# What is the equation of the normal line of #f(x)=12x^3-4x^2-5x# at #x=-2#?

We compute the slope of the tangent by evaluating the first derivative:

To calculate the equation of the normal line we use

To use this, we need a point on the line. We know that x = -2 is on the line, so we evaluate the original function at this point to get :

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The equation of the normal line of f(x)=12x^3-4x^2-5x at x=-2 is y = -23x - 26.

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