Answer
Vertex: $(2,1)$
x-intercept: none
y-intercept: $(0,9)$
Work Step by Step
$F(x) =2x^2-8x+9$
$a=2$, $b=-8$, $c=9$
The axis of symmetry is $x=-b/2a$, and the vertex of the graph is on the axis of symmetry.
$x=-b/2a$
$x=-(-8)/2*2$
$x=8/4$
$x=2$
$F(x) =2x^2-8x+9$
$F(2)=2*2^2-8*2+9$
$F(2)=2*4-16+9$
$F(2)=8-16+9$
$F(2)=1$
$(2, 1)$ is the vertex.
Since the vertex is above the x-axis and opens upward (due to the positive coefficient on $x^2$), there are no x-intercepts.
$x=0$
$F(x) =2x^2-8x+9$
$F(0) =2*0^2-8*0+9$
$F(0)=2*0-0+9$
$F(0)=0+9$
$F(0)=9$
$(0,9)$ is the y-intercept.
Vertex: $(2,1)$
x-intercept: none
y-intercept: $(0,9)$
