Exercise 3 Page 216

To draw the graph of the function, we will find some additional points and consider the end behavior. Let's use a table to find additional points.

$x$	$(x-1)(x-2)(x+2)$	$f(x)=(x-1)(x-2)(x+2)$
$\text{-}2$	$(\text{-}2-1)(\text{-}2-2)(\text{-}2+2)$	$0$
$\text{-}1$	$(\text{-}1-1)(\text{-}1-2)(\text{-}1+2)$	$6$
$0$	$(0-1)(0-2)(0+2)$	$4$
$1$	$(1-1)(1-2)(1+2)$	$0$

The points $(\text{-}2,0),$ $(\text{-}1,6),$ $(0,\text{-}4),$ and $(1,0)$ are on the graph of the function. Finally, let's apply the Distributive Property. This will simplify the equation and determine the leading coefficient and degree of the polynomial function. We can see now that the leading coefficient is $1,$ which is a positive number. Also, the degree is $3,$ which is an odd number. Therefore, the end behavior is down and up. Now, let's draw the graph!