Find the domain and range of $y=-x^2+4x-3$
My Attempt: $$y=f(x)=-x^2+4x-3$$ The given function is a polynomial of degree $2$ in $x$. $f(x)$ is defined for all $x\in R$, so the domain of $f=R$. Again, $$y=-x^2+4x-3$$ $$y=-(x^2-4x+3)$$ $$y=-(x-1)(x-3)$$ $$y=(x+1)(x-3)$$.
Write $f(x)=-x^2+4x-3=1-(x-2)^2$, then
$f(x)=1-(x-2)^2\leq -1$
and the range is $(-\infty,1]$ and the domain is all the set $\mathbb{R}=(-\infty,\infty)$ because $f$ is a polynomial function. It's minimum will be always at the middle point of the two roots: $x_1=-1$ and $x_2=3$ (because $f(x)$ is a polynomial function of degree 2) which is $x=2$, so $f(2)=1$ is the minimum value of $f(x)$ at $x=2$.
If you isolate $x$ from $y=1-(x-2)^2$ you find a two partial inverses (reciprocal binary relation, the inverse relation of $f$ is not a function) for $f(x)$:
$$ x = 2\pm\sqrt{1-y}$$
and then the range of $f$ is the domain of the functions:
$$f^{-1}(y)=2\pm\sqrt{1-y}.$$
which is $(-\infty,1]$.
Since $$-x^2+4x-3=1-(x-2)^2$$ it follows $f(x)\le 1$. So, the range of $f$ is $(-\infty, 1]$.
