Answer
$y=ax^2$ shows maximum curvature at $(0,0)$ and has no minimum curvature.
Work Step by Step
We have: $\dfrac{d f(x)}{dx}=2ax \\ \dfrac{d^2 f(x)}{dx^2}=2a$
Since, $\kappa (x)=\dfrac{[\dfrac{d^2 f(x)}{dx^2}]}{[1+(\dfrac{d f(x)}{dx}^2)]^{3/2}}$
$\implies \kappa (x)=\dfrac{[2a]}{[1+4a^2 \ x^2]^{3/2}}\\=-\dfrac{[24a^3x]}{[1+4a^2 \ x^2]^{5/2}}$
Next, $ \dfrac{d^2 \kappa}{dx^2}=\dfrac{d}{dx}[-\dfrac{[24a^3x]}{[1+4a^2x^2]^{5/2}}]$
or, $=\dfrac{-24a^3 \times [(1+4a^2 \ x^2)^{5/2}-20a^2 \ x^2 (4a^2 \ x^2+1)^{3/2}]}{(1+4a^2 \ x^2)^5}$
Consider $x=0$, and $ \dfrac{d^2 \kappa}{dx^2}=-24a^3 \lt 0$
Thus, $y=ax^2$ shows maximum curvature at $(0,0)$ and has no minimum curvature.
