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How can we straighten a curved relationship so a linear model works?

Apply squared, logarithmic and reciprocal transformations to linearise data, fit a least-squares line to the transformed data and use it to predict.

How to apply the squared, log and reciprocal transformations to straighten curved bivariate data, fit a line to the transformed variable, and back-substitute to predict in original units.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Why transform
Choosing the transformation from the shape
Fitting and predicting
Back-transforming a log model

What this dot point is asking

You must choose a transformation from the curve shape, fit a line to the transformed data, predict by substituting, and convert predictions back to original units.

Why transform

The least-squares line and the correlation coefficient $r$ only describe linear association. If the scatterplot is clearly curved, fitting a straight line gives poor predictions and a misleading $r$ . A transformation changes the scale of one variable so the relationship becomes linear, after which all the linear tools work.

Choosing the transformation from the shape

The three transformations in the course each suit a particular curve.

A practical method is the "circle of transformations": decide whether the curve bulges up or down and left or right, then pick the transformation that stretches the crowded part of the plot. In an exam, try the transformation suggested by the question and confirm the transformed plot or $r$ improves.

Fitting and predicting

After transforming, the least-squares line is written in terms of the transformed variable, for example $y = a + b(x^2)$ or $\log_{10} y = a + bx$ . To predict:

Substitute the given value, transforming it the same way as the data.
If the response variable was transformed, back-transform the result (square root, reciprocal, or $10^{\text{value}}$ ).

Worked example

Linearising a reciprocal relationship

The time $T$ (minutes) to fill a tank falls as the pump rate $r$ (litres per minute) rises, in a curve that drops steeply then levels. A least-squares line of $T$ against $\dfrac{1}{r}$ gives $T = 0.5 + 240\left(\dfrac{1}{r}\right)$ .

Confirm the transformation

A falling, levelling curve suits the reciprocal $\dfrac{1}{r}$ , which stretches the small- $r$ end where the curve is steep, straightening it.

Predict the time at r = 8

Substitute $\dfrac{1}{r} = \dfrac{1}{8} = 0.125$ : $T = 0.5 + 240(0.125) = 0.5 + 30 = 30.5$ minutes.

Predict the rate for a target time

For $T = 5$ minutes: $5 = 0.5 + 240\left(\dfrac{1}{r}\right)$ , so $240\left(\dfrac{1}{r}\right) = 4.5$ , giving $\dfrac{1}{r} = 0.01875$ and $r = 53.3$ litres per minute.

Interpret

The model predicts that filling in $5$ minutes needs a pump rate of about $53$ litres per minute. Because this rate is well beyond the data range, treat it as an extrapolation and report it with caution.

Back-transforming a log model

When $\log_{10} y$ is the response, predictions come out as a log value that must be undone with a power of ten. If $\log_{10} y = 2.0 + 0.3x$ gives $\log_{10} y = 3.5$ , then $y = 10^{3.5} = 3162$ . Forgetting this final step is the single most common error in log questions.

Exam-style practice questions

Practice questions written in the style of SCSA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WACE 20226 marksA scatterplot of

y

against

x

is curved, increasing and concave up. After applying the transformation

x^2

, the least-squares line of

y

x^2

y = 3.0 + 1.5(x^2)

. (a) Explain why a squared transformation was suitable. (b) Predict

y

when

x = 6

. (c) Predict the value of

x

(positive) when

y = 100

Show worked answer →

A concave-up increasing curve is straightened by stretching the $x$ axis with $x^2$ .

(a) The $x^2$ transformation spreads out large $x$ values, straightening a curve that rises ever more steeply, so the transformed plot of $y$ against $x^2$ is approximately linear. (2 marks)

(b) When $x = 6$ , $x^2 = 36$ , so $y = 3.0 + 1.5(36) = 3.0 + 54 = 57$ . (2 marks)

Markers reward justifying the transformation by the curve shape, substituting $x^2$ correctly, and solving back to original units.

WACE 20245 marksBacterial count

N

grows rapidly with time

t

(hours). A least-squares line fitted to

\log_{10} N

against

t

gives

\log_{10} N = 2.0 + 0.30t

. (a) Predict the count when

t = 5

. (b) State why a log transformation suited this data.

Show worked answer →

A log transformation linearises a quantity that multiplies by a roughly constant factor each period.

(a) When $t = 5$ , $\log_{10} N = 2.0 + 0.30(5) = 2.0 + 1.5 = 3.5$ , so $N = 10^{3.5} = 3162$ bacteria. (3 marks)

(b) Bacterial growth is exponential (a roughly constant percentage increase each hour), which plots as a steep curve. Taking $\log_{10} N$ converts the constant-ratio growth into a constant-difference, straight-line relationship with time. (2 marks)

Markers reward back-transforming with $10^{\log N}$ and explaining the log transformation by the exponential growth.