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=(x2−x−1(x−1))(x+2)
y=(x2−x−x+1)(x+2)
y=(x2−2x+1)(x+2)
y=x2(x+2)−2x(x+2)+1(x+2)
y=x3+2x2−2x(x+2)+1(x+2)
y=x3+2x2−2x2−4x+1(x+2)
y=x3+2x2−2x2−4x+x+2
y=x3−3x+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.