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Manager  G
Joined: 15 Dec 2015
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GMAT 1: 680 Q49 V34 GPA: 4
WE: Information Technology (Computer Software)
Does the curve y=b(x-2)^2+c lie completely above the x-axis?  [#permalink]

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Question Stats: 42% (01:38) correct 58% (01:55) wrong based on 126 sessions

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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
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 9705
Location: Pune, India
Re: Does the curve y=b(x-2)^2+c lie completely above the x-axis?  [#permalink]

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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.

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Joined: 14 Nov 2012
Posts: 23
GMAT 1: 740 Q51 V38 Does the curve y=b(x-2)^2+c lie completely above the x-axis?  [#permalink]

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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)
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Manager  G
Joined: 23 Jan 2018
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Location: India
WE: Information Technology (Computer Software)
Re: Does the curve y=b(x-2)^2+c lie completely above the x-axis?  [#permalink]

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chetan2u, Bunuel, VeritasKarishma, gmatbusters, amanvermagmat

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

Regards,
Arup
Manager  B
Joined: 29 Sep 2016
Posts: 113
Re: Does the curve y=b(x-2)^2+c lie completely above the x-axis?  [#permalink]

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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.

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  P
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Posts: 180
Re: Does the curve y=b(x-2)^2+c lie completely above the x-axis?  [#permalink]

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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 [ 72.92 KiB | Viewed 1214 times ]

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Manager  B
Joined: 07 May 2018
Posts: 61
Re: Does the curve y=b(x-2)^2+c lie completely above the x-axis?  [#permalink]

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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  P
Status: The darker the night, the nearer the dawn!
Joined: 16 Jun 2018
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Re: Does the curve y=b(x-2)^2+c lie completely above the x-axis?  [#permalink]

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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 = $$16b^2 - 4b(4b + c) = -4bc$$
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Giving Kudos is the best way to encourage and appreciate people. Re: Does the curve y=b(x-2)^2+c lie completely above the x-axis?   [#permalink] 04 May 2019, 11:04
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