Answer
Vertex: $(3,-5)$
Opens upward
x-intercepts: .8, 5.2
y-intercept: 4
Work Step by Step
$f(x)=x^2-6x+4$
$x=0$
$f(x)=x^2-6x+4$
$f(0)=0^2-6*0+4$
$f(0)=0-0+4$
$f(0)=4$
$y=x^2-6x+4$
$y=x^2-6x+(-6/2)^2-(-6/2)^2+4$
$y=x^2-6x+(-3)^2-(-3)^2+4$
$y=x^2-6x+9-9+4$
$y=(x-3)^2-5$
Vertex: $(3,-5)$
Coefficient of $x^2$ is positive, so graph opens upward
$y=0$
$y=(x-3)^2-5$
$0=(x-3)^2-5$
$0+5=(x-3)^2-5+5$
$5=(x-3)^2$
$\sqrt 5=\sqrt {(x-3)^2}$
$\sqrt 5 = (x-3)$
$\sqrt 5 = x-3$
$3+\sqrt 5 =x-3+3$
$3+\sqrt 5 =x$
$3 + 2.236 =x$
$5.236 = x$
$5.2 =x$
$-\sqrt 5 = x-3$
$-\sqrt 5 +3 = x-3+3$
$3-\sqrt 5 = x$
$3-2.236 = x$
$.764 =x$
$x=.8$
