## Precalculus (6th Edition) Blitzer

Step I: To check symmetry about the y-axis: Putting $x=-x$ in the given equation, if the equation remains the same, then it has symmetry about the y-axis. \begin{align} & {{x}^{3}}-{{y}^{2}}=5 \\ & {{\left( -x \right)}^{3}}-{{y}^{2}}=5 \\ & -{{x}^{3}}-{{y}^{2}}=5 \end{align} It is not the same as the provided equation, hence it is not symmetric about the y-axis. Step II: To check symmetry about the x-axis: Putting $y=-y$ in the given equation, if the equation remains the same, then it has symmetry about the x-axis. \begin{align} & {{x}^{3}}-{{y}^{2}}=5 \\ & {{x}^{3}}-{{\left( -y \right)}^{2}}=5 \\ & {{x}^{3}}-{{y}^{2}}=5 \end{align} It is the same as the provided equation, hence it has symmetry about the x-axis. Step III: To check symmetry about the origin: Putting $x=-x\text{ and }y=-y$ in the given equation, if the equation remains the same, then it has symmetry about the origin. \begin{align} & {{x}^{3}}-{{y}^{2}}=5 \\ & {{\left( -x \right)}^{3}}-{{\left( -y \right)}^{2}}=5 \\ & -{{x}^{3}}-{{y}^{2}}=5 \end{align} It is not the same as the provided equation, hence it is not symmetric about the origin. Therefore, the given equation has symmetry about the x-axis only.