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
[tex] \frac{2x^2+5x-12}{x+4}= \frac{(x+4)(2x-3)}{x+4}=2x-3[/tex]
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)
