Help plz

What is the smallest distance between the origin and a point on the graph below.

I can see that the graph is a concave up parabola'

So I know a smallest distance exists, there will be no maximum distance.

$y=\dfrac{1}{\sqrt{2}}\left(x^2-8\right)\\ \left ( x,\dfrac{1}{\sqrt{2}}\left(x^2-8\right)\right) \text{ is a general point on this graph}\\ \text{I need the distance of this point from (0,0)}\\ d^2=x^2+[\frac{1}{\sqrt{2}}\left(x^2-8\right)]^2\\ d^2=x^2+\frac{1}{2}\left(x^2-8\right)^2\\ $

d will be minimum when d^2 is minimum.  So to avoid confusion I will let  v=d^2

$v=x^2+\frac{1}{2}\left(x^2-8\right)^2\\$

$\frac{dv}{dx }=2x+(x^2-8)*2x\\ \frac{dv}{dx }=2x(1+x^2-8)\\ \frac{dv}{dx }=2x(x^2-7)\\ \frac{dv}{dx }=0\;\;when \;\;x=0\;\;or\;\;x=\pm\sqrt7\\ $

When x=0   v=32

When x= +/-sqrt7   v=7+0.5=7.5 

Clearly the min is when v=7.5

So the minimum distance is  $\sqrt{7.5}$

LaTex

y=\dfrac{1}{\sqrt{2}}\left(x^2-8\right)\\
\left ( x,\dfrac{1}{\sqrt{2}}\left(x^2-8\right)\right) \text{   is a general point on this graph}\\
\text{I need the distance of this point from (0,0)}\\
d^2=x^2+[\frac{1}{\sqrt{2}}\left(x^2-8\right)]^2\\
d^2=x^2+\frac{1}{2}\left(x^2-8\right)^2\\

v=x^2+\frac{1}{2}\left(x^2-8\right)^2\\

thank you very much