Transforming data to linearity in WACE Mathematics Applications Unit 3

Syllabus dot point

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

WACEMathematics ApplicationsUnit 38 min readUpdated 2 Jun 2026

Quick answer

When a scatterplot is curved, a straight line is the wrong model, so transform one variable to straighten it first. The squared transformation ( $x^2$ or $y^2$ ) and the reciprocal ( $1/x$ or $1/y$ ) and the logarithm ( $\log_{10}$ ) each straighten a particular curve shape. Fit the least-squares line to the transformed variable, then back-substitute to predict in the original units.

Jump to a section

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

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.

Straighten, then fit

Never fit a line to a curve. Transform first so the transformed scatterplot is approximately linear, then apply least-squares. Residual plots from the original fit will show a clear curved pattern, which is the signal that a transformation is needed.

Choosing the transformation from the shape

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

The three transformations

Squared ( $x^2$ or $y^2$ ): straightens a curve that increases at an increasing rate (concave up). Squaring $x$ stretches the right of the plot; squaring $y$ stretches the top.
Reciprocal ( $1/x$ or $1/y$ ): straightens a curve that decreases steeply then levels off, or rises towards a horizontal asymptote.
Logarithm ( $\log_{10} x$ or $\log_{10} y$ ): straightens exponential-type growth or decay, where the variable multiplies by a roughly constant factor.

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.

Common errors

Forgetting to back-transform: If the response variable was transformed, undo it (square root, reciprocal or $10^{\text{value}}$ ) to return to original units.
Transforming the prediction value differently from the data: Apply the same transformation to the input as was applied to build the model.
Fitting a line to obviously curved data: A transformation is needed first; a high $r$ on a curve does not justify a straight line.
Choosing the wrong transformation: Match the transformation to the curve shape, then check the transformed plot or $r^2$ improves before trusting it.

Exam-style questions

Questions in the style of SCSA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

WACE 2022-style6 marks

A 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 2024-style5 marks

Bacterial 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.