First of all do polynomial division to get your function in the nice form of:

#f(x) = 2x + 1 + 3/(x-2)#.

From this it is fairly clear where the asymptotes are, as #x# cannot equal #2# as #f(x)# would then be undefined. Similarly #f(x)# cannot equal #2x + 1#, as this would mean that #3/(x-2)# equals #0#, which it cannot as it implies #3=0#.

Then differentiate this function, applying the chain rule and product rules where necessary.

The derivative of #2x + 1# is of course #2#.

The derivative of #3/(x-2)# is done by letting #y=3(x-2)^-1#.

First apply the chain rule: say #t = x-2# and #w=t^-1#.

#:. (dt)/(dx) = 1# and #(dw)/(dt) = -t^-2#.

#:. (dw)/(dx) = -t^-2 = -(x-2)^-2#

The product rule isn't really necessary as it's pretty clear you will end up with:

#(dy)/(dx) = -3/(x-2)^2#.

#:. f'(x) = 2 -3/(x-2)^2#.

Letting #f'(x) = 0# next.

#:. 3 = 2(x-2)^2#

Simplify it down to arrive at #2x^2-8x+8 = 0#, which by the quadratic formula leaves us with: #x=(4-sqrt(6))/2# or #x=(sqrt(6)+4)/2#.

Solving for #f((4-sqrt(6))/2)# and #f((sqrt(6)+4)/2)# yields #5-sqrt(6)# and #2sqrt(6) + 5#, which are maximum and minimum respectively.

Finally calculate your intercepts, first by letting #f(x)=0#

#:. -3 = (2x+1)(x-2)#

#:. 2x^2-3x+1 =0#

#:. x = 1/2# or #x=1#.

Finally calculate #f(0)# which is simply #1-3/2#, therefore the #y#-intercept is where #y=-1/2#.