Answer:
[tex]y=x^3-3x^2[/tex]
Step-by-step explanation:
Given that on x=0 y=0, and x=3 y=0 the function has roots.
We can state that the function can be considered as:
[tex]\begin{gathered} y=ax(x-3)(x-b) \\ y=(ax^2-3ax)(x-b) \\ y=ax^3-abx^2-3ax^2+3abx \\ y=ax^3-x^2(ab+3a)+3abx \end{gathered}[/tex]
By the given table:
We can use two of the points given and create a system of 2 equations with two variables, and solve it for a and b:
[tex]\begin{gathered} x=1,\text{ y=-2} \\ -2=a-(ab+3a)+3ab\text{ (1)} \\ x=-1,\text{ y=-4} \\ -4=-a-(ab+3a)-3ab\text{ (2)} \end{gathered}[/tex]
Then, we can divide both equations to find b.
[tex]\begin{gathered} \frac{-2}{-4}=\frac{a-(ab+3a)+3ab}{-a-(ab+3a)-3ab} \\ \frac{1}{2}=\frac{a(1-(b+3)+3b)}{a(-1-(b+3)-3b)} \\ \frac{1}{2}=\frac{-2+2b}{-4-4b} \\ -4-4b=2(-2+2b) \\ -4-4b=-4+4b \\ 0=8b \\ b=\frac{0}{8} \\ b=0 \end{gathered}[/tex]
Now, knowing the value of b, we can substitute it into either equation and find a, to get the cubic function that models the data:
[tex]\begin{gathered} -2=a-3a \\ -2=-2a \\ a=\frac{-2}{-2} \\ a=1 \\ \end{gathered}[/tex]
Hence, the function that satisfies the data would be:
[tex]\begin{gathered} y=x(x-3)(x+0) \\ y=x^3-3x^2 \end{gathered}[/tex]
