Question 1046278
No restriction necessary on x, so the domain is ____________________________.
(This should be obvious).


Range?  Parabola concave downward, so vertex is a maximum.  Easiest to see if try to solve for the zeros and check the middle value between them.
{{{-2x^2+3=0}}}
{{{2x^2-3=0}}}
{{{2x^2=3}}}
{{{x^2=3/2}}}
{{{x=0+- sqrt(3)sqrt(2)/2}}}
{{{x=0+- sqrt(6)/2}}}
and the exact middle of these is 0.  This will mean that maximum value of the expression {{{-2x^2+3}}} occurs at x=0, and the range will be all values less than or equal to {{{-2*0^2+3}}}, or {{{highlight(y<=3)}}}.