We have vertex form for parabola equation as
![Y = a(X-h)^2 + k](https://tex.z-dn.net/?f=%20Y%20%3D%20a%28X-h%29%5E2%20%2B%20k%20)
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://tex.z-dn.net/?f=%20Y%20%3D%20a%28X-2%29%5E2%20%2B%201%20)
![5 = a(0 -2)^2 + 1](https://tex.z-dn.net/?f=%205%20%3D%20a%280%20-2%29%5E2%20%2B%201%20)
Simplify and solve for a as shown
![5 = a(-2)^2 + 1](https://tex.z-dn.net/?f=%205%20%3D%20a%28-2%29%5E2%20%2B%201%20)
![5 = a(4) + 1](https://tex.z-dn.net/?f=%205%20%3D%20a%284%29%20%2B%201%20)
5 -1 = a(4) + 1 - 1
4 = a(4)
![\frac{4}{4} = \frac{a(4)}{4}](https://tex.z-dn.net/?f=%20%5Cfrac%7B4%7D%7B4%7D%20%3D%20%5Cfrac%7Ba%284%29%7D%7B4%7D%20%20)
1 = a
Now plug 1 in a place in equation (1) as shown
![Y = 1(X-2)^2 + 1](https://tex.z-dn.net/?f=%20Y%20%3D%201%28X-2%29%5E2%20%2B%201%20)
![Y = (X-2)^2 + 1](https://tex.z-dn.net/?f=%20Y%20%3D%20%28X-2%29%5E2%20%2B%201%20)
So thats the vertex equation of parabola and final answer.