\(f(x)=x^2+k\) moves vertically.

\(f(x)=(x-r)^2\) moves horizontally.

Vertex form: \(f(x)=a(x-r)^2+k\), with vertex \(T(r,k)\).

For \(f(x)=ax^2+bx+c\), if the vertex is known then \(f(x)=a(x-r)^2+k\).

Since \(f(0)=c\), \(c=ar^2+k\), which determines \(a\).