Question 737927


{{{-2(x - 1)^2 = y + 5}}}...move {{{5}}} to the left

{{{-2(x -1)^2-5 = y }}}......now you have equation of parabola in vertex form {{{y=a(x-h)^2+k}}} where ({{{h=1}}},{{{k=-5}}}) are vertex coordinates, and coefficient {{{a=-2}}}
    If {{{a> 0}}}, the parabola opens upwards
    if {{{a< 0}}}, it opens downwards; since your {{{a=-2}}} =>  {{{a< 0}}} and parabola opens downwards


 to graph this function, you need several points to plot and draw a parabola through

table:

let first point be vertex

{{{x}}}|{{{y}}}
{{{1}}}|{{{-5}}}
{{{0}}}|{{{-7}}}
{{{2}}}|{{{-7}}}
{{{-2}}}|{{{-23}}}
{{{3}}}|{{{-13}}}


{{{drawing ( 600, 600, -10, 10, -30, 10,circle(1,-5,0.1),circle(0,-7,0.1),circle(2,-7,0.1),circle(-2,-23,0.1),circle(3,-13,0.1),graph( 600, 600, -10, 10, -30, 10,-2(x -1)^2-5)) }}}