Gradient of the curve help

  • #1
The gradient of the curve C is given by: dy/dx = (3x-1)2
the point P (1,4) lies on C.

a) Find an equation of the normal to C at P.
b) Find an equation for the curve C in the form y=f(x)
c) using dy/dx = (3x-1)2 show that there is no point on C at which the tangent is parallel to the line y=1-2x.

Well for a) i got: 4y= -x+17 (subbed x= 1 in dy/dx, got grad of normal and got eq)

But, i don't understand what part b) wants me to do. I think i could do c) by using the gradients and showing that they're not the same.

Some help on b) please!
 

Answers and Replies

  • #2
berkeman
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You have the derivative of the function y=f(x). Do you know how to get the original function back from its derivative? Do you know why they said to "find an equation" instead of "find the equation" in b)?
 
  • #3
do you mean i have to integrate dy/dx = (3x-1)2 for b)

to get : 3x3 - 3x2+x +c ?
 
Last edited:
  • #4
berkeman
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Integration is the correct operation. But your answer is not correct yet. You should be able to check your answer by differentiating it again to get back to the original equation for dy/dx.
 
  • #5
oh wait i remembered i need work out c, by using P(1,4)

Thanks for your help!
 
  • #6
berkeman
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Your welcome. Be sure to do the step where you check your answer using differentiation.
 

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