Answer
Vertex: $(2, 1)$
Opens upward
x-intercepts: none
y-intercept: 5
Work Step by Step
$f(x)=x^2-4x+5$
$y=x^2-4x+5$
$y-5=x^2-4x+5-5$
$y-5=x^2-4x$
$y-5+(-4/2)^2=x^2-4x+(-4/2)^2$
$y-5+(-2)^2=x^2-4x+(-2)^2$
$y-5+4=x^2-4x+4$
$y-1=(x-2)^2$
$y-1+1=(x-2)^2+1$
$y=(x-2)^2+1$
Vertex: $(2, 1)$
Opens upward
$x=0$
$f(x)=x^2-4x+5$
$f(0)=0^2-4*0+5$
$f(0)=0-0+5$
$f(0)=5$
$y=0$
$y=(x-2)^2+1$
$0=(x-2)^2+1$
$0-1=(x-2)^2+1-1$
$-1=(x-2)^2$
We can’t have the square root of a negative number, so there are no x-intercepts.
x-intercepts: none
y-intercept: 5
