Question 1157009
{{{f(x)= -(1/3)(x-1)^2-2}}}

compare to {{{F(x)= a(x-h)^2+k}}}

from given formula you see that {{{h=1}}} and {{{k=-2}}}, so vertex is at ({{{1}}},{{{-2}}})
the {{{vertical}}} line that goes through the vertex
{{{x= 1}}}

directrix:

{{{y= -(1/3)(x-1)^2-2}}}
{{{y+2= -(1/3)(x-1)^2}}}
{{{(y+2)/(-1/3)= (x-1)^2}}}
{{{-3(y+2))= (x-1)^2}}}.........compare to {{{4p(y-k)=(h-h)^2}}}
you see that {{{4p=-3}}}=>{{{p=-3/4}}}

{{{y=-2-(-3/4)}}}
{{{y=-8/4+3/4}}}
{{{y=-5/4}}}

table:

{{{x}}}|{{{y}}}
{{{1}}}|{{{-2}}}
{{{-1}}}|{{{-10/3}}}...{{{y= -(1/3)(-1-1)^2-2=-10/3}}}
{{{2}}}|{{{-7/3}}}...{{{y= -(1/3)(2-1)^2-2=-7/3}}}
{{{0}}}|{{{-7/3}}}...{{{y= -(1/3)(0-1)^2-2=-7/3}}}
{{{-2}}}|{{{-5}}}...{{{y= -(1/3)(-2-1)^2-2=-5}}}

{{{drawing( 600, 600, -10, 10, -10, 10, 
circle(1,-2,.12), locate(1,-1.5,v(1,-2)), 
circle(-1,-10/3,.12), locate(-2,-10/3,p(-1,-10/3)), 
circle(2,-7/3,.12), locate(2,-2,p(2,-7/3)),
 circle(0,-7/3,.12), locate(-0.9,-2,p(0,-7/3)), 
circle(-2,-5,.12), locate(-2,-5,p(-2,-5)), 
green(line(1,10,1,-10)),
 graph( 600, 600, -10, 10, -10, 10, -(1/3)(x-1)^2-2)) }}}