# What is the equation of the normal line of #f(x)=2x^3-8x^2+2x-1# at #x=-1#?

Find the derivative of the function through the power rule:

The slope of the tangent line is

Graphed are the function and its normal line:

graph{(2x^3-8x^2+2x-1-y)(y+13+(x+1)/24)=0 [-5, 7, -18.16, 2.12]}

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To find the equation of the normal line of f(x) at x=-1, we need to find the slope of the tangent line at that point and then determine the negative reciprocal of that slope to find the slope of the normal line.

First, we find the derivative of f(x) to get the slope of the tangent line. The derivative of f(x)=2x^3-8x^2+2x-1 is f'(x)=6x^2-16x+2.

Next, we substitute x=-1 into the derivative to find the slope of the tangent line at x=-1. f'(-1)=6(-1)^2-16(-1)+2=24.

The slope of the normal line is the negative reciprocal of the slope of the tangent line. So, the slope of the normal line is -1/24.

Now, we have the slope of the normal line and the point (-1, f(-1)) on the curve. We can use the point-slope form of a line to find the equation of the normal line.

Using the point-slope form, the equation of the normal line is y - f(-1) = (-1/24)(x - (-1)).

Simplifying this equation, we get y - f(-1) = (-1/24)(x + 1).

Therefore, the equation of the normal line of f(x)=2x^3-8x^2+2x-1 at x=-1 is y - f(-1) = (-1/24)(x + 1).

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