Graph

Question

Please explain.

(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.

Thanks

carsen · Answer

Hi

How did u arrive to this answer (the official answer is different). Can you post your approach please. Thanks boksana.

srijay007 · Answer

The diagram is not so clear and not to scale. 

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)

Using this we get

9=4a*4
or a=9/16

hence eq is (y)^2 = 9/4 * (x-2)^2
when, x=3 y=3/2=1.5

This is the answer I get, though its not in the answer choice. 

Did I commit any mistake?

carsen · Answer

Hello Vijay

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.

Thanks mate

srijay007 · Answer

-1/2 is imposs
Just looking at the graph you can tell that

DMJ · Answer

The answer is E

From the shape of the function we know its type: y=a*(x+b)^2

Since 0=a*(2+b)^2 ==> b=-2

Since y is little less than 3 for x=0

3>4a
a>3/4

If x = 3
y>3/4*(3-2)^2 ==> y is little more than Â¾.

DMJ · Answer

I made a silly mistake.

should be a<3/4 

If x =3

y<3/4*(3-2)^2 ==> y is between Â½ and Â¾. The answer is D

boksana · Answer

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.

Bhai · Answer

There you go Carsen.

Bhai · Answer

There you go Carsen.

AlexJS · Answer

E... and I just eyeballed it.

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