For the equation$3x^2-4y=12$, find the intercepts and check for symmetry.

Consider the given question,

$3x^2-4y=12$

Substitute $y=0$in the given equation,

localid="1645886422729" $$\begin{gathered} 3x^2-4\left(0\right)=12 \\ {x}^2=\frac{12}{3} \\ {x}^2=4 \\ x=\pm 2 \end{gathered}$$

Therefore, the x-intercepts are $\left(2,0\right),\left(-2,0\right)$.

Substitute $x=0$in the given equation,

$$\begin{gathered} 3{\left(0\right)}^2-4y=12 \\ -4y=12 \\ y=-\frac{12}{4} \\ y=-3 \end{gathered}$$

Therefore, the y-intercept is$\left(0,-3\right)$.

Replace y with $-y$ in the given equation,

$$\begin{gathered} 3x^2-4\left(-y\right)=12 \\ 3x^2+4y=12 \end{gathered}$$

As this equation is not equivalent to the original equation, then the graph is not symmetric with respect to the x-axis.

Replace xwith $-x$in the given equation,

$$\begin{gathered} 3{\left(-x\right)}^2-4y=12 \\ 3x^2-4y=12 \end{gathered}$$

As this equation is equivalent to the original equation, then the graph is not symmetric with respect to the y-axis.

Replace xwith $-x$ and y with $-y$in the given equation,

$$\begin{gathered} 3{\left(-x\right)}^2-4\left(-y\right)=12 \\ 3x^2+4y=12 \end{gathered}$$

As this equation is not equivalent to the original equation, then the graph is not symmetric with respect to the origin.