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According to the graph above, when x = 3, y most nearly ?

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According to the graph above, when x = 3, y most nearly ?  [#permalink]

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New post 30 Aug 2008, 13:08
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According to the graph above, when x = 3, y most nearly ?

(A) –1
(B) -1/2
(C) 0
(D) 1/2
(E) 1

Should we solely rely on the graph assuming that it is drawn to scale????
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Re: PS geometery ... little confused. [#permalink]

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New post 30 Aug 2008, 13:22
rao_1857 wrote:
According to the graph above, when x = 3, y most nearly ?

(A) –1
(B) -1/2
(C) 0
(D) 1/2
(E) 1

Should we solely rely on the graph assuming that it is drawn to scale????



What is the source of this question?

By observation answer should be 1/2 or 1
its more close to 1.
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Last edited by x2suresh on 30 Aug 2008, 13:26, edited 1 time in total.
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Re: PS geometery ... little confused. [#permalink]

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New post 30 Aug 2008, 13:24
rao_1857 wrote:
According to the graph above, when x = 3, y most nearly ?

(A) –1
(B) -1/2
(C) 0
(D) 1/2
(E) 1

Should we solely rely on the graph assuming that it is drawn to scale????



even if it is drawn to scale, it is hard to estimate between 1 and .5
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Re: PS geometery ... little confused. [#permalink]

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New post 30 Aug 2008, 20:28
i doubt if the Q is a correct one. Its a parabola and from the diagram, you can make out 2 coordinates, (2,0) and (0,3); that takes you no where.
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Re: PS geometery ... little confused. [#permalink]

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New post 31 Aug 2008, 00:27
nishchals wrote:
i doubt if the Q is a correct one. Its a parabola and from the diagram, you can make out 2 coordinates, (2,0) and (0,3); that takes you no where.


pls look closer....(0,3) is not on the parabola...if it were than the problem was simple.........in that situation, solve parabola's equation and you would get that y coord will be 3/4 when x=3. The figure shows only one conformed point with coordinates ie the vertex (2,0).

My guess for the answer is 1.
However I don't have mathematical proof to support it.
Logical proof is if (0,3) were on parabola....then (3, 3/4) is on parabola
Hence in the case of (0, little less than 3)......the point should be (3, little more than 3/4)....closest answer is 1
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Re: PS geometery ... little confused. [#permalink]

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New post 31 Aug 2008, 07:56
yeuapp .. I agree . there is not mathmatical way to solve this. by looking at it .. you can say its more close to 1.

src is 1000 ps.
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Re: PS geometery ... little confused. [#permalink]

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New post 16 Apr 2009, 00:17
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proof:
vertex form of equation :
y = a (x-h)^2 + k

From the figure: upward parabola = a is +ve
vertex is at (2,0) => h = 2 k =0

y = a (x- 2)^2

point (0, 2.8) lies on this parabola => a = 0.7

y = 0.7 (x-2)^2

when x = 3, y =0.7 => ~1

HTH
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Re: PS geometery ... little confused. [#permalink]

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New post 16 Apr 2009, 05:11
My guess is 1/2. but I don;t know how to solve it :) It's just seems to me that way.
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Re: PS geometery ... little confused. [#permalink]

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New post 23 Apr 2009, 23:59
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alpha_plus_gamma wrote:
proof:

vertex form of equation :
y = a (x-h)^2 + k

From the figure: upward parabola = a is +ve
vertex is at (2,0) => h = 2 k =0

y = a (x- 2)^2

point (0, 2.8) lies on this parabola => a = 0.7

y = 0.7 (x-2)^2

when x = 3, y =0.7 => ~1

HTH


Hi alpha_plus_gamma ,

This is the bang on solution. This question features at many places.
I wanted to ask one thing. How to figure out the starting eqn.

