Answer
Please see the graph.
Work Step by Step
$y=x^2-3x+2$
Axis of symmetry: $x=-b/2a$
$x=-b/2a$
$x=-(-3)/2(1)$
$x=3/2$
$y=x^2-3x+2$
$y=(3/2)^2-3*(3/2)+2$
$y=9/4-9/2+2$
$y=-9/4+2$
$y=-1/4$
$(1.5,-.25)$
$y=x^2-3x+2$
$y=x^2-2x-x+2$
$y=x(x-2)-1(x-2)$
$y=(x-2)(x-1)$
If either $x=2$ or $x=1$, then that part of the equation equals zero, and then the entire formula equals zero. Thus, we have the two points $(1,0)$ and $(2,0)$ on the curve.
