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In the xy-plane, a parabola intersects with axis-y at point (0,y). Is y<0.

I. The vertex of parabola is (2,-5) II. The parabola intersects with axis-x at point (-2,0) and (6,0)

Though it's possible to solve this question algebraically the easiest way will be to visualize it and draw on a paper.

(1) The vertex of parabola is (2,-5) --> the vertex is in the IV quadrant: if the parabola is downward it'll have negative y-intercept, but if it's upward then it can have positive as well as negative y-intercept. Not sufficient.

(2) The parabola intersects with axis-x at point (-2,0) and (6,0) --> now if the vertex is above x-axis then parabola will have positive y-intercept and if its vertex is below x-axis it'll have negative y-intercept. Not sufficient.

(1)+(2) As from (1) the vertex is below x-axis then from (2) we'll have that parabola must have negative y-intercept. Sufficient.

You can look at the diagram below to see that a parabola passing through the given three points must have negative y-intercept only.

Attachment:

MSP139819db6ebe95a9e2a900005889a632a09g5628.gif [ 3.25 KiB | Viewed 6667 times ]

Thanks Bunuel for the explanation, I was thinking since option B gives us (-2,0) and (6,0) vertices, with this we can assume that the parabola is directed upwards and it's Y intersect will be -ve and hence sufficient, but after reading your explanation I understand it better 1+ Kudos to you.

Re: In the xy-plane, a parabola intersects with axis-y at point [#permalink]

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28 Oct 2013, 00:40

Hi Bunuel

What is wrong in the following approach

1) Insufficient The vertex of parabola is (2,-5) --> the vertex is in the IV quadrant: if the parabola is downward it'll have negative y-intercept, but if it's upward then it can have positive as well as negative y-intercept. Not sufficient

I was thinking from option 2 we can find the product of the roots and sum of the roots product of roots = c/a = -12

and sum of roots = -b/a = 4

hence come up with an equation y = x^2-4x-12

and from the question stem we have that it intercepts at 0,y

Putting x = 0 we get y = -12 (negative) and B alone is sufficient .

1) Insufficient The vertex of parabola is (2,-5) --> the vertex is in the IV quadrant: if the parabola is downward it'll have negative y-intercept, but if it's upward then it can have positive as well as negative y-intercept. Not sufficient

I was thinking from option 2 we can find the product of the roots and sum of the roots product of roots = c/a = -12

and sum of roots = -b/a = 4

hence come up with an equation y = x^2-4x-12

and from the question stem we have that it intercepts at 0,y

Putting x = 0 we get y = -12 (negative) and B alone is sufficient .

How did you get y=x^2-4x-12 from c/a=-12 and -b/a=4? You cannot solve c/a=-12 and -b/a=4 to get unique values of a, b, and c.

For example if a=2, c=-12, and b=-4 you'll get 2x^2-24x-8=0:

Attachment:

MSP2141ga1ih773ii7e3di00002h2000h2fiibia6c.gif [ 3.51 KiB | Viewed 5203 times ]

If a=-1, c=12, and b=4 you'll get -x^2+4x+12=0:

Attachment:

MSP24971d0ia65iig79i4gg000037eb4cg662i7adih.gif [ 3.44 KiB | Viewed 5203 times ]

Or in other words infinitely many parabolas have x intercepts at -2 and 6. You cannot get unique equation only from that info.
_________________

Re: In the xy-plane, a parabola intersects with axis-y at point [#permalink]

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15 Oct 2014, 01:53

Bunuel wrote:

sr2013 wrote:

Hi Bunuel

What is wrong in the following approach

1) Insufficient The vertex of parabola is (2,-5) --> the vertex is in the IV quadrant: if the parabola is downward it'll have negative y-intercept, but if it's upward then it can have positive as well as negative y-intercept. Not sufficient

I was thinking from option 2 we can find the product of the roots and sum of the roots product of roots = c/a = -12

and sum of roots = -b/a = 4

hence come up with an equation y = x^2-4x-12

and from the question stem we have that it intercepts at 0,y

Putting x = 0 we get y = -12 (negative) and B alone is sufficient .

How did you get y=x^2-4x-12 from c/a=-12 and -b/a=4? You cannot solve c/a=-12 and -b/a=4 to get unique values of a, b, and c.

