#f(x) = (x^2 - 3x + 9)/(x - 3) = (x^2 - 3x)/(x - 3) + 9/(x - 3)#

#= (x(x - 3))/(x - 3) + 9/(x - 3) = x + 9/(x - 3)#

So you actually wanna sketch #f(x) = x + 9/(x - 3)#

Domain:

The domain of this function is all Real x except 3

because #9/(x - 3) # would become unreal!

This brings us to Asymptotes:

They're two Asymptotes:

No 1: #y = x# because as #x -> oo # # f(x) -> x#

No 2: #x= 3# is a vertical asymptote because as #f(x) -> oo # # f(x)-> 3#

Next, turning points:

To find turning points (maxima and minima),

We let# y = f(x) => y = x + 9/(x - 3)#

We make #x# the subject of the equation,

#=> yx - 3y = x^2 - 3x + 9#

#=> x^2 -(3 + y)x + 9+ 3y = 0#

Since all the values of #x# are real, #b^2 - 4ac >= 0#

#=> (3 + y)^2 - 4(9 + 3y) >= 0#

#=> y^2 - 6y - 27 >=0 => (y - 9)(y + 3)>= 0#

Hence, #y<=-3 uu y >= 9#

Putting the values of #y# in #f(x)# to get #x#

we have the points #(0, -3)# maximum and #(6, 9)# as minimum

To find the intercepts :

#y#- intercept is when #x = 0# we have #(0, -3)#

#x#- intercept is when #y = 0# but #y# cannot be #0#(see the range above)

Table of variation:

So now we're all set to sketch that graph;

Here's what i got: