Does "normal" mean perpendicular to the tangent?
If that's the case:
slope_normal(x) * slope_tangent(x) = -1
16 * slope_tangent(x) = -1
where slope_tangent(x) = y'(x)
16 * y'(x) = -1
Solve for x.
-O
Ok so I am at this point where dx/dy (x^4) = -1/16.
from here I got 4x^3 = -1/16
i multipled by 4 * 16 to get -1/64 and got the third root of this to get me -1/4 which is my x. How do i get the y vaule? i tried subbing it in into x^4 and i got the wrong answer