Answer
Vertex: $(-1/2, 1/4)$
x-intercepts: none
y-intercept: $(0, 1)$
Work Step by Step
$f(x)=3x^2+3x+1$
$a=3$, $b=3$, $c=1$
Vertex is at $x=-b/2a$
$x=-b/2a$
$x=-3/2*3$
$x=-3/6$
$x=-1/2$
$x=-1/2$
$f(x)=3x^2+3x+1$
$f(-1/2)=3(-1/2)^2+3(-1/2)+1$
$f(-1/2)=3*1/4-3/2+1$
$f(-1/2)=3/4-3/2+1$
$f(-1/2)=1/4$
$x=0$
$f(x)=3x^2+3x+1$
$f(0)=3*0^2+3*0+1$
$f(0)=3*0+0+1$
$f(0)=0+1$
$f(0)=1$
$f(x)=0$
The vertex of the parabola is at $(-1/2, 1/4)$, and the coefficient of the function is positive. Thus, the graph of the function opens up, and there are no x-intercepts.
