Answer: \(A\approx7\), \(b\approx2\), and \(x\approx2.2\) when \(y=200\).

Start by taking the key information from the graph or diagram, such as intercepts, turning points, amplitudes, periods, or intersection points. Then use the relevant algebra to justify the result.

Taking natural logarithms of \(y=Ab^{x^2}\) gives

\(\ln y=\ln A+x^2\ln b.\)

So a graph of \(\ln y\) against \(x^2\) should be a straight line with gradient \(\ln b\) and intercept \(\ln A\).

Using the line of best fit, the gradient is approximately \(0.69\), so

\(\ln b\approx0.69\), and hence \(b\approx2\).

The intercept is approximately \(1.95\), so

\(\ln A\approx1.95\), and hence \(A\approx7\) to 1 significant figure.

When \(y=200\), \(\ln y=\ln200\approx5.30\).

Using \(\ln y\approx1.95+0.69x^2\),

\(x^2\approx\frac{5.30-1.95}{0.69}\approx4.8\).

Thus \(x\approx2.2\) to 2 significant figures.

This gives the required answer: \(A\approx7\), \(b\approx2\), and \(x\approx2.2\) when \(y=200\).