y=3x2-2x+5,
dy/dx=6x-2 which is the slope of the tangent.
When y=6, 6=3x2-2x+5, 3x2-2x-1=0=(3x+1)(x-1). So x=-⅓ and 1 when y=6.
This gives us dy/dx=-4 at (-⅓,6) and dy/dx=4 at (1,6), so there are two tangent lines and two normal lines.
If we represent the tangent equations as y=m1x+a1 and y=m2x+a2, where m1=-4 and m2=4, then we have:
y=-4x+a1 and y=4+a2, and we find a1 and a2 by plugging in the coords of the tangents:
6=4/3+a1, a1=6-4/3=14/3; 6=4+a2, a2=6-4=2, giving us y=-4x+14/3 and y=4x+2.
The slopes of the normals are -1/(-4)=¼, -1/4=-¼.
The normals must pass through the same points as the tangents (-⅓,6) and (1,6):
y=x/4+a3 and y=-x/4+a4 are the general equations of the normals. Plug in the coords:
6=-1/12+a3, a3=73/12; 6=-1/4+a4, a4=25/4, so y=x/4+73/12, y=-x/4+25/4 are the normals.
We can write the equations:
Tangents: 12x+3y=14, y=4x+2
Normals: 12y-3x=73, x+4y=25