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Question 1136446: Find the equation of the normal line to the curve.
1. y=x^3-4x and that is parallel to the line x+8y-8=0
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! 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
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