jmc

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

The greater zero is at $x=\sqrt{\frac{5}{2}}+2$, so $n=\boxed{2.5}$. The graph of the parabola is below: