## Algebra 1

$y=x^2-3x+2$ 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.