How do you graph `y=1/[x(x-2)]`?

Step 1. Find the `y`-intercepts.

`y = f(x) = 1/(x(x-2))`

`f(0) = 1/(0(0-2)) = 1/0`

There is no `y`-intercept.

Step 2. Find the `x`-intercepts.

`0 = 1/(x(x-2))`

`0 = 0`

There is no `x`-intercept.

Step 3. Find the vertical asymptotes.

Set the denominator equal to zero and solve for `x`.

`x(x-2) = 0`

`x = 0` or `x-2 = 0`

`x = 0` or `x=2`

There are vertical asymptotes at `x = 0` and `x = 2`.

Step 4. Find the horizontal asymptote.

The degree of the denominator is greater than the degree of the numerator, so

The horizontal asymptote is at `y=0` (the `x`-axis).

Step 5. Draw your axes and the asymptotes.

The vertical asymptotes divide the graph into three regions of `x`s.

(a) The left hand region has the `x`- and `y`-axes as asymptotes.

`f(-1) = 1/(-1)(-1-2) = 1/(-1)(-3)) = 1/(3) ≈ 0.33`.

The point at (`-1,0.33`) is in the second quadrant, so we have a "hyperbola" above the horizontal asymptote.

(b) The right hand region has `x =2` and the `x`-axis as asymptotes.

`f(3) = 1/(3(3-2)) = 1/(3×1) = 1/3 ≈ 0.33`.

The point at (`3,0.33`) is in the second quadrant, so we have a "hyperbola" above the horizontal asymptote.

So we have a mirror-image hyperbola in the first quadrant.

(c) In the middle region, we have

`f(1) = 1/(1(1-2)) = 1/(1(-1)) = 1/(-1) = -1` and

`f(1.5) = 1/(1.5(1.5-2)) = 1/(1.5(-0.5)) = 1/(-0.75) ≈ -1.33`

`f(0.5) = 1/(0.5(0.5-2)) = 1/(0.5(-1.5)) = 1/(-0.75) ≈-1.33`

The points at (`0.5, -1.33`), (`1,-1`), and (`1.5,-1.33`) are all below the `y`-intercept, so we have an "inverted parabola" between the vertical asymptotes.

And we have our graph.