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The graph above shows two lines, one on the left of the y-ax

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The graph above shows two lines, one on the left of the y-ax [#permalink]

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The graph above shows two lines, one on the left of the y-axis and one on the right., that represent the curve f(x) . What is the value of f(4) . f(-4) ?

(1) a = 3
(2) b = 5
[Reveal] Spoiler: OA

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Re: The graph above shows two lines, one on the left of the y-ax [#permalink]

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New post 05 Aug 2013, 12:49
Asifpirlo wrote:
The graph above shows two lines, one on the left of the y-axis and one on the right., that represent the curve f(x) . What is the value of f(4) . f(-4) ?

(1) a = 3
(2) b = 5

[Reveal] Spoiler:
need discussions


to determine the equation of a line we need atleast 2 points.

now in this case we need to know the value of \(f(4) * f(-4)\)

by statement 1 we can get f(0) = 3...(0,3)==>this alone is insufficient to determine the equation of line
by statement 2 we can get f(8) = 5...(8,5)==>this alone is insufficient to determine the equation of line

now combining we can get the equation of line of right side and hence will be able to calculate f(4)...but for calculating f(-4) we need the equation of line of left side which is still unknown....hence insufficient

hence E
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Re: The graph above shows two lines, one on the left of the y-ax [#permalink]

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New post 05 Aug 2013, 13:17
blueseas wrote:
Asifpirlo wrote:
The graph above shows two lines, one on the left of the y-axis and one on the right., that represent the curve f(x) . What is the value of f(4) . f(-4) ?

(1) a = 3
(2) b = 5

[Reveal] Spoiler:
need discussions


to determine the equation of a line we need atleast 2 points.

now in this case we need to know the value of \(f(4) * f(-4)\)

by statement 1 we can get f(0) = 3...(0,3)==>this alone is insufficient to determine the equation of line
by statement 2 we can get f(8) = 5...(8,5)==>this alone is insufficient to determine the equation of line

now combining we can get the equation of line of right side and hence will be able to calculate f(4)...but for calculating f(-4) we need the equation of line of left side which is still unknown....hence insufficient

hence E


blueseas,
Do you think (0,3) is sufficient to form an equation of the line. how? its not went through the origin.
i think (0,3) and (8,5) both together are sufficient to form the equation at the right side.

And we dont know the value of C thats why we are unable to form the equation of the left side. thats why f(-4) is not possible to evaluate.
thus answer is (E).
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Re: The graph above shows two lines, one on the left of the y-ax [#permalink]

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New post 05 Aug 2013, 13:21
Asifpirlo wrote:

blueseas,
Do you think (0,3) is sufficient to form an equation of the line. how? its not went through the origin.
i think (0,3) and (8,5) both together are sufficient to form the equation at the right side.

And we dont know the value of C thats why we are unable to form the equation of the left side. thats why f(-4) is not possible to evaluate.
thus answer is (E).


YES I AGREE THAT ...AND THAT IS WHAT I EXPLAINED.....i think i have written the same thing...please highlite the area which you feel is wrong
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Re: The graph above shows two lines, one on the left of the y-ax [#permalink]

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New post 05 Aug 2013, 14:25
blueseas wrote:
Asifpirlo wrote:

blueseas,
Do you think (0,3) is sufficient to form an equation of the line. how? its not went through the origin.
i think (0,3) and (8,5) both together are sufficient to form the equation at the right side.

And we dont know the value of C thats why we are unable to form the equation of the left side. thats why f(-4) is not possible to evaluate.
thus answer is (E).


YES I AGREE THAT ...AND THAT IS WHAT I EXPLAINED.....i think i have written the same thing...please highlite the area which you feel is wrong



by statement 1 we can get f(0) = 3...(0,3)==>this alone is insufficient to determine the equation of line

you wrote right, but i didn't see insufficient, i saw that sufficient.. sorry brother.... :)
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Re: The graph above shows two lines, one on the left of the y-ax [#permalink]

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New post 03 Jun 2014, 03:49
To find F(X) i.e. y = mx+c
since a=3 hence at x=0, c=3
also, slope = -(y2- y1)/(x2-x1)
to get y2 - y1 we need both the points a and b ... this give the equation for one line in the first quardrant but the other equation is not possible
hence the answer is E.

