Question 175623
If the general equation is
{{{y = ax^2 + bx + c}}}
The vertex is at {{{-(b/2a)}}}
1) {{{y = -(x-2)^2 + 1}}}
{{{y = -(x^2 - 4x + 4) + 1}}}
{{{y = -x^2 + 4x - 3}}}
{{{-(b/(2a)) = -(4/(2*(-1)))}}}
{{{4/2 = 2}}}
When {{{x=2}}},{{{y=1}}}, so the 
vertex is at (2,1)
It's a max because of the - sign 
in front of (x-2)^2
Graphing the 2 equations:
{{{ graph( 600, 600, -10, 10, -5, 10,-x^2 + 4x - 3,-x^2 + 7) }}}