## Precalculus (6th Edition) Blitzer

Step I: To check even or odd. In the provided equation, if $g\left( -x \right)=g\left( x \right)$ then it is an even function, and if $g\left( -x \right)=-g\left( x \right)$ , then it is odd. \begin{align} & g\left( x \right)={{x}^{2}}+x \\ & g\left( -x \right)={{\left( -x \right)}^{2}}+\left( -x \right) \\ & ={{x}^{2}}-x \end{align} It is neither an even nor odd function. Step II: To check symmetry about the y- axis: In the provided equation put $x=-x$ , if the equation remains the same, then it has symmetry about the y-axis. \begin{align} & y={{x}^{2}}+x \\ & y={{\left( -x \right)}^{2}}+\left( -x \right) \\ & y={{x}^{2}}-x \\ \end{align} It is not the same as the given equation, hence it is not symmetric about the y-axis. Step III: To check symmetry about the x- axis: In the provided equation put $y=-y$ , if the equation remains the same, then it has symmetry about the x- axis. \begin{align} & y={{x}^{2}}+x \\ & \left( -y \right)={{x}^{2}}+x \\ & -y={{x}^{2}}+x \end{align} It is not the same as the given equation, hence it is not symmetric about the x-axis. Step IV: To check symmetry about the origin: In the provided equation put $x=-x,\text{ and }y=-y$ , if the equation remains the same, then it has symmetry about the origin. \begin{align} & y={{x}^{2}}+x \\ & \left( -y \right)={{\left( -x \right)}^{2}}+\left( -x \right) \\ & -y={{x}^{2}}-x \\ & y=-{{x}^{2}}+x \end{align} It is not the same as the given equation, hence it is not symmetric about the origin. Therefore, the provided function is neither an even nor odd function.