Answer: \(A\approx2.09\), \(b\approx1.36\); \(y\approx5.66\times10^5\); \(x\approx0.456\).

Taking logarithms to base \(10\),

\(\lg y=\lg A+bx^2.\)

So the graph of \(\lg y\) against \(x^2\) has gradient \(b\) and intercept \(\lg A\).

Using the two given points,

\(b=\frac{6.88-5.25}{4.83-3.63} =\frac{1.63}{1.20} =1.358\ldots.\)

So

\(b\approx1.36.\)

Now use \(5.25=\lg A+3.63b\):

\(\lg A=5.25-3.63(1.358\ldots)=0.31925\ldots.\)

Therefore

\(A=10^{0.31925\ldots}=2.085\ldots\approx2.09.\)

(b) When \(x=2\),

\(y=A\times10^{b(2)^2}.\)

Using the values above,

\(y\approx2.08569\times10^{1.35833\ldots\times4} =565696\ldots.\)

So

\(y\approx5.66\times10^5.\)

(c) When \(y=4\),

\(\lg4=\lg A+bx^2.\)

So

\(x^2=\frac{\lg4-\lg A}{b}.\)

Hence

\(x=\sqrt{\frac{\lg4-0.31925\ldots}{1.35833\ldots}} =0.456\ldots.\)

The positive value is

\(x\approx0.456.\)