Question 744398
The vertex form of the equation of a parabola is
{{{y=a(x-h)^2+k}}} with a coefficient {{{a}}} and using the coordinates (h,k) of the vertex.
 
A SLOW AND EASY WAY:
If you expand the square in {{{y=a(x-h)^2+k}}} you get
{{{y=a(x^2-2hx+h^2)+k}}} --> {{{y=a^x^2-2ahx+ah^2+k}}}
If you compare that to {{{y=-3x^2+12x-8}}} you realize that:
the coefficient of {{{x^2}}} is {{{a=-3}}}, and
the coefficient of {{{x}}} is {{{-2ah=12}}}
That gives you the values of {{{a}}} and {{{h}}}:
{{{a=-3}}} was clear, and substituting that value in {{{-2ah=12}}} you get
{{{-2*(-3)*h=12}}} --> {{{6h=12}}} --> {{{h=12/6}}} --> {{{h=2}}}
Plugging those values into
{{{y=a(x-h)^2+k}}} you get
{{{y=-3(x-2)^2+k}}} --> {{{y=-3(x-2)^2+k}}} --> {{{y=-3(x^2-4x+4)+k}}} --> {{{y=-3x^2+12x-12+k}}}
Comparing that last equation to {{{y=-3x^2+12x-8}}}, you realize that the independent term (the part without any {{{x}}} ) is
{{{-12+k=-8}}} --> {{{k=-8+12}}} --> {{{k=4}}}
Plugging the values found for {{{a}}}, {{{h}}}, and {{{k}}} into {{{y=a(x-h)^2+k}}} you get
{{{highlight(y=-3(x-2)^2+4)}}}.
 
THE FAST AND STEEP WAY that makes you look so smart:
{{{y=-3x^2+12x-8}}} --> {{{y=-3(x^2-4x)-8}}} (taking out (-3) as common factor from {{{-3x^2+12x}}}
{{{y=-3(x^2-4x)-8}}} --> {{{y=-3(x^2-4x+4)+3*4-8}}} (adding {{{3*4}}} and {{{-3(4)}}} which cancel out)
{{{y=-3(x^2-4x+4)+3*4-8}}} --> {{{y=-3(x^2-4x+4)+12-8}}} --> {{{y=-3(x^2-4x+4)+4}}} --> {{{highlight(y=-3(x-2)^2+4)}}}.
 
THE MEMORIZED FORMULA WAY:
{{{y=ax^2+bx+c}}} --> {{{y=a(x+b/2a)^2+(4ac-b^2)/4a}}}
For {{{y=-3x^2+12x-8}}}, {{{a=-3}}}, {{{b=12}}} and {{{c=-8}}} so
{{{b/2a=12/(2*(-3))=12/(-6)=-2}}} and
{{{(4ac-b^2)/4a=(4*(-3)*(-8)-12^2)/(4(-3))=(96-144)/(-12)=-148/(-12)=4}}} so
the vertex form is {{{highlight(y=-3(x-2)^2+4)}}}.
 
NOTE:
If your teacher favors the fast and steep way, show your teacher that way (even if you figured it out a different way on a separate piece of paper).