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

Equation of normal is

graph{(x-12y+49)(y+2x^3-4x^2-2x)=0 [-9.71, 10.29, -1.28, 8.72]}

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To find the equation of the normal line at a given point, 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 + 2x is f'(x) = -6x^2 + 8x + 2.

To find the slope at x = -1, we substitute -1 into the derivative: f'(-1) = -6(-1)^2 + 8(-1) + 2 = -6 + (-8) + 2 = -12.

The slope of the tangent line at x = -1 is -12.

The negative reciprocal of -12 is 1/12.

Therefore, the equation of the normal line at x = -1 is y = (1/12)(x - (-1)) + f(-1).

Simplifying this equation, we have y = (1/12)(x + 1) + f(-1).

To find f(-1), we substitute -1 into the original function: f(-1) = -2(-1)^3 + 4(-1)^2 + 2(-1) = -2 + 4 - 2 = 0.

Therefore, the equation of the normal line at x = -1 is y = (1/12)(x + 1) + 0, which simplifies to y = (1/12)(x + 1).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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