ANSWER
B. they all resemble a linear function
Step-by-step explanation
We have to find the zeros of this function. Since it is a 3rd-degree polynomial it is expected that there will be 3 zeros, but one or the three of them can have a multiplicity higher than 2, in which case we will find one or two zeros only.
In this case, the zeros of the function are x = -1.5, x = -1, and x = 2, as we can see in the graph,
So we can write the function as,
![y=2(x+1.5)(x+1)(x-2)](https://img.qammunity.org/2023/formulas/mathematics/college/9yhlypxa5bobd1uc450vsgmvlvno2dwpfb.png)
If we replace each zero in the function, but only for the factors not corresponding to that zero, we will get only one factor. For example, replacing x = 2,
![y=2(2+1.5)(2+1)(x-2)=2\cdot3.5\cdot3(x-2)=21(x-2)](https://img.qammunity.org/2023/formulas/mathematics/college/pttr49vwf2i2kc0no3bcqzq3iupjt0gxq1.png)
Which is a linear function. And this happens also for both remaining zeros.
Hence, all three zeros resemble a linear function.