Question

find an equation of each normal line to y=x^3-4x that is parallel to the line x+8y-8=0​

Answer 1

x+8y=8

y=(-1/8)x+1. Has a slope of -1/8

The normal line to the curve has a slope of -1/8 as well

It is perpendicular to the line tangent to the curve, and that has a negative reciprocal of -1/8 slope, or 8.

so we want the tangent line to the curve y=x^3-4x, where the slope is 8

Take the derivative, which is 3x^2-4, and set it equal to 8

3x^2=12

x^2=4

x=-2 and 2.

At x=both -2 and 2, y=0, so the tangent of the line to the curve at (-2, 0) and (2, 0) has a slope of 8. From the point slope formula y-y1=m(x-x1), m slope and (x1, y1) point, y=8(x-2) or y=8x-16 and y=8(x+2) or y=8x +16. These are equations of the tangent line to the curve at those two points.

The normal line to the curve has slope -1/8 and goes through (-2, 0), and its equation is y=(-1/8)(x+2) or

y=-(1/8)x-1/4.

There is a second one that has equation y=(-1/8)(x-2) or y=(-1/8)(x)+1/4