Answer
Vertex: $(2, 1)$
Opens upward
x-intercepts: none
y-intercepts: 5
Work Step by Step
$f(x)=x^2-4x+5$
$f(x)=x^2-4x+5$
$y=x^2-4x+5$
$y-1=x^2-4x+5-1$
$y-1=x^2-4x+4$
$y-1=(x-2)^2$
$y-1+1=(x-2)^2+1$
$y=(x-2)^2+1$
The vertex is at $(2, 1)$. Since this vertex is above the x-axis and has a positive coefficient for $x^2$, the graph opens up. Also, there are no x-intercepts.
$x=0$
$f(x)=x^2-4x+5$
$f(0)=0^2-4*0+5$
$f(0)=0-0+5$
$f(0)=5$
x-intercepts: none
y-intercepts: 5
