Cambridge O Level Additional Mathematics · Syllabus 4037 · 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 O Level Additional Mathematics.
Straight-Line Form in context
Straight-line form is the rearrangement of a non-linear relationship into \(Y=mX+c\), where \(X\) and \(Y\) are new variables built from \(x\) and \(y\), so that a curved relationship plots as a line whose gradient and intercept reveal hidden constants. For the power model \(y=Ax^{n}\), taking logarithms gives \(\ln y=\ln A+n\ln x\), so plotting \(\ln y\) against \(\ln x\) gives gradient \(n\) and intercept \(\ln A\). For the exponential model \(y=Ab^{x}\), \(\ln y=\ln A+x\ln b\), so plotting \(\ln y\) against \(x\) gives gradient \(\ln b\). Because the intercept is \(\ln A\), the constant \(A\) is recovered by exponentiating.
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.
Questions students ask about Straight-Line Form
How do you turn \(y=Ax^{n}\) into a straight line?
Take natural logarithms of both sides: \(\ln y=\ln A+n\ln x\), which has the form \(Y=mX+c\) with \(Y=\ln y\) and \(X=\ln x\). Plot \(\ln y\) against \(\ln x\); the gradient is \(n\) and the intercept is \(\ln A\), so \(A=\mathrm{e}^{\text{intercept}}\). For \(y=Ab^{x}\), plot \(\ln y\) against \(x\): the gradient is \(\ln b\), so \(b=\mathrm{e}^{\text{gradient}}\). Write the straight-line form first and name \(X\) and \(Y\) before you plot; this needs \(x>0\) and \(y>0\).
Why can't I read \(A\) straight off the intercept of a \(\ln y\) against \(\ln x\) graph?
Because the straight-line form is \(\ln y=n\ln x+\ln A\), so the constant term is \(\ln A\), not \(A\). An intercept of \(1.7\) means \(\ln A=1.7\), so \(A=\mathrm{e}^{1.7}\approx5.47\). The same trap catches the gradient of a \(\ln y\) against \(x\) graph, which is \(\ln b\) rather than \(b\). Keep the intercept unrounded until you exponentiate, because exponentiating magnifies any rounding error.