We are about to learn integrals, but this homework does not expect an understanding of them. I'm having a hard time understanding what I am to do.

The problem asks me to find cubic function in the form of $\displaystyle f(x)=ax^3+cx^2+d$ that has a local maximum value of 9 at -3 and a local minimum value of 7 at 0.

So I knew that the zeros for $\displaystyle f'(x)$ were equal to -3 and 0. Using that I came up with $\displaystyle x(x+3)$. I realize though that any coefficient attached to the first x would still give a zero of 0. So how do I find that coefficient?

Taking the antiderrivitive of my function, then adding the constant 7 gave: $\displaystyle f(x) = \frac{1}{3}x^3+\frac{3}{2}x^2+7$. However with this, $\displaystyle f(-3) = 11.5$, not 9. What am I missing?

Thanks in advance!

EDIT:

I wanted to add that I tried also working from the second derivative, too. Thinking that the inflection point ought to be halfway between 0 and -3, I made $\displaystyle f''(x) = x + \frac{3}{2}$ which makes $\displaystyle f'(x) = \frac{1}{2}x^2 + \frac{3}{2}x$ which still has zeros of 0 and -3. The anti derivative of that gives $\displaystyle f(x) = \frac{1}{6}x^3 + \frac{3}{4}x^2 +7$ after adding the constant. We're really close, but $\displaystyle f(-3) = 9.25$ which isn't 9.

Just wanted whoever to know that I'm giving it the old college try