Resources For Teachers For Tutors For Students & Parents Pricing
Create Free Account
Year 11 Methods (Unit 1 & 2) Variation

Fitting Data

20 practice questions 1 video lesson Theory + worked examples

Learn how to fit data to a variation model for Queensland Year 11 Mathematical Methods (QCAA). Given a table of values, you decide which relationship it follows before making any predictions.

You will learn to test ratios and products to choose between direct, square, inverse and square-root models, find the constant of variation, and use the rule to predict — the modelling skill that turns measurements into a reliable formula.

Practice 20 questions
Practice questions

Every question with a fully worked solution.

Start practising
Watch 1 video(s)
  • Fitting Data - Video - Proportion: Inverse Proportion to Square From Table Watch
Create a free accountTrack your progress and save your work as you go.
Create free account

Theory

In Year 11 Mathematical Methods (QCAA, Unit 1), fitting data means choosing a variation model that matches a table of values, finding the constant of variation \(k\), and using it to predict. This page shows how to test ratios and products to decide between direct, square, inverse and square-root models.

Fitting data is deciding which simple relationship links two columns of numbers, then finding its constant. The candidates are the variation models: direct \(y=kx\), square \(y=kx^2\), inverse \(y=\dfrac{k}{x}\) and square-root \(y=k\sqrt{x}\).

You choose the model with a constancy test: work out a candidate ratio or product for every row. Whichever stays constant identifies the model, and that constant value is \(k\). For example, if \(\dfrac{y}{x}\) is constant the model is direct; if \(xy\) is constant it is inverse.

As a first guide, if \(y\) increases with \(x\) try a direct power model; if \(y\) decreases as \(x\) increases, try an inverse model.

Find the constant, then predict. Once the model type is known, \(k\) comes from any one row, and the finished rule predicts new values.
Direct-variation fit to dataData points (2,6),(4,12),(5,15) lying on the line y=3x through the origin. x y
Data on a line through the origin: direct variation \(y=3x\).
Inverse-variation fit to dataData points (2,30),(3,20),(5,12) lying on the hyperbola y=60/x. x y
Data on a hyperbola: inverse variation \(y=\dfrac{60}{x}\).

The constancy tests that identify each model:

\[\text{direct: }\dfrac{y}{x}=k\qquad \text{square: }\dfrac{y}{x^2}=k\qquad \text{inverse: }xy=k\qquad \text{root: }\dfrac{y}{\sqrt{x}}=k\]
yx=k

Once \(k\) is found, the fitted rule is one of:

\[y=kx\qquad y=kx^2\qquad y=\dfrac{k}{x}\qquad y=k\sqrt{x}\]
y=kx2
One test at a time: compute the candidate ratio for every row; the model is the one whose value is the same all the way down.

How to fit a variation model to data

  1. Look at the trend: is \(y\) increasing (direct power) or decreasing (inverse) as \(x\) increases?
  2. Test for constancy: compute \(\dfrac{y}{x}\), \(\dfrac{y}{x^2}\), \(xy\) or \(\dfrac{y}{\sqrt{x}}\) for each row until one is constant — that value is \(k\).
  3. Write and use the rule: state \(y=kx^n\) (or \(\dfrac{k}{x}\)) and substitute to predict.
Example 1 — A direct model
Fit a model to the data \((2,\,6),(4,\,12),(5,\,15)\), then predict \(y\) when \(x=10\).
Solution

Trend — \(y\) rises with \(x\); test \(\dfrac{y}{x}\):

\(\dfrac{6}{2}\)\(=\)\(3\)
\(\dfrac{12}{4}\)\(=\)\(3\)
\(\dfrac{15}{5}\)\(=\)\(3\)

The ratio \(\dfrac{y}{x}=3\) is constant, so the model is direct with \(k=3\): \(y=3x\).

Predict — \(x=10\):

\(y\)\(=\)\(3(10)\)
\(=\)\(30\)

\(y=3x\); when \(x=10\), \(y=30\).

