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Re: PS geometery ... little confused. [#permalink]

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31 Aug 2008, 00:27

nishchals wrote:

i doubt if the Q is a correct one. Its a parabola and from the diagram, you can make out 2 coordinates, (2,0) and (0,3); that takes you no where.

pls look closer....(0,3) is not on the parabola...if it were than the problem was simple.........in that situation, solve parabola's equation and you would get that y coord will be 3/4 when x=3. The figure shows only one conformed point with coordinates ie the vertex (2,0).

My guess for the answer is 1. However I don't have mathematical proof to support it. Logical proof is if (0,3) were on parabola....then (3, 3/4) is on parabola Hence in the case of (0, little less than 3)......the point should be (3, little more than 3/4)....closest answer is 1

Re: PS geometery ... little confused. [#permalink]

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24 Apr 2009, 00:58

bang on alpha, for simplification:

This is an equation of parabola .. y = a (x-h)^2 + k where h & k are values of vertex A<0 if parabola opens downwards a>0 if parabola opens upwards here a is +ve vertex is at (2,0) => h = 2 k =0 y = a (x- 2)^2

To find a, check other point on parabola , here (0, 2.8) 2.8= a(0+2)^2 2.8= a(4) a= .7 now y = 0.7 (x-2)^2 put x= 3 then y = .7

Re: PS geometery ... little confused. [#permalink]

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16 Aug 2010, 04:50

Expert's post

fatihaysu wrote:

I really dont understand anything

It should be like that but i dont know

X1= 1/2 x2= 1

y1=2 y2=1

then

(y2-y1)/(x2-x1)=M which is -2 in this case

then

-2= (y-y1)/(x-x1)

y= -2x + 3 is equa.

ıf we put there as X y= -6 +3 then

y= -3 which is my answer....ı am confused where i am tripping up

Frankly I don't quite understand what is tested here...

But your approach is wrong because your method for finding equation of a graph is for straight line (linear equation) and obviously the curve shown on the diagram is not straight line.

Is it a graph of quadratic equation? Are the intersection points with axes (0,3) and (2,0)? If the answer on both questions is YES then the equation of the graph would be \(y=\frac{3}{4}x^2-3x+3\) and for \(x=1\) --> \(y=\frac{3}{4}x^2-3x+3=\frac{3}{4}\approx{1}\).

I wouldn't worry about this question at all.

P.S. Check the link about Coordinate Geometry in my signature for more on this issue. _________________

Re: PS geometery ... little confused. [#permalink]

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16 Aug 2010, 06:26

Bunuel wrote:

fatihaysu wrote:

I really dont understand anything

It should be like that but i dont know

X1= 1/2 x2= 1

y1=2 y2=1

then

(y2-y1)/(x2-x1)=M which is -2 in this case

then

-2= (y-y1)/(x-x1)

y= -2x + 3 is equa.

ıf we put there as X y= -6 +3 then

y= -3 which is my answer....ı am confused where i am tripping up

Frankly I don't quite understand what is tested here...

But your approach is wrong because your method for finding equation of a graph is for straight line (linear equation) and obviously the curve shown on the diagram is not straight line.

Is it a graph of quadratic equation? Are the intersection points with axes (0,3) and (2,0)? If the answer on both questions is YES then the equation of the graph would be \(y=\frac{3}{4}x^2-3x+3\) and for \(x=1\) --> \(y=\frac{3}{4}x^2-3x+3=\frac{3}{4}\approx{1}\).

I wouldn't worry about this question at all.

P.S. Check the link about Coordinate Geometry in my signature for more on this issue.

It was important for me that why my approach is wrong,i know now...

Re: According to the graph above, when x = 3, y most nearly ? [#permalink]

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13 Feb 2016, 19:04

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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