Answer
Vertex: $(1, -11)$
Opens downward
x-intercept: none
y-intercept: -12
Work Step by Step
$f(x)=-x^2+2x-12$
$y=-x^2+2x-12$
$y=-(x^2-2x)-12$
$y=-(x^2-2x)+(-2/2)^2-(-2/2)^2-12$
$y=-(x^2-2x)+(-1)^2-(-1)^2-12$
$y=-(x^2-2x)+1-1-12$
$y=-(x^2-2x+1)-11$
$y=-(x-1)^2-11$
Vertex: $(1, -11)$
Opens downward
$x=0$
$f(x)=-x^2+2x-12$
$f(0)=-0^2+2*0-12$
$f(0)=-0+0-12$
$f(0)=-12$
There are no x-intercepts since the graph is below the x-axis and opens downward.
y-intercept: -12
