Question 913239
Convert to vertex form,
{{{h(x)=2x^2+6x+25}}}
{{{h(x)=2(x^2+3x)+25}}}
{{{h(x)=2(x^2+3x+(3/2)^2)+25-2(3/2)^2)}}}
{{{h(x)=2(x+3/2)^2+50/2-9/2}}}
{{{h(x)=2(x+3/2)^2+41/2}}}
The vertex is ({{{-3/2}}},{{{41/2}}}).
The axis of symmetry is {{{x=-3/2}}}
The parabola opens upwards.
{{{drawing(300,300,-10,10,-10,10,blue(line(-3/2,50,-3/2,-50)),circle(-3/2,41/2,0.4),graph(300,300,-10,10,-5,45,2x^2+6x+25))}}}