Question 1157011

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

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

from given formula you see that {{{h=-1}}} and {{{k=-3}}}, so vertex is at ({{{-1}}},{{{-3}}})
the {{{vertical}}} line that goes through the vertex
{{{x= -1}}}

directrix:
{{{y= 2(x+1)^2-3}}}
{{{(1/2)(y+3)= (x+1)^2}}}...since {{{4p=1/2}}}=>{{{p=1/8}}}

{{{y=-3-p}}}
{{{y=-3-1/8}}}
{{{y=-25/8}}}

table:

{{{x}}}|{{{y}}}
{{{-1}}}|{{{-3}}}
{{{1}}}|{{{5}}}
{{{-3}}}|{{{5}}}
{{{0}}}|{{{-1}}}
{{{-2}}}|{{{-1}}}

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