Answer
A) (i) The x axis (a, -b) This means that you take the equation and subistitute all y with (-) if after solving it, is the same as the original it means it is symmetric.
(ii) y axis (-a,b) this means that you take the equation and subtitute all x with (-). If after solving it, is the same as the original it means it is symmetric.
(iii) origin (-a,-b) this means that you take the equation and substitite all x and y by (-). If after solving it, is the same as the original it means it is symmetric
B) x-axis symmetry.
Work Step by Step
A) How do you test whether the graph pf an equation is symmetric with respect to the (i) x axis (ii) y axs and (iii) origin?
(i) The x axis (a, -b) This means that you take the equation and subistitute all y with (-) if after solving it, is the same as the original it means it is symmetric.
(ii) y axis (-a,b) this means that you take the equation and subtitute all x with (-). If after solving it, is the same as the original it means it is symmetric.
(iii) origin (-a,-b) this means that you take the equation and substitite all x and y by (-). If after solving it, is the same as the original it means it is symmetric.
B) What type of symmetry does the graph of the equation $xy^{2}+y^{2}x^{2}=3x$ have?
X axis $$x(-y)^{2}+(-y)^{2}x^{2}=3x$$ $$xy^{2}+y^{2}x^{2}=3x$$
It is symmetric respecting x axis.
Y axis $$(-x)y^{2}+y^{2}(-x)^{2}=3(-x)$$ $$-xy^{2}+y^{2}x^{2}=-3x$$
No symmetry.
Origin $$(-x)(-y)^{2}+(-y)^{2}(-x)^{2}=3(-x)$$ $$(-x)y^{2}+y^{2}x^{2}=-3x$$
No symmetry.