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# On the x-y coordinate plane there is a parabola, y=x(6-x). Which of th

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Math Revolution GMAT Instructor
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On the x-y coordinate plane there is a parabola, y=x(6-x). Which of th  [#permalink]

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14 Jan 2016, 19:00
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55% (hard)

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69% (02:40) correct 31% (02:14) wrong based on 140 sessions

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On the x-y coordinate plane there is a parabola, y=x(6-x). Which of the following points is in the region that surround with this parabola and x-axis?

I. (1, 4) II. (3, 6) III. (5, 4)

A. I only
B. II only
C. III only
D. I & II
E. I, II &III

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Re: On the x-y coordinate plane there is a parabola, y=x(6-x). Which of th  [#permalink]

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15 Jan 2016, 08:34
y = x(6 - x)

x 0 1 2 3 4 5 6
y 0 5 8 9 5 5 0

From the above values of x and y we can see that all the three points lie between the x axis and the parabola.

Answer can be determined by plotting a graph too.
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Picture.jpg [ 3.77 MiB | Viewed 1895 times ]

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Re: On the x-y coordinate plane there is a parabola, y=x(6-x). Which of th  [#permalink]

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18 Jan 2016, 19:11
On the x-y coordinate plane there is a parabola, y=x(6-x). Which of the following points is in the region that surround with this parabola and x-axis?

I. (1, 4)II. (3, 6) III. (5, 4)

A. I only B. II only C. III only D. I &II E.I, II &III

Attachment:

6-2nd question to GC.jpg [ 1.81 MiB | Viewed 1863 times ]

==>If you look at the attached picture, I, II &III are all included in the graph. Therefore, the answer is E.
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On the x-y coordinate plane there is a parabola, y=x(6-x). Which of th  [#permalink]

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22 Jan 2016, 05:36
1
MathRevolution wrote:
On the x-y coordinate plane there is a parabola, y=x(6-x). Which of the following points is in the region that surround with this parabola and x-axis?

I. (1, 4) II. (3, 6) III. (5, 4)

A. I only
B. II only
C. III only
D. I & II
E. I, II &III

Plot the graph of equation y=x(6-x)

Since the graph will be an inverted parabola open downwards (i.e. y will be less than x(6-x) for points to be inside the parabola) therefore any point for which y<x(6-x) with positive value of y will be surrounded by the graph and x-axis (as shown in figure)

All points are surrounded as shown in figure and satisfied by (red highlighted equation). Hence,

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Re: On the x-y coordinate plane there is a parabola, y=x(6-x). Which of th  [#permalink]

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10 Apr 2016, 19:51
is there a faster way to solve it???
my approach
we have -x^2, so parabola is downwards
when x=0, y=0, so (0;0) is a point on the parabola.
another point of intersection with x axis is x=6.
now..
x=5; y=5
x=4; y=8
x=3; y=9
x=2; y=8
x=1; y=5

everything below is inside the parabola
1. 1;4 - is inside, because 1;5 is on the parabola.
2. 3;6 - is inside, because 3;9 is on the parabola.
3. 5;4 - is inside, because 5;5 is on the parabola.

E
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Re: On the x-y coordinate plane there is a parabola, y=x(6-x). Which of th  [#permalink]

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10 Apr 2016, 21:42
mvictor wrote:
is there a faster way to solve it???
my approach
we have -x^2, so parabola is downwards
when x=0, y=0, so (0;0) is a point on the parabola.
another point of intersection with x axis is x=6.
now..
x=5; y=5
x=4; y=8
x=3; y=9
x=2; y=8
x=1; y=5

everything below is inside the parabola
1. 1;4 - is inside, because 1;5 is on the parabola.
2. 3;6 - is inside, because 3;9 is on the parabola.
3. 5;4 - is inside, because 5;5 is on the parabola.

E

Yes, you know that since co-efficient of x^2 is negative, the parabola will be downward facing. So the parabola and the x axis will enclose some points. We don't even need to draw out the parabola.

Just put the relevant points in y = x(6 - x)
x = 1, y = 5 lies on the parabola. So (1, 4) will lie below it. Hence, it will be enclosed by the parabola and x axis.
x = 3, y = 9 lies on the parabola. So (3, 6) will lie below it. Hence, it will be enclosed by the parabola and x axis.
x = 5, y = 5 lies on the parabola. So (5, 4) will lie below it. Hence, it will be enclosed by the parabola and x axis.

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Re: On the x-y coordinate plane there is a parabola, y=x(6-x). Which of th  [#permalink]

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16 Oct 2018, 02:25
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Re: On the x-y coordinate plane there is a parabola, y=x(6-x). Which of th   [#permalink] 16 Oct 2018, 02:25
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