Question 372620
The parabola is already in vertex form,{{{x=c(y-h)^2+d}}}
{{{x=2y^2-3}}}
{{{x=2(y-0)^2-3}}}
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Vertex:({{{-3}}},{{{0}}})
The vertex lies on the axis of symmetry, {{{y=0}}}
The parabola opens to the right.{{{2>0}}}
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{{{drawing(300,300,-5,9,-7,7,grid(1),circle(-3,0,0.2), graph(300,300,-5,9,-7,7,0,sqrt((x+3)/2)),graph(300,300,-5,9,-7,7,0,-sqrt((x+3)/2)))}}}