Answer
$x$-intercept: $-4$
$y$-intercepts: $-2$ and $2$
symmetric with the $x$-axis only
Work Step by Step
Step $1$. For $y^2=x+4$, to get the $x$-intercept/s, let $y=0$ then solve for $x$:
$$\begin{align*}
0^2&=x+4\\
0&=x+4\\
-4&=x
\end{align*}$$
Thus, the $x$-intercept is $-4$.
Step $2$. To get the $y$-intercept, let $x=0$ then solve for $y$.
$$\begin{align*}
y^2&=0+4\\
y^2&=4\\
y&\pm2
\end{align*}$$
Thus, the $y$-intercepts are $-2$ and $2$.
Step $3$. Check for symmetry with respect to the $x$-axis by replacing $(x, y)$ with $(x,-y)$:
$$\begin{align*}
(-y)^2&=x+4\\
y^2&=x+4
\end{align*}$$
Since the resulting equation is the same as the original, then the graph of the equation is symmetric with respect to the $x$-axis.
Check for symmetry with respect to the $y$-axis by replacing $(x, y)$ with $(-x,y)$:
$$\begin{align*}
y^2&=-x+4\\
\end{align*}$$
Since the resulting equation is not the same as the original, then the graph of the equation is NOT symmetric with respect to the $y$-axis.
Check for symmetry with respect to the origin by replacing $(x, y)$ with $(-x,-y)$:
$$\begin{align*}
(-y)^2&=-x+4\\
y^2&=-x+4
\end{align*}$$
Since the resulting equation is not the same as the original, then the graph of the equation is NOT symmetric with respect to the origin.
