Answer
Vertex: $(2, 3)$
Opens upward
x-intercept: none
y-intercept: 15
Work Step by Step
$f(x)=3x^2-12x+15$
$f(x)=3(x^2-4x)+15$
$f(x)=3(x^2-4x)+15+3*(-4/2)^2-3*(-4/2)^2$
$f(x)=3(x^2-4x)+15+3*(-2)^2-3*(-2)^2$
$f(x)=3(x^2-4x)+15+(3*4)-3*4$
$f(x)=3(x^2-4x+4)+15-3*4$
$f(x)=3(x-2)^2+15-12$
$f(x)=3(x-2)^2+3$
Vertex: $(2, 3)$
Opens upward
$x=0$
$f(x)=3x^2-12x+15$
$f(0)=3*0^2-12*0+15$
$f(0)=3*0-0+15$
$f(0)=0+15$
$f(0)=15$
Since the graph is above the x-axis and opens upward, there are no x-intercepts.
y-intercept: 15
