Find the slopes of the tangent and the normal to the curve `y=2x^2+3sinx` at `x=0`

Explanation

$$ \begin{aligned} &y=2 x^{2}+3 \sin x \\ &\frac{d y}{d x}=4 x+3 \cos x \end{aligned} $$
Thus slope of tangent at `x=0` is `4(0)+3 \cos 0=3`


Hence slope of normal at the same point is `-\frac{1}{m}=-\frac{1}{3}`