A straight line is not the final answer to a linear-law question. It is an intermediate representation. The real task is to understand what its axes mean, identify what its gradient and intercept represent, and recover the constants in the original relationship.
Suppose a graph of y/x against 1/x2 has gradient 3 and vertical intercept 4. It is tempting to write y = 3x + 4. That would describe the wrong variables. The actual line is y/x = 3/x2 + 4, so the original relationship is y = 4x + 3/x, with x ≠ 0. The gradient became the coefficient of 1/x, not the coefficient of x.
This guide teaches the complete route: original relationship → transformed variables → straight-line equation → original constants → verification. You will work with powers, reciprocals and logarithms; see why some transformations fail; and practise returning from a graph to the relationship it represents.
Curriculum scope: transformation of relationships into straight-line form, including power and exponential relationships, is included in the coordinate-geometry content of the 2026 O-Level Additional Mathematics 4049 syllabus and 2027 SEC G3 Additional Mathematics K341 syllabus. This is a G3/O-Level worked guide, not a claim that every transformation below is separately named in the syllabus.
All numerical data and questions below are original teaching examples. Tables labelled exact are constructed to obey the stated relationship. The later measurement example is explicitly hypothetical and approximate; it is not a report of a real experiment.
Choose a section: foundations · transformation map · worked examples · measurement and fitting · mistake clinic · practice · full solutions.
The three meanings that must remain separate
In an ordinary straight-line equation y = mx + c, x is the horizontal coordinate and y is the vertical coordinate. The gradient m tells us the change in the vertical coordinate per unit change in the horizontal coordinate. The intercept c is the vertical coordinate where the horizontal coordinate is zero.
In a transformed graph, the horizontal coordinate may instead be x2, 1/x or log x. The vertical coordinate may be y2, y/x or log y. The ordinary line rule still works, but it applies to these new coordinates.
Write X for the horizontal transformed quantity and Y for the vertical transformed quantity. Then Y = mX + c keeps the line structure visible while you work out what X and Y mean. Capital letters here are labels for transformed quantities, not new independent measurements.
Three things are therefore different: the original variables x and y; the transformed coordinates X and Y; and the model constants a, b, k or n. A successful solution translates between these three layers without silently swapping their meanings.
What does “Y against X” mean?
A graph of Y against X places X on the horizontal axis and Y on the vertical axis. Thus “log y against log x” means horizontal log x and vertical log y. Reversing those axes generally changes both the gradient and the intercept.
Before using any two points, state the axes in words or symbols. For example: “Let X = 1/x2 and Y = y/x.” This short line prevents an entire page of otherwise accurate calculations from being attached to the wrong variables.
A small foundation check
Find the gradient of the line through (2,7) and (6,19). The vertical change is 12 and the horizontal change is 4, so m = 3. Substituting into Y = 3X + c gives 7 = 6 + c, hence c = 1.
Next, rearrange y = ax + b/x by dividing both sides by x, with x ≠ 0. You obtain y/x = a + b/x2. Finally, explain why log(ab) = log a + log b is valid for positive a and b, while log(a + b) cannot generally be split that way.
If one of these steps is uncertain, repair it before adding more graph types. Use the Coordinate Geometry guide for lines and gradients, or the Logarithms and Exponentials guide for log laws. Linear law depends on both; it does not replace either.
A transformation map: identify the quantities before the constants
| Original relationship | Choose X and Y | Gradient and intercept |
|---|---|---|
| y = ax2 + b | X = x2; Y = y | m = a; c = b |
| y = a/x + b | X = 1/x; Y = y | m = a; c = b |
| xy = ax2 + b | X = x2; Y = xy | m = a; c = b |
| y = ax + b/x | X = 1/x2; Y = y/x | m = b; c = a |
| y = axn | X = log x; Y = log y | m = n; c = log a |
| y = kbx | X = x; Y = log y | m = log b; c = log k |
| 1/y = a/x + b | X = 1/x; Y = 1/y | m = a; c = b |
These rows are consequences of algebra, not a list to memorise without inspection. Write the transformed equation explicitly each time. In particular, the row involving y/x reverses the apparent roles of a and b: the constant term is a and the coefficient of X is b.
Reciprocal transformations require nonzero denominators. The real-log transformations shown here require positive logarithm arguments. For the power model, take x > 0 and a > 0, so y > 0. For the exponential model, take k > 0 and b > 0. A zero or negative value cannot simply be entered into the same real-log table.
The symbol log may use a stated base such as 10, provided the same base is used consistently. Natural logarithms, written ln, use base e. A slope that is itself a logarithm must be inverted using the matching base when recovering a constant.
