# What is the equation of the tangent line of #f(x) =(x^2)/(x-4)# at #x=5#?

Determine f(x)'s value at x= 5 first.

Let's now determine f(x)'s derivative at x=5.

We can create a tangent line equation for x=5 using all of this information, and the solution is provided here.

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To find the equation of the tangent line of f(x) at x=5, we need to find the slope of the tangent line and a point on the line.

First, we find the derivative of f(x) using the quotient rule:

f'(x) = [(x-4)(2x) - (x^2)(1)] / (x-4)^2

Simplifying this expression, we get:

f'(x) = (2x^2 - 8x - x^2) / (x-4)^2

Combining like terms, we have:

f'(x) = (x^2 - 8x) / (x-4)^2

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

f'(5) = (5^2 - 8(5)) / (5-4)^2 = (25 - 40) / 1 = -15

So, the slope of the tangent line at x=5 is -15.

To find a point on the line, we substitute x=5 into f(x):

f(5) = (5^2) / (5-4) = 25 / 1 = 25

Therefore, a point on the tangent line is (5, 25).

Using the point-slope form of a line, we can write the equation of the tangent line:

y - 25 = -15(x - 5)

Simplifying this equation, we get:

y - 25 = -15x + 75

Finally, rearranging the equation, we have:

y = -15x + 100

Therefore, the equation of the tangent line of f(x) at x=5 is y = -15x + 100.

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