Solution:
When a and b are the zeros of a function, this implies that
[tex]\begin{gathered} x=a \\ \Rightarrow(x-a) \\ x=b \\ \Rightarrow(x-b) \end{gathered}[/tex][tex]A(x-a)(x-b)=0[/tex]This implies the values of x when the function equals zero.
Given that the zeros of the cubic function f(x) are -6, 1 and 7, this implies that
[tex]f(x)=A(x+6)(x-1)(x-7)\text{ ---- equation 1}[/tex]Given that
[tex]f(8)=9[/tex]we substitute 9 for f(x) and 8 for x.
Thus,
[tex]\begin{gathered} 9=A(8+6)(8-1)(8-7) \\ 9=A\times14\times7\times1 \\ 9=98A \\ \Rightarrow A=\frac{9}{98} \end{gathered}[/tex]Substitute the value of A into equation 1.
[tex]f(x)=\frac{9}{98}(x+6)(x-1)(x-7)[/tex]Expand the resulting equation.
[tex]\begin{gathered} f(x)=\frac{9}{98}\mleft(x^2+5x-6\mright)\mleft(x-7\mright) \\ \frac{9}{98}(x-7)=\mleft(\frac{9x}{98}-\frac{9}{14}\mright) \\ t\text{hus, we have} \\ f(x)=(\frac{9x}{98}-\frac{9}{14})(x^2+5x-6) \\ =(\frac{9x}{98}\times x^2)+(\frac{9x}{98}\times\: 5x)+\frac{9x}{98}\mleft(-6\mright)-\frac{9}{14}x^2-(\frac{9}{14}\times\: 5x)-\frac{9}{14}\mleft(-6\mright) \\ \therefore f(x)=\frac{9x^3}{98}-\frac{9x^2}{49}-\frac{369x}{98}+\frac{27}{7} \end{gathered}[/tex]Hence, the equation for f(x) is
[tex]f(x)=\frac{9x^3}{98}-\frac{9x^2}{49}-\frac{369x}{98}+\frac{27}{7}[/tex]