Last visit was: 18 Nov 2025, 20:47 It is currently 18 Nov 2025, 20:47
Close
GMAT Club Daily Prep
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.
Close
Request Expert Reply
Confirm Cancel
User avatar
Sumithra
Joined: 24 Oct 2006
Last visit: 19 Sep 2012
Posts: 168
Own Kudos:
66
 [18]
Posts: 168
Kudos: 66
 [18]
6
Kudos
Add Kudos
12
Bookmarks
Bookmark this Post
User avatar
Sumithra
Joined: 24 Oct 2006
Last visit: 19 Sep 2012
Posts: 168
Own Kudos:
66
 [1]
Posts: 168
Kudos: 66
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
Sumithra
Joined: 24 Oct 2006
Last visit: 19 Sep 2012
Posts: 168
Own Kudos:
66
 [1]
Posts: 168
Kudos: 66
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
Fig
Joined: 01 May 2006
Last visit: 02 Feb 2025
Posts: 1,031
Own Kudos:
253
 [4]
Posts: 1,031
Kudos: 253
 [4]
2
Kudos
Add Kudos
2
Bookmarks
Bookmark this Post
Sumithra
Would appreciate if someone could shed some light on how to view an equation/inequality in xy-plane, for both straight lines and curves?


For lines, curves or any functions represented on an XY Plan, the concepts remain similar.

o y > f(x)
On the XY plan, the matching values of y are "above" the draw of y = f(x).

To know where is the concerned region, u can use 1 value of x : x=0 (if definied on 0), calculate f(0) and then look where is a value of y such that y > f(0). U will be "above" the f(x).

Ex 1 : For the line f(x) = 3*x+1, y > f(x) is in green (fig 1). f(0) = 1 so u know that the point (0,2) is in good region. Now, we can draw the region limited by the line and containing (0,2).

Ex 2 : For the line f(x) = -2*x-1, y > f(x) is in green (fig 2). f(0) = -1 so u know that the point (0,0) is in good region.

Ex 3 : For the line f(x) = x^2 -2*x -1, y > f(x) is in green (fig 3). f(0) = -1 so u know that the point (0,0) is in good region.



o y < f(x)
On the XY plan, the matching values of y are "under" the draw of y = f(x).

To know where is the concerned region, u can use 1 value of x : x=0 (if definied on 0), calculate f(0) and then look where is a value of y such that y < f(0). U will be "under" the f(x).

Ex 4 : For the line f(x) = -x+1, y < f(x) is in green (fig 4). f(0) = 1 so u know that the point (0,0) is in good region.

Ex 5 : For the line f(x) = 2*x^2 +4*x -1, y < f(x) is in green (fig 5). f(0) = -1 so u know that the point (0,-2) is in good region.
Attachments

Fig1_y_above_3x+1.gif
Fig1_y_above_3x+1.gif [ 2.88 KiB | Viewed 20304 times ]

Fig2_y_above_-2x-1.gif
Fig2_y_above_-2x-1.gif [ 2.76 KiB | Viewed 20270 times ]

Fig3_Y_above_Xpow2-2X-1.gif
Fig3_Y_above_Xpow2-2X-1.gif [ 2.66 KiB | Viewed 20275 times ]

User avatar
Fig
Joined: 01 May 2006
Last visit: 02 Feb 2025
Posts: 1,031
Own Kudos:
Posts: 1,031
Kudos: 253
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Fig 4 and 5 :)
Attachments

Fig4_Y_under_-X+1.gif
Fig4_Y_under_-X+1.gif [ 2.73 KiB | Viewed 20241 times ]

Fig5_Y_under_2Xpow2+4X-1.gif
Fig5_Y_under_2Xpow2+4X-1.gif [ 2.6 KiB | Viewed 20250 times ]

User avatar
svas
Joined: 15 Jan 2007
Last visit: 20 Aug 2007
Posts: 16
Own Kudos:
4
 [1]
Posts: 16
Kudos: 4
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
hey things like distance between (a,b) (c,d) : its a right angle triangle. simple pythagoras formula that is applied there.

all who gave gmat already, Did you try to remember all of these ? I am trying to rely on doing these right there. am i being over-ambitious ?
User avatar
Sumithra
Joined: 24 Oct 2006
Last visit: 19 Sep 2012
Posts: 168
Own Kudos:
Posts: 168
Kudos: 66
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Thanks Fig for taking pains to explain. Now, I understand better.
avatar
Dr MBA
Joined: 13 Jul 2007
Last visit: 16 Sep 2008
Posts: 3
Own Kudos:
3
 [1]
Posts: 3
Kudos: 3
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
wow.... all tis stuff is great... made my day.......... hopefully also my GMAT


Better to understand a little than to misunderstand a lot
User avatar
beckee529
Joined: 11 Jun 2007
Last visit: 23 Feb 2012
Posts: 394
Own Kudos:
1,925
 [1]
Posts: 394
Kudos: 1,925
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Thanks for putting in the time to provide this guide, this will be very helpful with my studies!
avatar
getbig
Joined: 09 Dec 2008
Last visit: 05 Jan 2009
Posts: 1
Posts: 1
Kudos: 0
Kudos
Add Kudos
Bookmarks
Bookmark this Post
thanks for the collection of formulas... I've been looking for something like this.
User avatar
Ibodullo
Joined: 13 Jan 2009
Last visit: 26 Jan 2023
Posts: 120
Own Kudos:
Given Kudos: 9
Concentration: Finance
Schools:Harvard Business School, Stanford
Posts: 120
Kudos: 33
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Thanks for that one!
User avatar
irajeevsingh
Joined: 27 Jun 2008
Last visit: 05 Aug 2009
Posts: 72
Own Kudos:
Given Kudos: 11
Posts: 72
Kudos: 42
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Thanks for collating it at one place.
User avatar
amit2k9
Joined: 08 May 2009
Last visit: 18 Jun 2017
Posts: 535
Own Kudos:
Given Kudos: 10
Status:There is always something new !!
Affiliations: PMI,QAI Global,eXampleCG
Posts: 535
Kudos: 636
Kudos
Add Kudos
Bookmarks
Bookmark this Post
thanks for the approach.However,Can somebody please explain me how to plot,
y>f(x),where f(x) = x^2-2x -1.

Thanks in advance.
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 18 Nov 2025
Posts: 105,355
Own Kudos:
Given Kudos: 99,964
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,355
Kudos: 778,091
Kudos
Add Kudos
Bookmarks
Bookmark this Post
amit2k9
thanks for the approach.However,Can somebody please explain me how to plot,
y>f(x),where f(x) = x^2-2x -1.

Thanks in advance.

Plot y=x^2-2x-1. y>x^2-2x-1 would be above the graph:
Attachment:
MSP5724200423hdf0492bch00005382e4a5b3eah3fg.gif
MSP5724200423hdf0492bch00005382e4a5b3eah3fg.gif [ 3.1 KiB | Viewed 7871 times ]
Does this make sense?
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 38,587
Own Kudos:
Posts: 38,587
Kudos: 1,079
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Moderator:
Math Expert
105355 posts