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In the xyplane, a parabola intersects with axisy at point [#permalink]
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10 Dec 2010, 11:03
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In the xyplane, a parabola intersects with axisy at point (0,y). Is y < 0 ? (1) The vertex of parabola is (2,5) (2) The parabola intersects with axisx at point (2,0) and (6,0)
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Re: #32 parabola intersts Y axis [#permalink]
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aalriy wrote: In the xyplane, a parabola intersects with axisy at point (0,y). Is y<0.
I. The vertex of parabola is (2,5) II. The parabola intersects with axisx 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 yintercept, but if it's upward then it can have positive as well as negative yintercept. Not sufficient. (2) The parabola intersects with axisx at point (2,0) and (6,0) > now if the vertex is above xaxis then parabola will have positive yintercept and if its vertex is below xaxis it'll have negative yintercept. Not sufficient. (1)+(2) As from (1) the vertex is below xaxis then from (2) we'll have that parabola must have negative yintercept. Sufficient. You can look at the diagram below to see that a parabola passing through the given three points must have negative yintercept only. Attachment:
MSP139819db6ebe95a9e2a900005889a632a09g5628.gif [ 3.25 KiB  Viewed 5765 times ]
Answer: C. Hope it's clear.
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Re: #32 parabola intersts Y axis [#permalink]
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20 Oct 2011, 22:12
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.



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Re: In the xyplane, a parabola intersects with axisy at point [#permalink]
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27 Jun 2013, 22:30



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Re: In the xyplane, a parabola intersects with axisy 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 yintercept, but if it's upward then it can have positive as well as negative yintercept. 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^24x12
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 .



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Re: In the xyplane, a parabola intersects with axisy at point [#permalink]
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28 Oct 2013, 01:13
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 yintercept, but if it's upward then it can have positive as well as negative yintercept. 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^24x12
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^24x12 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^224x8=0: Attachment:
MSP2141ga1ih773ii7e3di00002h2000h2fiibia6c.gif [ 3.51 KiB  Viewed 4303 times ]
If a=1, c=12, and b=4 you'll get x^2+4x+12=0: Attachment:
MSP24971d0ia65iig79i4gg000037eb4cg662i7adih.gif [ 3.44 KiB  Viewed 4304 times ]
Or in other words infinitely many parabolas have x intercepts at 2 and 6. You cannot get unique equation only from that info.
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Re: In the xyplane, a parabola intersects with axisy 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 yintercept, but if it's upward then it can have positive as well as negative yintercept. 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^24x12
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^24x12 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^224x8=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.



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In the xyplane, a parabola intersects with axisy at point [#permalink]
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15 Oct 2014, 02:59
kd1989 wrote: 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 yintercept, but if it's upward then it can have positive as well as negative yintercept. 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^24x12
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^24x12 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^224x8=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. P.S. Please read this: rulesforpostingpleasereadthisbeforeposting133935.html#p1096628 (Writing Mathematical Formulas on the Forum)
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In the xy plane,a parabola intersects with axisy at point (0,y). [#permalink]
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12 May 2016, 19:49
In the xy plane,a parabola intersects with axisy 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)



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Re: In the xy plane,a parabola intersects with axisy at point (0,y). [#permalink]
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12 May 2016, 20:55
ruchi857 wrote: In the xy plane,a parabola intersects with axisy 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 xaxis, but  a) if it opens upwards the intersect with yaxis can be either below xaxis or above it.. b) if it opens downwards the intersect with yaxis will be below xaxis .. 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 xaxis.. 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 yaxis will be below x axis, as the vertex will be below xaxis.. b) if it opens downwards the intersect with yaxis will be above xaxis, as the vertex will be above xaxis .. combined statement I tells us that the vertex is below the xaxis and statement II tells us that if vertex is below x axis the intersect is also below xaxis.. so y<0.. Suff C
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Re: In the xyplane, a parabola intersects with axisy at point [#permalink]
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Re: In the xyplane, a parabola intersects with axisy at point
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