Fitting y equals three xDirect-variation data with the prediction (10,30) on y=3x. x y
y=3x
Example 2 — A square model
Fit a model to \((1,\,2),(2,\,8),(3,\,18)\), then predict \(y\) when \(x=5\).
Solution

Test \(\dfrac{y}{x}\) first — not constant:

\(\dfrac{2}{1}\)\(=\)\(2\)
\(\dfrac{8}{2}\)\(=\)\(4\)
\(\dfrac{18}{3}\)\(=\)\(6\)

Test \(\dfrac{y}{x^2}\):

\(\dfrac{2}{1^2}\)\(=\)\(2\)
\(\dfrac{8}{2^2}\)\(=\)\(2\)
\(\dfrac{18}{3^2}\)\(=\)\(2\)

The ratio \(\dfrac{y}{x^2}=2\) is constant, so \(y=2x^2\).

Predict — \(x=5\):

\(y\)\(=\)\(2(5)^2\)
\(=\)\(50\)

\(y=2x^2\); when \(x=5\), \(y=50\).

Fitting y equals two x squaredDirect square-variation data with the prediction (5,50) on y=2x^2. x y
y=2x2
Example 3 — An inverse model
Fit a model to \((2,\,30),(3,\,20),(5,\,12)\), then predict \(y\) when \(x=10\).
Solution

Trend — \(y\) falls as \(x\) rises; test the product \(xy\):

\((2)(30)\)\(=\)\(60\)
\((3)(20)\)\(=\)\(60\)
\((5)(12)\)\(=\)\(60\)

The product \(xy=60\) is constant, so the model is inverse with \(k=60\): \(y=\dfrac{60}{x}\).

Predict — \(x=10\):

\(y\)\(=\)\(\dfrac{60}{10}\)
\(=\)\(6\)

\(y=\dfrac{60}{x}\); when \(x=10\), \(y=6\).

Fitting y equals sixty over xInverse-variation data with the prediction (10,6) on y=60/x. x y
y=60x
Example 4 — A square-root model
Fit a model to \((1,\,4),(4,\,8),(9,\,12)\), then predict \(y\) when \(x=25\).
Solution

\(y\) rises slowly; test \(\dfrac{y}{\sqrt{x}}\):

\(\dfrac{4}{\sqrt{1}}\)\(=\)\(4\)
\(\dfrac{8}{\sqrt{4}}\)\(=\)\(4\)
\(\dfrac{12}{\sqrt{9}}\)\(=\)\(4\)

The ratio \(\dfrac{y}{\sqrt{x}}=4\) is constant, so \(y=4\sqrt{x}\).

Predict — \(x=25\):

\(y\)\(=\)\(4\sqrt{25}\)
\(=\)\(4(5)\)
\(=\)\(20\)

\(y=4\sqrt{x}\); when \(x=25\), \(y=20\).

Fitting y equals four root xSquare-root-variation data with the prediction (25,20) on y=4sqrt(x). x y
y=4x

Common pitfalls

Testing only one model. If \(\dfrac{y}{x}\) is not constant, do not give up — try \(\dfrac{y}{x^2}\), \(xy\) or \(\dfrac{y}{\sqrt{x}}\) before deciding.
Direct versus inverse. A rising table needs a direct power model; a falling table needs an inverse model. Check the trend first.
Reading \(k\) from the wrong test. The constant \(k\) is the value of the test that stayed constant, not the first ratio you happened to compute.

Frequently asked questions

How do you fit a variation model to data?

Decide whether \(y\) rises or falls with \(x\), test a candidate ratio or product for each row until one is constant, and take that constant as \(k\).

How do you know if data shows direct or inverse variation?

If \(\dfrac{y}{x}\) is constant it is direct (\(y=kx\)); if the product \(xy\) is constant it is inverse (\(y=\dfrac{k}{x}\)).

How do you find the constant of variation k from a table?

Once the model type is known, substitute any one data pair; for a direct model \(k=\dfrac{y}{x}\), for an inverse model \(k=xy\).

What if y over x is not constant?

Try other tests: \(\dfrac{y}{x^2}\) for a square model, \(\dfrac{y}{\sqrt{x}}\) for a square-root model, or \(xy\) for an inverse model.

How do you use a fitted model to make a prediction?

Write the finished rule with the value of \(k\), then substitute the new \(x\)-value to compute the predicted \(y\).

More in Variation
Direct Variation
20 questions · Video
Inverse Variation
20 questions · Video