The general vertex form is
color(white)("XXX")y=color(green)m(x-color(red)a)^2+color(blue)b
with vertex at (color(red)a,color(blue)b)
Given
color(white)("XXX")y=2x^2+8x+4
Extract the color(green)m factor (in this case color(green)2 from the first two terms
color(white)("XXX")y=color(green)2(x^2+4x)+4
Supposing that the (x^2+4x) are part of a squared binomial (x-color(red)a)^2=(x^2-2color(red)ax+color(red)a^2)
then -2color(red)ax must be equal to 4x
rarr color(red)a=4/(-2)=color(red)(-2)
and in order to "complete the square" an extra color(red)a^2=(-2)^2=color(magenta)4 will need to be added inside the brackets to the (x^2+4x) we already have.
Note that if we insert this color(magenta)(+4) because of the factor color(green)m=color(green)2
we will really be adding color(green)2 xx color(magenta)4=color(brown)8 to the expression.
To maintain equality if we add color(brown)8 we will also need to subtract it:
color(white)("XXX")y=color(green)2(x^2+4x+color(magenta)4)+4color(brown)(-8)
Simplifying
color(white)("XXX")y=color(green)2(x+2)^2-4
Adjusting to match the sign requirements of the vertex form:
color(white)("XXX")y=color(green)2(x-color(red)(""(-2)))^2+color(blue)(""(-4))
The graph below of the original equation supports this result.
graph{2x^2+8x+4 [-8.386, 2.71, -5.243, 0.304]}
