Answer
The $x-$intercepts are $3,\,-3$, and the $y-$intercept is $9$.
The graph of the equation ${{x}^{2}}+y=9$ is symmetric about the $y-$axis.
Work Step by Step
To determine the $x$-intercepts, substitute $y=0$ in the equation ${{x}^{2}}+y=9$, and solve for $x$.
By substituting $y=0$ in ${{x}^{2}}+y=9$,
${{x}^{2}}+0=9$.
By simplifying,
$\Rightarrow {{x}^{2}}=9$.
By taking the square root,
$\Rightarrow x=\pm \,3$.
Therefore, the $x$- intercepts are $3$ and $-3$.
To determine the $y$-intercepts, substitute $x=0$ in the equation ${{x}^{2}}+y=9$, and solve for $y$.
By substituting $x=0$ in ${{x}^{2}}+y=9$,
${{\left( 0 \right)}^{2}}+y=9$.
By simplifying,
$\Rightarrow 0+y=9$.
$\Rightarrow y=9$.
Therefore, the $y$-intercept is $9$.
Now, test for symmetry:
For the $x-$axis: Replace $y$ with $-y$ in the equation ${{x}^{2}}+y=9$. It gives ${{x}^{2}}-y=9$, which is not equal to ${{x}^{2}}+y=9$.
For the $y-$axis: Replace $x$ with $-x$ in the equation ${{x}^{2}}+y=9$. It gives ${{\left( -x \right)}^{2}}+y=9$, which is equivalent to ${{x}^{2}}+y=9$. So, the graph of the equation ${{x}^{2}}+y=9$ is symmetric with respect to the $y-$axis.
For the origin: Replace $x$ with $-x$ and $y$ with $-y$ in the equation ${{x}^{2}}+y=9$. It gives ${{\left( -x \right)}^{2}}+\left( -y \right)=9$, which is not equal to ${{x}^{2}}+y=9$.
Hence, the graph of the equation ${{x}^{2}}+y=9$ is symmetric with respect to the $y-$axis.
