Question 740130
The standard form of a parabola's equation is generally expressed:

    {{{y = ax^2 + bx + c}}}
        The role of '{{{a}}}'
            If {{{a> 0}}}, the parabola opens upwards
            if {{{a< 0}}}, it opens downwards.
        The axis of symmetry
            The axis of symmetry is the line {{{x = -b/2a}}} 



to  put your parabola into standard form, solve for {{{y}}}


{{{x^2-4x-8y+2=0 }}}


{{{x^2-4x+2=8y }}}


{{{(1/8)x^2-(4/8)x+2/8=y }}}


{{{y=(1/8)x^2-(1/2)x+1/4 }}}


{{{a=1/8}}}=> {{{a> 0}}}, the parabola opens upwards


The vertex form of a parabola's equation is generally expressed as :

{{{y= a(x-h)^2+k}}}

    ({{{h}}},{{{k}}}) is the {{{vertex}}}

in vertex form will be:

{{{y= a(x-h)^2+k}}}

{{{y=((1/8)x^2-(1/2)x+______)+1/4 }}}....complete the square


{{{y=(1/8)(x-2)^2-1/4}}}

({{{h=2}}},{{{k=-1/4}}}) is the {{{vertex}}}


{{{ graph( 600, 600, -5, 10, -5, 10,(1/8)x^2-(1/2)x+1/4) }}}