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Drawing a graph that earns the marks

Covers AS91168

The graph is the evidence

The graph is not decoration — it is the evidence for your conclusion, and it carries its own marks.

Plot the independent variable on the horizontal axis and the dependent variable on the vertical axis. Always.
Label both axes with the quantity and its unit, e.g. "length $L$ (m)", "period² $T^2$ (s²)".
Choose a scale that spreads the data across at least half of each axis, and uses easy intervals (1, 2, 5, 10 per square — never 3 or 7).
Plot every point accurately with a small cross or dot, and draw a line of best fit — a single smooth line with roughly equal numbers of points either side. Never join the dots.

Reading a gradient properly

Take your two points from on the line of best fit, not from your data table. The line is the average of all your data; a raw point is just one measurement.
Choose two points that are far apart on the line — the further apart, the smaller the effect of any error in reading them.
Use the standard gradient calculation:

$\text{gradient} = \frac{\text{rise}}{\text{run}} = \frac{y_2 - y_1}{x_2 - x_1}$

Give the gradient a unit, found by dividing the vertical axis unit by the horizontal axis unit.
- For $T^2$ (s²) against $L$ (m), the gradient's unit is s² m⁻¹.

Identifying an anomalous result

An anomalous point lies clearly away from the trend followed by all the others.
Do not delete it. Circle it, ignore it when drawing the line of best fit, and discuss it — a plausible cause for an anomaly is one of the best Excellence discussion points available to you.

What a good graph looks like

Feature	Why it matters
Axes labelled with quantity and unit	Without units, no gradient can be interpreted
Sensible scale, data filling the space	Squashed data hides the shape of the relationship
Points plotted accurately	The line of best fit is only as good as the points
Single smooth line of best fit	Joining the dots implies every reading is exactly right
Gradient triangle drawn on the line	Shows the marker which two points you used

Worked ExampleFinding a gradient and its unit

A student plots $T^2$ (s²) against $L$ (m) for a pendulum and draws a line of best fit. Two points on the line are $(0.20, 0.82)$ and $(0.90, 3.62)$ . Find the gradient, with its unit, and state the equation of the relationship.

Step 1 — Check the points are on the line

Both points are read from the line of best fit, not from the data table — so they represent the averaged trend of all the readings.

Step 2 — Calculate the gradient

$\text{gradient} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3.62 - 0.82}{0.90 - 0.20} = \frac{2.80}{0.70} = 4.0$

Step 3 — Find the unit

The vertical axis is in s² and the horizontal axis is in m, so the gradient's unit is

$\frac{\text{s}^2}{\text{m}} = \text{s}^2\text{ m}^{-1}$

Step 4 — State the relationship

The line passes through the origin, so there is no intercept term:

$T^2 = 4.0L$

Taking the square root of both sides gives the relationship between the original variables:

$T = 2.0\sqrt{L}$

$\textbf{gradient = 4.0 s}^{\textbf{2}}\textbf{ m}^{\textbf{−1}}\textbf{, T = 2.0√L}$

Drawing a graph that earns the marks

Covers AS91168

The graph is the evidence

The graph is not decoration — it is the evidence for your conclusion, and it carries its own marks.

Plot the independent variable on the horizontal axis and the dependent variable on the vertical axis. Always.
Label both axes with the quantity and its unit, e.g. "length $L$ (m)", "period² $T^2$ (s²)".
Choose a scale that spreads the data across at least half of each axis, and uses easy intervals (1, 2, 5, 10 per square — never 3 or 7).
Plot every point accurately with a small cross or dot, and draw a line of best fit — a single smooth line with roughly equal numbers of points either side. Never join the dots.

Reading a gradient properly

Take your two points from on the line of best fit, not from your data table. The line is the average of all your data; a raw point is just one measurement.
Choose two points that are far apart on the line — the further apart, the smaller the effect of any error in reading them.
Use the standard gradient calculation:

$\text{gradient} = \frac{\text{rise}}{\text{run}} = \frac{y_2 - y_1}{x_2 - x_1}$

Give the gradient a unit, found by dividing the vertical axis unit by the horizontal axis unit.
- For $T^2$ (s²) against $L$ (m), the gradient's unit is s² m⁻¹.

Identifying an anomalous result

An anomalous point lies clearly away from the trend followed by all the others.
Do not delete it. Circle it, ignore it when drawing the line of best fit, and discuss it — a plausible cause for an anomaly is one of the best Excellence discussion points available to you.

What a good graph looks like

Feature	Why it matters
Axes labelled with quantity and unit	Without units, no gradient can be interpreted
Sensible scale, data filling the space	Squashed data hides the shape of the relationship
Points plotted accurately	The line of best fit is only as good as the points
Single smooth line of best fit	Joining the dots implies every reading is exactly right
Gradient triangle drawn on the line	Shows the marker which two points you used

Worked ExampleFinding a gradient and its unit

Step 1 — Check the points are on the line

Both points are read from the line of best fit, not from the data table — so they represent the averaged trend of all the readings.

Step 2 — Calculate the gradient

$\text{gradient} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3.62 - 0.82}{0.90 - 0.20} = \frac{2.80}{0.70} = 4.0$

Step 3 — Find the unit

The vertical axis is in s² and the horizontal axis is in m, so the gradient's unit is

$\frac{\text{s}^2}{\text{m}} = \text{s}^2\text{ m}^{-1}$

Step 4 — State the relationship

The line passes through the origin, so there is no intercept term:

$T^2 = 4.0L$

Taking the square root of both sides gives the relationship between the original variables:

$T = 2.0\sqrt{L}$

$\textbf{gradient = 4.0 s}^{\textbf{2}}\textbf{ m}^{\textbf{−1}}\textbf{, T = 2.0√L}$