find an equation of the normal line to the curve y=radx-3 that is parallel to the line 6x+3y-4=0

Answer 1

Let the point of contact be P(a,b)
or P(a, √(a-3) )
If it is to be a normal at P, then the tangent at P must be perpendicular to 6x + 3y - 4 = 0
that is,
the tangent must have a slope of +1/2

dy/dx = (1/2)(x-3)^(-1/2)
= 1/( 2√(x-3) )

1/(2√(x-3) ) = 1/2
2√(x-3) = 2
√(x-3) = 1
x-3 = 1
x = 4

so point P is (4,1)

so the normal must have a slope of -2
equation:
y = -2x + b, with (4,1) on it ....
4 = -2(1) + b
b = 6

normal equation : y = -2x + 6

checking by looking at graph
http://www.wolframalpha.com/input/?i=+plot+y+%3D+√%28x-3%29%2C+y+%3D+-2x%2B6%2C+6x+%2B+3y+-+4+%3D+0

Answer 2

actually, the normal line is
y-1 = -2(x-4)
y = -2x+9