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 their is no point on C at which the tangent is parralel to the line y= 1-2x

Posted Tue 4th December, 2012 @ 11:09 by

SabahEdited by

Sabah on Tue 4th December, 2012 @ 11:10

1) sub x=1 to find the gradient at p.

(3(1)-1)^2 =4

use grad times grad of normal =-1

therefore gradient of -1/4

then use y-y1=m(x-x1) use values of x and y at p

y-4=-1/4(x-1)

tidy up =y=-x/4 +17/4

Answered Sun 9th December, 2012 @ 16:40 by

Edward Pinches
thanx bro :D :D glad some1 finally decided 2 help me ^_^

Answered Mon 10th December, 2012 @ 17:13 by

Sabah