Answer
The given function is an even function and has symmetry about the y-axis only.
Work Step by Step
Step I:
To check if even or odd.
In the given equation, if $f\left( -x \right)=f\left( x \right)$ , then it is an even function, and if $f\left( -x \right)=-f\left( x \right)$ , then it is odd.
$\begin{align}
& f\left( x \right)={{x}^{4}}-2{{x}^{2}}+1 \\
& f\left( -x \right)={{\left( -x \right)}^{4}}-2{{\left( -x \right)}^{2}}+1 \\
& ={{x}^{4}}-2{{x}^{2}}+1 \\
& =f\left( x \right).
\end{align}$
It is an even function.
Step II: 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.
$\begin{align}
& f\left( x \right)={{x}^{4}}-2{{x}^{2}}+1 \\
& f\left( -x \right)={{\left( -x \right)}^{4}}-2{{\left( -x \right)}^{2}}+1 \\
& ={{x}^{4}}-2{{x}^{2}}+1
\end{align}$
It is the same as the provided equation above, hence it has symmetry about the y-axis.
Step III:
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.
$\begin{align}
& y={{x}^{4}}-2{{x}^{2}}+1 \\
& \left( -y \right)={{x}^{4}}-2{{x}^{2}}+1 \\
& -y={{x}^{4}}-2{{x}^{2}}+1 \\
& y=-\left( {{x}^{4}}-2{{x}^{2}}+1 \right)
\end{align}$
It is not the same as the provided equation above, hence it is not symmetric about the x-axis.
Step IV:
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.
$\begin{align}
& y={{x}^{4}}-2{{x}^{2}}+1 \\
& \left( -y \right)={{\left( -x \right)}^{4}}-2{{\left( -x \right)}^{2}}+1 \\
& -y={{x}^{4}}-2{{x}^{2}}+1 \\
& y=-\left( {{x}^{4}}-2{{x}^{2}}+1 \right)
\end{align}$
It is not the same as the provided equation above, hence it has no symmetry about the origin.
Therefore, the given function is an even function and has symmetry about the y-axis only.
