<s>[INST] 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?
 [/INST] $-4x^2 -2x -4 = -(4x^2 + 2x + 4)$
The $4x^2 +2x + 4$ must come from some $(2x+b)^2$, to get the right square, and this has linear term $4b$, which should equal $2x$ so $b=\frac{1}{2}$.
Now $(2x + \frac12)^2 = 4x^2 + 2x + \frac14$, so we need an extra $3\frac34$ to get $4$, like we need. So in all 
$$-4x^2 -2x -4 = -\left((2x+\frac12)^2 + 3\frac34\right)$$
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