Worked example 1: a reciprocal graph
A graph of y against 1/x has equation Y = −6X + 4. Find y in terms of x and calculate y when x = 2.
Here X = 1/x and Y = y. Substituting the definitions of the axes gives y = −6/x + 4, with x ≠ 0. When x = 2, y = −3 + 4 = 1.
The transformed graph is straight, but the graph of y against the original x is not the line y = −6x + 4. This distinction is the purpose of the transformation: a nonlinear relationship in one choice of coordinates may become linear in another.
The intercept 4 belongs to X = 0. Because X = 1/x, no finite nonzero x makes X exactly zero. An intercept can still help identify the model constant even when it does not correspond to an attainable original input. Do not describe it automatically as “the value at x = 0”.
Worked example 2: squaring can hide a sign
A graph of y2 against x is the straight line Y = 4X + 9. Find the possible values of y when x = 4.
Since X = x and Y = y2, the original relation is y2 = 4x + 9. At x = 4, y2 = 25, so y = 5 or y = −5.
If the question states that y is a nonnegative length, the contextual condition selects 5. Without such a condition, taking only the positive square root would lose a valid branch. The transformation has made y and −y share the same vertical coordinate.
The original real-valued relation also requires 4x + 9 ≥ 0, hence x ≥ −9/4. A line can be drawn beyond that boundary in the transformed plane, but a negative value of Y would not represent a real y2. Coordinate convenience does not remove the original restrictions.
Worked example 3: recover constants whose positions have changed
The variables satisfy y = ax + b/x. A graph of y/x against 1/x2 passes through (1,7) and (4,16). Find a and b. These ordered pairs are transformed coordinates, not original (x,y) pairs.
Divide the original relation by x, with x ≠ 0:
y/x = a + b/x2.
Let Y = y/x and X = 1/x2. Then Y = bX + a. The gradient is (16 − 7)/(4 − 1) = 3. Using (1,7), the intercept is 7 − 3(1) = 4. Therefore b = 3 and a = 4.
The original relationship is y = 4x + 3/x. At x = 2, this gives y = 8 + 1.5 = 9.5. To check through the transformed line, X = 1/4 and Y = 3(1/4) + 4 = 4.75. Multiplying Y by x gives y = 9.5 again.
This two-route check is valuable because it tests the axis interpretation as well as the arithmetic. A student who writes a = 3, b = 4 can obtain a perfectly neat answer while failing the actual relationship.
Worked example 4: a power relationship becomes a log–log line
The exact data below follow y = axn, with a > 0 and x > 0. Find a and n.
| x | y | Useful comparison |
|---|---|---|
| 1 | 3 | The value at x = 1 gives a. |
| 2 | 12 | Doubling x multiplies y by 4. |
| 4 | 48 | The same factor repeats. |
| 8 | 192 | The same factor repeats again. |
Taking logarithms gives log y = log a + n log x. Thus a graph of log y against log x has gradient n and intercept log a.
Using the points corresponding to x = 2 and x = 4, the gradient is
n = (log 48 − log 12)/(log 4 − log 2) = log 4/log 2 = 2.
At x = 1, log x = 0 and y = 3, so the intercept is log 3. Consequently, a = 3 and the original model is y = 3x2.
The intercept is log 3, not 3. With base-10 logs it is approximately 0.4771. Raising 10 to the unrounded intercept recovers a. With natural logs, the intercept would instead be ln 3 and exponentiating with e would recover the same a. The power n is unchanged because both axes use the same log base.
A direct ratio check provides another route: 48/12 = (4/2)n, so 4 = 2n and n = 2. This confirms the result for these exact data. It does not mean the graph method should be omitted when a question specifically asks for a transformed graph.
Worked example 5: an exponential model uses different axes
The exact relationship is y = kbx. It passes through original-coordinate points (0,5) and (2,45), with k > 0 and b > 0. Find k and b.
At x = 0, b0 = 1, so k = 5. At x = 2, 45 = 5b2, giving b2 = 9. The positive-base condition gives b = 3. Therefore y = 5 × 3x.
The straight-line version is log y = x log b + log k. Its horizontal axis is x, not log x. The gradient is log 3 and the intercept is log 5. To recover b and k from a base-10 graph, take 10 to the respective gradient and intercept.
This is the key contrast: a power model becomes a line on log–log axes, while this exponential model becomes a line with log y on the vertical axis and x on the horizontal axis. Both contain logarithms, but they represent different relationships.
