To find the slope of the normal to the curve \(y = 2x^2 + 3\sin x\) at \(x = 0\), we first need to determine the derivative of \(y\) with respect to \(x\), which gives us the slope of the tangent line at any point on the curve.
The derivative is calculated as follows:
\(\frac{dy}{dx} = \frac{d}{dx}(2x^2 + 3\sin x) = 4x + 3\cos x\)
Substituting \(x = 0\) into this derivative gives the slope of the tangent line at \(x = 0\):
\(\frac{dy}{dx}\bigg|_{x=0} = 4(0) + 3\cos(0) = 3\)
The slope of the normal line is the negative reciprocal of the slope of the tangent line. Therefore, the slope of the normal at \(x = 0\) is:
\(-\frac{1}{3}\)
Hence, the correct answer is \(-\frac{1}{3}\).