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

Do the rest calculations

#### Explanation:

Given function:

$f \left(x\right) = - {x}^{2} + 8 x - \frac{4}{x}$

$f ' \left(x\right) = - 2 x + 8 + \frac{4}{x} ^ 2$

Now, setting $f \left(x\right) = 0$, we get

$- {x}^{2} + 8 x - \frac{4}{x} = 0$

$\setminus \frac{- {x}^{3} + 8 {x}^{2} - 4}{x} = 0$

${x}^{3} - 8 {x}^{2} + 4 = 0 \setminus \setminus \quad \left(x \setminus \ne 0\right)$

Solving above equations, we get

$x = - 0.679 , 0.742 , 7.936$

Hence, the points where tangents are drawn are

$\left(- 0.679 , 0\right)$, $\left(0.742 , 0\right)$ & $\left(7.936 , 0\right)$

Now, one can find out the equations of tangents at above three points by finding slopes at respective points

Hope it helps