We have vertex form for parabola equation as
![Y = a(X-h)^2 + k](https://img.qammunity.org/2019/formulas/mathematics/middle-school/cltur6u62yizq5xrecpbigih87dvfobyw9.png)
where (h,k) is the vertex.
As the turning point given here is (2,1) so thats the vertex.
On comparing (2,1) with (h,k), we can see
h = 2, k = 1
Plugging 2 in h place and 1 in k place in
we get
------------------------ (1)
Now we need to find value of a.
For that we will use point (0,5) given on parabola.
On comparing (0,5) with point (X,Y) we get X = 0, Y = 5
so plug 0 in X place and 5 in Y place in equation (1)
![Y = a(X-2)^2 + 1](https://img.qammunity.org/2019/formulas/mathematics/middle-school/e5ve3c18ku1zq10q5eqxfgx5s6tknin7l1.png)
![5 = a(0 -2)^2 + 1](https://img.qammunity.org/2019/formulas/mathematics/middle-school/qknpnlas8vb0j83asvmyav4an7coba0afd.png)
Simplify and solve for a as shown
![5 = a(-2)^2 + 1](https://img.qammunity.org/2019/formulas/mathematics/middle-school/ouedt2wabkwvi22fwtv04ixzd6e2koptkf.png)
![5 = a(4) + 1](https://img.qammunity.org/2019/formulas/mathematics/middle-school/u9zcuykaa0po4flft5dlw0cp8xc2gy11co.png)
5 -1 = a(4) + 1 - 1
4 = a(4)
![(4)/(4) = (a(4))/(4)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/l1b6v5w0hgjod345o2csokas2ijy67w2s6.png)
1 = a
Now plug 1 in a place in equation (1) as shown
![Y = 1(X-2)^2 + 1](https://img.qammunity.org/2019/formulas/mathematics/middle-school/4rwc96duz5a2gst1lytgkrpds2jffgokj7.png)
![Y = (X-2)^2 + 1](https://img.qammunity.org/2019/formulas/mathematics/middle-school/7lzgjh2tvjdetegsxi4a464j26jymqgg91.png)
So thats the vertex equation of parabola and final answer.