Answer
Vertex: $(2,0)$
x-intercept: $(2,0)$
y-intercept: $(0,4)$
Work Step by Step
$h(x)=x^2-4x+4$
$a=1$, $b=-4$, $c=4$
The axis of symmetry is $x=-b/2a$, and the vertex of the graph is on the axis of symmetry.
$x=-b/2a$
$x=-(-4)/2*1$
$x=4/2$
$x=2$
$h(x)=x^2-4x+4$
$h(2)=2^2-4*2+4$
$h(2)=4-8+4$
$h(2)=0$
$(2, 0)$ is the vertex.
$x=0$
$h(x)=x^2-4x+4$
$h(0)=0^2-4*0+4$
$h(0)=0-0+4$
$h(0)=4$
$(0,4)$ is the y-intercept.
$y=0$
$0=x^2-4x+4$
$0=(x-2)^2$
$\sqrt 0 = \sqrt {(x-2)^2}$
$0= x-2$
$0+2=x-2+2$
$2=x$
$(2,0)$ is the x-intercept.
Vertex: $(2,0)$
x-intercept: $(2,0)$
y-intercept: $(0,4)$