Sudeep
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Re: PS geometery ... little confused. [#permalink]

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New post 24 Apr 2009, 00:58
bang on alpha, for simplification:

This is an equation of parabola .. y = a (x-h)^2 + k where h & k are values of vertex
A<0 if parabola opens downwards
a>0 if parabola opens upwards
here a is +ve
vertex is at (2,0) => h = 2 k =0
y = a (x- 2)^2

To find a, check other point on parabola , here (0, 2.8)
2.8= a(0+2)^2
2.8= a(4) a= .7
now
y = 0.7 (x-2)^2 put x= 3 then y = .7
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Re: PS geometery ... little confused. [#permalink]

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New post 24 Apr 2009, 09:57
If drawn to scale, it's closer to 1 than to 0.5.
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Re: PS geometery ... little confused. [#permalink]

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New post 16 Aug 2010, 03:47
I really dont understand anything

It should be like that but i dont know

X1= 1/2
x2= 1

y1=2
y2=1

then

(y2-y1)/(x2-x1)=M which is -2 in this case

then

-2= (y-y1)/(x-x1)

y= -2x + 3 is equa.

ıf we put there as X y= -6 +3 then

y= -3 which is my answer....ı am confused where i am tripping up
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Re: PS geometery ... little confused. [#permalink]

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New post 16 Aug 2010, 04:08
Bunuel can you look at it pleasee
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Re: PS geometery ... little confused. [#permalink]

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New post 16 Aug 2010, 04:50
Expert's post
fatihaysu wrote:
I really dont understand anything

It should be like that but i dont know

X1= 1/2
x2= 1

y1=2
y2=1

then

(y2-y1)/(x2-x1)=M which is -2 in this case

then

-2= (y-y1)/(x-x1)

y= -2x + 3 is equa.

ıf we put there as X y= -6 +3 then

y= -3 which is my answer....ı am confused where i am tripping up


Frankly I don't quite understand what is tested here...

But your approach is wrong because your method for finding equation of a graph is for straight line (linear equation) and obviously the curve shown on the diagram is not straight line.

Is it a graph of quadratic equation? Are the intersection points with axes (0,3) and (2,0)? If the answer on both questions is YES then the equation of the graph would be \(y=\frac{3}{4}x^2-3x+3\) and for \(x=1\) --> \(y=\frac{3}{4}x^2-3x+3=\frac{3}{4}\approx{1}\).

I wouldn't worry about this question at all.

P.S. Check the link about Coordinate Geometry in my signature for more on this issue.
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Re: PS geometery ... little confused. [#permalink]

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New post 16 Aug 2010, 06:26
Bunuel wrote:
fatihaysu wrote:
I really dont understand anything

It should be like that but i dont know

X1= 1/2
x2= 1

y1=2
y2=1

then

(y2-y1)/(x2-x1)=M which is -2 in this case

then

-2= (y-y1)/(x-x1)

y= -2x + 3 is equa.

ıf we put there as X y= -6 +3 then

y= -3 which is my answer....ı am confused where i am tripping up


Frankly I don't quite understand what is tested here...

But your approach is wrong because your method for finding equation of a graph is for straight line (linear equation) and obviously the curve shown on the diagram is not straight line.

Is it a graph of quadratic equation? Are the intersection points with axes (0,3) and (2,0)? If the answer on both questions is YES then the equation of the graph would be \(y=\frac{3}{4}x^2-3x+3\) and for \(x=1\) --> \(y=\frac{3}{4}x^2-3x+3=\frac{3}{4}\approx{1}\).

I wouldn't worry about this question at all.

P.S. Check the link about Coordinate Geometry in my signature for more on this issue.


It was important for me that why my approach is wrong,i know now...

Thank you Bunuel
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Re: According to the graph above, when x = 3, y most nearly ?  [#permalink]

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Re: According to the graph above, when x = 3, y most nearly ?   [#permalink] 13 Feb 2016, 19:04
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According to the graph above, when x = 3, y most nearly ?

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