For the curve Y=3X-3X2 find:
A
i: its points of intersection with the X and Y axis.
ii: Its max and min turning points.
B Sketch the curve
C Calulate the area inclosed by:
i: The curve and X axis.
ii: The x axis, the lines though the turning points perpendicular to the X axis and the part of the curve between the turning points
~ Hell I dont even know where to start!
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Started by Sammie, Jan 14 2007 04:57 PM
6 replies to this topic
#1
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Posted 14 January 2007 - 04:57 PM
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#2
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Posted 14 January 2007 - 05:01 PM
At the points of intersection with the x-axis, y=0, and vice versa. Does this help? And to find its max and min turning points, use differentiation to find maximas and minimas. Sketching the curve is easy, all you do is plot the values, trace the line, as you know where max and mins are.
Calculating the area between the curve and x-axis is easy, your limits are the two points you find, and it's a basic definite integral question. If I didn't have a headache, I'd be able to understand that last part. I'll let someone else explain it.
Calculating the area between the curve and x-axis is easy, your limits are the two points you find, and it's a basic definite integral question. If I didn't have a headache, I'd be able to understand that last part. I'll let someone else explain it.
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#3
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Posted 14 January 2007 - 11:03 PM
QUOTE(Sammie @ Jan 14 2007, 04:57 PM) 
For the curve Y=3X-3X2 find:
A
i: its points of intersection with the X and Y axis.
ii: Its max and min turning points.
B Sketch the curve
C Calulate the area inclosed by:
i: The curve and X axis.
ii: The x axis, the lines though the turning points perpendicular to the X axis and the part of the curve between the turning points
~ Hell I dont even know where to start!
Y=3X-3X2 are you sure this is the correct equation as this will only give you a maximum turning point as it is a quadratic. Can't really make sense of c(ii) if it only has one turning point
#4
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Posted 15 January 2007 - 01:08 PM
Sorry its Y=3X-3X3
~head desk~
~head desk~
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Posted 15 January 2007 - 09:47 PM
QUOTE(Sammie @ Jan 15 2007, 01:08 PM)
Sorry its Y=3X-3X3
~head desk~
That makes more sense now
Do as the wedge effect suggests to get a sketch of the graph.
This should give you a sketch like attachment 1
C(i)
This allows you to set up two definite integrals
0-13X-3X3dx and 103X-3X3dxFrom this you should get values of -0.75 and 0.75
This will give an area of 1.5
C(ii)
(see attachment 2)
I think this is the area that you are supposed to find.
Again you must set up 2 definite integrals, this time the limits will be the x-coordinates of the stationary points.
From these you should get an answer of -5/12 and 5/12 which will give you an area of 10/12 or 5/6
Cant seem to send the second attachment. I will try and explain:
Imagine a vertical line from each turning point and where these meet the x-axis.
The area you have to find is the area that the function makes with the x-axis between these two x-values of the turning points.
Hope this helps
Attached Thumbnails
#6
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Posted 17 January 2007 - 06:25 PM
Oh thank you thank you thank you
~ I may have to ask you to marry me now ~tackle hugs~
MAkes so much more sence now!
~ I may have to ask you to marry me now ~tackle hugs~
MAkes so much more sence now!
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Posted 18 January 2007 - 12:27 AM
Marriage is a bit steep! 
But it is nice to be of help sometimes.
But it is nice to be of help sometimes.
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