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

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To find the equation of the line normal to f(x) at x=-3, we need to determine the slope of the tangent line at that point. The slope of the tangent line is equal to the derivative of the function at that point. Taking the derivative of f(x)=3x^2 -x +1, we get f'(x) = 6x - 1. Evaluating f'(-3), we find that the slope of the tangent line at x=-3 is -19.

Since the line normal to a given line has a slope that is the negative reciprocal of the slope of the original line, the slope of the line normal to f(x) at x=-3 is 1/19.

Using the point-slope form of a line, where (x1, y1) is the given point on the line, we can write the equation of the line normal to f(x) at x=-3 as y - f(-3) = (1/19)(x - (-3)). Simplifying this equation, we get y + 1 = (1/19)(x + 3).

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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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