GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 26 Jun 2019, 05:10

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

Does the curve y=b(x-2)^2+c lie completely above the x-axis?

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

 
Manager
Manager
User avatar
G
Joined: 15 Dec 2015
Posts: 115
GMAT 1: 660 Q46 V35
GPA: 4
WE: Information Technology (Computer Software)
Does the curve y=b(x-2)^2+c lie completely above the x-axis?  [#permalink]

Show Tags

New post 31 Jul 2017, 11:47
1
1
10
00:00
A
B
C
D
E

Difficulty:

  85% (hard)

Question Stats:

38% (01:34) correct 62% (01:56) wrong based on 117 sessions

HideShow timer Statistics


Does the curve \(y=b(x-2)^2+c\) lie completely above the x-axis?

Statement 1: b>0,c<0
Statement 2: b>2,c<2
Most Helpful Expert Reply
Veritas Prep GMAT Instructor
User avatar
D
Joined: 16 Oct 2010
Posts: 9369
Location: Pune, India
Re: Does the curve y=b(x-2)^2+c lie completely above the x-axis?  [#permalink]

Show Tags

New post 18 Feb 2019, 07:50
2
3
ArupRS wrote:
chetan2u, Bunuel, VeritasKarishma, gmatbusters, amanvermagmat

Dear experts: Does GMAT test the concepts of conic sections?

Regards,
Arup


A circle is fair play though you may not see other conic sections.

But this question requires absolutely nothing but basic understanding of co-ordinate geometry.

"completely above x axis" means y is not 0 and not negative. y must be positive.

\(y = b(x - 2)^2 + c\)

Statement 1: b>0,c<0

c is negative.
b is positive. (x - 2)^2 can not be negative. It will be 0 (when x = 2) or positive.
So at x = 2, the first term will be 0 but c will be negative. This means y will be negative.
So the curve will not lie completely above x axis.

Statement 2: b>2,c<2

c could be positive or negative.
b is positive so first term would be 0 or positive.
Depending on the value of c, the graph could lie completely above x axis (say if c = 1) or below too (if c = -1)
Not sufficient.

Answer (A)
_________________
Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
General Discussion
Intern
Intern
avatar
B
Joined: 14 Nov 2012
Posts: 23
GMAT 1: 740 Q51 V38
Reviews Badge
Does the curve y=b(x-2)^2+c lie completely above the x-axis?  [#permalink]

Show Tags

New post 28 Oct 2017, 13:38
3
1
DH99 wrote:
Does the curve \(y=b(x-2)^2+c\) lie completely above the x-axis?

Statement 1: b>0,c<0
Statement 2: b>2,c<2


The question is basically about: whether y >0 or not ?
We need to check the sign of the discriminant \((b^2 - 4ac)\)
If the discriminant >= 0, then the curve will not lie completely above the x-axis.

\(y=b(x-2)^2+c\)
=> \(y = b(x^2 - 4x +4) + c\)
=> \(y = bx^2 - 4bx + 4b + c\)

The discrimimant = \(16b^2 - 4b(4b + c) = -4bc\)

Statement 1: b>0,c<0
=> The discrimimant = -4bc > 0 (Sufficient)

Statement 2: b>2,c<2
With c < 2, it is unsure if c < 0 or c > 0 (Insufficient)
Manager
Manager
avatar
S
Joined: 23 Jan 2018
Posts: 173
Location: India
WE: Information Technology (Computer Software)
Re: Does the curve y=b(x-2)^2+c lie completely above the x-axis?  [#permalink]

Show Tags

New post 17 Feb 2019, 23:59
chetan2u, Bunuel, VeritasKarishma, gmatbusters, amanvermagmat

Dear experts: Does GMAT test the concepts of conic sections?

Regards,
Arup
Manager
Manager
avatar
B
Joined: 29 Sep 2016
Posts: 99
Re: Does the curve y=b(x-2)^2+c lie completely above the x-axis?  [#permalink]

Show Tags

New post 20 Mar 2019, 17:56
VeritasKarishma wrote:
ArupRS wrote:
chetan2u, Bunuel, VeritasKarishma, gmatbusters, amanvermagmat

Dear experts: Does GMAT test the concepts of conic sections?

Regards,
Arup


A circle is fair play though you may not see other conic sections.

But this question requires absolutely nothing but basic understanding of co-ordinate geometry.

"completely above x axis" means y is not 0 and not negative. y must be positive.

\(y = b(x - 2)^2 + c\)

Statement 1: b>0,c<0

c is negative.
b is positive. (x - 2)^2 can not be negative. It will be 0 (when x = 2) or positive.
So at x = 2, the first term will be 0 but c will be negative. This means y will be negative.
So the curve will not lie completely above x axis.

Statement 2: b>2,c<2

c could be positive or negative.
b is positive so first term would be 0 or positive.
Depending on the value of c, the graph could lie completely above x axis (say if c = 1) or below too (if c = -1)
Not sufficient.

Answer (A)


Hi Vertiaskarishma
I am not very clear.

y = (positive or zero) + c

Since there is no limitation/restriction of c given. If absolute of c is a small number, y can be positive.
If absolute of c is a large number, y can be negative.

Pls let me know where am I going wrong.
Manager
Manager
User avatar
G
Status: The darker the night, the nearer the dawn!
Joined: 16 Jun 2018
Posts: 104
GMAT ToolKit User Reviews Badge
Re: Does the curve y=b(x-2)^2+c lie completely above the x-axis?  [#permalink]

Show Tags

New post 04 Apr 2019, 22:46
DHAR wrote:
Does the curve \(y=b(x-2)^2+c\) lie completely above the x-axis?

