At what point on the curve y = x^4 does the normal have a slope of 16?
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