Answer
no x-intercepts
y-intercept is $(0,5)$
vertex is $(-1/2,9/2)$
Work Step by Step
$f(x)=2x^2+2x+5$
$y=2x^2+2x+5$
$y-5=2x^2+2x$
$(y-5)/2=(2x^2+2x)/2$
$y/2-5/2=x^2+x$
$y/2-5/2+(1/2)^2=x^2+x+(1/2)^2$
$y/2-5/2+1/4=x^2+x+1/4$
$y/2-9/4=(x+1/2)^2$
$y/2-9/4=(x+1/2)^2$
$0/2-9/4=(x+1/2)^2$
$-9/4=(x+1/2)^2$
Since we would be taking the square root of a negative number, we know that there are no x-intercepts.
y-intercept:
$f(x)=2x^2+2x+5$
$f(0)=2*0^2+2*0+5$
$f(0)=0+0+5$
$f(0)=5$
Vertex:
$y=2x^2+2x+5$
$y-5=2x^2+2x$
$(y-5)/2=(2x^2+2x)/2$
$y/2-5/2=x^2+x$
$y/2-5/2+(1/2)^2=x^2+x+(1/2)^2$
$y/2-5/2+1/4=x^2+x+1/4$
$y/2-9/4=(x+1/2)^2$
$2*(y/2-9/4=(x+1/2)^2)$
$y-9/2=2(x+1/2)^2$
Vertex of $(-1/2,9/2)$
