Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: PS geometery ... little confused. [#permalink]
30 Aug 2008, 23:27

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

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]
16 Aug 2010, 03:50

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]
16 Aug 2010, 05:26

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...

Thank you Bunuel _________________

This is not finished here...Watch me.....

gmatclubot

Re: PS geometery ... little confused.
[#permalink]
16 Aug 2010, 05:26

Low GPA MBA Acceptance Rate Analysis Many applicants worry about applying to business school if they have a low GPA. I analyzed the low GPA MBA acceptance rate at...

In out-of-the-way places of the heart, Where your thoughts never think to wander, This beginning has been quietly forming, Waiting until you were ready to emerge. For a long...