Also notice that x = 0 is perfectly acceptable in this exponential model. It would not be an acceptable horizontal value in a log x transformation. Do not transfer a restriction from one model to another without checking where the logarithm actually appears.
Worked example 6: read a transformed equation back into a power
A line satisfies log10 y = (3/2)log10 x + log10 2. Express y in terms of x.
The slope is 3/2 and the intercept is log10 2. Using the power law for logarithms, the right side becomes log10(2x3/2). Since the logarithm is one-to-one on positive inputs, y = 2x3/2, with x > 0.
Do not write y = (3/2)x + 2. That would treat the coefficients of transformed coordinates as though they belonged to an untransformed line. Equally, the exponent is not 103/2: the gradient in a log–log graph is already the power n, not log n.
Worked example 7: reverse the axes carefully
Take y = 3x2, with x > 0. The familiar form is log y = 2 log x + log 3. Now suppose the vertical axis is log x and the horizontal axis is log y.
Rearrange for the new vertical quantity:
log x = (1/2)log y − (1/2)log 3.
The new gradient is 1/2, not 2, and the intercept is −(1/2)log 3, not log 3. This is not a second power model. It is the same relationship viewed with the two axes exchanged.
A useful repair habit is to begin every solution with “vertical quantity = gradient × horizontal quantity + intercept”. Then substitute the actual labels. This is safer than trying to remember whether a particular symbol usually denotes a horizontal or vertical variable.
Worked example 8: an additive offset changes the log transformation
Consider y = 2 + 3x2, with x > 0. Can we write log y = log 2 + log 3 + 2 log x? No. A logarithm of a sum cannot be split into a sum of logarithms in that way.
Because the offset 2 is known, subtract it first: y − 2 = 3x2. Then log(y − 2) = log 3 + 2 log x. This gives a line when plotting log(y − 2) against log x, provided y > 2.
If the offset were unknown, subtracting 2 merely because it creates a convenient-looking line would not be justified. The transformation must follow the stated model and known information. Unknown constants cannot be inserted into transformed data as though they had already been found.
There is also a simpler non-log route here: plot y against x2. That yields gradient 3 and intercept 2 directly. Multiple valid transformations may exist. Choose the one requested, or the simplest one that identifies the desired constants without creating unnecessary restrictions.
From an exact exercise to hypothetical measured data
Exact constructed tables and measured tables should not be treated identically. A set of exact model values can lie precisely on a line after transformation. Measurements may contain rounding, noise or model mismatch. A nearly straight transformed plot can support a proposed relationship over the observed range, but it does not establish an exact universal law.
Suppose these hypothetical observations are being compared with a model y = ax2 + b:
| x | X = x2 | Observed y |
|---|---|---|
| 1 | 1 | 12.2 |
| 2 | 4 | 41.8 |
| 3 | 9 | 92.3 |
| 4 | 16 | 161.7 |
For illustration, suppose a drawn approximate line on the transformed graph passes through the convenient line points (2,22) and (14,142). These are points on the selected line, not two entries copied from the measurement table. The gradient is (142 − 22)/(14 − 2) = 10, and the intercept is 22 − 10(2) = 2.
The resulting approximate model is y ≈ 10x2 + 2. At x = 2.5, it estimates y ≈ 10(6.25) + 2 = 64.5. This chosen line is an illustrative fit, not a claimed least-squares optimum.
For x = 1, 2, 3 and 4, the model predicts 12, 42, 92 and 162. Subtracting these predictions from the observations gives differences 0.2, −0.2, 0.3 and −0.3. These small differences help describe agreement for this invented example, but no uncertainty model or real measurement precision has been supplied.
Why choose two well-separated points on the fitted line?
When the purpose is to estimate the gradient of a drawn line, use points on that line and read them as accurately as possible. Widely separated points reduce the influence of a small reading error relative to the total horizontal and vertical changes. Using two nearby noisy observations instead can estimate a local fluctuation rather than the intended fitted trend.
Follow the actual instructions of a school or examination question. A question may specify which line to draw, how to estimate its gradient or whether to use algebra instead. Do not assume every graph problem calls for statistical fitting.
Interpolation, extrapolation and model limits
The estimate at x = 2.5 lies within the observed x-range 1 to 4. An estimate at x = 20 would be outside it. Even a convincing fit within the measured range does not establish that the same relationship continues indefinitely.
Transformations also change how errors look. Taking a logarithm compresses large values relative to small ones. Fitting a straight line to log-transformed data is therefore not automatically the same optimisation as fitting a curve to the original y-values. That distinction is a model-checking boundary, not an extra computation required in every A-Math exercise.
