Answer
$x$-intercepts: $\quad-2$ and $2$
$y$-intercept: $\quad-3$;
symmetric with the $y$-axis
Work Step by Step
Step $1$. To find the $x$-intercept, set $y=0$ then solve for $x$ to obtain:
$$\begin{align*}
3x^2-4y&=12\\
3x^2-4(0)&=12\\
3x^2&=12\\
x^2&=4\\
x&=\pm\sqrt{4}\\
x&=\pm2
\end{align*}$$
Thus, the $x$-intercepts are $\pm2$
Step $2$. To find the $y$-intercept, set $x=0$ then solve for $y$ to obtain:
$$\begin{align*}
3x^2-4y&=12\\
3(0^2)-4y&=12\\
-4y&=12\\
y&=-3
\end{align*}$$
Thus, the $y$-intercept is $-3$.
Step $3$. To test for symmetry, replace $(x,y)$ with $(x,-y)$ (x-axis symmetry), or $(-x,y)$ (y-axis symmetry), or $(-x,-y)$ (origin symmetry), and see if the equation remain unchanged.
Replace $(x, y)$ with $(x, -y)$ to obtain:
$\begin{align*}
3x^2-4(-y)&=12\\
3x^2+4y&=12
\end{align*}$
The equation is different from the original so there is no symmetry with respect to $x$-axis.
Replace $(x, y)$ with $(-x, y)$ to obtain:
$\begin{align*}
3(-x)^2-4y&=12\\
3x^2-4y&=12
\end{align*}$
The equation is the same as the original so there is symmetry with respect to $y$-axis.
Replace $(x, y)$ with $(-x, -y)$ to obtain:
$\begin{align*}
3(-x)^2-4(-y)&=12\\
3x^2+4y&=12
\end{align*}$
The equation is different from the original so there is no symmetry with resspect to the origin.
We can see that the equation has only $y$-axis symmetry.
