The only thing I know I should do is integrate the dy/dx equation, but then I don't really know what to do afterwards :/
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creativebuzz
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- 28-12-2014 21:59
Last edited by creativebuzz; 28-12-2014 at 22:15. -
morgan8002
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- 28-12-2014 22:10
I think you've uploaded the wrong image. The stuff you've typed does not match the image(text is integration, image is surds).
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creativebuzz
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- 28-12-2014 22:15
(Original post by morgan8002)
I think you've uploaded the wrong image. The stuff you've typed does not match the image(text is integration, image is surds). -
morgan8002
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- 28-12-2014 22:24
To find the equation of a tangent, you need to find the gradient of the tangent and have a point on the line. Does this help?
The first step is to find the equation of the tangent. Integrating comes later in the question. -
creativebuzz
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- 28-12-2014 22:29
(Original post by morgan8002)
To find the equation of a tangent, you need to find the gradient of the tangent and have a point on the line. Does this help?
The first step is to find the equation of the tangent. Integrating comes later in the question. -
morgan8002
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- 28-12-2014 22:32
(Original post by creativebuzz)
I managed to get the gradient but I didn't know what the y coordinate would be so I could form and equation?
You need to use all of the information that the question gives you, because they will never give you more information than you need. -
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- 28-12-2014 22:37
F'(x) = 3(x)^2 - 4x -1
F(x) (which is equal to y) is just the integral of f'(x)
Then set up an equation involving y and sub in x=2 to get the y value innit
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morgan8002
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- 28-12-2014 22:40
(Original post by Fermions)
F'(x) = 3(x)^2 - 4x -1
F(x) (which is equal to y) is just the integral of f'(x)
Then set up an equation involving y and sub in x=2 to get the y value innit
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- 28-12-2014 22:41
(Original post by morgan8002)
You are left with an arbitrary constant of integration, so you have to use the other information given in the question to find the value of the constant and thus the equation of the line.
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creativebuzz
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- 28-12-2014 22:52
Last edited by creativebuzz; 28-12-2014 at 22:53. -
morgan8002
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- 28-12-2014 22:53
(Original post by creativebuzz)
That is exactly where I got stuck -
creativebuzz
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- 28-12-2014 22:56
(Original post by morgan8002)
So have you managed to find the equation of the tangent?
1) find gradient = 3
2) then the eq of the tangent: y - 0 = 3(x -0)
to give me y=3x
but I wasn't confident in my answer :/ -
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- 28-12-2014 22:56
To do this you must first find the equation of the tangent. You would try to find the gradient l. You will find this to be 3 i think. You would use 3 and also the point (0,0) to form an equation of the tanget. You will find this to be y=3x. Next you would sub in 2 to get a y value of 6. Then you would integrate the given equation and subsitute in 2 as the x value and 6 as the y value to find the constant.
Hope this helps
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morgan8002
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- 28-12-2014 22:59
(Original post by creativebuzz)
Well my initial method was to
1) find gradient = 3
2) then the eq of the tangent: y - 0 = 3(x -0)
to give me y=3x
but I wasn't confident in my answer :/ -
morgan8002
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- 28-12-2014 23:00
(Original post by ImNervous)
To do this you must first find the equation of the tangent. You would try to find the gradient l. You will find this to be 3 i think. You would use 3 and also the point (0,0) to form an equation of the tanget. You will find this to be y=3x. Next you would sub in 2 to get a y value of 6. Then you would integrate the given equation and subsitute in 2 as the x value and 6 as the y value to find the constant.
Hope this helps
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creativebuzz
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- 28-12-2014 23:04
(Original post by ImNervous)
To do this you must first find the equation of the tangent. You would try to find the gradient l. You will find this to be 3 i think. You would use 3 and also the point (0,0) to form an equation of the tanget. You will find this to be y=3x. Next you would sub in 2 to get a y value of 6. Then you would integrate the given equation and subsitute in 2 as the x value and 6 as the y value to find the constant.
Hope this helps
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Updated: December 28, 2014
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