Answer
vertex:$(2,1)$ and defines at opens downwards.
and Domain is: $(- \infty, \infty)$ and range is: $(-\infty, 1]$
Work Step by Step
Since, we have $y=-x^2+4x-3$
This equation can be written as:
$y=-(x^2-4x+4)-3+4$
or, $y=-(x-2)^2+1$
Standard form of a horizontal parabola is $x=a(y-k)^2+h$
Here, vertex:$(h,k)$
Compare the above equation with the standard form.
Thus, we have vertex:$(2,1)$ and defines at opens downwards.
and Domain is: $(- \infty, \infty)$ and range is: $(-\infty, 1]$
