Answer:
[tex]\text{D. }f(x)=(x-3)^2(x-2)(x-1)[/tex]
Step-by-step explanation:
The equation below shows three zeroes at [tex]x=1[/tex], [tex]x=2[/tex], [tex]x=3[/tex]. By definition, these values of [tex]x[/tex] must yield a y-value of zero. Since the function bounces off of [tex]x=3[/tex], there must be two occurrences of it.
Thus, the equation of this line, in factored form, is [tex]\boxed{f(x)=(x-3)^2(x-2)(x-1)}[/tex]
