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

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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) = 5x^3 + x^2 - 5x - 2 is f'(x) = 15x^2 + 2x - 5.

To find the slope of the tangent line at x = 2, we substitute x = 2 into the derivative: f'(2) = 15(2)^2 + 2(2) - 5 = 60 + 4 - 5 = 59.

The slope of the tangent line at x = 2 is 59.

To find the slope of the normal line, we take the negative reciprocal of the slope of the tangent line: -1/59.

Now, we have the slope of the normal line and the point (2, f(2)) 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(2) = (-1/59)(x - 2).

Simplifying this equation will give us the final answer.

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