Answer
Vertex: $(0,1)$
Opens upward
x-intercept: none
y-intercept: 1
Work Step by Step
$f(x)=x^2+1$
$a=1$, $b=0$, $c=1$
Vertex: $-b/2a=x$
$x=-b/2a$
$x=-0/2*1$
$x=0*1$
$x=0$
$f(x)=x^2+1$
$f(0)=0^2+1$
$f(0)=0+1$
$f(0)=1$
Vertex: $(0,1)$
Opens upward
The y-intercept is found when $x=0$, and since the vertex has an x-coordinate of 0, we already have the y-intercept (1).
There are no x-intercepts. The graph sits on the x-axis and opens up.
y-intercept: 1
