Solution:
Given:
![f(x)=x(x+3)(x+2)^2](https://img.qammunity.org/2023/formulas/mathematics/college/1qybzpu171qjtfp718sragvu8yfyiamp47.png)
To get the zeros, the zeros exist when f(x) = 0
![\begin{gathered} 0=x(x+3)(x+2)^2 \\ x=0 \\ Multiplicity=1\text{ \lparen odd multiplicity\rparen} \\ \\ \\ x+3=0 \\ Hence,\text{ }x=-3 \\ Multiplicity=1\text{ \lparen odd multiplicity\rparen} \\ \\ \\ (x+2)^2=0 \\ (x+2)(x+2)=0 \\ x=0-2 \\ x=-2\text{ \lparen twice\rparen} \\ Multiplicity=2\text{ \lparen even multiplicity\rparen} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/iifq4649ao2xo08e1tqwwshorek26wuu72.png)
To get the behavior of the graph, the rule below applies;
The graph of a polynomial function will touch the x-axis at zeros with even multiplicities. The graph will cross the x-axis at zeros with odd multiplicities.
The graph of the function showing the x-intercepts and the behavior of the zeros is shown;
Therefore, in conclusion:
![x=0,cross;x=-3,cross;x=-2,touch](https://img.qammunity.org/2023/formulas/mathematics/college/mcmoj2j3nm4g5cg1e7323w34crpnqb360u.png)