Please Bunuel - post your explanation.
Thanks

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Re: The graph above shows two lines, one on the left of the y-ax [#permalink]

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New post 09 Jun 2014, 03:01
blueseas wrote:
Asifpirlo wrote:
The graph above shows two lines, one on the left of the y-axis and one on the right., that represent the curve f(x) . What is the value of f(4) . f(-4) ?

(1) a = 3
(2) b = 5

[Reveal] Spoiler:
need discussions


to determine the equation of a line we need atleast 2 points.

now in this case we need to know the value of \(f(4) * f(-4)\)

by statement 1 we can get f(0) = 3...(0,3)==>this alone is insufficient to determine the equation of line
by statement 2 we can get f(8) = 5...(8,5)==>this alone is insufficient to determine the equation of line

now combining we can get the equation of line of right side and hence will be able to calculate f(4)...but for calculating f(-4) we need the equation of line of left side which is still unknown....hence insufficient

hence E


@Blueseas

Can you elaborate further , I know the answer is E but just to clear my coordinate concepts I wanted to know even if we know both the points how can we get f(4).

using y = mx+ c
where = 2/8
c= 3
and putting x= 4 I get y = 3
does that mean f(4) = 3

How can both f(0) and f(4) be 3 from the figure it seems f(4) > f(0)?
if you can elaborate.

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Re: The graph above shows two lines, one on the left of the y-ax [#permalink]

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New post 10 Jun 2014, 04:28
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qlx wrote:
blueseas wrote:
Asifpirlo wrote:
The graph above shows two lines, one on the left of the y-axis and one on the right., that represent the curve f(x) . What is the value of f(4) . f(-4) ?

(1) a = 3
(2) b = 5

[Reveal] Spoiler:
need discussions


to determine the equation of a line we need atleast 2 points.

now in this case we need to know the value of \(f(4) * f(-4)\)

by statement 1 we can get f(0) = 3...(0,3)==>this alone is insufficient to determine the equation of line
by statement 2 we can get f(8) = 5...(8,5)==>this alone is insufficient to determine the equation of line

now combining we can get the equation of line of right side and hence will be able to calculate f(4)...but for calculating f(-4) we need the equation of line of left side which is still unknown....hence insufficient

hence E


@Blueseas

Can you elaborate further , I know the answer is E but just to clear my coordinate concepts I wanted to know even if we know both the points how can we get f(4).

using y = mx+ c
where = 2/8
c= 3
and putting x= 4 I get y = 3
does that mean f(4) = 3

How can both f(0) and f(4) be 3 from the figure it seems f(4) > f(0)?
if you can elaborate.


in the equation y=mx+c , or f(x) = mx+c as you have rightly calculated m=1/4 and c=4

thus we have y=(1/4)x + 3

now f(4)= 1 + 3 = 4

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Re: The graph above shows two lines, one on the left of the y-ax [#permalink]

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New post 10 Jun 2014, 14:04
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manpreetsingh86 wrote:
qlx wrote:

@Blueseas

Can you elaborate further , I know the answer is E but just to clear my coordinate concepts I wanted to know even if we know both the points how can we get f(4).

using y = mx+ c
where = 2/8
c= 3
and putting x= 4 I get y = 3
does that mean f(4) = 3

How can both f(0) and f(4) be 3 from the figure it seems f(4) > f(0)?
if you can elaborate.


in the equation y=mx+c , or f(x) = mx+c as you have rightly calculated m=1/4 and c=4

thus we have y=(1/4)x + 3

now f(4)= 1 + 3 = 4


Hi

Don't know what i was thinking, when I managed to get f(4) = 3

Manpreet thank you so much really appreciate it, this query was bothering me for a while
you certainly deserve Kudos .

Looking forward to more assistance from you in the future too!

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Re: The graph above shows two lines, one on the left of the y-ax [#permalink]

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New post 08 Sep 2015, 20:46
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Re: The graph above shows two lines, one on the left of the y-ax [#permalink]

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New post 22 Sep 2017, 23:17
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).

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Re: The graph above shows two lines, one on the left of the y-ax   [#permalink] 22 Sep 2017, 23:17
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