Finding the equation of a cubic function? I am given two points on a cubic curve: a local maximum at $(-2,3)$ and a local minimum at $(1,0)$. I am to find an $f(x)=ax^3+bx^2+cx+d$.
So far, I've been able to ascertain only a little:
$f'(x) = A(x+2)(x-1) = 3ax^2 + 2bx +c$
therefore
$$
\begin{eqnarray}
2b &=& 3a \\
c &=& -4b = -6a \\
\end{eqnarray}
$$
However, I am unsure how to proceed. The textbook chapter did not cover this; it described ways to use $f(x)$ and it's derivatives to analyze the appearance of a graph, but nothing about finding an equation from part of it's derivative.
 A: You have four linear equations and four unknown quantities: $$f(-2)=3,\ \  f(1)=0,\ \  f'(-2)=0, \mbox{ and } f'(1)=0$$ Thus
$$f(-2)=-8a+4b-2c+d=3$$$$f(1)=a+b+c+d=0$$$$f'(-2)= 12a-4b+c=0$$$$f'(1)=3a+2b+c=0$$
A: First, you need to require your points to lie on the curve. Clearly this will depend on the values of $a,b,c,d$. So since $(1,0)$ is on the curve, this implies that $f(1) = a+b+c+d= 0$
This is your first equation; you will need other 3 to find $a,b,c,d$
Another one you get from the other point; next, you impose that $f'(x)=0$ in those points. This gives you sufficient conditions to solve the problem 
A: You're almost there.
You got the derivative: $f'(x) = 3ax^2 + 2bx + c$.  You have the relations $2b = 3a, c = -4b = -6a$ by setting the derivative to zero at those two points.
Now, go back to your original equation.  The relations allow you to eliminate two of the four unknown constants.  But you have the $y$ coordinates for those two values, and hence can solve for the two remaining unknowns.  Once you know those two, you can find the other two with the relations you have already.
So, let's get rid of $b$ and $c$ to leave $a$ and $d$:
$$f(x) = ax^3 + \frac{3a}{2}x^2 + (-6a)x + d.$$
Now, evaluate this for $(-2, 3)$:
$$3 = a(-2)^3 + \frac{3a}{2}(-2)^2 + (-6a)(-2) + d.$$
One equation with two unknowns.
Repeat by evaluating the function at $(1,0)$.  Then you'll be able to solve for $a$ and $d$, and then you can solve for $b$ and $c$.
Can you proceed from here?
A: if a cubic poly has a local max at $x = -2$ and local min a t $x = 0,$ then it has an inflection point at $x = \frac 12.$ so you can write  $$y = a(x+\frac 12)^3 + b.$$ now setting $x = 0, y = 0 \to a = -8b.$ and subbing $x = -2, y = 3 \to 3 = -\frac{27} 8 a + b $ solving these two gives $$a = -\frac 6 7, b = \frac 3{28}, y = -\frac 6 7\left(x + \frac 12\right)^3 +\frac 3{28} $$
