IA modelling, step by step · Reciprocal
Reciprocal model, step by step: y = a/x + b and y = a/(x − h) + k
Inverse relationships — time against speed, cost per person against group size — curve towards two asymptotes. Take the vertical asymptote from the context, plot y against 1/(x − h), and the gradient and intercept of that line give a and k.
Example data, invented for this guide. The context is realistic, but the numbers were made up to show the method. Use your own measured or sourced data in your IA.
When to use it
The shape of the data
- y falls steeply at first, then levels off towards a horizontal line (or rises towards one).
- x × y (or (x − h)(y − k)) is roughly constant.
- A plot of y against 1/x is close to straight.
The context
- Time for a fixed distance or task at different speeds or rates.
- Cost per person when a fixed cost is shared.
- Physics: Boyle's law, lens formulas, intensity-type relationships.
Course fit: AA SL and HL (the reciprocal function and rational functions); AI SL (inverse variation, axⁿ with n = −1); AI HL (linearising). Everyone can justify it well because the asymptotes usually have a meaning.
The example data
A cyclist logs the same commute on different days: average speed while moving (from a bike computer) and total door-to-door time, which includes waiting at traffic lights.
| i | x: Average moving speed (km/h) | y: Journey time (min) |
|---|---|---|
| 1 | 12 | 66.5 |
| 2 | 15 | 53.6 |
| 3 | 18 | 46.4 |
| 4 | 20 | 41.8 |
| 5 | 22 | 39.1 |
| 6 | 25 | 34.5 |
| 7 | 28 | 32 |
| 8 | 30 | 30.3 |
Example 1: y = a/x + b (journey time and speed)
Time = distance ÷ speed suggests a/x, and time spent stopped at lights adds a constant b. The vertical asymptote is x = 0 (speed zero), so plot y against 1/x.
Step 1 · Asymptotes from the context, then linearise
The vertical asymptote is taken at x = 0. If y = a/(x − h) + k, then plotting y against X = 1/x gives a straight line with gradient a and intercept k (the horizontal asymptote).
Step 2 · Tabulate the sums
| i | 1/x | y | 1/x² | 1/xy | y² |
|---|---|---|---|---|---|
| 1 | 0.083333 | 66.5 | 0.0069444 | 5.5417 | 4422.25 |
| 2 | 0.066667 | 53.6 | 0.0044444 | 3.5733 | 2872.96 |
| 3 | 0.055556 | 46.4 | 0.0030864 | 2.5778 | 2152.96 |
| 4 | 0.05 | 41.8 | 0.0025 | 2.09 | 1747.24 |
| 5 | 0.045455 | 39.1 | 0.0020661 | 1.7773 | 1528.81 |
| 6 | 0.04 | 34.5 | 0.0016 | 1.38 | 1190.25 |
| 7 | 0.035714 | 32 | 0.0012755 | 1.1429 | 1024 |
| 8 | 0.033333 | 30.3 | 0.0011111 | 1.0100 | 918.09 |
| Σ | 0.410058 | 344.200 | 0.0230280 | 19.0929 | 15856.6 |
n = 8 points. The transformed values are rounded here; keep them unrounded in your calculator or spreadsheet.
Step 3 · Gradient
m = (nΣXY − ΣX ΣY) / (nΣX² − (ΣX)²)
m = (8 × 19.0929 − 0.410058 × 344.200) / (8 × 0.0230280 − 0.410058²) = 11.6014 / 0.0160770 = 721.6
Here X = 1/x and Y = y.
Step 4 · Intercept
X̄ = ΣX/n = 0.051257, Ȳ = ΣY/n = 43.025
c = Ȳ − m X̄ = 43.025 − 721.6 × 0.051257 = 6.037
The line of best fit always passes through the mean point (X̄, Ȳ).
Step 5 · Correlation
r = (nΣXY − ΣX ΣY) / √[(nΣX² − (ΣX)²)(nΣY² − (ΣY)²)] = 11.6014 / √(0.0160770 × 8378.84) = 0.9996
r² = 0.9991: a very strong positive linear correlation between 1/x and y.
Step 6 · Back-substitute
y = 721.6/x + 6.037
Asymptotes: x = 0 (vertical) and y = 6.037 (horizontal). Check that both make sense in the context — the horizontal one is the value y approaches as x grows large.
