How do you sketch the graph of `f(x) = (x^2 - 3x + 9)/(x - 3)`?

`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: