## Precalculus (6th Edition) Blitzer

Step I: To check even or odd. In the provided equation, if $h\left( -x \right)=h\left( x \right)$ then it is an even function, and if $h\left( -x \right)=-h\left( x \right)$ , then it is odd. \begin{align} & h\left( x \right)={{x}^{2}}+2{{x}^{4}} \\ & h\left( -x \right)={{\left( -x \right)}^{2}}+2{{\left( -x \right)}^{4}} \\ & ={{x}^{2}}+2{{x}^{4}} \\ & =h\left( x \right) \end{align} It is an even 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} & h\left( x \right)={{x}^{2}}+2{{x}^{4}} \\ & h\left( -x \right)={{\left( -x \right)}^{2}}+2{{\left( -x \right)}^{4}} \\ & ={{x}^{2}}+2{{x}^{4}} \\ & =h\left( x \right) \end{align} It is the same as the given equation, hence it has symmetry 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} & h\left( x \right)={{x}^{2}}+2{{x}^{4}} \\ & y={{x}^{2}}+2{{x}^{4}} \\ & -y={{x}^{2}}+2{{x}^{4}} \\ & y=-{{x}^{2}}-2{{x}^{4}} \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}}+2{{x}^{4}} \\ & \left( -y \right)={{\left( -x \right)}^{2}}+2{{\left( -x \right)}^{4}} \\ & -y={{x}^{2}}+2{{x}^{4}} \\ & y=-{{x}^{2}}-2{{x}^{4}} \end{align} It is not the same as provided equation, hence it is not symmetric about the origin. Therefore, the provided function is even, has symmetry about the y-axis only.