To find f(x) we will do an integration
[tex]\begin{gathered} f^{\prime}(x)=x^2-3x+12\text{ } \\ f(x)=\int (x^2-3x+12) \end{gathered}[/tex][tex]\int (x^2-3x+12)=\frac{x^3}{3}-\frac{3x^2}{2}+12x+c[/tex]
To find c substitute x by 0 and y by 9 because f(0) = 9
[tex]\begin{gathered} f(x)=\frac{1}{3}x^3-\frac{3}{2}x^2+12x+c \\ f(0)=\frac{1}{3}(0)^3-\frac{3}{2}(0)^2+12(0)+c=9 \\ c=9 \end{gathered}[/tex]
The function f(x) is
[tex]f(x)=\frac{1}{3}x^3-\frac{3}{2}x^2+12x+9[/tex]
Answer D
