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# On the graph above, when x = 1/2, y = 2; and when x = 1, y =

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Re: On the graph above, when x = 1/2, y = 2; and when x = 1, y = [#permalink]
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(E) 1

its symmetric with respect to x=2.
when x=1, 1 unit to the left of x(2), y is 1.
when x=3, 1 unit to the right of x(2), y should again be 1 as it is symmetric.
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Re: On the graph above, when x = 1/2, y = 2; and when x = 1, y = [#permalink]
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Bunuel wrote:
Attachment:
Graph.png

On the graph above, when x = 1/2, y = 2; and when x = 1, y = 1. The graph is symmetric with respect to the vertical line at x = 2. According to the graph, when x = 3, y =

(A) -1
(B) -1/2
(C) 0
(D) 1/2
(E) 1

Half of the entire graph is a mirror image of the other side. The center coordinate is (2,0) according to the graph. Thus, y when x=1 and y when x-3 are equal.

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Re: On the graph above, when x = 1/2, y = 2; and when x = 1, y = [#permalink]
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Bunuel wrote:
Attachment:
Graph.png
On the graph above, when x = 1/2, y = 2; and when x = 1, y = 1. The graph is symmetric with respect to the vertical line at x = 2. According to the graph, when x = 3, y =

(A) -1
(B) -1/2
(C) 0
(D) 1/2
(E) 1

Practice Questions
Question: 7
Page: 153
Difficulty: 600

graph is symmetric with respect to the vertical line

so if x = 1, y = 1 & x=2 , y=0 then x=3 , y=1 ...
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Re: On the graph above, when x = 1/2, y = 2; and when x = 1, y = [#permalink]
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Bunuel wrote:
Attachment:
Graph.png
On the graph above, when x = 1/2, y = 2; and when x = 1, y = 1. The graph is symmetric with respect to the vertical line at x = 2. According to the graph, when x = 3, y =

(A) -1
(B) -1/2
(C) 0
(D) 1/2
(E) 1

We are given that the graph is SYMMETRIC along the vertical line x = 2. We are also given that when x = 1, y = 1. We are asked to determine the value of y when x is 3. Keep in mind that when x = 1, it is ONE UNIT away from the line x = 2, and when x = 3, it is also ONE UNIT away from the line x = 2.

Since the y-coordinate is 1, when x is 1, the y-coordinate is ALSO 1 when x is 3. Again, this is because the graph is symmetric with the line x = 2.

Another way to look at this problem is to see that the point (1,1) is actually being reflected over the line x = 2.

We see that the point (1,1) is exactly 1 unit from the line x = 2. In fact the point (2,1) is on the line x = 2 which is exactly 1 unit from (1,1). However, (1,1) is exactly 1 unit on the LEFT of (2,1), so by symmetry, the point that is exactly 1 unit on the RIGHT of (2,1) is (3,1). Thus the y-coordinate is 1.

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Re: On the graph above, when x = 1/2, y = 2; and when x = 1, y = [#permalink]
Bunuel wrote:

On the graph above, when x = 1/2, y = 2; and when x = 1, y = 1. The graph is symmetric with respect to the vertical line at x = 2. According to the graph, when x = 3, y =

(A) -1
(B) -1/2
(C) 0
(D) 1/2
(E) 1

Bunuel
What if x=4?
y would be 3?
Thanks__
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Re: On the graph above, when x = 1/2, y = 2; and when x = 1, y = [#permalink]
Y would be 1 as it is symmetrical around 2 so y at X=1 equal to y at x=3

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Re: On the graph above, when x = 1/2, y = 2; and when x = 1, y = [#permalink]
Bunuel wrote:

On the graph above, when x = 1/2, y = 2; and when x = 1, y = 1. The graph is symmetric with respect to the vertical line at x = 2. According to the graph, when x = 3, y =

(A) -1
(B) -1/2
(C) 0
(D) 1/2
(E) 1

Bunuel
What if x=4?
y would be 3?
Thanks__

_____________________________
Yes.
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Re: On the graph above, when x = 1/2, y = 2; and when x = 1, y = [#permalink]
BrentGMATPrepNow

I was just curious and tried to come up with the equation of the parabola but couldn't figure it out. Isn't (2, 0) the vertex of the parabola?
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Re: On the graph above, when x = 1/2, y = 2; and when x = 1, y = [#permalink]
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Vegita wrote:
BrentGMATPrepNow

I was just curious and tried to come up with the equation of the parabola but couldn't figure it out. Isn't (2, 0) the vertex of the parabola?

It LOOKS like (2, 0) is the vertex, but it isn't.
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Re: On the graph above, when x = 1/2, y = 2; and when x = 1, y = [#permalink]
According to the given information, we know that the graph is symmetric with respect to the vertical line at x = 2. This means that if we reflect any point on the graph across the line x = 2, we will obtain another point on the graph.

Given that when x = 1/2, y = 2, we can observe that this point is located 1.5 units to the left of the line x = 2. Therefore, if we reflect this point across the line x = 2, the corresponding y-coordinate will be the same, but the x-coordinate will be 1.5 units to the right of the line x = 2.

So, the reflected point will have an x-coordinate of 2 + 1.5 = 3.5.

However, we are asked to find the value of y when x = 3, not x = 3.5. Since the graph is symmetric with respect to the line x = 2, the y-coordinate will also be symmetric. Thus, the value of y when x = 3 will be the same as the value of y when x = 3.5.

Given that when x = 1, y = 1, we can conclude that when x = 3, y will also be (E) 1.
Re: On the graph above, when x = 1/2, y = 2; and when x = 1, y = [#permalink]
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