Remmber you can't divide by 0 if you can canel out something from both that is a hole
exampe if you had y=[(x-2)(x+3)]/(x-2), then the equation is y=x+3 with a hole at x=2 (since x=2 to make x-2=0 true)
so
y=2x-3 is theh graph zero at y=0 or x=3/2 (1.5,0) is the zero set factored out thing to zero x+4=0 x=-4 at x=-4 input into later one y=2x-3 y=2(-4)-3 y=-8-3 y=-11 hole at (-4,-11) the hole is at (-4,-11) zero at (3/2,0)