Answer:
![f(x)=5(x-2)^{2}+6](https://tex.z-dn.net/?f=f%28x%29%3D5%28x-2%29%5E%7B2%7D%2B6)
Step-by-step explanation:
we have
![f(x)=5x^{2} -20x+26](https://tex.z-dn.net/?f=f%28x%29%3D5x%5E%7B2%7D%20-20x%2B26)
This is the equation of a vertical parabola open upward
The vertex is a minimum
The equation of a vertical parabola in vertex form is equal to
![f(x)=a(x-h)^2+k](https://tex.z-dn.net/?f=f%28x%29%3Da%28x-h%29%5E2%2Bk)
where
(h,k) is the vertex
Convert the given function to vertex form
Factor the leading coefficient
![f(x)=5(x^{2} -4x)+26](https://tex.z-dn.net/?f=f%28x%29%3D5%28x%5E%7B2%7D%20-4x%29%2B26)
Complete the square
![f(x)=5(x^{2} -4x+4)+26-20](https://tex.z-dn.net/?f=f%28x%29%3D5%28x%5E%7B2%7D%20-4x%2B4%29%2B26-20)
![f(x)=5(x^{2} -4x+4)+6](https://tex.z-dn.net/?f=f%28x%29%3D5%28x%5E%7B2%7D%20-4x%2B4%29%2B6)
Rewrite as perfect squares
------> equation in vertex form
The vertex is the point (2,6)