What is the equation of the normal line of #f(x)=-2x^3-4x^2+6x-1# at #x=2#?

To find the equation of the normal line at a given point on a curve, we need to determine the slope of the tangent line at that point and then find the negative reciprocal of that slope.

To find the slope of the tangent line, we take the derivative of the function f(x).

The derivative of f(x) = -2x^3 - 4x^2 + 6x - 1 is f'(x) = -6x^2 - 8x + 6.

To find the slope of the tangent line at x = 2, we substitute x = 2 into the derivative: f'(2) = -6(2)^2 - 8(2) + 6 = -24.

The slope of the tangent line at x = 2 is -24.

To find the slope of the normal line, we take the negative reciprocal of the slope of the tangent line: -1/(-24) = 1/24.

Now, we have the slope of the normal line, and we also have the point of tangency, which is (2, f(2)).

To find f(2), we substitute x = 2 into the original function: f(2) = -2(2)^3 - 4(2)^2 + 6(2) - 1 = -23.

Therefore, the point of tangency is (2, -23).

Using the point-slope form of a line, we can write the equation of the normal line:

y - y1 = m(x - x1), where (x1, y1) is the point of tangency and m is the slope of the normal line.

Substituting the values, we get:

y - (-23) = (1/24)(x - 2).

Simplifying, we have:

y + 23 = (1/24)x - 1/12.

Rearranging the equation, we get the equation of the normal line:

y = (1/24)x - 1/12 - 23.

Simplifying further, we have:

y = (1/24)x - 575/24.

Therefore, the equation of the normal line of f(x) = -2x^3 - 4x^2 + 6x - 1 at x = 2 is y = (1/24)x - 575/24.