Hello, hejmh!
The general cubic function is: .$\displaystyle f(x) \:=\:ax^3 + bx^2 + cx + d$
. . Its derivative is: .$\displaystyle f'(x) \;=\;3ax^2 + 2bx + c$
We need to determine the four coefficients: .$\displaystyle a,b,c,d.$
We know five facts about the function:
. . $\displaystyle \begin{Bmatrix} f(0) &=& 10 \\ f(2) &=& 0 \\ f(\text{-}3) &=& 0 \\ \\[-3mm] f'(0) &=& 0 \\ f'(2) &=& 0 \end{Bmatrix}$
Set up a system of equations and solve.
Edit: Ah, I see that DrSteve has a nicer form of the cubic.
Because, as Dr. Steve said, x= 2 is a double root- the x-axis is tangent to the graph at x= 2.
Another way of seeing the same thing- moving from one side of 2 to the other, say from x= 1.9 to x= 2.1, changes the sign of x- 2. x- 2 must be to an even power since y does not change sign.
NO, it isn't. $\displaystyle f(0)= a(0+ 3)(0- 2)^2= 12a$, not 0.2- You see in the figure f(0)=10 but in this solution f(0)=0
According to your graph $\displaystyle f'(0)$ may or may not be $\displaystyle 0$. It turns out that it is not. If you combine my post above with HallsofIvy's you will see that
$\displaystyle f(x)=\frac{5}{6}(x+3)(x-2)^2$.
The derivative of this function is not 0 at $\displaystyle x=0$.
Soroban's method should work if you just leave out the condition $\displaystyle f'(0)=0$. You will get a system of 3 equations in 3 unknowns which should have a unique solution (I'm pretty sure that the 3 equations in 2 unknowns you have above is inconsistent). You'll probably want to do Gauss-Jordan Reduction to solve this. It's quite a bit more work than the original solution I posted.