Differentiation doubt

[spiderman]

Hey guys,
Pls could you'll help me with this sum:

-: Fin the equation to the normal to the curve y=3x^2 -2x-1 which is parallel to the line y=x-3.

[astarmathsandphysics]

Hey guys,
Pls could you'll help me with this sum:

-: Fin the equation to the normal to the curve y=3x^2 -2x-1 which is parallel to the line y=x-3.

dy/dx=6x-2
since it is perpendicular  to y=x-3 the gradient=dy/dx=-1/1 s0 6x-2=-1 so x=1/6
y=3(1/6)^2-2(1/6)-1=-5/4 so the point is (1/6,-5/4)
y+5/4=-1(x-1/6) so 12y+15=-12x+2 so 12x+12y=-13
Didn't read the question then put x=-1/6 somewhere

[nid404]

the answer to this is different
It's 12y=12x-17

In y=x-3
m=1
therefore gradient will be same for point on the curve 3x2- 2x-1

6x-2=1
x=1/2

Using this too.....i get a wrong answer

Even m confused wid this one...

[astarmathsandphysics]

Corrected nmy post

[astarmathsandphysics]

Do you mean the normal is parallel to y=x-2 or the tangent is parallel to y=x-3

[nid404]

Do you mean the normal is parallel to y=x-2 or the tangent is parallel to y=x-3

tangent parallel to y=x-3....i think. we have to find normal to tangent on the curve(when the curve's gradient is same as y=x-3)

[astarmathsandphysics]

It works out worse for that.