Question 1155967


 {{{f(x) = -3x^2 + 6x-1}}}....complete the square 

{{{f(x) = (-3x^2 +6x)-1}}}

{{{f(x) = -3(x^2-2x)-1}}}

{{{f(x) = -3(x^2-2x+b^2)-(-3)b^2-1}}}

{{{f(x) = -3(x^2-2x+1^2)+3*1^2-1}}}

{{{f(x) = -3(x-1)^2+3-1}}}

{{{f(x) = -3(x-1)^2+2}}}

=> {{{h=1}}} and {{{k=2}}}

=> vertex is at  ({{{1}}},{{{2}}})


{{{drawing( 600,600, -10, 10, -10, 10,
circle(1,2,.12), locate(1,2,V(1,2)),
 graph( 600,600, -10, 10, -10, 10, -3(x-1)^2+2)) }}}