Answer
$y=ax^2$ shows the maximum curvature at $(0,0)$ and has no minimum curvature.
Work Step by Step
We have: $f(x)=ax^2$
This implies that $\dfrac{d f(x)}{dx}=2ax$ and $\dfrac{d^2 f(x)}{dx^2}=2a$
We know that $\kappa (x)=\dfrac{[\dfrac{d^2 f(x)}{dx^2}]}{[1+(\dfrac{d f(x)}{dx}^2)]^{3/2}}$
So, $\kappa (x)=\dfrac{[2a]}{[1+4a^2x^2]^{3/2}}$
and $\dfrac{d \kappa}{dx}=-\dfrac{[24a^3x]}{[1+4a^2x^2]^{5/2}}$
and,$ \dfrac{d^2 \kappa}{dx^2}=\dfrac{d}{dx}[-\dfrac{[24a^3x]}{[1+4a^2x^2]^{5/2}}]=\dfrac{-24a^3[(1+4a^2x^2)^{5/2}-20a^2x^2 (4a^2x^2+1)^{3/2}]}{(1+4a^2x^2)^5}$
Set $x=0$, so $ \dfrac{d^2 \kappa}{dx^2}=-24a^3 \lt 0$
This means that the maximum occurs at $x=0, y=0$. Therefore, $y=ax^2$ shows the most curvature at $(0,0)$ and has no minimum curvature.
