Answer
Vertex: $(3,-18)$
Opens downward
x-intercept: none
y-intercept: 0
Work Step by Step
$f(x)=-2x^2+12x$
$f(x)=-2(x^2-6x)$
$f(x)=-2(x^2-6x+(-6/2)^2)-2(-6/2)^2$
$f(x)=-2(x^2-6x+(-3)^2)-2(-3)^2$
$f(x)=-2(x^2-6x+9)-2(9)$
$f(x)=-2(x-3)-18$
Vertex: $(3,-18)$
Opens downward
$x=0$
$f(x)=-2x^2+12x$
$f(0)=-2*0^2+12*0$
$f(0)=-2*0+0$
$f(0)=0+0$
$f(0)=0$
Since the vertex is below the x-axis and the graph opens down, there are no x-intercepts.
Y-intercept: 0
