You are right that $y = ax^2 + bx + c.$ This is because its axis is the $y$-axis. You have three unknowns, namely $a$, $b$ and $c$. Luckily your are given three points on the curve. We are going to use these points to create three simultaneous equations in $a$, $b$ and $c$.
The first point is $(x,y) = (4,11)$ and so we substitute $x=4$ and $y=11$ into $y = ax^2 + bx + c.$ The resulting equation is $11 = 16a + 4b + c.$ The second point is $(x,y) = (3,4)$ and so we substitute $x=3$ and $y=4$ into $y = ax^2 + bx + c.$ The resulting equation is $4 = 9a + 3b + c.$ The third point is $(x,y) = (0,3)$ and so we substitute $x=0$ and $y=3$ into $y = ax^2 + bx + c.$ The resulting equation is $3 = c.$ We need to solve the following system of simultaneous equations:
$ \begin{array}{cCCcccc} 16a & + & 4b & + & c & = & 11 \ ,\\ 9a & + & 3b & + & c & = & 4 \ , \\ & & & & c & = & 3 \ . \end{array}$
Clearly, $c=3$, and so we can substitute this into the first two equations. Rearranging gives:
$ \begin{array}{ccccc} 16a & + & 4b & = & 8 \ , \\ 9a & + & 3b & = & 1 \ . \end{array} $
The first equation has a common factor of $4$ and so we may divide both sides by $4$ to give:
$ \begin{array}{ccccc} 4a & + & b & = & 2 \ , \\ 9a & + & 3b & = & 1 \ . \end{array} $
Let's solve these equations by elimination. First, let's multiply both sides of the first equation by $3$:
$ \begin{array}{ccccc} 12a & + & 3b & = & 6 \ , \\ 9a & + & 3b & = & 1 \ . \end{array} $
Subtracting the second from the first gives $3a = 5$, which implies that $a=5/3.$ Next, we may substitute $a = 5/3$ into either equation; I will chose the first one. If $a=5/3$ and $12a + 3b = 6$ then $20 + 3b = 6$ and so $3b = -14$ and $b=-14/3.$ It follows that $a = 5/3,$ $b=-14/3$ and $c=3$. Using this:
$y = \frac{5}{3}x^2 - \frac{14}{3}x + 3 \, . $