Maths
posted by Annie .
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'=6x5 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^25x, when x is 1/6, then y= 3/3630/36=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.