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

1 Answer

Do the rest calculations

Explanation:

Given function:

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

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

Now, setting f(x)=0, we get

-x^2+8x-4/x=0

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

x^3-8x^2+4=0\ \quad (x\ne 0)

Solving above equations, we get

x=-0.679, 0.742, 7.936

Hence, the points where tangents are drawn are

(-0.679, 0), (0.742, 0) & (7.936, 0)

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

Hope it helps