Question 23313
There are several ways to find the vertex of this.  I think the easiest way is to use a formula for the vertex that comes from the quadratic formula.  Remember that?   {{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}   Well, this formula is just the first part of this without the radical.  It turns out the vertex of a parabola {{{y = ax^2 +bx +c}}} (which opens up or down!) is always at {{{x=-b/(2a) }}}


In your case, {{{y = -3x^2 + 6x}}}, a=-3 and b=6, so {{{x= -6/(2(-3)) = 1}}}.


Now, if x=1, then
{{{y = -3x^2 + 6x}}}
{{{y=-3 +6= 3}}}


Vertex is at (1,3).


Check it out with the graph:
{{{graph (500,500, -10,10,-10,10, -3x^2 + 6x) }}}

Does this graph look like the vertex is at (1,3), and does the graph open down?


R^2 at SCC.