Answer
Vertex: $(1,4)$
Opens upward
x-intercepts: none
y-intercept: 7
Work Step by Step
$f(x)=3x^2-6x+7$
$x=0$
$f(x)=3x^2-6x+7$
$f(0)=3*0^2-6*0+7$
$f(0)=3*0-0+7$
$f(0)=0+7$
$f(0)=7$
$f(x)=3x^2-6x+7$
$y=3x^2-6x+7$
$y=3(x^2-2x)+7$
$y=3(x^2-2x+(-2/2)^2)+7-3(-2/2)^2$
$y=3(x^2-2x+(-1)^2)+7-3(-1)^2$
$y=3(x^2-2x+1)+7-3(1)$
$y=3(x-1)^2+4$
Vertex: $(1, 4)$
Coefficient of $x^2$ is positive, so graph opens up
Since the graph opens up, and the vertex of the graph is above the x-axis, there are no x-intercepts.
