Answer
Vertex: $(-1, 3)$
Opens upward
x-intercept: none
y-intercept: 5
Work Step by Step
$f(x)=2x^2+4x+5$
$f(x)=(2x^2+4x)+5$
$f(x)=2(x^2+2x)+5$
$f(x)=2(x^2+2x+(2/2)^2)+5-2*(2/2)^2$
$f(x)=2(x^2+2x+(1^2)+5-2*(1)^2$
$f(x)=2(x^2+2x+1)+5-2*1$
$f(x)=2(x+1)^2+5-2$
$f(x)=2(x+1)^2+3$
Vertex: $(-1, 3)$
Opens upward
$x=0$
$f(x)=2x^2+4x+5$
$f(0)=2*0^2+4*0+5$
$f(0)=2*0+0+5$
$f(0)=0+5$
$f(0)=5$
Since the vertex of the graph has a minimum y-value of 3 and opens up, there are no lower points. Thus, there are no x-intercepts.
y-intercept: 5
