```Question 549395
The quadratic function, or equation of a parabola (whichever way you want to see it)
{{{f(x)=-3x^2-60x-294}}} or {{{y=-3x^2-60x-294}}}
can be transformed into the vertex form in steps, using facts of algebra you learned before.
You can also calculate the coordinates of the vertex from formulas in your textbook, and then plug those coordinates into another formula from your textbook to get the vertex form.
The steps, go like this:
I want all the terms with x, and only the terms with x on the right side.
{{{y=-3x^2-60x-294}}} --> {{{y+294=-3x^2-60x}}} (I added 294 to each side)
I don't want a coefficient, or factor, or minus sign, directly in front of {{{x^2}}}
{{{y+294=-3x^2-60x}}} --> {{{y+294=-3(x^2+20x)}}}, extracting common factor (-3).
At this point, it's time to "complete the square".
The right side parenthesis reminds me of the square of {{{(x-10)}}}:
{{{(x-10)^2=x^2-20x+100}}}
so I decide to add to both sides
{{{-300=-3*100}}} or, if you prefer,
I'll subtract {{{300=3*100}}} from both sides to get
{{{y+294-300=-3(x^2+20x)-3*100}}} --> {{{y+294-300=-3(x^2+20x+100)}}}
Now I can write what's in parentheses as the square I know it is
{{{y-6=-3(x+10)^2}}} (I also simplified the left hand side, at the same time).
NOTE: (That is not yet what most call the vertex form, but it shows that it is the parabola that results from modifying {{{y=x^2}}} by flipping it upside down, stretching it vertically to 3 times its size, shifting it to the left 10 units and shifting it up 6 units).
LAST STEP: Add 6 to both sides of the equation.
{{{y-6=-3(x+10)^2}}} ---> {{{y=-3(x+10)^2+6}}}```