1. The problem statement, all variables and given/known data Find the equation of each of the normal lines to the curve y = x^3 - 4x that is parallel to the line x + 8y - 8 = 0. 2. Relevant equations Differentiation, y - y1 = m(x - x1) 3. The attempt at a solution Well, clearly I start by differentiating y = x^3 - 4x to gey dy/dx = 3x^2 - 4. Then, this is the slope of the tangent to the graph y. This slope is equal to the slope of the line x + 8y - 8 = 0. Which is -8. So, I then evaluate for x and sub back into dy/dx, giving me the slope of the tangent. The slope of the normal line is the negative reciprocal of this. If you solve for x using 3x^2 - 4 = -8, you get an imaginary value though. Where have I gone wrong?