Find the equation of the normal to the curve y = 3x^2 - 5x that is perpendicular to the line y = x + 4.
Find the slope of the tangent first.
the perpendicular to the line x+4 has a slope of -1.
so the normal to the curve has the same slope.
y'=6x-5 and then the perpendicular to this must have a slope of -1/6x
so -1/6x=-1 or x= 1/6
y=mx+b slope is -1
y=-x+b but the point of tangencey is x=1/6
y=-1/6+b
Now for y, in the curve y=3x^2-5x, when x is 1/6, then y= 3/36-30/36=-27/36
so solve for b.
All this depends on my intrepretation of normal to the curve that is perpendicular to the line...
That is most unusual wording.