Answer: \(y=e^{571/40-(109/50)x^2}-2\).

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.

Let

\(Y=\ln(y+2)\quad\text{and}\quad X=x^2\).

Since \(Y\) is plotted against \(X\), the straight line has equation

\(Y=mX+c\).

The two points are \((2.25,9.37)\) and \((4.75,3.92)\), so the gradient is

\(m=\frac{3.92-9.37}{4.75-2.25}\).

Thus

\(m=\frac{-5.45}{2.50}=-2.18=-\frac{109}{50}\).

Use \((2.25,9.37)\) to find \(c\):

\(9.37=-2.18(2.25)+c\).

So

\(c=9.37+4.905=14.275=\frac{571}{40}\).

Therefore

\(\ln(y+2)=\frac{571}{40}-\frac{109}{50}x^2\).

Exponentiate both sides:

\(y+2=e^{571/40-(109/50)x^2}\).

Hence

\(y=e^{571/40-(109/50)x^2}-2\).