hals
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The gradient of the curve is given by
dy/dx = ax + b
The curve passes through the points (0,0) , (1,2) and (-4,4)

a) find the values of a and b
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16Characters....
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(Original post by hals)
The gradient of the curve is given by
dy/dx = ax + b
The curve passes through the points (0,0) , (1,2) and (-4,4)

a) find the values of a and b
What have you done so far?
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hals
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(Original post by 16Characters....)
What have you done so far?
I've tried working out the equation for the gradient at each point by replacing x and I've tried integrating dy/dx to get y then substituting x and y values but i didn't get the right answers.

Fyi the answers are : a=6 and b=-1
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16Characters....
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(Original post by hals)
I've tried working out the equation for the gradient at each point by replacing x and I've tried integrating dy/dx to get y then substituting x and y values but i didn't get the right answers.

Fyi the answers are : a=6 and b=-1
Integration should work. What did you get when you integrated?
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Naruke
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(Original post by hals)
I've tried working out the equation for the gradient at each point by replacing x and I've tried integrating dy/dx to get y then substituting x and y values but i didn't get the right answers.

Fyi the answers are : a=6 and b=-1
Note:

  • Integration is the inverse process of Differentiation. Did you know that? You could view differentiation and integration like this: If  f(x) = \displaystyle \frac{d...}{dx} \Rightarrow f^{-1}(x) = \int ...  dx . What I'm really trying to imply is that  (f \circ f^{-1})(x) = x

Can you see how you could apply that to your problem? if you integrated both sides, you'd just get  y = f(x) and then it would just be a case of plugging in the points that were given to you and solving them simultaneously to find a and b.
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starshine909
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never liked maths in the first place
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RDKGames
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(Original post by hals)
The gradient of the curve is given by
dy/dx = ax + b
The curve passes through the points (0,0) , (1,2) and (-4,4)

a) find the values of a and b
You don't need to work out any gradients. Simply integrate and sub into the new equation your (x,y) coordinates that the curve passes through and that should leave you with simultaneous equations to solve in a and b
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hals
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(Original post by 16Characters....)
Integration should work. What did you get when you integrated?
I integrated dy/dx and got
y = 1/2ax^2 + bx
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Mr M
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(Original post by hals)
I integrated dy/dx and got
y = 1/2ax^2 + bx
Don't forget the constant of integration (it will go away again in a minute but you should include it).
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