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

Find the slope of the tangent line.

Write the equation in point-slope form:

In slope-intercept form:

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To find the equation of the normal line of f(x) = √(x^2 - x) at x = 2, we need to determine the slope of the tangent line at x = 2 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:

f'(x) = (1/2) * (2x - 1) / √(x^2 - x)

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

f'(2) = (1/2) * (2(2) - 1) / √(2^2 - 2) = (1/2) * (3) / √(4 - 2) = 3/4

Since the slope of the tangent line is 3/4, the slope of the normal line will be the negative reciprocal of 3/4, which is -4/3.

Now, we have the slope (-4/3) and the point (2, f(2)) on the normal line. We can use the point-slope form of a line to find the equation of the normal line:

y - f(2) = (-4/3)(x - 2)

Simplifying this equation will give us the final answer.

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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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