# `x^2=-6y` Graph the equation. Identify the focus, directrix, and axis of symmetry of the parabola.

*print*Print*list*Cite

`x^2=-6y`

Take note that one of the vertex form of a parabola is

`(x - h)^2 = 4p(y-k)`

where

(h,k) is the vertex and,

p is the distance between the vertex and the focus and also the same distance between the vertex and the directrix.

So to graph it, first find the vertex.

Rewriting the given equation in exact form as above, it becomes:

`(x-0)^2=-6(y-0)`

So the vertex is (0,0).

Then, determine the other points of the parabola. To do so, isolate the y.

`x^2=-6y`

`x^2/(-6)=y`

`-x^2/6=y`

Then, assign values to x. And solve for the y values.

`x=-6` , `y=-(-6)^2/6=-36/6=-6`

`x=6` , `y=-6^2/6=-36/6=-6`

Plot these three points (-6,-6), (0,0) and (6,-6). And connect them.

**Therefore, the graph of the given equation is:**

To determine the focus and the directrix, consider the coefficient of the unsquared portion of the given equation and set it equal to 4p.

`4p=-6`

And, solve for p.

`p=-6/4`

`p=-3/2`

So the focus of the parabola is 3/2 units from the vertex. Since the parabola is going down, the coordinates of the focus is:

`(h, k+p)=(0,0+(-3/2))=(0,-3/2)`

And the equation of directrix is:

`y=k-p`

`y= 0 - (-3/2)`

`y= 3/2`

**Therefore, the focus is `(0,-3/2)` and the directrix is `y=3/2` .**

Take note that the axis of symmetry of a parabola is line that passes the vertex and the focus. And it is perpendicular to the directrix. (See attachment.)

**Therefore, the equation of its axis of symmetry is `x=0` .**