#x#-intercepts of a curve #y=f(x)=(x^2+4)/(2x)# appear when #y=0# and #y#-intercepts of a curve when #x=0#. As we have #x# in denominator, #y#-intercepts appear only at #x=+-oo# and as we have #x^2+4#, which is always greater than or equal to #4#, it is never zero. Hence, there are no #x# or #y#-intercepts of the function.
Function #f(x)# is increasing when #f'(x)>0# and is decreasing when #f'(x)<0#. When #f'(x)=0#, we have an local extrema and we have a local maxima, when #f''(x)<0# and a local minima when #f''(x)>0#.
In the interval, where #f''(x)<0#, the function is concave and when #f''(x)>0#, the function is convex.
As #f(x)=(x^2+4)/(2x)#, we can find #f'(x)# using quotient rule.
#f'(x)=(2x(2x)-2(x^2+4))/(2x)^2=(2x^2-8)/(4x^2)=(x^2-4)/(2x^2)#
hence we have local extrema when #x^2-4=0# or #x=+-2#
further #f''(x)=(2x^2(2x)-4x(x^2-4))/(4x^4)=(16x)/(4x^4)=4/x^3#
Observe that #f''(x)>0# when #x>0# and hence function is convex in the interval #(0,oo)# and as when #x<0# i.e. in the interval #(-oo,0)#, #f''(x)<0#, the function is concave.
As at #x=-2#, #f''(x)=4/(-8)=-1/2<0# we have a local maxima.
and at #x=2#, #f''(x)=4/8=1/2>0# we have a local minima.
The graph appears as shown below.
graph{(x^2+4)/(2x) [-20, 20, -10, 10]}
