euation of the line?

Question

Find the equation of the tangent and normal line to the curve y = 3x² – 2x + 5 where y = 6

Answer 1

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