## Precalculus (6th Edition) Blitzer

The graph of the equation ${{x}^{3}}-{{y}^{2}}=5$ symmetric to the x-axis only.
Consider the equation, \begin{align} & {{x}^{3}}-{{y}^{2}}=5 \\ & {{x}^{2}}={{y}^{2}}+5 \\ \end{align} Check symmetry about the y-axis: An equation is symmetric about the y-axis if $-x$ is substituted in the function and the result is an equivalent equation then the graph of the equation is symmetric with respect to the y-axis. Substitute $x=-x$ in the equation, \begin{align} & {{x}^{3}}={{y}^{2}}+5 \\ & {{\left( -x \right)}^{3}}={{y}^{2}}+5 \\ & -{{x}^{3}}={{y}^{2}}+5 \end{align} Therefore, the equation is not symmetric about the y-axis Now, check symmetry about the x-axis: An equation is symmetric about the x-axis, if $-y$ is substituted in the function and it leads to an equivalent equation than the graph is symmetric with respect to the x-axis. Substitute $y=-y$ , \begin{align} & {{x}^{3}}={{\left( -y \right)}^{2}}+5 \\ & ={{y}^{2}}+5 \end{align} Therefore, the equation is symmetric about the x-axis. Now, check symmetry about the origin: An equation is symmetric about origin if $x=-x,y=-y$ and it leads to an equivalent equation, this implies that the function is symmetric about the origin. Substitute $x=-x,y=-y$ \begin{align} & {{\left( -x \right)}^{3}}={{\left( -y \right)}^{2}}+5 \\ & -{{x}^{3}}={{y}^{2}}+5 \end{align} Therefore, the equation is not symmetric about the origin Hence, the equation is symmetric about the x-axis.