Parabolic Representation The points $(-1/2, 0)$, $\left(0, \frac{\sqrt3}{2}\right)$, $(1/2, 0)$ are the vertices of an equilateral triangle. A parabolic equation that contains the three points is 
$$y=
(-2 \sqrt3)x^2+\frac{\sqrt{3}}{2} \,.$$ How can the other two parabolas, that are congruent to the first parabola and contain the same three points, be represented? Can this be done with parametric equations or do I need a higher level of understanding?   
 A: Template for rotated formula
Suppose you take your parabola and rotate it by $\varphi$ and translate it by $(a,b)$. That means a point $(x,y)$ lies on your new parabola if its preimage $$\begin{pmatrix}\cos\varphi(x-a)+\sin\varphi(y-b)\\-\sin\varphi(x-a)+\cos\varphi(y-b)\end{pmatrix}$$ lies on the original parabola. In other words if
$$-\sin\varphi(x-a)+\cos\varphi(y-b)=(-2\sqrt3)
\bigl(\cos\varphi(x-a)+\sin\varphi(y-b)\bigr)^2+\frac{\sqrt3}2$$
Special case
In this special case, with the three points forming a equilateral triangle, you know that $\varphi=\pm120°$ for reasons of symmetry. You also know that the center of rotation is the center of the triangle, at $(0,1/\sqrt6)$. So you can either compute $a$ and $b$ from that, or you do the transformation in three steps: one translation to move the center of rotation to the origin, then the rotation, and then the translation back to the original position.
General case
But in order to make this answer more widely applicable, I'll also consider the more generic case where your points are not in such a special location, so that you don't know the rotation angles up front.
In that case you can plug your three points into the generic rotated equation above, and try to find $\varphi,a,b$ from that. But the trigonometric functions make this a bit ugly, so I'd suggest applying the tangent half-angle substitution first.
$$\bigl(-2t(x-a)+(1-t^2)(y-b)\bigr)(1+t^2)=(-2\sqrt3)
\bigl((1-t^2)(x-a)+2t(y-b)\bigr)^2+\frac{\sqrt3}2(1+t^2)^2$$
Now insert the three points you want to have on your parabola, and feed the resulting polynomial equations to the computer algebra system of your choice to get three solutions. One is the original parabola with $a=b=t=0$, while the other two have $a=\pm\tfrac14,b=\tfrac14\sqrt3,t=\pm\sqrt3$. This leads to the following equation for the parabola:
$$-8\sqrt3x^2 \pm48xy -24\sqrt3y^2 + 32y + 2\sqrt3 = 0$$
You can also undo the tangent half-angle substitution of $2\arctan(t)=\pm120°$ to see that this is a rotation by that angle, the thing I took for granted in the special case section above.
Parametric description
If you want a parametric description of your curve, use the original $x$ coordinate as a parameter (which I call $u$ to avoid confusion), use your original equation to derive $y$ from that, then apply the rotation and translation to that vector:
$$\begin{pmatrix}x\\y\end{pmatrix}=
\begin{pmatrix}\cos120° & \mp\sin120° \\
\pm\sin120° & \cos120°\end{pmatrix}\cdot
\begin{pmatrix}u\\(-2\sqrt3)u^2+\tfrac{\sqrt3}2\end{pmatrix}+
\begin{pmatrix}\pm\tfrac14\\\tfrac14\sqrt3\end{pmatrix}$$
Of course that is only one possible parametrization, and depending on the application others might be more suitable. But that is already true for your original (unrotated) parabola.
A: Since you are given the three points, there is a straight vectorial approach that you can follow.
Premised that
the parabola $y=x^2$,  the parabola  with vertex in $O=(0,0)$ and passing through
 $U_{-1}=(-1,1)$ and  $U_{1}=(1,1)$, can be written as:
$$
\mathop {OP}\limits^ \to   \cdot \mathbf{u}_y  = \left( {\mathop {OP}\limits^ \to   \cdot \mathbf{u}_x } \right)^2 
$$
where $\mathop {OP}\limits^ \to   = (x,y)$ is the position vector, and
$\mathbf{u}_x  = \left( {1,0} \right)\quad \mathbf{u}_y  = \left( {0,1} \right)$
are the unit vectors on the axes, and that we have
$$
\mathop {OU_1 }\limits^ \to   = \mathbf{u}_y  + \mathbf{u}_x \quad \mathop {OU_{ - 1} }\limits^ \to   = \mathbf{u}_y  - \mathbf{u}_x 
$$
then
in the case of the three symmetric points under consideration
$$
A = \left( { - 1/2,\;0} \right)\quad B = \left( {1/2,\;0} \right)\quad C = \left( {0,\;\sqrt 3 /2} \right)
$$
let's consider the vectors 
$$
\begin{gathered}
  \mathbf{v} = \frac{1}
{2}\left( {\mathop {AB}\limits^ \to   - \mathop {AC}\limits^ \to  } \right) = \frac{1}
{2}\left( {\left( {x_B  - x_C } \right),\;\left( {y_B  - y_C } \right)} \right) \hfill \\
  \mathbf{w} = \frac{1}
{2}\left( {\mathop {AB}\limits^ \to   + \mathop {AC}\limits^ \to  } \right) = \frac{1}
{2}\left( {\left( {x_B  + x_C  - 2x_A } \right),\;\left( {y_B  + y_C  - 2y_A } \right)} \right) \hfill \\ 
\end{gathered} 
$$
as shown in the figure

we want that
$$
\left\{ \begin{gathered}
  \mathop {AB}\limits^ \to   \cdot \mathbf{w} = \left\| \mathbf{w} \right\|^{\,2} \quad  \Rightarrow \quad 1 \hfill \\
  \mathop {AB}\limits^ \to   \cdot \mathbf{v} = \left\| \mathbf{v} \right\|^{\,2} \quad  \Rightarrow \quad 1 \hfill \\ 
\end{gathered}  \right.
$$
therefore the parabola with vertex in $A$ e passing through $B$ and $C$ can be written as:
$$
\frac{{\mathop {AP}\limits^ \to   \cdot \mathbf{w}}}
{{\left\| \mathbf{w} \right\|^{\,2} }} = \left( {\frac{{\mathop {AP}\limits^ \to   \cdot \mathbf{v}}}
{{\left\| \mathbf{v} \right\|^{\,2} }}} \right)^2 
$$
i.e.
$$
\left( {\left( {x + 1/2} \right),\;y} \right) \cdot \frac{1}
{2}\left( {\begin{array}{*{20}c}
   {3/2}  \\
   {\sqrt 3 /2}  \\
 \end{array} } \right)\frac{4}
{3} = \left( {\left( {\left( {x + 1/2} \right),\;y} \right) \cdot \frac{1}
{2}\left( {\begin{array}{*{20}c}
   {1/2}  \\
   { - \sqrt 3 /2}  \\
 \end{array} } \right)4} \right)^2 
$$
thus
$$
2\left( {6x + 3 + 2\sqrt 3 y} \right) = 3\left( {2x + 1 - 2\sqrt 3 y} \right)^2 
$$
or
$$
12x^2  + 36y^2  - 24\sqrt 3 xy - 16\sqrt 3 y - 3 = 0
$$
