[tex]\begin{gathered} ((x+3)(x-2)(x-6))+12 \\ \text{ All the points that has (x,y) will be transformed into }x-y\text{ thus} \\ (-3,0)\rightarrow(x-(-3))=x+3 \\ (2,0)\rightarrow(x-2) \\ (6,0)\rightarrow(x-6) \\ \text{and lastly, the (0,12) is the y intercept for which f(0)=12} \\ \text{therefore it is a constant +12} \end{gathered}[/tex]