Where quantities have physical units, keep the unit convention fixed. Strictly, logarithms operate on dimensionless numbers; school tables often use the numerical values expressed in stated units. Changing that unit convention changes those numerical values and can change a log intercept. A defensible model records the units rather than treating its constants as context-free numbers.
Mistake clinic: when a correct line gives a wrong model
Error: “The gradient is a because the original formula uses a first.” Letters do not determine roles. In y/x = a + b/x2, the gradient against 1/x2 is b. Repair the line by matching the coefficient of the actual horizontal quantity.
Error: “The intercept of a log y graph is the original constant.” In log y = n log x + log a, the intercept is log a. If the base is 10 and the intercept is −1, a = 0.1, not −1. A negative logarithm can represent a positive number smaller than one.
Error: “The transformed point (2,7) means x = 2, y = 7.” That is true only when the axes are the original variables. If the axes are x2 and xy, the point instead means x2 = 2 and xy = 7. Recovering original coordinates may involve multiple possibilities or extra conditions.
Error: “Use log on every expression until it looks linear.” A transformation is justified by a valid identity. The log of a sum is not the sum of its logs, and real logs of nonpositive inputs are not available. Repair the algebra before entering values into the calculator.
Error: “A good-looking line proves the model.” Exact exercise data and measurements have different evidential roles. For measurements, inspect departures from the line and the range of observation. Treat the model as a supported description within stated limits, not as proof of causation or permanent validity.
Error: “Round every logarithm immediately.” Early rounding can shift a gradient or an antilog. Retain calculator precision for calculations and round displayed tables or final answers according to the question. Distinguish a graph-reading estimate from an exact algebraic answer.
Independent practice: twelve questions
For each transformation question, write the vertical quantity, the horizontal quantity and the corresponding Y = mX + c equation. For a numerical question, end by substituting back into the original relation or checking against the supplied points.
1. Express y = 5x2 − 2 in straight-line form using X = x2. State Y, m and c.
2. A graph of y against 1/x has gradient 8 and intercept −3. The original relation is y = a/x + b. Find a, b and y at x = 4.
3. For y = ax + b/x, a graph of y/x against 1/x2 has gradient −2 and intercept 5. Find a, b and y at x = 2.
4. Given log10 y = 2 log10 x − 1, express y in terms of x for x > 0.
5. Given log10 y = (log10 2)x + log10 7, recover the exponential model and find y at x = 3.
6. Given ln y = −0.4x + ln 9, recover y and find its exact value when x = 5.
7. Exact data (1,5), (2,20) and (3,45) obey y = axn, with a > 0 and x > 0. Find a, n and y at x = 4.
8. A graph of log10 y against log10 x passes through exact transformed-coordinate points (0,0.6) and (2,3.6). Find a and n in y = axn.
9. The vertical axis is log10 x and the horizontal axis is log10 y. The line passes through (0,1) and (2,2). Find y in terms of x.
10. A graph of y2 against x passes through (1,9) and (4,21). Find p and q in y2 = px + q. Find both possible y-values when x = 1.
11. A student writes log y = log 2 + n log x from y = 2 + xn. Explain the error and give a suitable log transformation for x > 0.
12. A graph of xy against x2 passes through (2,7) and (6,19). Find y in terms of x, for x ≠ 0, and calculate y at x = 2.
Full solutions
1–3: identify the axes and the constant roles
1. Choose Y = y. Then Y = 5X − 2, so m = 5 and c = −2. The line is straight in X, while the original relation is quadratic in x.
2. With X = 1/x and Y = y, the equation is Y = aX + b. Hence a = 8 and b = −3. At x = 4, y = 8/4 − 3 = −1. A negative output is allowed in this non-log example.
3. Divide by x: y/x = a + b/x2. Thus Y = bX + a, giving b = −2 and a = 5. At x = 2, y = 5(2) − 2/2 = 9. Verify through Y = −2X + 5: X = 1/4 gives Y = 4.5 and y = xY = 9.
4–6: undo the correct logarithm
4. The intercept is log10 a = −1, so a = 10−1 = 0.1. The exponent is 2. Therefore y = 0.1x2.
5. Compare with log y = x log b + log k. Here b = 2 and k = 7, giving y = 7 × 2x. At x = 3, y = 7 × 8 = 56.
6. Exponentiating with base e gives y = 9e−0.4x. At x = 5, the exact value is 9e−2, approximately 1.218. Using 10−0.4x would invert the wrong logarithm.
7–9: power models and reversed coordinates
7. At x = 1, a = 5. Comparing the first two points gives 20/5 = 2n, so n = 2. The third point checks the model: 5(3)2 = 45. Thus y = 5x2 and y at x = 4 is 80.
