Find all School-related info fast with the new School-Specific MBA Forum

It is currently 26 Jul 2014, 09:21

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.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

If the parabola represented by f(x) = ax^2 + bx + c passes

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
1 KUDOS received
Manager
Manager
avatar
Joined: 29 Jul 2009
Posts: 121
Followers: 2

Kudos [?]: 11 [1] , given: 23

If the parabola represented by f(x) = ax^2 + bx + c passes [#permalink] New post 02 Mar 2010, 07:10
1
This post received
KUDOS
3
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

24% (03:34) correct 76% (02:11) wrong based on 42 sessions
If the parabola represented by f(x) = ax^2 + bx + c passes through points (-3,0) , (0,3) and (5,0), which of the following must be true?

I f(-1) > f(2)
II f(1) > f(0)
III f(2) > f(1)

A. only I
B. only II
c. only III
D. I and II
E. I and III
[Reveal] Spoiler: OA

Last edited by Bunuel on 03 Apr 2014, 05:24, edited 3 times in total.
Edited the OA.
Intern
Intern
avatar
Joined: 22 Nov 2009
Posts: 32
Followers: 0

Kudos [?]: 6 [0], given: 1

Re: Co ordinate geometry!! [#permalink] New post 02 Mar 2010, 09:54
apoorvasrivastva wrote:

If the parabola represented by f(x) = ax^2 + bx + c passes through points (-3,0) , (0,3) and (5,0), which of the following must be true?

I f(-1) > f(2)
II f(1) > f(0)
III f(2) > f(1)


f(x)=y

Substituting (x,y) as (-3,0) , (0,3) and (5,0), we get the following equations:

0 = 9a-3b+c
3 = c
0 = 25a+5b+c

Solving: a=-1/5, b=2/5, c=3
f(x) = -x^2/5 + x/5 +3

The options are on f(-1), f(0), f(1) and f(2).
f(-1) = -1/5 + 2/5 + 3 = 16/5 = 3.2
f(0) = 3
f(1) = 1/5 + 2/5 + 3 = 18/5 = 3.6
f(2) = -4/5 + 4/5 + 3 = 3

I f(-1) > f(2) true
II f(1) > f(0) true
III f(2) > f(1) false

So, option D, I and II

(not sure if there is an easier way to solve this!)
_________________

kudos +1 ?

Expert Post
2 KUDOS received
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 18746
Followers: 3251

Kudos [?]: 22454 [2] , given: 2619

Re: Co ordinate geometry!! [#permalink] New post 02 Mar 2010, 17:32
2
This post received
KUDOS
Expert's post
1
This post was
BOOKMARKED
apoorvasrivastva wrote:
If the parabola represented by f(x) = ax^2 + bx + c passes through points (-3,0) , (0,3) and (5,0), which of the following must be true?

I f(-1) > f(2)
II f(1) > f(0)
III f(2) > f(1)

A. only I
B. only II
c. only III
D. I and II
E. I and III

please suggest me as to how do i crack this question in 2 mints..i am completely lost on this :(

OA is
[Reveal] Spoiler:
D


Answer to this question cannot be D, it should be B (II only).

To solve this question you don't need to calculate a, b, and c. We should put the given three points on the XY-plane and we'll get: parabola f(x)=ax^2+bx+c will have the vertex at x=1, halfway between x=-3 and x=5, as f(-3)=0=f(5). Plus, as f(0)=3 the parabola would be downward. As the parabola is downward, value of f(x) at x=1, f(1), is the the greatest value of f(x). As we move from x=1 to either of direction the value of f(x) will decrease. So any f(m) would be greater than f(n) if m is closer to x=1 than n. For example f(9)>f(10)>f(-12) or f(6)=f(-4) or f(-5)=f(7)>f(-100).

Hence
I. f(-1)>f(2) is false, as x=2 is 1 farther from x=1 and x=-1 is 2 farther than x=1. (True inequality would be f(-1)<f(2)).

II. f(1)>f(0) is true as f(1) is vertex and more than any f(x).

III. f(2)>f(1) is false as no f(x) is more than f(1). (The correct would be f(2)<f(1)).

Hope it's clear.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Manager
Manager
avatar
Joined: 29 Jul 2009
Posts: 121
Followers: 2

Kudos [?]: 11 [0], given: 23

Re: Co ordinate geometry!! [#permalink] New post 02 Mar 2010, 23:01
@bunnuel i have dited my post for the st I ..it was a typo error i am sorry abt that

it is f(-1) > f(-2)

i am not clear as to how did u get the vertex as x=1 :(
Expert Post
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 18746
Followers: 3251

Kudos [?]: 22454 [0], given: 2619

Re: Co ordinate geometry!! [#permalink] New post 02 Mar 2010, 23:18
Expert's post
apoorvasrivastva wrote:
@bunnuel i have dited my post for the st I ..it was a typo error i am sorry abt that

it is f(-1) > f(-2)

i am not clear as to how did u get the vertex as x=1 :(


Intersection points of parabola with x-axis are (-3,0) and (5,0):

-----(-3)--0----(5)---, as parabola is symmetric, the x coordinate of the vertex must be halfway between x=-3 and x=5 --> x=\frac{-3+5}{2}=1. As parabola is downward, f(x) naturally will have it's max values at vertex x=1, f(1).

As for the typo in the stem. If I is saying: f(-1) > f(-2), then it's true as x=-1 is closer to x=1, than x=-2, which means that f(-1)>f(-2).
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Intern
Intern
avatar
Joined: 04 Mar 2010
Posts: 3
Followers: 0

Kudos [?]: 0 [0], given: 1

Re: Co ordinate geometry!! [#permalink] New post 04 Mar 2010, 20:22
amazing explanation Bunuel. I would normally have found a, b and c and then solved. Thanks for showing me the way to think more than the way you solved the problem.
SVP
SVP
User avatar
Joined: 09 Sep 2013
Posts: 1750
Followers: 165

Kudos [?]: 33 [0], given: 0

Premium Member
Re: If the parabola represented by f(x) = ax^2 + bx + c passes [#permalink] New post 23 Dec 2013, 20:20
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.
_________________

GMAT Books | GMAT Club Tests | Best Prices on GMAT Courses | GMAT Mobile App | Math Resources | Verbal Resources

Re: If the parabola represented by f(x) = ax^2 + bx + c passes   [#permalink] 23 Dec 2013, 20:20
    Similar topics Author Replies Last post
Similar
Topics:
If the parabola represented by f(x)=ax2+bx+c passes through peergmatclub 2 21 Jun 2008, 14:26
2 If the parabola represented by f(x) = ax^2 + bx + c passes vd 10 03 Jun 2008, 21:41
Does the line represented by y=ax^2+bx+c intersect with getzgetzu 5 21 Apr 2006, 02:16
f(x)=ax^2+bx+c. Point (1,6) is on the curve of f(x). Is laxieqv 2 04 Mar 2006, 09:31
Does the line represented by ax^2+bx+c intersect the x-axis? joemama142000 2 21 Feb 2006, 16:11
Display posts from previous: Sort by

If the parabola represented by f(x) = ax^2 + bx + c passes

  Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.