Cambridge IGCSE Additional Mathematics · Syllabus 0606 · Straight-Line Graphs
Straight-Line Form
What is Straight-Line Form?
Straight-line form is the rearrangement of a non-linear relationship into the shape Y = mX + c, where X and Y are new variables built from the original x and y. Plotting Y against X then produces a straight line whose gradient and vertical intercept reveal the unknown constants in the original relationship. For a power model y = Ax to the n, taking natural logarithms gives ln y = ln A + n ln x, so plotting ln y against ln x yields a line of gradient n and intercept ln A. For an exponential model y = A b to the x, taking natural logarithms gives ln y = ln A + x ln b, so plotting ln y against x yields a line of gradient ln b and intercept ln A. Because the intercept is ln A rather than A, the constant A is recovered by exponentiating the intercept, and both transformations require the quantities inside the logarithms to be strictly positive.
This definition is part of the Straight-Line Graphs chapter in Cambridge IGCSE Additional Mathematics.
Common mistakes with Straight-Line Form
- 13. Reading \(A\) directly from an intercept that is \(\ln A\) The error A plot of \(\ln y\) against \(\ln x\) has intercept \(1.7\); concluding \(A=1.7\). Why it fails The straight-line form is \(\ln y=n\ln x+\ln A\), so the constant term is \(\ln A\). An intercept of \(1.7\) therefore says \(\ln A=1.7\), giving \(A=\mathrm{e}^{1.7}\approx5.47\) — more than three times the value reported. Fix Write “intercept \(=\ln A\)” before substituting any number, so the exponentiation is unavoidable.
- 14. Reading \(b\) directly from a gradient that is \(\ln b\) The error A plot of \(\ln y\) against \(x\) has gradient \(1.099\); concluding \(b=1.099\). Why it fails For \(y=Ab^{x}\), the straight-line form is \(\ln y=x\ln b+\ln A\), so the gradient is \(\ln b\). The actual base is \(b=\mathrm{e}^{1.099}=3\). Reporting \(1.099\) describes a relationship that barely grows, where the true one triples at every step. Fix Sanity-check against the data. If \(y\) roughly triples as \(x\) increases by \(1\), then \(b\) must be near \(3\), so a \(b\) close to \(1\) is immediately suspect.