Answer:
Analyzed ans Sketched.
Step-by-step explanation:
We are given the function:
[tex]y = 8 \ln (x)/x^2[/tex]
The first derivative of y:
[tex]y' = \frac{8x-16x\ln(x)}{x^4} =\frac{8-16\ln(x)}{x^3} =0[/tex]
The root [tex]x = \sqrt e[/tex] is absolute maximum.
The second derivative of y:
[tex]y''=\frac{-16x^2-24x^2+48x^2\ln(x)}{x^6} =\frac{48\ln(x)-40}{x^4} =0[/tex]
The root [tex]x = e^{5/6}[/tex] is point where concavity changes from down to up.
x = 0 is vertical asymptote.
y = 0 is horizontal asymptote.
The sketch is given in the attachment.