Judging by the picture, I don't think the focus point is correct, as it outside of the parabola.
I derived the coordinates of the parabola by first determining the a, b, and c value.
$$a=3, b=(-6), c=2$$
And then using the formula for the focus point:
$$F=(b/(2a), 1/(4a) + d/(4a))$$
Calculations
What is the correct focus point?
EDIT:
Better by differentiation with symbolic coefficients
$$ y= ax^2+bx+c;\, y^{'}=2ax+b ;\, y''=2a= 1/R $$ where $R$ is the radius of curvature, as $y^{'}=0$ in the curvature expression
The extremum point is
$$ (-b/2a, c-b^2/4a)$$ to whose $y$ coordinate we add $R/2=1/ 4a$ (sign automatically valid due to $a$ symbolically assigned ) to get $y$ coordinate of focal point as by virtue of a property of parabola its focal point bisects vertex-center of curvature line and this can be easily proved if we take parabola $ x^2= 4 f y.$ This gives us the formula for the focus.
$$ \boxed{(-b/2a,\, c+(1-b^2)/4a ) } $$
When $(a,b,c)=(3,-6,2)$ is plugged in, focal point coordinates are:
$$(1,-\frac{11}{12}). $$