Linearisation means rewriting a relationship so that two chosen quantities, call them Y and X, satisfy Y = mX + c. If the plotted points then lie on a straight line, the relationship is confirmed, and the gradient and intercept give the constants.
This appears in straight lines and linearisation and is the main reason Additional Mathematics asks about gradients and intercepts at all.
How do you decide what to plot?
Compare the given formula with Y = mX + c and match the pattern. The expression that the gradient multiplies becomes X, and the constant on its own becomes c. You may need to multiply or divide through first.
| Relationship | Plot Y against X | Gradient | Intercept |
|---|---|---|---|
| y = ax² + b | y against x² | a | b |
| y = a/x + b | y against 1/x | a | b |
| y = ax³ + b | y against x³ | a | b |
| y = ax + b/x | xy against x² | a | b |
For the last row, multiply both sides by x: xy = ax² + b. Relationships such as y = axn and y = abx need logarithms, which is covered in linearising an exponential relationship.
Worked example
Data for y = px² + q are recorded in this table.
| x | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| y | 7 | 13 | 23 | 37 |
Step 1, choose the axes: plot y against x², so X = x².
Step 2, new row: x² = 1, 4, 9, 16.
Step 3, check it is straight: the gradient from X = 1 to 4 is (13 − 7) ÷ (4 − 1) = 2. From X = 4 to 9 it is (23 − 13) ÷ (9 − 4) = 2. From X = 9 to 16 it is (37 − 23) ÷ (16 − 9) = 2. The gradient is constant, so the points lie on a line.
Step 4, constants: p = 2. Substituting (1, 7): 7 = 2(1) + q, so q = 5.
Result: y = 2x² + 5. Check: at x = 3, 2(9) + 5 = 23.
The mistake to watch for
A common slip is to calculate the gradient with the original x values even though the graph uses x².
Mistaken working: using x = 2 and x = 4, m = (37 − 13) ÷ (4 − 2) = 12
The student subtracted x values (2 and 4) instead of X values (4 and 16).
The correction is to add the transformed row to your table and read every gradient from that row only. With X = 4 and 16 the gradient is (37 − 13) ÷ (16 − 4) = 24 ÷ 12 = 2, which agrees with the correct answer.
Check yourself
1. The relationship is y = a/x + b. State what to plot on each axis, and what the gradient and intercept represent.
Show answer
Plot y on the vertical axis and 1/x on the horizontal axis. The gradient is a and the intercept is b.
2. Data follow y = ax³ + b. When x = 1, y = 3. When x = 2, y = 17. Find a and b.
Show answer
X = x³ gives X = 1 and X = 8. Gradient a = (17 − 3) ÷ (8 − 1) = 14 ÷ 7 = 2. Then 3 = 2(1) + b, so b = 1.
a = 2, b = 1. Check: 2(8) + 1 = 17.
3. For y = ax + b/x, a graph of xy against x² passes through (1, 9) and (9, 17). Find a and b.
Show answer
Gradient = (17 − 9) ÷ (9 − 1) = 8 ÷ 8 = 1, so a = 1. Intercept: 9 = 1(1) + b, so b = 8.
a = 1, b = 8, so y = x + 8/x. Check at x = 3: xy = 9 + 8 = 17.
Where this leads next
Once you can linearise, recover model constants from a gradient and intercept when only a graph or two readings are given. The non-calculator working trainer is useful for practising exact fraction gradients, and the practice set mixes these forms together.
If the “what goes on each axis” decision is where you stall, our teachers can drill it with you in online one-to-one Additional Mathematics tuition.