Question 1078317
<pre>
{{{y}}}{{{""=""}}}{{{expr(-1/12)x^2-2x-1}}}

First get into standard form, which is

{{{(x-h)^2}}}{{{""=""}}}{{{4p(y-k)}}}

Then the vertex is (h,k), the focus is (h,k+p),
and the equation of the directrix is y=k-p
-----------------------------------------------
Instead of doing your homework for you, I'll
do one exactly like it.  Then you can use it 
as a model to do yours step by step

{{{y}}}{{{""=""}}}{{{expr(-1/20)x^2-2x-10}}}

Clear of fractions by multiplying through by 20

{{{20y}}}{{{""=""}}}{{{-x^2-40x-200}}}

Get the x terms on the left side
and all the other terms on the right

{{{x^2+40x}}}{{{""=""}}}{{{-20y-200}}}

Complete the square on the left:

1. Multiply the coefficient of x, which is 40, by 1/2, getting 20
2. Square what you got, 20<sup>2</sup>=400
3. Add that, 400, to both sides of the equation

{{{x^2+40x+400}}}{{{""=""}}}{{{-20y-200+400}}}

Factor the left side as (x+20)(x+20) and write it as (x+20)<sup>2</sup>
Combine the -200+400 on the right as +200

{{{(x+20)^2}}}{{{""=""}}}{{{-20y+200}}}

Factor out the coefficient of y, which is -20 on the right,
remembering that when you factor out a negative, you get
a sign change inside the parentheses:

{{{(x+20)^2}}}{{{""=""}}}{{{-20(y-10)}}}

Compare that to

{{{(x-h)^2}}}{{{""=""}}}{{{4p(y-k)}}}

where the vertex is (h,k), the focus is (h,k+p),
and the equation of the directrix is y=k-p

So h = -20, 4p = -20, k=10.

And since 4p =-20, p = -5.

So the vertex is (-20,10), the focus is (-20,10-5) or (-20,5)

(-20,5)  <-- focus

and the equation of the directrix is y=10-(-5) or y = 15 

{{{drawing(400,400,-50,10,-30,30,

graph(400,400,-50,10,-30,30,(-1/20)x^2-2x-10),
line(-100,15,100,15),circle(-20,5,.4),locate(-20,5,"(-20,5)") )}}}

Now you can do yours exactly the same way step by step.

Edwin</pre>