Answer: \(A\approx20\), \(b\approx1.5\), and \(y\approx130\) when \(x=3.5\).

Transform the given relationship into a straight-line form. The gradient and intercept of the straight line then give the constants in the original model.

Since \(y=Ax^b\), take natural logarithms of both sides:

\(\ln y=\ln A+b\ln x\).

This has the straight-line form

\(Y=C+bX\),

where \(Y=\ln y\), \(X=\ln x\), and \(C=\ln A\).

The values used for the straight-line graph are approximately:

A line of best fit gives a vertical intercept of about \(3\). Hence

\(\ln A\approx3\),

so

\(A\approx e^3\approx20\) to 2 significant figures.

The gradient of the line gives \(b\). From the graph,

\(b\approx1.5\).

For \(x=3.5\),

\(\ln x=\ln3.5\approx1.25\).

Reading from the straight-line graph gives approximately \(\ln y\approx4.9\), so

\(y\approx e^{4.9}\approx130\).

This is consistent with the mark scheme range \(120\leq y\leq150\).