Answer: \(y=\ln(2x^2+0.52)\).

The key point is to identify the important features of the graph before using it to answer the question.

Since \(e^y\) is plotted against \(x^2\) and gives a straight line, write

\(e^y=mx^2+c.\)

The line passes through \((2.24,5)\) and \((4.74,10)\), so

\(m=\frac{10-5}{4.74-2.24} =\frac5{2.50}=2.\)

Use the point \((2.24,5)\):

\(5=2(2.24)+c.\)

Thus

\(c=5-4.48=0.52.\)

So

\(e^y=2x^2+0.52.\)

Taking natural logarithms,

\(y=\ln(2x^2+0.52).\)