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# Point P lies on the equation y= x^2 - 1 and Point Q lies on the equati

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Joined: 10 Nov 2017
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Point P lies on the equation y= x^2 - 1 and Point Q lies on the equati  [#permalink]

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Updated on: 17 Dec 2018, 04:49
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Difficulty:

85% (hard)

Question Stats:

34% (01:39) correct 66% (01:33) wrong based on 187 sessions

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Point P lies on the equation $$y= x^2 - 1$$ and Point Q lies on the equation $$y= -x^2+1$$. Find the difference between the minimum y coordinate value of Point P and the maximum y coordinate value of Point Q.

A. -2
B. -1
C. 0
D. 1
E. 2

Originally posted by Arun1994 on 03 Dec 2018, 05:51.
Last edited by Arun1994 on 17 Dec 2018, 04:49, edited 1 time in total.
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Joined: 15 Aug 2018
Posts: 51
GMAT 1: 740 Q47 V45
GPA: 3.5
Re: Point P lies on the equation y= x^2 - 1 and Point Q lies on the equati  [#permalink]

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06 Dec 2018, 09:04
I would propose zero.
The "B" value in the quadratic equation must be zero. Thus, I assume that both extreme values will have a zero in the numerator, leaving us with a value of zero for both functions.
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Re: Point P lies on the equation y= x^2 - 1 and Point Q lies on the equati  [#permalink]

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07 Dec 2018, 03:56
Arun1994 wrote:
Point P lies on the equation $$y= X^2 - 1$$ and Point Q lies on the equation $$y= -X^2+1$$. Find the difference between the minimum y coordinate value of Point P and the maximum y coordinate value of Point Q.

A. -2
B. -1
C. 0
D. 1
E. 2

I am not able to solve the problem and I dont know the anwser. Can someone help me please ?

y=x^2-1 [Curve---P]
=> y+1=x^2 { This is an equation of upward parabola with vertex at y=-1 at symetric about y axis }
so, y (min)=-1

y=-x^2+1 [Curve----Q]
=> y-1= -x^2 ( This is an equation of downward parabola with vertex at y=1 at symetric about y axis }
so, y(max)=1

Req. difference= 1-(-1)=2 {i assume by difference they r asking about distance between points}

Req. Ans-E
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Re: Point P lies on the equation y= x^2 - 1 and Point Q lies on the equati  [#permalink]

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17 Dec 2018, 04:04
get y intercepts of both equations by putting x =0
so minimum value= -1 ...max (y)=1
difference=minimum-maximum =-1-1=-2
option A
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Joined: 10 May 2018
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Re: Point P lies on the equation y= x^2 - 1 and Point Q lies on the equati  [#permalink]

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07 Jan 2019, 09:11
Does it asks max(y) - min(y) or min(y) - max(y)..I got confused

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Re: Point P lies on the equation y= x^2 - 1 and Point Q lies on the equati   [#permalink] 07 Jan 2019, 09:11
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