The graph of the function `y= (x^2-4x+4)/(x^3-6x^2+9x-4)` is:

(a) x and y intercepts

To determine the x intercepts, set y=0.

`0= (x^2-4x+4)/(x^3-6x^2+9x-4)`

Then, simplify the equation by multiplying both sides by the denominator.

`(x^3-6x^2+9x-4)*0= (x^2-4x+4)/(x^3-6x^2+9x-4)*(x^3-6x^2+9x-4)`

`0=x^2-4x+4`

Then, factor right side and solve for x.

`0=(x-2)^2`

`0=x-2`

`2=x`

**Hence, the x-intercept is `(2,0)` .**

To solve for the y-intercept, set x=0.

`y=(x^2-4x+4)/(x^3-6x^2+9x-4)`

`y=(0^2-4(0)+4)/(0^3-6(0)^2+9(0)-4)`

`y=4/(-4)`

`y=-1`

**Thus, the y-intercept is `(0,-1)` .**

(b) Vertical Asymptotes / Holes

To determine if the function has vertical asymptotes and holes, factor the numerator and denominator.

`y=(x^2-4x+4)/(x^3-6x^2+9x-4)`

`y=(x-2)^2/((x-4)(x-1)^2)`

Notice that the numerator and denominator has no common factor.**Hence, the function has no holes.**

Then, set each factor in the denominator equal to zero to solve for the vertical asymptotes.

For the first factor in the denominator:

`x-4=0`

`x=4`

For the second factor:

`(x-1)^2=0`

`x-1=0`

`x=1`

**Therefore, the vertical asymptotes are `x=1` and `x=4` .**

(c) Long range behavior.

Base on the graph, the range of the function is all real numbers.

(d)Points at which the graph crosses its Horizontal or Slant Asymptote (if any)

Since the degree of the numerator is less than the degree of the denominator, the function has no slant asymptote. It has only horizontal asymptote at `y=0` .

So to determine if the graph crosses the horizontal asymptote y=0, refer to the x-intercept.

**Hence, the graph passes the horizontal asymptote at `(2,0)` .**