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