jmc

Rewrite the equation of the parabola as $y=a(x-h{)}^2+k$, where $a$, $h$, and $k$ are constants and $(h,k)$ are the coordinates of the vertex. If the parabola has a vertical line of symmetry at $x=1$, then the $x$-coordinate of the vertex is $x=1$, so $h=1$. The equation of the parabola becomes $y=a(x-1{)}^2+k$. Plugging in the two given points into this equation, we have the two equations $$\begin{array}{rl}3& =a(-1-1{)}^2+k\Rightarrow 3=4a+k\\ -2& =a(2-1{)}^2+k\Rightarrow -2=a+k\end{array}$$ Subtracting the second equation from the first yields $5=3a$, so $a=5/3$. Plugging this value into the second equation to solve for $k$, we find that $k=-11/3$. So the equation of the parabola is $y=\frac{5}{3}(x-1{)}^2-\frac{11}{3}$. To find the zeros of the parabola, we set $y=0$ and solve for $x$: $$\begin{array}{rlr}0& =\frac{5}{3}(x-1{)}^2-\frac{11}{3}& \\ \frac{11}{3}& =\frac{5}{3}(x-1{)}^2& \\ \frac{11}{5}& =(x-1{)}^2& \\ x& =\pm \sqrt{\frac{11}{5}}+1& \end{array}$$

The greater zero is at $x=\sqrt{\frac{11}{5}}+1$, so $n=\boxed{2.2}$. The graph of the parabola is below: