Choosing a Model: Linear, Quadratic or Exponential
Theory
Is a pattern growing by a constant amount, by a steadily changing amount, or by a constant factor? This Year 10 maths page shows how to choose a model from a table of values: constant first differences mean a linear model, constant second differences a quadratic model, and a constant ratio an exponential model. It also shows how to write the rule, make predictions and state the domain where the model makes sense. Curriculum: VC2M10A15 / AC9M10A04.
A model is a rule that describes how one quantity depends on another. When the \(x\)-values in a table go up in equal steps, the pattern in the \(y\)-values shows which kind of model fits.
A linear model \(y = mx + c\) has constant first differences: \(y\) changes by the same amount each step. Its graph is a straight line.
A quadratic model \(y = ax^2 + bx + c\) has first differences that change by the same amount each step, so its second differences (the differences of the differences) are constant. Its graph is a parabola.
An exponential model \(y = a \times b^x\) has a constant ratio: each \(y\)-value is the previous one multiplied by \(b\). It describes growth (\(b \gt 1\)) or decay (\(0 \lt b \lt 1\)), such as compound interest, populations and depreciation.
The domain of a model is the set of input values for which it makes sense. A model can give impossible values, such as a negative height or a negative amount of water, outside that domain.
| Model | Test (equal steps in \(x\)) | Rule |
|---|---|---|
| linear | first differences constant | \(y = mx + c\) |
| quadratic | second differences constant | \(y = ax^2 + bx + c\) |
| exponential | ratio of consecutive values constant | \(y = a \times b^x\) |
When \(x\) goes up by \(1\) each step:
In each case the value at \(x = 0\) gives the constant: \(c\) for linear and quadratic models, \(a\) for an exponential model.
How to choose a model from a table
- Check the steps: make sure the \(x\)-values go up by equal amounts.
- First differences: if they are all equal, the model is linear.
- Second differences: if the first differences are not equal but the second differences are, the model is quadratic.
- Ratios: divide each \(y\)-value by the one before. If the ratio is constant, the model is exponential.
- Write the rule (see Formulas) and check it against every point in the table, not just the first two.
Using a model
- Predict by substituting. Predictions far outside the data (extrapolation) are less reliable.
- Solve for an input: set the rule equal to the value and solve, or use a table of values when the unknown is a power.
- Domain: state the inputs for which the model makes sense, such as \(0 \le t \le\) the time when a tank is empty or a ball lands.
- Compare models by working out their values at the same input; an exponential model with \(b \gt 1\) eventually overtakes any linear model.
| \(x\) | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) |
|---|---|---|---|---|---|
| A | \(7\) | \(4\) | \(1\) | \(-2\) | \(-5\) |
| B | \(2\) | \(3\) | \(6\) | \(11\) | \(18\) |
| C | \(80\) | \(40\) | \(20\) | \(10\) | \(5\) |
A: first differences \(-3, -3, -3, -3\), all equal. Linear, starting at \(7\) and falling by \(3\) each step: \(\boldsymbol{y = 7 - 3x}\).
B: first differences \(1, 3, 5, 7\); second differences \(2, 2, 2\). Quadratic with \(2a = 2\), so \(a = 1\). The value at \(x = 0\) is \(2\), and \(y = x^2 + 2\) fits every point: \(\boldsymbol{y = x^2 + 2}\).
C: first differences \(-40, -20, -10, -5\) are not equal, and neither are their differences. Ratios: \(\dfrac{40}{80} = \dfrac{20}{40} = \dfrac{10}{20} = \dfrac{5}{10} = \dfrac{1}{2}\). Exponential decay: \(\boldsymbol{y = 80 \times \left(\tfrac{1}{2}\right)^x}\).
| \(x\) | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) |
|---|---|---|---|---|---|
| \(y\) | \(3\) | \(2\) | \(5\) | \(12\) | \(23\) |
(ii) Predict \(y\) when \(x = 10\).
(i) First differences \(-1, 3, 7, 11\); second differences \(4, 4, 4\).
| \(2a\) | \(=\) | \(4 \quad\Rightarrow\quad a = 2\) |
| \(x = 0:\ c\) | \(=\) | \(3\) |
| \(x = 1:\ 2 + b + 3\) | \(=\) | \(2 \quad\Rightarrow\quad b = -3\) |
So \(\boldsymbol{y = 2x^2 - 3x + 3}\). Check at \(x = 4\): \(32 - 12 + 3 = 23\). \(\checkmark\)
| \(\text{(ii) } y\) | \(=\) | \(2(10)^2 - 3(10) + 3\) |
| \(\) | \(=\) | \(200 - 30 + 3 = \mathbf{173}\) |
| \(t\) | \(0\) | \(1\) | \(2\) | \(3\) |
|---|---|---|---|---|
| \(V\) | \(1500\) | \(1200\) | \(960\) | \(768\) |
(ii) Write a rule for \(V\) and find the value after \(5\) years.
(i) The value drops by \(\$300\), \(\$240\), \(\$192\): not a constant amount, so not linear. The ratios are
| \(\dfrac{1200}{1500} = \dfrac{960}{1200} = \dfrac{768}{960}\) | \(=\) | \(0.8\) |
a constant ratio, so the model is exponential: each year the laptop keeps \(80\%\) of its value.
| \(\text{(ii) } V\) | \(=\) | \(\boldsymbol{1500 \times 0.8^t}\) |
| \(t = 5:\ V\) | \(=\) | \(1500 \times 0.8^5 = \mathbf{\$491.52}\) |
| \(t\) | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) |
|---|---|---|---|---|---|
| \(d\) | \(0\) | \(0.4\) | \(1.6\) | \(3.6\) | \(6.4\) |
(ii) When does the ball reach the bottom? State the domain of the model.
(i) First differences \(0.4, 1.2, 2.0, 2.8\); second differences \(0.8, 0.8, 0.8\), so quadratic with \(2a = 0.8\), \(a = 0.4\). At \(t = 0\), \(c = 0\); at \(t = 1\), \(0.4 + b = 0.4\), so \(b = 0\).
| \(d\) | \(=\) | \(\boldsymbol{0.4t^2}\) |
(ii) Set \(d = 10\):
| \(0.4t^2\) | \(=\) | \(10\) |
| \(t^2\) | \(=\) | \(25\) |
| \(t\) | \(=\) | \(5\ (t \ge 0)\) |
The ball reaches the bottom after \(\mathbf{5}\) seconds. After that it is off the ramp, so the model only applies for \(\boldsymbol{0 \le t \le 5}\).
Common pitfalls
Frequently asked questions
How do you tell if a table is linear, quadratic or exponential?
Work out the first differences. If they are constant the table is linear. If not, find the second differences; if those are constant it is quadratic. If instead each value is the previous one times the same number, it is exponential.
What are second differences?
They are the differences between consecutive first differences. For \(1, 4, 9, 16\) the first differences are \(3, 5, 7\) and the second differences are \(2, 2\).
How do you find a in a quadratic model from a table?
When \(x\) goes up by \(1\) each step, the constant second difference equals \(2a\), so \(a\) is half the second difference.
Why does exponential growth overtake linear growth?
A linear model adds the same amount each step, but an exponential model multiplies by the same factor, so its increases keep getting bigger. For example \(100 \times 1.1^n\) passes \(100 + 20n\) at \(n = 15\).
What is the domain of a model?
The input values for which the model makes sense. A model for water in a tank only works until the tank is empty; a model for a thrown ball only works until it lands.