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Graph

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carsen
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Graph [#permalink] New post   11 Jul 2004, 18:58
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



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Re: Graph [#permalink] New post   11 Jul 2004, 20:12
Hi

How did u arrive to this answer (the official answer is different). Can you post your approach please. Thanks boksana.
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Re: Graph [#permalink] New post   11 Jul 2004, 20:34
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?
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Re: Graph [#permalink] New post   11 Jul 2004, 20:44
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
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Re: Graph [#permalink] New post   11 Jul 2004, 20:54
-1/2 is imposs
Just looking at the graph you can tell that
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Re: Graph [#permalink] New post   12 Jul 2004, 09:01
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 ¾.
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Re: Graph [#permalink] New post   12 Jul 2004, 09:33
I made a silly mistake.

should be a<3/4

If x =3

y<3/4*(3-2)^2 ==> y is between ½ and ¾. The answer is D
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Re: Graph [#permalink] New post   12 Jul 2004, 13:00
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.
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Re: Graph [#permalink] New post   12 Jul 2004, 14:19
carsen wrote:
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


There you go Carsen.
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Carsen.JPG
Carsen.JPG [ 14.31 KiB | Viewed 2022 times ]

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Bhai
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Re: Graph [#permalink] New post   12 Jul 2004, 14:25
carsen wrote:
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


There you go Carsen.
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Carsen.JPG
Carsen.JPG [ 14.31 KiB | Viewed 2016 times ]

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AlexJS
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Re: Graph [#permalink] New post   12 Jul 2004, 18:58
E... and I just eyeballed it. ;)



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Re: Graph [#permalink]
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