What is not reversing a transformation on y?

If the response was logged, the equation predicts log y, not y. You must antilog at the end, or your answer is in the wrong units.

§-Syllabus dot point

QLDGeneral MathematicsSyllabus dot point

Topic 1: Bivariate data analysis - how do we straighten a curved relationship so a least-squares line can be fitted?

Apply a square, logarithmic or reciprocal transformation to one variable to linearise a non-linear association, fit a least-squares line to the transformed data, use the transformed equation to predict, and choose the transformation that best straightens the scatter

A focused answer to the QCE General Mathematics Unit 3 dot point on data transformation. Covers when to transform, the square, log and reciprocal transformations, how to fit and use a least-squares line on transformed data, and how to predict by back-substituting, with arithmetic-verified worked examples for IA2 and the external assessment.

Generated by Claude Opus 4.87 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

QCAA wants you to handle bivariate data whose scatter is clearly curved, where fitting a straight line directly would be wrong. The fix is to transform one of the variables (square it, take its logarithm, or take its reciprocal) so the relationship straightens out, then fit a least-squares line to the transformed data and use that line to predict. You also have to choose which transformation does the best straightening. This is the natural follow-on from residual analysis in Unit 3 Topic 1 and is regularly tested in IA1, IA2 and the external assessment.

The answer

Why transform at all

Least-squares regression only describes a straight-line relationship. When a scatterplot or residual plot shows a smooth curve, a straight line fitted to the raw data gives biased predictions. Rather than abandon regression, you change the scale of one variable so the curve becomes a line. This is called linearising the data.

The three transformations

In General Mathematics you choose from three transformations, applied to either the explanatory variable $x$ or the response variable $y$ .

The squared transformation ( $x^2$ or $y^2$ ) stretches the upper end of a variable. It straightens data that curves upward more and more steeply.
The logarithmic transformation ( $\log x$ or $\log y$ ) compresses the upper end. It straightens data that rises quickly then flattens, or data that grows by a roughly constant percentage.
The reciprocal transformation ( $1/x$ or $1/y$ ) strongly compresses large values and is used for data that drops steeply and then levels off towards an asymptote.

Choosing the transformation

You pick the transformation that makes the transformed scatterplot look most like a straight line. In practice you compare residual plots or the value of $r^2$ for the candidate transformations and select the one with the most random residuals and the highest $r^2$ . The transformation can be applied to either axis; sometimes squaring $x$ works while sometimes taking $\log y$ works, so test rather than guess.

Fitting and predicting

Once transformed, treat the new variable exactly like ordinary data: fit the least-squares line on CAS. The fitted equation is written in terms of the transformed variable, for example

\hat{y} = a + b x^2 \qquad \text{or} \qquad \widehat{\log y} = a + b x.

To predict, substitute the value into the transformed equation, then undo any transformation on the response variable. If you transformed $y$ to $\log y$ , you must take the antilog ( $10$ to the power) at the end to return to the original units.

Predicting from a squared transformation

A curved scatter of $y$ against $x$ is straightened by squaring the explanatory variable. CAS fits the least-squares line $\hat{y} = 4 + 0.5 x^2$ . Predict $y$ when $x = 6$ .

Step 1: Apply the transformation to the input value

The equation is fitted in terms of $x^2$ , not $x$ , so you must square the given value of $x$ before substituting. This is the most common place to lose marks.

x^2 = 6^2 = 36

Verify: $6 \times 6 = 36$ .

Step 2: Substitute the transformed value into the fitted equation

Replace $x^2$ with $36$ in the equation:

\hat{y} = 4 + 0.5 \times 36

Step 3: Evaluate to find the prediction

0.5 \times 36 = 18, \qquad \hat{y} = 4 + 18 = 22

Verify: $4 + 18 = 22$ .

Step 4: Check whether a back-transformation is needed

Here only the explanatory variable $x$ was transformed, not the response $y$ , so the equation already predicts $y$ directly. No antilog or reciprocal step is needed at the end. If instead the model had been $\widehat{\log y} = 4 + 0.5x$ , you would compute the right-hand side first, then raise $10$ to that power to recover $y$ in its original units.

Final answer: The predicted value of $y$ when $x = 6$ is $22$ .

Common traps

Forgetting to transform the input before predicting: With $\hat{y} = 4 + 0.5x^2$ you must square the value of $x$ first. Substituting $x = 6$ as $6$ rather than $36$ gives a wrong prediction.
Not reversing a transformation on y: If the response was logged, the equation predicts $\log y$ , not $y$ . You must antilog at the end, or your answer is in the wrong units.
Choosing the transformation by guesswork: The correct transformation is the one that straightens the scatter best, judged by the residual plot and $r^2$ , not by which looks tidiest in the equation.
Transforming both variables at once unnecessarily: General Mathematics transformations change one variable. Changing both makes the model hard to interpret and is rarely required.
Squaring a negative input wrongly: Squaring removes the sign, so a negative $x$ becomes positive after transforming; keep track of this when the data includes negatives.

Exam-style practice questions

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

QCAA 20224 marksA curved association between

y

and

x

is straightened by squaring the explanatory variable. The least-squares line fitted to the transformed data is

\hat{y} = 12 + 0.8x^2

. (a) Determine the predicted value of

y

when

x = 5

. (b) State why a

\log y

transformation might be chosen instead for a different data set.

Show worked answer →

A 4 mark technique question: apply the transformation, substitute, evaluate, and give the reasoning.

(a) Transform the input first: $x^2 = 5^2 = 25$ . Substitute into the fitted line: $\hat{y} = 12 + 0.8 \times 25 = 12 + 20 = 32$ . The predicted value of $y$ is $32$ .

(b) A $\log y$ transformation compresses large $y$ values, so it straightens data that rises quickly then flattens, or that grows by a roughly constant percentage (multiplicative growth). The squared transformation instead suits data that curves upward ever more steeply.

Markers reward squaring the input before substituting, the correct prediction, and a transformation choice tied to the shape of the data.

QCAA 20236 marksResearchers model bacterial count

N

(thousands) against time

t

(hours). A

\log_{10} N

transformation straightens the scatter, and the least-squares line is

\widehat{\log_{10} N} = 0.6 + 0.25t

. (a) Determine the predicted count when

t = 8

hours, giving your answer to the nearest thousand. (b) Justify why the residual plot for the transformed data is examined before trusting this prediction.

Show worked answer →

A 6 mark extended question needing back-transformation and an interpretation.

(a) Substitute $t = 8$ : $\widehat{\log_{10} N} = 0.6 + 0.25 \times 8 = 0.6 + 2 = 2.6$ . The equation predicts $\log_{10} N$ , so reverse the transformation by raising $10$ to that power: $N = 10^{2.6} \approx 398.1$ . To the nearest thousand the count is about $398{,}000$ bacteria (since $N$ is in thousands, $398.1 \times 1000$ ).

(b) The residual plot for the transformed data should show random scatter about zero. If it shows a remaining pattern, the chosen transformation has not fully straightened the relationship, so predictions, especially extrapolated ones, would be biased. A random residual plot and a high $r^2$ justify trusting the linear fit.

Markers reward the substitution, raising $10$ to the power to undo the log, the rounded count, and linking the residual plot to the validity of prediction.