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

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

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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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Doc1.doc [31.5 KiB]

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

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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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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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untitled.JPG [ 6.71 KiB | Viewed 3682 times ]

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

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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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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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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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16 Apr 2009, 00:17
1
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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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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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23 Apr 2009, 23:59
1
KUDOS
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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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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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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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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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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16 Aug 2010, 04:50
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 $$x=1$$ --> $$y=\frac{3}{4}x^2-3x+3=\frac{3}{4}\approx{1}$$.

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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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 $$x=1$$ --> $$y=\frac{3}{4}x^2-3x+3=\frac{3}{4}\approx{1}$$.

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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13 Feb 2016, 19:04
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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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