Question 785270
In the graph of {{{y=3+x^2}}} all the points are above the x-axis {{{y>=3>0}}}, so it could not be symmetric with respect to the x- axis, or the origin.
Since {{{y=3+x^2=3+(-x)^2}}}, for each point (x,y) in the graph there is a corresponding (-x,y) point at the same distance to the other side of the y-axis, so it is symmetric with respect to the y-axis.
To be symmetric with respect to the y- axis, changing {{{x}}} to {{{-x}}} should yield an equivalent equation, and that is true for {{{y=3+x^2}}}.


To be symmetric with respect to the x- axis, changing {{{y}}} to {{{-y}}} should yield an equivalent equation.
In this case, if {{{y=3+x^2}}} is true, {{{-y=3+x^2}}} is definitely not true.


To be symmetric with respect to the origin, replacing {{{-y}}} for {{{y}}} AND {{{-x}}} for {{{x}}} AT THE SAME TIME should yield an equivalent equation.
In this case, if {{{y=3+x^2}}} is true, {{{-y=3+(-x)^2}}}<-->{{{-y=3+x^2}}} is definitely not true.