Use the general equation $y=a(x-h)^2+k$, where $(h,k)$ are the coordinates of the vertex.
You already know that the $y$-coordinate of the vertex is $k=-7$. The $x$-coordinate of the vertex, $h$, is given to you indirectly: it is by symmetry the midpoint of the $x$-intercepts. So, find the value of $h$ using this and then substitute the known values of $h$ and $k$ into the general equation.
This still leaves you the unknown $a$. But, you know that the point $(3,0)$ is on the parabola, and you can substitute $x=3$ and $y=0$ into your equation and solve for $a$.
One uses Pascal's theorem for hexagons inscribed in conics. The hexagons do not need to be convex and embedded, but the order of the points (following cyclicity) is important.
You have two points on the parabola $P_1$ and $P_2$ and the axis $L$ of the parabola. Then the point at infinity $P_{\infty}$ of on $L$ is also on the parabola. Reflect points $P_1$ and $P_2$ with respect to $L$ and obtain the points $P'_1$ and $P'_2$ respectively, which also lie on the parabola.
Big Step 1. Construct the tip of the tip of the parabola $P_0$, i.e. the point where the parabola intersects the parabola's axis $L$, together with the line $b$ through $P_0$ orthogonal to $L$. The line $b$ is the tangent to the parabola at point $P_0$.
Given the five points $P_1, \, P_2, \, P'_1, \, P'_2$ and $P_{\infty}$ and the line $L$ you can construct the sixth $P_0$ using Pascal's theorem for the hexagon $P_{\infty}P_2P'_2P'_1P_1P_0$ (again, the order is importan).
Denote by $P_{\infty}P_2$ the line through $P_2$ parallel to $L$. Construct $Q_1 = P_{\infty}P_2 \cap P_1P_1'$;
Construct $Q_0 = L \cap P_1'P_2'$;
Line $Q_0Q_1$ is Pascal's line for hexagon $P_{\infty}P_2P'_2P'_1P_1P_0$.
Construct $Q_2 = Q_0Q_1 \cap P_2P_2'$;
Then construct point $P_0 = P_1Q_2 \cap L$ which is the sought point (Pascal's theorem). Draw line $b$ through $P_0$ orthogonal to axis $L$.
Big Step 2. Construct that tangent $t_2$ to the parabola at point $P_2$. To do that one can use the degenerate version of Pascal's theorem where the hexagon is $P_0P_0P_1P_2P_2P_2'$ where the line defined by the degenerate edge $P_0P_0$ is tangent line $b$ at $P_0$ and the line defined by the degenerate edge $P_2P_2$ is tangent line $t_2$ at $P_2$.
As already constructed, point $Q_2 = P_0P_1 \cap P_2P_2'$;
Construct $\hat{Q} = P_1P_2 \cap P_0P_2'$
Line $\hat{Q}Q_2 $ is the Pascal line for degenerate hexagon $P_0P_0P_1P_2P_2P_2'$;
Construct $M = \hat{Q}Q_2 \cap b$;
Construct line $t_2 = MP_2$ which is the tangent to the parabola at point $P_2$ (Pascal's theorem, degenerate version).
Concluding Big Step 3.
Draw line $m$ passing through point $M$ and orthogonal to line $t_2$.
Construct the point $F = m \cap L$. This is the focus of the parabola.
Reflect point $F$ with respect to line $b$ and obtain the point $S$ on $L$ such that $SP_0 = FP_0$, i.e. $P_0$ is the midpoint of segment $FS$.
Construct the line $\Gamma$ through $S$ orthogonal to axis $L$. This is the directrix.
Best Answer
You have to have more information than that. You can construct infinite number of parabolas with arbitrary x-axis intersects (zero, one or two).
For example you can write $y = (x-x_0)(x-x_1)$ toget a parabola that intersects at $x_1$ and $x_2$, but there are more ways than than. You could for example take $x = y^2+3+\sqrt7$ which is also a parabola - just that it doesn't happen to be facing upwards or downwards.
If you on the other hand requires the parabola to be expressable as $y = P(x)$ where $P$ is an polynomial with rational coefficioents then the solution is unique. The polinomial is uniquely determined up to rational scaling.
Suppose that the coefficient of the $x^2$ term is one, then we have
$$P(x) = (x+3+\sqrt7)(x-c) = x^2 + (3 + \sqrt7-c)x+(3+\sqrt7)cx$$
Now we require $3+\sqrt7-c$ to be rational which makes $c = q+\sqrt7$ for some rational $q$, but we also require $(3+\sqrt7)c = (3+\sqrt7)(q+\sqrt7) = 3q + (3+q)\sqrt7 + 49$ to be rational. But since $\sqrt7$ is irrational we must have that $3+q=0$ and consequently that $c = q+\sqrt7 = -3+\sqrt 7$.