As we can see on the picture, the graphs crosses the x-axis at x=0 and it bounces of the x-axis at x=2. In the first case, the polynomial has a factor of degree 1 at x=0. In the second case, the factor has a degree 2 at x=2. So, we can write the following:
[tex]f(x)=A(x)(x-2)^2[/tex]where A is a constant to be determined. We can find this constant by noticing that when x=3, f(3)=3, then we have
[tex]3=A(3)(3-2)^2[/tex]which gives
[tex]\begin{gathered} 3=A\cdot3\cdot1^2 \\ 3=3A \\ \text{then} \\ A=1 \end{gathered}[/tex]Therefore, we get
[tex]f(x)=x(x-2)^2[/tex]Now, lets expand this result. By
[tex]\begin{gathered} f(x)=x(x^2-4x+4) \\ \end{gathered}[/tex]then, the answer is
[tex]f(x)=x^3-4x^2+4x[/tex]