How do I complete the square of $y= -4x^2-2x-4$? $y = -4x^2 - 2x - 4$ 
I just can't figure this out, do I divide the second number and the third number by $4$ then by $2$ and then add the product to the second one and subtract it from the third one?

First, you need to isolate x^2 from the equation by factor the -4 out, which leave the equation like this:
$$y= -4(x^2+ \frac{1}{2} x ) -4 $$
We know that the equation of square binomials is:
$(x+b)^2= x^2+2xb+b^2$
To write the binomial $x^2 + \frac{1}{2}x$ to a square binomials, we need to find b^2
We got :
$2xb = \frac{1}{2} x $ 
=> $2b= \frac{1}{2}$ 
=> $b= \frac{1}{4}$ 
=> $b^2= \frac{1}{16}$
To balance the equation we need to add and subtract b^2 in the parentheses
$$y= -4(x^2+ \frac{1}{2} x +\frac{1}{16} -\frac{1}{16}) -4$$
Distribute the $-\frac{1}{16}$ out and combine like term
$$y= -4(x^2+ \frac{1}{2} x +\frac{1}{16}) +\frac{1}{4} -4$$
$$y= -4(x^2+ \frac{1}{2} x +\frac{1}{16}) -\frac{15}{4}$$
Now convert the trinominal inside of the parentheses to square binomial 
$$y= -4(x^2+ 2\frac{1}{4} x +(\frac{1}{4})^2) -\frac{15}{4}$$
$$y= -4(x + \frac{1}{4})^2 -\frac{15}{4}$$
