Year11guy
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The question says sketch the curve: y=x^3-6x^2+9x
I drew it with solutions 0 and 3,3.
The next question asks:
Using your answer to (b) (the last question) sketch on a separate diagram the curve with equation: y=(x-2)^3-6(x-2)^2+9(x-2) showing the coordinates of the points at which the curve meets the x axis. I don't know where to begin or what the answer has to do with the previous question.
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TenOfThem
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(Original post by Year11guy)
The question says sketch the curve: y=x^3-6x^2+9x
I drew it with solutions 0 and 3,3.
The next question asks:
Using your answer to (b) (the last question) sketch on a separate diagram the curve with equation: y=(x-2)^3-6(x-2)^2+9(x-2) showing the coordinates of the points at which the curve meets the x axis. I don't know where to begin or what the answer has to do with the previous question.
You have drawn f(x) and now need to draw f(x-2)

Have you done Graph Transformations?
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BabyMaths
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(Original post by Year11guy)
The question says sketch the curve: y=x^3-6x^2+9x
I drew it with solutions 0 and 3,3.
The next question asks:
Using your answer to (b) (the last question) sketch on a separate diagram the curve with equation: y=(x-2)^3-6(x-2)^2+9(x-2) showing the coordinates of the points at which the curve meets the x axis. I don't know where to begin or what the answer has to do with the previous question.
You do notice the similarity don't you?
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Year11guy
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(Original post by TenOfThem)
You have drawn f(x) and now need to draw f(x-2)

Have you done Graph Transformations?

(Original post by BabyMaths)
You do notice the similarity don't you?
So it's the same graph but just 2 units to the right? Would you get the same answer from expanding because that's what I began doing.
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BabyMaths
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(Original post by Year11guy)
So it's the same graph but just 2 units to the right? Would you get the same answer from expanding because that's what I began doing.
Yes, you could expand and so on but that would take considerably longer and you wouldn't be using your last answer as required by the question.
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TenOfThem
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(Original post by Year11guy)
So it's the same graph but just 2 units to the right? Would you get the same answer from expanding because that's what I began doing.
You would get the same answer, yes

But you would be wasting time
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Year11guy
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(Original post by Year11guy)
So it's the same graph but just 2 units to the right? Would you get the same answer from expanding because that's what I began doing.
(Original post by TenOfThem)
You would get the same answer, yes

But you would be wasting time
How would I find the equations of the asymptotes of y=3/x+2
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TenOfThem
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(Original post by Year11guy)
How would I find the equations of the asymptotes of y=3/x+2
http://www.thestudentroom.co.uk/show....php?t=2873737

What do you know about asymptotes?
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Year11guy
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(Original post by TenOfThem)
http://www.thestudentroom.co.uk/show....php?t=2873737

What do you know about asymptotes?
y=3/(x+2)

its a line that is approached but not touched by the curve
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TenOfThem
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(Original post by Year11guy)
y=3/(x+2)

its a line that is approached but not touched by the curve
Can I ask the context of your question - are you actually in Year 11
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Year11guy
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(Original post by TenOfThem)
Can I ask the context of your question - are you actually in Year 11
Lol, this was made when I was in year 11, I'm doing AS
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TenOfThem
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(Original post by Year11guy)
Lol, this was made when I was in year 11, I'm doing AS
Remember to label threads as sixth form then


So - do you know what value x cannot take?
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Year11guy
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(Original post by TenOfThem)
Remember to label threads as sixth form then


So - do you know what value x cannot take?
0?
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TenOfThem
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(Original post by Year11guy)
0?
Why would that be?

If x = 0, y = 3/2

I see no problem with that
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Year11guy
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(Original post by TenOfThem)
Why would that be?

If x = 0, y = 3/2

I see no problem with that
-2 sorry
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TenOfThem
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(Original post by Year11guy)
-2 sorry
So you know one asymptote x=-2


Now you need the y asymptote

Can you see a value that y cannot =
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Year11guy
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(Original post by TenOfThem)
So you know one asymptote x=-2


Now you need the y asymptote

Can you see a value that y cannot =
Does it have to be a whole number? Is there a way to calculate it?
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TenOfThem
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(Original post by Year11guy)
Does it have to be a whole number? Is there a way to calculate it?
There are a number of ways - have you been shown any

Simply spotting in this case
Splitting the numerator
Dividing by x
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Year11guy
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(Original post by TenOfThem)
There are a number of ways - have you been shown any

Simply spotting in this case
Splitting the numerator
Dividing by x
I haven't really learned about this. I was shown the spotting one. Which is the short/easiest?My exam board is edexcel if there's a specific one I must know
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TenOfThem
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(Original post by Year11guy)
I haven't really learned about this. I was shown the spotting one. Which is the short/easiest?My exam board is edexcel if there's a specific one I must know
You can use any

Clearly "spotting" is the shortest - so if you have been shown that use that
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