To do so, we need to find the values of x for which g(x)=0.
g(x)=0 ⇔ (x+1)(x-1)(x+2)=0
Since the function is already written in factored form, we will use the Zero Product Property.
We found that the zeros of the function are - 2, - 1, and 1. We can stop here and recognize that the only graph with those zeros is graph B. We can also continue by graphing the function with a table of values as shown below.
Extra
Graphing the function using a table of values
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-1)(x+2)
g(x)=(x+1)(x-1)(x+2)
- 3
( - 3+1)( - 3-1)( - 3+2)
- 8
- 1.5
( - 1.5+1)( - 1.5-1)( - 1.5+2)
0.625
0
( 0+1)( 0-1)( 0+2)
- 2
0.5
( 0.5+1)( 0.5-1)( 0.5+2)
- 1.875
The points ( - 3, - 8), ( - 1.5, 0.625), ( 0, - 2), and ( 0.5, - 1.875) 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!
