1 #-# If a graph represented by #f(x,y)=0# is symmetric with respect to #x#-axis, we should have #f(x,y)=f(x,-y)#.
Here, in #6x^2=y-1#, we have #f(x,y)=6x^2-y+1=0# and #f(x,-y)=6x^2-(-y)+1=6x^2+y+1# and hence
#f(x,y)!=f(x,-y)# and hence it is **not symmetric w.r.t. #x#-axis.
2 #-# If a graph represented by #f(x,y)=0# is symmetric with respect to #y#-axis, we should have #f(x,y)=f(-x,y)#.
In #6x^2=y-1#, we have #f(-x,y)=6(-x)^2-y+1=6x^2-y+1# and hence
#f(x,y)=f(-x,y)# and hence it is symmetric w.r.t. #y#-axis.
3 #-# If a graph represented by #f(x,y)=0# is symmetric with respect to line #y=x#, we should have #f(x,y)=f(y,x)#.
In #6x^2=y-1#, we have #f(y,x)=6y^2-x+1# and hence
#f(x,y)!=f(y,x)# and hence it is not symmetric w.r.t. line #y=x#.
4 #-# If a graph represented by #f(x,y)=0# is symmetric with respect to line #y=-x#, we should have #f(x,y)=f(-y,-x)#.
In #6x^2=y-1#, we have #f(-y,-x)=6(-y)^2-(-x)+1=6y^2+x+1# and hence
#f(x,y)!=f(-y,-x)# and hence it is not symmetric w.r.t. line #y=-x#.
graph{(6x^2-y+1)(x-y)(x+y)=0 [-5.394, 4.606, -0.64, 4.36]}
