How do you graph `y=1/(x^2-x-6)`?

**Step 1. ** Find the `y`-intercepts.

Set `x=0` and solve for `y`.

`f(0) =-1/6`

The `y`-intercept is at (`0,-1/6`).

Step 2. Find the `x`-intercepts.

Set `y=0` and solve for `x`.

`0 = 1/(x^2-x-6)`

`0 = -1`

This is an impossibility, so there is no `x`-intercept.

Step 3. Find the vertical asymptotes.

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

`x^2-x-6= 0`

`(x-3)(x+2) = 0`

`x =3` and `x=-2`

There are vertical asymptotes at `x = -2` and `x =3`.

Step 4. Find the horizontal asymptote.

The degree of the denominator is less 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.

Step 6. Sketch the graph in each region defined by the asymptotes.

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

`f(-3) = 1/6`.

The point at (`-3,1/6`) is in the second quadrant, so we have a "hyperbola" above the horizontal asymptote.

(b) The right hand region has `x =3` and the `x`-axis as asymptotes.

`f(4) = 1/6`.

The point at (`4,1/6`) is in the first 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(0) =-1/6` and

`f(1) = -1/6`

`f(0.5) = -1/6.25`

The points at (`0,-1/6`), (`0.5,-1/6.25`), and (`1,-1/6`) are all below the `x`-axis, so we have an "inverted parabola" between the vertical asymptotes.