure than interpolation and should not be treated as an observed value.

Recognise a linear relationship

A relationship of the form y=mx+cy=mx+c is linear because equal changes in xx produce equal changes in yy; its graph is a straight line.

y=mx+c

Term Meaning on a graph of yy against xx
mm gradient: the change in yy for each unit change in xx
cc vertical-axis intercept: the value of yy when x=0x=0
m>0m>0 the line rises from left to right
m<0m<0 the line falls from left to right
m=0m=0 yy is constant, so the line is horizontal

For y=3x+2y=3x+2, increasing xx by 11 increases yy by 33, and the line crosses the yy-axis at 22. The points (0,2)(0,2), (1,5)(1,5) and (2,8)(2,8) therefore lie on the same straight line.

A straight-line relationship is directly proportional only when c=0c=0, so the line passes through the origin. A straight line with a non-zero intercept is linear but not directly proportional.

Plot two variables from data

Plotting two variables turns paired data into a graph so that a pattern, relationship or anomalous result can be seen.

Step Action
1 place the independent variable on the horizontal axis and the dependent variable on the vertical axis
2 label both axes with the variable and unit
3 choose simple, even scales that use most of the grid
4 plot every coordinate accurately as a small cross
5 draw a suitable line or smooth curve of best fit through the overall trend

Continuous measurements, such as temperature and time, may take values between recorded points and usually support a line or curve. Discrete values occur as separate counts or categories and should remain distinct. A best-fit construction should balance the scatter rather than join each point in turn.

Do not force an anomalous point into the trend or omit it without evidence. Keep it plotted, identify it as a possible anomaly, and use the pattern of the remaining data when judging the best fit.

Determine the gradient and intercept of a line

The gradient measures how quickly yy changes with xx, while the intercept is the value of yy where the line crosses the vertical axis.

m= rac{\Delta y}{\Delta x}= rac{y_2-y_1}{x_2-x_1},\qquad c=y-mx

Choose two widely separated points on the straight best-fit line, not necessarily measured data points. Form a large gradient triangle, calculate vertical change divided by horizontal change, then substitute one point into c=ymxc=y-mx or read the intercept at x=0x=0.

ext{using }(2,5) ext{ and }(8,17):\quad m= rac{17-5}{8-2}=2,\quad c=5-(2 imes2)=1,\quad y=2x+1

Gradient units are yy-axis units divided by xx-axis units. Preserve the sign: a falling line has a negative gradient. A small triangle magnifies reading error, so use much of the line where possible.

Use a tangent to find an instantaneous rate

For a curved graph, the gradient changes from point to point. The gradient of a tangent at one point measures the instantaneous rate of change there.

Step Action
1 mark the point at which the rate is required
2 draw a straight tangent that follows the curve's local direction and touches it at that point
3 choose two widely separated points on the tangent
4 calculate Δy/Δx\Delta y/\Delta x and state the derived rate unit

ext{instantaneous rate}= ext{tangent gradient}= rac{\Delta y}{\Delta x}

On a hydrogen-volume against time graph, the tangent gradient has units such as cm3s1\mathrm{cm^3\,s^{-1}}. A steep tangent means a faster rate; as the curve becomes less steep, the instantaneous rate decreases.

A chord joining two points on the curve gives an average rate over an interval, not the rate at one instant. Calculate the gradient from points on the tangent, not from nearby points on the curve.