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Re: #32 parabola intersts Y axis [#permalink]
10 Dec 2010, 12:52

4

This post received KUDOS

Expert's post

aalriy wrote:

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 3568 times ]

Re: #32 parabola intersts Y axis [#permalink]
20 Oct 2011, 21:12

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.

Re: In the xy-plane, a parabola intersects with axis-y at point [#permalink]
27 Oct 2013, 23:40

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 .

Re: In the xy-plane, a parabola intersects with axis-y at point [#permalink]
28 Oct 2013, 00:13

Expert's post

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 [ 3.51 KiB | Viewed 2114 times ]

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

Attachment:

MSP24971d0ia65iig79i4gg000037eb4cg662i7adih.gif [ 3.44 KiB | Viewed 2116 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]
15 Oct 2014, 00:53

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.

In the xy-plane, a parabola intersects with axis-y at point [#permalink]
15 Oct 2014, 01:59

Expert's post

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 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.

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