We want to find the of the given .
$y=(x−1)_{2}(x+2) $
Zeros of the Function
To do so, we need to find the values of
$x$ for which
$y=0.$
$y=0⇔(x−1)_{2}(x+2)=0 $
Since the function is already written in , we will use the .
$(x−1)_{2}(x+2)=0$
$(x+1)_{2}=0x+2=0 (I)(II) $
$x+1=0x+2=0 (I)(II) $
$x=1x+2=0 (I)(II) $
$x=1x=2 (I)(II) $
We found that the zeros of the function are
$2,$ and
$1.$ Graph
To draw the graph of the function, we will find some additional points and consider the . Let's use a table to find additional points.
$x$

$(x−1)_{2}(x+2)$

$y=(x−1)_{2}(x+2)$

$1.5$

$(1.5−1)_{2}(1.5+2)$

$3.125$

$1$

$(1−1)_{2}(1+2)$

$4$

$0$

$(0−1)_{2}(0+2)$

$2$

$0.5$

$(0.5−1)_{2}(0.5+2)$

$0.625$

The points
$(1.5,3.125),$ $(1,4),$ $(0,2),$ and
$(0.5,0.625)$ are on the graph of the function. Finally, let's apply the . This will simplify the equation and determine the leading and of the polynomial function.
$y=(x−1)_{2}(x+2)$
Rewrite $(x−1)_{2}$ as $(x−1)(x−1)$
$y=(x(x−1)−1(x−1))(x+2)$
$y=(x_{2}−x−1(x−1))(x+2)$
$y=(x_{2}−x−x+1)(x+2)$
$y=(x_{2}−2x+1)(x+2)$
$y=x_{2}(x+2)−2x(x+2)+1(x+2)$
$y=x_{3}+2x_{2}−2x(x+2)+1(x+2)$
$y=x_{3}+2x_{2}−2x_{2}−4x+1(x+2)$
$y=x_{3}+2x_{2}−2x_{2}−4x+x+2$
$y=x_{3}−3x+2$
We can see now that the 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.
The function correpsonds to graph $D.$