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# C1 Integration and Differentiation watch

1. The curve C has equation y = f(x)
Given that dy/dx = 3x^2 - 20x + 29

And that C passes thru the point P(2,6)

And C goes through the point (4,0)
Find an equation of the the tangent to C at P (the answer is y = x+4)

The tangent to C at the point Q is parallel to the tangent at P, so calculate the xcoord of Q.

So far I have worked out the gradient of Q's tangent is 1..

what next?

thanks
2. Find the value of the gradient at the point P, then the tangent has equation
y = mx + c, where m is the gradient you found.
Subsitute in the co-ordinates of P for y and x and solve to find c
3. (Original post by koldtoast)
The curve C has equation y = f(x)
Given that dy/dx = 3x^2 - 20x + 29

And that C passes thru the point P(2,6)

And C goes through the point (4,0)
Find an equation of the the tangent to C at P (the answer is y = x+4)

The tangent to C at the point Q is parallel to the tangent at P, so calculate the xcoord of Q.

So far I have worked out the gradient of Q's tangent is 1..

what next?

thanks
Plug 2 (x-coordinate of P) into your expression for dy/dx to get m. dy/dx=3(2)^2 - 20(2) + 29 = 1. Hence, we have y = x + c. So, 6 = 2 + c, --> c=4. So y=x+4
4. (Original post by koldtoast)
The curve C has equation y = f(x)
Given that dy/dx = 3x^2 - 20x + 29

And that C passes thru the point P(2,6)

And C goes through the point (4,0)
Find an equation of the the tangent to C at P (the answer is y = x+4)

The tangent to C at the point Q is parallel to the tangent at P, so calculate the xcoord of Q.
dy/dx = 3x^2 - 20x + 29
---> y = x^3 - 10x^2 + 29x + k
When x = 2, y = 6:
---> 6 = 8 - 40 + 58 + k
---> 6 = 26 + k
---> k = - 20

---> Curve C: y = x^3 - 10x^2 + 29x - 20

At P: Gradient = 12 - 40 + 29 = 1
---> Tangent at P has gradient = 1 and passes through P(2, 6).
Equation Of Tangent: y - 6 = 1(x - 2)
---> y - 6 = x - 2
---> y = x + 4

As the tangent at P is parallel to the tangent at Q, therefore the tangents at P and Q have the same gradient and thus the curve has the same gradient at these 2 points.

At Q when dy/dx = 1:
---> 1 = 3x^2 - 20x + 29
---> 3x^2 - 20x + 28 = 0
---> x = {20(+/-)Sqrt[400 - 4(3)(28)]}/6
= [20(+/-)8]/6
---> x Co-ord Of Q = 28/6 = 28/6 = 14/3 OR:
---> x Co-ord Of Q = 12/6 = 2
5. Thanks nima..so x = 14/3 or 2 is a perfectly valid answer? I wouldn't have to work out which of the x values it is?
6. (Original post by koldtoast)
Thanks nima..so x = 14/3 or 2 is a perfectly valid answer? I wouldn't have to work out which of the x values it is?
its 14/3

the 2 relates to the P

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