Constants and predictions from your own graph

Once your line of best fit is drawn, its gradient and vertical intercept give the constants of the law, and the line itself gives predictions. A prediction is a value of the original variables, so every reading from the line has to be changed back.

Why: the line is the law in disguise. For \(y=ax^n\), \(\lg y=n\lg x+\lg a\), so the gradient is \(n\) and the intercept is \(\lg a\). Undoing the logarithm with \(a=10^{c}\) recovers the constant, and the same undoing turns any point of the line into an \((x,\,y)\) pair.

Method

1. Read the gradient from a gradient triangle on your line and the intercept where the line meets the vertical axis.
2. Match them to the linear form of the law and solve for the constants. Undo logarithms: \(\lg a=c\) gives \(a=10^{c}\); \(\ln A=c\) gives \(A=\mathrm{e}^{c}\).
3. To predict \(y\) from \(x\): change \(x\) into \(X\), read \(Y\) from the line (or substitute into its equation), then change \(Y\) back into \(y\).
4. To find \(x\) for a given \(y\): change \(y\) into \(Y\), read \(X\), then change back.
5. Give answers to \(3\) significant figures with units. A value outside the range of the data is an extrapolation and is less reliable.

Example

Readings follow \(y=ax^n\) for \(2\le x\le12\). The line of best fit of \(\lg y\) against \(\lg x\) passes through \((0.3,\,1.25)\) and \((1.1,\,2.45)\).

\(n=\dfrac{2.45-1.25}{1.1-0.3}=\dfrac{1.2}{0.8}=1.5\); intercept \(\lg a=1.25-1.5(0.3)=0.8\), so \(a=10^{0.8}\approx6.31\).

When \(x=5\): \(\lg x=0.699\), \(\lg y=1.5(0.699)+0.8=1.849\), so \(y=10^{1.849}\approx70.6\).

When \(y=100\): \(\lg y=2\), \(\lg x=\dfrac{2-0.8}{1.5}=0.8\), so \(x=10^{0.8}\approx6.31\), inside the range of the data.