# What is the equation of the normal line of #f(x)= x^2 − 5x + 1# at #x = 5#?

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To find the equation of the normal line of f(x) = x^2 - 5x + 1 at x = 5, we need to determine the slope of the tangent line at x = 5 and then find the negative reciprocal of that slope to obtain the slope of the normal line.

First, we find the derivative of f(x) with respect to x, which gives us f'(x) = 2x - 5.

Next, we substitute x = 5 into f'(x) to find the slope of the tangent line at x = 5: f'(5) = 2(5) - 5 = 5.

Since the slope of the normal line is the negative reciprocal of the slope of the tangent line, the slope of the normal line is -1/5.

Now, we have the slope of the normal line (-1/5) and the point of tangency (5, f(5)).

Using the point-slope form of a line, we can write the equation of the normal line as y - f(5) = (-1/5)(x - 5).

Simplifying this equation gives us the equation of the normal line of f(x) = x^2 - 5x + 1 at x = 5 as y = (-1/5)x + 6.

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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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