Original area of triangle divided by 10 is the answer to your question.
Answer:
In geometry, a median of a triangle is a line segment joining a vertex to the midpoint of the opposite side, thus bisecting that side. Every triangle has exactly three medians, one from each vertex, and they all intersect each other at the triangle's centroid.
Check the picture below, so the parabola looks more or less like so, hmmm with a vertex at (-1 , -4), so, using those values from the table
![~~~~~~\textit{vertical parabola vertex form} \\\\ y=a(x- h)^2+ k\qquad \begin{cases} \stackrel{vertex}{(h,k)}\\\\ \stackrel{"a"~is~negative}{op ens~\cap}\qquad \stackrel{"a"~is~positive}{op ens~\cup} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}](https://tex.z-dn.net/?f=~~~~~~%5Ctextit%7Bvertical%20parabola%20vertex%20form%7D%20%5C%5C%5C%5C%20y%3Da%28x-%20h%29%5E2%2B%20k%5Cqquad%20%5Cbegin%7Bcases%7D%20%5Cstackrel%7Bvertex%7D%7B%28h%2Ck%29%7D%5C%5C%5C%5C%20%5Cstackrel%7B%22a%22~is~negative%7D%7Bop%20ens~%5Ccap%7D%5Cqquad%20%5Cstackrel%7B%22a%22~is~positive%7D%7Bop%20ens~%5Ccup%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D)
![\stackrel{vertex}{\stackrel{h}{-1}~~,~~\stackrel{k}{-4}}\qquad \implies y=a[x-(-1)]^2-4\implies y=a(x+1)^2-4 \\\\\\ \textit{we also know that} \begin{cases} x=2\\ y=14 \end{cases}\implies 14=a(2+1)^2-4\implies 18=9a \\\\\\ \cfrac{18}{9}=a\implies 2=a~\hspace{10em}\boxed{y=2(x+1)^2-4}](https://tex.z-dn.net/?f=%5Cstackrel%7Bvertex%7D%7B%5Cstackrel%7Bh%7D%7B-1%7D~~%2C~~%5Cstackrel%7Bk%7D%7B-4%7D%7D%5Cqquad%20%5Cimplies%20y%3Da%5Bx-%28-1%29%5D%5E2-4%5Cimplies%20y%3Da%28x%2B1%29%5E2-4%20%5C%5C%5C%5C%5C%5C%20%5Ctextit%7Bwe%20also%20know%20that%7D%20%5Cbegin%7Bcases%7D%20x%3D2%5C%5C%20y%3D14%20%5Cend%7Bcases%7D%5Cimplies%2014%3Da%282%2B1%29%5E2-4%5Cimplies%2018%3D9a%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B18%7D%7B9%7D%3Da%5Cimplies%202%3Da~%5Chspace%7B10em%7D%5Cboxed%7By%3D2%28x%2B1%29%5E2-4%7D)