Step 7 · Residuals and goodness of fit
For each point, the residual is the observed value minus the model's value: e = y − ŷ.
| i | x | y (data) | ŷ (model) | y − ŷ | (y − ŷ)² |
|---|---|---|---|---|---|
| 1 | 12 | 66.5 | 66.17 | 0.328 | 0.108 |
| 2 | 15 | 53.6 | 54.14 | −0.545 | 0.297 |
| 3 | 18 | 46.4 | 46.13 | 0.273 | 0.0747 |
| 4 | 20 | 41.8 | 42.12 | −0.318 | 0.101 |
| 5 | 22 | 39.1 | 38.84 | 0.262 | 0.0688 |
| 6 | 25 | 34.5 | 34.90 | −0.402 | 0.161 |
| 7 | 28 | 32 | 31.81 | 0.191 | 0.0365 |
| 8 | 30 | 30.3 | 30.09 | 0.209 | 0.0437 |
| Σ | 0.8905 |
- Sum of squared residuals: SSR = Σ(y − ŷ)² = 0.8905
- Total sum of squares about the mean ȳ = 43.02: SST = Σ(y − ȳ)² = 1047
- Coefficient of determination: R² = 1 − SSR/SST = 1 − 0.8905/1047 = 0.9991
- Root-mean-square error: RMSE = √(SSR/n) = 0.334 — a typical size of a residual, in the units of y.
Example 2: y = a/(x − h) + k (a lens, asymptotes from the context)
For a lens of focal length f, no sharp image forms when the object is at the focal point, so the vertical asymptote is at u = f = 10 cm: h = 10 from the context. Plot v against 1/(u − 10).
A lamp is placed at different distances from a converging lens labelled f = 10 cm, and a screen is moved until the image is sharp.
| i | x: Object distance u (cm) | y: Image distance v (cm) |
|---|---|---|
| 1 | 15 | 30.4 |
| 2 | 20 | 19.8 |
| 3 | 25 | 16.9 |
| 4 | 30 | 14.8 |
| 5 | 40 | 13.4 |
| 6 | 50 | 12.4 |
| 7 | 60 | 12.1 |
Step 1 · Asymptotes from the context, then linearise
The vertical asymptote is taken from the context at x = h = 10. If y = a/(x − h) + k, then plotting y against X = 1/(x − 10) gives a straight line with gradient a and intercept k (the horizontal asymptote).
Step 2 · Tabulate the sums
| i | 1/(x − 10) | y | (1/(x − 10))² | (1/(x − 10))y | y² |
|---|---|---|---|---|---|
| 1 | 0.2 | 30.4 | 0.04 | 6.08 | 924.16 |
| 2 | 0.1 | 19.8 | 0.01 | 1.98 | 392.04 |
| 3 | 0.066667 | 16.9 | 0.0044444 | 1.1267 | 285.61 |
| 4 | 0.05 | 14.8 | 0.0025 | 0.74 | 219.04 |
| 5 | 0.033333 | 13.4 | 0.0011111 | 0.44667 | 179.56 |
| 6 | 0.025 | 12.4 | 0.000625 | 0.31 | 153.76 |
| 7 | 0.02 | 12.1 | 0.0004 | 0.242 | 146.41 |
| Σ | 0.495000 | 119.800 | 0.0590806 | 10.9253 | 2300.58 |
n = 7 points. The transformed values are rounded here; keep them unrounded in your calculator or spreadsheet.
Step 3 · Gradient
m = (nΣXY − ΣX ΣY) / (nΣX² − (ΣX)²)
m = (7 × 10.9253 − 0.495000 × 119.800) / (7 × 0.0590806 − 0.495000²) = 17.1763 / 0.168539 = 101.9
Here X = 1/(x − 10) and Y = y.
Step 4 · Intercept
X̄ = ΣX/n = 0.070714, Ȳ = ΣY/n = 17.114
c = Ȳ − m X̄ = 17.114 − 101.9 × 0.070714 = 9.908
The line of best fit always passes through the mean point (X̄, Ȳ).
Step 5 · Correlation
r = (nΣXY − ΣX ΣY) / √[(nΣX² − (ΣX)²)(nΣY² − (ΣY)²)] = 17.1763 / √(0.168539 × 1752.02) = 0.9996
r² = 0.9991: a very strong positive linear correlation between 1/(x − 10) and y.
Step 6 · Back-substitute
y = 101.9/(x − 10) + 9.908
Asymptotes: x = 10 (vertical) and y = 9.908 (horizontal). Check that both make sense in the context — the horizontal one is the value y approaches as x grows large.
Step 7 · Residuals and goodness of fit
For each point, the residual is the observed value minus the model's value: e = y − ŷ.
| i | x | y (data) | ŷ (model) | y − ŷ | (y − ŷ)² |
|---|---|---|---|---|---|
| 1 | 15 | 30.4 | 30.29 | 0.110 | 0.0121 |
| 2 | 20 | 19.8 | 20.10 | −0.299 | 0.0893 |
| 3 | 25 | 16.9 | 16.70 | 0.198 | 0.0393 |
| 4 | 30 | 14.8 | 15.00 | −0.203 | 0.0413 |
| 5 | 40 | 13.4 | 13.30 | 0.0953 | 0.00909 |
| 6 | 50 | 12.4 | 12.46 | −0.0554 | 0.00307 |
| 7 | 60 | 12.1 | 11.95 | 0.154 | 0.0238 |
| Σ | 0.2179 |
- Sum of squared residuals: SSR = Σ(y − ŷ)² = 0.2179
- Total sum of squares about the mean ȳ = 17.11: SST = Σ(y − ȳ)² = 250.3
- Coefficient of determination: R² = 1 − SSR/SST = 1 − 0.2179/250.3 = 0.9991
- Root-mean-square error: RMSE = √(SSR/n) = 0.176 — a typical size of a residual, in the units of y.
