Answer: \(A\approx0.5\), \(b\approx4\), \(y\approx35\) when \(x=1.75\), and \(x\approx1.63\) when \(y=20\).

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.

The given relationship is

\(y=Ab^{x^2}.\)

Taking natural logarithms gives

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

This has the form

\(Y=c+mX,\)

where \(Y=\ln y\) and \(X=x^2\). Therefore a graph of \(\ln y\) against \(x^2\) is a straight line with gradient \(\ln b\) and intercept \(\ln A\).

Using the data, the transformed values are approximately:

These points lie close to a straight line with gradient about

\(\ln4\)

and intercept about

\(\ln0.5.\)

Thus

\(A\approx0.5 \quad\text{and}\quad b\approx4.\)

Using \(y\approx0.5(4^{x^2})\), when \(x=1.75\),

\(x^2=3.0625.\)

So

\(y\approx0.5(4^{3.0625})\approx35.\)

When \(y=20\),

\(20\approx0.5(4^{x^2}).\)

Hence

\(4^{x^2}\approx40.\)

Taking logarithms,

\(x^2\ln4\approx\ln40.\)

Therefore

\(x\approx\sqrt{\frac{\ln40}{\ln4}}\approx1.63.\)