[tex]f(x)=x^2-4x+3[/tex]
a=1, b=-4, c=3
If the vertex has coordinates (2;-1)(p=2,q=-1) we can write vertex form of a parabola equation:
[tex]f(x)=a(x-p)^2+q[/tex]
[tex]f(x)=1(x-2)^2-1[/tex]
We need to put (x-2) at the place of (x) in f(x) equation to get g(x)
[tex]g(x)=1[(x-2)-2]^2-1[/tex]
[tex]g(x)=(x-2-2)^2-1[/tex]
[tex]g(x)=(x-4)^2-1[/tex]
So:
p=4, q=-1
Vertex of the parabola defined by g(x)=f(x-2) has the vertex at [tex]\boxed{(4;-1)}[/tex]
:)
