91Ó°ÊÓ

The given points are $(2,0),(4,0),\text{and}(0,1)$.

Use the factor theorem, is follows:

$f\left(x\right)=\left(x-2\right)\left(x-4\right)$

Add a constant $k$, we get

$f\left(x\right)=\left(x-2\right)\left(x-4\right)+k$

Substitute $\left(0,1\right)$for $\left(x,y\right)$in the above function.

$\begin{array}{rcl}1& =& \left(0-2\right)\left(0-4\right)+k\\ k& =& -7\end{array}$

Substitute $-7$for $k$in $f\left(x\right)=\left(x-2\right)\left(x-4\right)+k$.

$\begin{array}{rcl}f\left(x\right)& =& \left(x-2\right)\left(x-4\right)+-7\\ & =& x^2-6x+1\end{array}$

Perform the test for symmetry with respect to the $x$-axis.

Replace $y$with $-y$.

$\begin{array}{rcl}-y& =& x^2-6x+1\\ y& =& -x^2+6x-1\end{array}$

The above equation is not same as the given equation.

Perform the test for symmetry with respect to the $y$-axis.

Replace $x$with $-x$.

$\begin{array}{rcl}y& =& {\left(-x\right)}^2-6\left(-x\right)+1\\ & =& x^2+6x+1\end{array}$

The above equation is not same as the given equation.

Perform the test for symmetry with respect to the origin.

Replace $x$with $-x$and $y$with $-y$.

$\begin{array}{rcl}-y& =& {\left(-x\right)}^2-6\left(-x\right)+1\\ -y& =& x^2+6x+1\\ y& =& -x^2-6x-1\end{array}$

The above equation is not same as the given equation.

Therefore, the equation is neither symmetry about the$x$axis nor$y$-axis.

The equation is $y=x^2-6x+1$. It is neither symmetric to $x$-axis nor$y$-axis.