Answer
since $y"$>0, the curve is concave up for all $x$.
Work Step by Step
when $y$=$x^2-4x+3$,
then $y'$=$2x-4$=$2(x-2)$ and
$y"$=$2$.
The curve rises on $(2,\infty)$ and falls on $(-\infty,2)$.At $x$=$2$ there is a minimum.since $y"$>0, the curve is concave up for all x.
