The graph gives you m and c for Y = mX + c. Your job is to translate back: the plotted Y and X are expressions in x and y, so m and c must be placed into the original relationship, not into y = mx + c.
This is the second half of straight lines and linearisation. It follows transforming data into straight-line form.
What does the conversion look like?
Write three things in order: what was plotted, the line’s equation, and then the original relationship after rearranging.
Suppose a graph of y/x against x is straight. Then Y = y/x, X = x, and Y = mX + c becomes y/x = mx + c. Multiply by x to get y = mx² + cx.
The same pattern works for every form in the previous lesson. The steps are always: identify Y and X, find m and c, write Y = mX + c with numbers, then undo the transformation.
Worked example
Variables x and y satisfy y = px² + qx. A graph of y/x against x is a straight line through (1, 5) and (3, 11). Find p and q, then y when x = 4.
Step 1, plotted quantities: Y = y/x and X = x. Dividing the model by x gives y/x = px + q.
Step 2, gradient: m = (11 − 5) ÷ (3 − 1) = 6 ÷ 2 = 3. So p = 3.
Step 3, intercept: 5 = 3(1) + c, so c = 2. So q = 2.
Step 4, original model: y = 3x² + 2x.
Step 5, prediction: when x = 4, y = 3(16) + 2(4) = 48 + 8 = 56.
Check: at x = 3, y = 27 + 6 = 33, and y/x = 33 ÷ 3 = 11, which matches the point (3, 11).
The mistake to watch for
A common slip is to treat the plotted Y as if it were y.
Mistaken working: gradient 3, intercept 2, so y = 3x + 2
This ignores that the vertical axis showed y/x, not y.
It even fits the first point, since 3(1) + 2 = 5, which makes it look right. The second point exposes it: the graph point (3, 11) means y/x = 11, so y = 33, but 3(3) + 2 = 11. The correction is to write “Y = y/x” at the start of your working and substitute it back at the end.
Check yourself
1. A graph of y against x² is a line through (0, 4) and (2, 10). Find the formula for y in terms of x.
Show answer
Gradient = (10 − 4) ÷ (2 − 0) = 3. The intercept is 4, read at X = 0.
y = 3x² + 4. Check: when x² = 2, y = 6 + 4 = 10.
2. A graph of xy against x is a line through (2, 7) and (5, 16). Find y in terms of x.
Show answer
Gradient = (16 − 7) ÷ (5 − 2) = 3. Then 7 = 3(2) + c, so c = 1. So xy = 3x + 1.
y = 3 + 1/x. Check at x = 5: 3 + 0.2 = 3.2, and xy = 5 × 3.2 = 16.
3. A graph of y against x² has gradient 1.5 and passes through (4, 11). Find y = px² + q, then find the positive x for which y = 59.
Show answer
p = 1.5. Substitute: 11 = 1.5(4) + q, so q = 5. The formula is y = 1.5x² + 5.
For y = 59: 1.5x² = 54, so x² = 36 and x = 6. Check: 1.5(36) + 5 = 59.
Where this leads next
The next question is how far such a model can be trusted, which is the focus of assessing an extrapolated linear model. You can also test your conversions in the mixed practice set, and the non-calculator working trainer keeps fractional gradients exact.
Some students do the graph work well and lose the final mark at the conversion step. Our teachers look for that pattern in online one-to-one Additional Mathematics tuition.