8. The gradient is (3.6 − 0.6)/(2 − 0) = 1.5. The intercept is 0.6, so n = 3/2 and a = 100.6, approximately 3.981. Since the transformed coordinates were specified as exact, retaining 100.6 preserves the exact constant.
9. The line has gradient (2 − 1)/(2 − 0) = 1/2 and intercept 1. Therefore log x = (1/2)log y + 1. Rearranging gives log y = 2 log x − 2, so y = 0.01x2, with x > 0. Reading the original line as log y against log x would produce the wrong model.
10–12: restrictions and returning to the original variables
10. The gradient is (21 − 9)/(4 − 1) = 4, and the intercept is 9 − 4(1) = 5. Therefore p = 4 and q = 5. At x = 1, y2 = 9, so y = 3 or −3. A nonnegative context would select 3, but none was supplied.
11. The student incorrectly split a logarithm of a sum. First write y − 2 = xn. For the stated positive x, this is positive, so log(y − 2) = n log x. A graph of log(y − 2) against log x passes through the origin for the exact model.
12. Let X = x2 and Y = xy. The line has gradient 3 and intercept 1, hence xy = 3x2 + 1. Dividing by x gives y = 3x + 1/x. At x = 2, y = 6 + 1/2 = 13/2.
A changed retest: can you read the representation independently?
Retest A: A graph of P against 1/t has gradient 12 and intercept 2. Write the original relation. It is P = 12/t + 2, with t ≠ 0. The intercept does not represent a finite t = 0 value.
Retest B: A relation is log10 C = −0.2t + 0.7. It becomes C = 100.7(10−0.2)t. The starting coefficient is positive, and the exponential base lies between zero and one.
Retest C: A graph of y/x2 against 1/x2 has gradient 5 and intercept 3. The transformed relation is y/x2 = 5/x2 + 3, so y = 3x2 + 5. The transformation itself used x ≠ 0. Although the recovered polynomial can be evaluated at zero, the transformed graph did not supply a coordinate there.
Use these after a delay rather than immediately after copying the solutions. The question is whether you can identify the representation and return from it without a tutor confirming every step.
A practical teaching and correction sequence
Begin with a line whose axes are ordinary X and Y. Ask the student to identify m and c. Then change only the labels to x2 and y. Next change both labels to 1/x2 and y/x. The geometry of the line has not changed; its mathematical interpretation has.
After this is secure, contrast a power model with an exponential model. Ask the student to explain why log x appears in one horizontal axis but not the other. This explanation is stronger evidence than remembering a table row.
When an answer is wrong, locate the first failed transition. Was the original algebra rearranged incorrectly? Were the axes reversed? Was a log intercept mistaken for a constant? Was an inverse transformation applied with the wrong base? Or was a valid negative branch discarded? Each error needs a different short repair.
A useful correction has four lines: the original relationship, the definitions of X and Y, the line equation, and the recovered relationship. Add one substitution check. This compact record exposes the student’s decisions without requiring a long paragraph around every calculation.
For students who can solve familiar forms but hesitate in mixed questions, remove the chapter heading and present two possible transformations. Ask which one would produce a line and why. Method selection must eventually happen before the calculator is touched.
What a complete answer should leave behind
A complete linear-law answer identifies its axes, uses valid algebra, states the gradient and intercept accurately, recovers the correct original constants and respects the original domain. Where data are approximate, it also distinguishes a fitted estimate from an exact identity.
The powerful idea is that changing representation can make a relationship easier to inspect. The accompanying responsibility is to return to the original variables without losing their meaning. A straight line is useful because of what it reveals, not merely because it is straight.
Continue through the Additional Mathematics Hub for neighbouring topics and the correct examination route. The Additional Mathematics Tuition pillar remains the separate destination for families considering teaching support. A marked graph question showing the student’s transformed axes is especially useful evidence for that conversation.
Source and scope review: 6 September 2026. The official links above establish the G3/O-Level curriculum boundary. Derivations, numerical examples, invented data, practice questions and solutions are original teaching material. This guide does not claim that a fitted graph proves causation, that every model extends beyond its observed range, or that completing the exercises guarantees a grade.
Continue the linear-law route across eduKate
For another worked teaching route, continue to the eduKateSengkang Coordinate Geometry, Circles and Linearising Relationships guide.
When linear law is part of a wider modelling pattern involving exponentials, logarithms and transformed relationships, use the Bukit Timah Tutor Exponentials, Logarithms, Linear Law and Modelling Synthesis Guide.