Statement 1: b>0,c<0
Statement 2: b>2,c<2


The curve is of the standard-form: \(Ax^2 + Bx + C\)
Discriminant (D): \(B^2 - 4AC\)

The equation holds the lowest/highest point at coordinates \((x,y)\): \(( -B/2A , -D/4A)\)

We need to confirm whether the quadratic equation opens down or up - Refer the picture.

Given:
    "completely above x axis" means y is not 0 and not negative. y must be positive.
    \(Y = b(x-2)^2+c\)

    \(Y = bx^2 - 4bx + 4b + c\)
    Comparing with the standard-form:
      A: \(b\)
      B: \(-4b\)
      C: \(4b+c\)

Thus, the Y coordinate having the lowest/highest point: \((-D/4A)\)
Resultant-Discriminant (D): \(- [(-4b)^2 - 4*b*(4b+c) ]/4b\) = \(c\) - Rest of the term cancels out.

Hence, only the value of \(c\) is vital to determine the coordinates.
    If \(b > 0\): Parabola opens UP - b is the co-efficient of \(x^2\).
      \(c > 0\) ---------> The lowest Y-coordinate is +ve
      \(c < 0\) ---------> The lowest Y-coordinate is -ve

    If \(b < 0\): Parabola opens DOWN
      \(c > 0\) ---------> The highest Y-coordinate is +ve
      \(c < 0\) ---------> The highest Y-coordinate is -ve

Thus, Only statement-1 is SUFFICIENT.
Statement-2 can yield +ve, -ve OR zero, depending on the respective values of \(c\)
Attachments

The coefficient of x2 determines the direction in which the parabola opens.For positive coefficients, the parabola opens up..jpg
The coefficient of x2 determines the direction in which the parabola opens.For positive coefficients, the parabola opens up..jpg [ 72.92 KiB | Viewed 589 times ]


_________________
---------------------------------------------------------------------------------
“The trouble is, you think you have time.” – Buddha
Giving Kudos is the best way to encourage and appreciate people.
Manager
Manager
avatar
B
Joined: 07 May 2018
Posts: 53
Re: Does the curve y=b(x-2)^2+c lie completely above the x-axis?  [#permalink]

Show Tags

New post 04 May 2019, 03:50
elainetianfong wrote:
DH99 wrote:
Does the curve \(y=b(x-2)^2+c\) lie completely above the x-axis?

Statement 1: b>0,c<0
Statement 2: b>2,c<2


The question is basically about: whether y >0 or not ?
We need to check the sign of the discriminant \((b^2 - 4ac)\)
If the discriminant >= 0, then the curve will not lie completely above the x-axis.

\(y=b(x-2)^2+c\)
=> \(y = b(x^2 - 4x +4) + c\)
=> \(y = bx^2 - 4bx + 4b + c\)

The discrimimant = \(16b^2 - 4b(4b + c) = -4bc\)

Statement 1: b>0,c<0
=> The discrimimant = -4bc > 0 (Sufficient)

Statement 2: b>2,c<2
With c < 2, it is unsure if c < 0 or c > 0 (Insufficient)



Could you please explain how you got the discriminant?

Regards,

Ritvik
Manager
Manager
User avatar
G
Status: The darker the night, the nearer the dawn!
Joined: 16 Jun 2018
Posts: 104
GMAT ToolKit User Reviews Badge
Re: Does the curve y=b(x-2)^2+c lie completely above the x-axis?  [#permalink]

Show Tags

New post 04 May 2019, 11:04
menonrit wrote:
elainetianfong wrote:
DH99 wrote:
Does the curve \(y=b(x-2)^2+c\) lie completely above the x-axis?

Statement 1: b>0,c<0
Statement 2: b>2,c<2


The question is basically about: whether y >0 or not ?
We need to check the sign of the discriminant \((b^2 - 4ac)\)
If the discriminant >= 0, then the curve will not lie completely above the x-axis.

\(y=b(x-2)^2+c\)
=> \(y = b(x^2 - 4x +4) + c\)
=> \(y = bx^2 - 4bx + 4b + c\)

The discrimimant = \(16b^2 - 4b(4b + c) = -4bc\)

Statement 1: b>0,c<0
=> The discrimimant = -4bc > 0 (Sufficient)

Statement 2: b>2,c<2
With c < 2, it is unsure if c < 0 or c > 0 (Insufficient)



Could you please explain how you got the discriminant?

Regards,

Ritvik


The curve is of the standard-form: \(Ax^2+Bx+C\)
Discriminant (D): \(B^2−4AC\)
Discriminant (D) = Square of Co-effecient-of \(x\) - 4*( Co-effecient-of \(x^2\) )*( Constant term )

A: Co-effecient-of \(x^2\)
B: Square of Co-effecient-of \(x\)
C: Constant term

Analogically, in the equation: y = b\(x^2\) − 4b\(x\) + 4b+c
    A: Co-effecient-of \(x^2\) ---------> \(b\)
    B: Square of Co-effecient-of \(x\) --------> \((-4b)^2\)
    C: Constant term ----------> \(4b+c\)

Thus, the discrimimant = \(16b^2 - 4b(4b + c) = -4bc\)
_________________
---------------------------------------------------------------------------------
“The trouble is, you think you have time.” – Buddha
Giving Kudos is the best way to encourage and appreciate people.
GMAT Club Bot
Re: Does the curve y=b(x-2)^2+c lie completely above the x-axis?   [#permalink] 04 May 2019, 11:04
Display posts from previous: Sort by

Does the curve y=b(x-2)^2+c lie completely above the x-axis?

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  


Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne