We have #f(x) = g(x) / (h(x)) = (2x^3 -x^2 - 2x +1) / (x^2+3x+2)#.

Function #h(x)# quite easily factorises down to #(x+1)(x+2)#.

Divide #g(x)# by #x+1# to factorise the cubic, and it can be found through polynomial division that #g(x) = (x+1)(2x^2-3x+1)#, the latter quadratic of which factorises down to #(2x-1)(x-1)#.

Hence #f(x) = ((2x-1)(x-1)(x+1)) / ((x+1)(x+2)) = ((2x-1)(x-1)) / (x+2)#

By further polynomial division, that is #(2x^2-3x+1)/(x+2)#, it can be found that #f(x) = 2x - 7 + 15/(x+2)#.

Clearly #x# cannot equal #-2# as it leaves the fraction undefined, so there is an asymptote at #x=-2#.

Secondly #f(x)# cannot equal #2x-7#, as this would suggest that #15/(x+2) = 0 rArr 15 = 0#, so there is an oblique asymptote at #y=2x-7#.

To solve for #x#-intercept(s) return to the factorised function, from which it can be seen easily that for #f(x)# to equal zero, #(2x-1)(x-1) = 0#, therefore #x=1/2# or #1# for the #x#-intercepts.

Let #x = 0# to find #y#-intercepts. Thus #f(x) = ((-1)(-1))/2#, therefore #y=1/2#.

Lastly, the 'holes'. Two #x+1# expressions were cancelled early on, hence there is a hole at #x=-1#. So, a discontinuity is found at #(-1, f(-1)) -= (-1, 6)#.

To find stationary points, calculate #f'(x)# and let it equal zero:

#f'(x) = x - 15/((x+2)^2) = 0#.

#:. x(x+2)^2 = 15#

#:. x^3 + 4x^2 + 4x -15 = 0#

So you end up with a stationary point at approximately #(1.34, f(1.34))#.

Hope this helps!