For example if a=2, c=-12, and b=-4 you'll get 2x^2-24x-8=0:

Attachment:

MSP2141ga1ih773ii7e3di00002h2000h2fiibia6c.gif

If a=-1, c=12, and b=4 you'll get -x^2+4x+12=0:

Attachment:

MSP24971d0ia65iig79i4gg000037eb4cg662i7adih.gif

Or in other words infinitely many parabolas have x intercepts at -2 and 6. You cannot get unique equation only from that info.

Hi Bunuel

I have tried solving it using standard expressions for parabola For parabola y= ax2+ bx+ c, standard vertex is located at point (-\frac{b}{2a}, c-\frac{b^2}{4a}).

From a) we know the value of vertex as (2,-5) by putting the value in standard vertex we can get c=-4 it is also given in the question stem that parabola intersects y axis at (0,y) from this we can get the value of y as -4. which is sufficient to answer the question.

Please let me know whats wrong with this approach.

1) Insufficient The vertex of parabola is (2,-5) --> the vertex is in the IV quadrant: if the parabola is downward it'll have negative y-intercept, but if it's upward then it can have positive as well as negative y-intercept. Not sufficient

I was thinking from option 2 we can find the product of the roots and sum of the roots product of roots = c/a = -12

and sum of roots = -b/a = 4

hence come up with an equation y = x^2-4x-12

and from the question stem we have that it intercepts at 0,y

Putting x = 0 we get y = -12 (negative) and B alone is sufficient .

How did you get y=x^2-4x-12 from c/a=-12 and -b/a=4? You cannot solve c/a=-12 and -b/a=4 to get unique values of a, b, and c.

For example if a=2, c=-12, and b=-4 you'll get 2x^2-24x-8=0:

Attachment:

MSP2141ga1ih773ii7e3di00002h2000h2fiibia6c.gif

If a=-1, c=12, and b=4 you'll get -x^2+4x+12=0:

Attachment:

MSP24971d0ia65iig79i4gg000037eb4cg662i7adih.gif

Or in other words infinitely many parabolas have x intercepts at -2 and 6. You cannot get unique equation only from that info.

Hi Bunuel

I have tried solving it using standard expressions for parabola For parabola y= ax2+ bx+ c, standard vertex is located at point (-\frac{b}{2a}, c-\frac{b^2}{4a}).

From a) we know the value of vertex as (2,-5) by putting the value in standard vertex we can get c=-4 it is also given in the question stem that parabola intersects y axis at (0,y) from this we can get the value of y as -4. which is sufficient to answer the question.

Please let me know whats wrong with this approach.

The post you are quoting has an answer to your question. You cannot solve for c.

Re: In the xy-plane, a parabola intersects with axis-y at point [#permalink]

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19 Mar 2016, 23:20

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In the xy plane,a parabola intersects with axis-y at point (0,y). is y<0?

1. The vertex of the parabola is (2,-5) 2. The parabola intersects with axis X at point (-2,0) and (6,0)

A parabola has a vertex and a similar curve on both sides.. the vertex can be MIN or MAX value depending on the way parabola opens up...

1. The vertex of the parabola is (2,-5) we know vertex lies below the x-axis, but -- a) if it opens upwards- the intersect with y-axis can be either below x-axis or above it.. b) if it opens downwards- the intersect with y-axis will be below x-axis .. Insuff

2. The parabola intersects with axis X at point (-2,0) and (6,0) we know vertex would lie at (-2+6)/2 = 2 as x, BUT we cannot determine if VERTEX is above or below x-axis.. whereever vertex lies, the intersect will lie on that point .. since the curve moves from 2 to -2 in that Quadrant -

a) if it opens upwards- the intersect with y-axis will be below x axis, as the vertex will be below x-axis.. b) if it opens downwards- the intersect with y-axis will be above x-axis, as the vertex will be above x-axis ..

combined- statement I tells us that the vertex is below the x-axis and statement II tells us that if vertex is below x- axis the intersect is also below x-axis.. so y<0.. Suff C
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Re: In the xy-plane, a parabola intersects with axis-y at point [#permalink]

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04 Jul 2017, 11:20

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.
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