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

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Answer 1 · 2018-07-28T13:36:43

Do the rest calculations

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