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(I do not know the method to upload the way BHAI does.) Can someone edit this and show the graph, instead of the download option). Or perhaps, let me know how to show the graph on the screen.

Anyways here's my approach:
Eq of a parabola with vertex at (x0,y0) is
(y-y0)^2=4a(x-x0)^2 (x,y) any pint on the parabola. 4a is called the latus rectum (it is just a constant)

Based on the diagram looks like the parabola passes thru (0,3)
vertex of the parabola here is (2,0)

thanks man. I really do not know how to solve this. this question came from a frnd of mine in china. He says the OA is (-1/2). No idea, how that is. Perhaps, the OA is wrong or may be it needs a different approach.

I've looked at the pic. It's obvious that y can be only positive, so then scan answer choices. Between 1 and 1/2, 1/2 suits better.
But formally, you would have write down eq. y=ax^2+bx+c, then by using given points find coefficients a, b and c.

(I do not know the method to upload the way BHAI does.) Can someone edit this and show the graph, instead of the download option). Or perhaps, let me know how to show the graph on the screen.

(I do not know the method to upload the way BHAI does.) Can someone edit this and show the graph, instead of the download option). Or perhaps, let me know how to show the graph on the screen.