<PRE><B>Question 453215</B>
The general form of a parabola is given by the equation:

  {{{y=Ax^2 + Bx + C  }}}where {{{A}}}, {{{B}}}, and {{{C}}} are arbitrary {{{real }}}{{{constants}}}.

The {{{vertex}}} is  at: ({{{ -B/2A }}}, {{{C - B^2/4A}}} )

you are given:

{{{y = -x^2 + 5x - 6}}} which is already in general form of a parabola

so, {{{A=-1}}}, {{{B=5}}}, and {{{C=-6}}}

now find the coordinates of the vertex:


({{{ -B/2A }}}, {{{C - B^2/4A}}} )


{{{ -B/2A=-5/2(-1)=-5/-2=5/2=2.5 }}}, 

{{{C - B^2/4A=(-6)-25/4(-1)=-6+25/4=-6+6.25=0.25}}} 

so, the vertex is at: ({{{2.5}}},{{{0.25}}})


{{{ graph( 500, 500, -5, 5, -5, 5, -x^2 + 5x - 6) }}}


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