What the model tells you
- Example 1: a ≈ 722 min·km/h. Since time (min) = 60 × distance ÷ speed, a/60 ≈ 12.0 km is the length of the commute — a check against a map is excellent validation.
- Example 1: b ≈ 6.0 min is the horizontal asymptote: however fast you ride, the journey never takes less than about 6 minutes (the time stopped at lights).
- Example 2: the lens formula 1/u + 1/v = 1/f rearranges to v = f²/(u − f) + f, so theory predicts a = f² = 100 and k = f = 10 for this lens. The fit gives a = 102 and k = 9.91: close, and comparing them with theory is strong reflection.
The same on a GDC
Enter and plot the data first, then fit. The key sequences are for current operating systems; menus differ slightly between versions.
TI-84 Plus CE
- Data: [stat] → 1: Edit… Type the x values in L1 and the y values in L2. To plot: [2nd] [y=] (STAT PLOT) → Plot1: On, Type: scatter, Xlist: L1, Ylist: L2, then [zoom] → 9: ZoomStat.
- Make the transformed list: in L3's header type 1/L1 (or 1/(L1 − 10)) [enter].
- [stat] → CALC → 4: LinReg(ax+b), Xlist L3, Ylist L2. The gradient is a, the intercept is b (or k).
- There is no built-in reciprocal regression on the TI-84, so the linearisation is the method.
TI-Nspire CX
- Data: Add a Lists & Spreadsheet page; name column A xs and column B ys and type the data. Add a Data & Statistics page (or a Graphs page with menu → Graph Entry/Edit → Scatter Plot) and choose xs and ys.
- In Lists & Spreadsheet add a column recip with formula =1/xs (or =1/(xs − 10)).
- Linear Regression (mx+b) with X List recip, Y List ys: m is a, b is the asymptote.
Casio fx-CG50
- Data: [MENU] → Statistics. Type the x values in List 1 and the y values in List 2. To plot: [F1] (GRAPH) → [F6] (SET): Graph Type Scatter, XList List1, YList List2; [EXIT], then [F1] (GRAPH1).
- Put 1 ÷ List 1 (or 1 ÷ (List 1 − 10)) in List 3: move to List 3's header cell and type it (List is in [OPTN] → [F1] (LIST)).
- [F2] (CALC) → [F6] (SET): XList List3, YList List2; then [F3] (REG) → [F1] (X) → [F1] (ax+b).
In Desmos
Free at desmos.com/calculator. In a regression, ~ means “fit this model”; subscripts are typed with an underscore (x_1 shows as x₁).
- Data in x₁, y₁.
y_1 ~ a/x_1 + b— linear in a and b, so this is the same least-squares answer as the 1/x line.- With a vertical asymptote from context:
y_1 ~ a/(x_1 - 10) + k. - Draw the asymptotes as dashed lines (
x = 10,y = k) to show them on your graph.
More on technology in the IA: using Desmos, GeoGebra and Excel.
How to write it up in your IA
- Explain where each asymptote comes from in the context before fitting.
- Show the transformed table (1/x) and the straight-line plot.
- Show the least-squares line once and how a and the asymptote come from it.
- Interpret a, h and k in context; check them against theory or a measurement if you can.
- State the domain (the model is undefined at x = h) and discuss extrapolation near the asymptote.
These are the points to cover, not sentences to copy. Write every explanation in your own words, about your own data.
Common mistakes
- Fitting y = a/x when the context clearly has a non-zero asymptote (b ≠ 0).
- Including a data point at x = h (division by zero).
- Confusing an inverse relationship with exponential decay: exponential decay reaches half its height in a fixed step, a reciprocal does not.
- Not stating units for a, which often has a physical meaning.
Is this good enough for Criterion E?
- The asymptotes are justified from the context.
- The linearisation (y against 1/(x − h)) is shown with a table and graph.
- a and k are found and interpreted in context, with units.
- The fit is checked with residuals and compared with at least one other model (for example exponential decay).
- Domain and extrapolation limits are stated.
- HL: h is estimated from the data as well (by minimising SSR for several values of h) and compared with the context value.
SL or HL? Criterion E asks for mathematics that fits your course. At SL, fitting with technology is fine when you explain the method and justify every choice. At HL, show more of the mathematics yourself — the last item in the list is an example. See Criterion E and Criterion D.
Frequently asked questions
How do I choose h for y = a/(x − h) + k?
From the context whenever possible — the x value where the quantity becomes undefined or infinite (a focal length, a minimum group size, a threshold). If the context gives no value, try several values of h and choose the one with the smallest SSR, and say so.
Free: the IA checklist an examiner uses
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