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?
 A: $-4x^2 -2x -4 = -(4x^2 + 2x + 4)$
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 
$$-4x^2 -2x -4 = -\left((2x+\frac12)^2 + 3\frac34\right)$$
A: 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 
$$y= -4(x^2+ 2\frac{1}{4} x +(\frac{1}{4})^2) -\frac{15}{4}$$
$$y= -4(x + \frac{1}{4})^2 -\frac{15}{4}$$
A: There is a general formula you can use it. Every quadratic function can be writen as:
$$f(x)= a(x-p)^2+q$$wher $p=-{b\over 2a}$ and $q= {-D\over 4a}$ and $D=b^2-4ac$.
A: The simplest strategy is to multiply both sides by $4$:
$$
4y=-(\underbrace{16x^2+8x}_{\text{to complete}}+16)
$$
Then notice that $(4x+1)^2=16x^2+8x+1$, so we have
$$
4y=-((4x+1)^2+15)
$$
and then
$$
y=-\frac{1}{4}(4x+1)^2-\frac{15}{4}
$$
In general, if you have a polynomial like $ax^2+bx+c$, you can easily complete the square by multiplying by $4a$:
$$
\underbrace{4a^2x^2+4abx}_{\text{to complete}}+4ac=
4a^2x^2+4abx+b^2-b^2+4ac=(2ax+b)^2+(4ac-b^2)
$$
Then you can divide back by $4a$. In this particular case, it's not necessary to multiply by $16$, as $4$ suffices.
A: In general, if you want to complete the square on $ax^2+bx+c$, here's the formula and derivation
Suppose that
$$ax^2+bx+c=A(x+B)^2+C$$
Then:
$$ax^2+bx+c=Ax^2+2ABx+AB^2+C$$
Therefore:
$$A=a$$
$$b=2aB$$
$$c=aB^2+C$$
Therefore:
$$B=\frac{b}{2a}$$
$$C=-a(\frac{b}{2a})^2+c=-\frac{b^2}{4a}+c$$
Plug it all in:
$$ax^2+bx+c=a(x+\frac{b}{2a})^2+c-\frac{b^2}{4a}$$
Which is the complete square.
A: Method 1:
$y = -4x^2 -2x -4$
$\frac {y}{-4} = x^2 + \frac 12x + 1$
$-\frac {y}{4}- 1 = x^2 + 2*\frac 14 x$
$-1-\frac y4  +(\frac 14)^2= x^2 + 2*\frac 14 x+ (\frac 14)^2$
$-1+\frac 1{16}-\frac y4  = (x + \frac 14)^2$
$(x+  \frac 14)^2 = -\frac {15}{16} - \frac y4 = \frac {-15 - 4y}{16}$
$x + \frac 14 = \pm \sqrt {\frac {-15-4y}{16} }= \frac {\sqrt {-15 - 4y}}4$
$x = \frac {\sqrt {-15 - 4y}}4 -\frac 14 = \frac {\sqrt{-15 - 4y} - 1}4$
Method 2:
$y = -4x^2 - 2x -4 \iff y+4 = -4x^2 -2x$ 
We want $-4x^2 - 2x$ to be the first to summands of $(mx + n)^2 = m^2x + 2mnx + n^2$.  But $m^2$ is positive while $-4$ is negative.
So we multiple everything by $-1$.
$y+4 = -4x^2 -2x \iff -y - 4 = 4x^2 + 2x$.  Now we want $4 = m^2$ and $2 = 2mn$.
Or in other words $m = \pm \sqrt 4$ and $n = \frac 1m$. Or in other words $m = \pm 2$ and $n = \frac 12$.  
$-y -4 = 4x^2 + 2x = m^2x^2 + 2mnx \iff$
$-y-4 + n^2 = -y -4 + (\frac 12)^2= 4x^2 + 2x + (\frac 12)^2 = m^2x^2 + 2mnx +n^2 = (mx + n)^2 = (2x + \frac 12)^2$
$\iff -y -4 +\frac 14 = -y -\frac {15}4 = (2x + \frac 12)^2$
$\iff 2x +\frac 12 =\pm \sqrt {-y -\frac {15}{4}}$
$\iff 2x =-\frac 12 \pm \sqrt {-y-\frac {15}4}=-\frac 12 \pm \frac {\sqrt{-4y-15}}2 = \frac {-1 \pm\sqrt {-4y - 15}}2$
$\iff x = \frac {-1 \pm \sqrt{-4y -15}}4$ 
