What is the equation of the tangent lines of #f(x) =-x^2+8x-4/x# at # f(x)=0#?

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When we solve the aforementioned equations, we obtain

Thus, the locations of the drawn tangents are

Now, by determining the slopes at each of the three points above, one can determine the equations of the tangents at those locations.

I hope it's useful.

To find the equation of the tangent lines of f(x) = (-x^2 + 8x - 4)/x at f(x) = 0, we need to determine the derivative of the function and substitute the x-value where f(x) = 0 into the derivative. The derivative of f(x) is given by f'(x) = (2x^2 - 8x + 4)/x^2.

To find the x-value where f(x) = 0, we set the numerator of the function equal to zero: -x^2 + 8x - 4 = 0. Solving this quadratic equation, we find x = 2 ± √3.

Now, we substitute these x-values into the derivative f'(x) to find the slopes of the tangent lines. When x = 2 + √3, f'(x) = (2(2 + √3)^2 - 8(2 + √3) + 4)/(2 + √3)^2. Simplifying this expression, we get f'(x) = 2 + √3.

Similarly, when x = 2 - √3, f'(x) = (2(2 - √3)^2 - 8(2 - √3) + 4)/(2 - √3)^2. Simplifying this expression, we get f'(x) = 2 - √3.

Therefore, the equations of the tangent lines at f(x) = 0 are y = (2 + √3)(x - (2 + √3)) and y = (2 - √3)(x - (2 - √3)).