<s>[INST] determining the cubic function from a graph This question may be a bit below the regular level for this forum, but I really struggle finding the solution. All of my attempts seem to end up describing $y = 1$, probably because all the points I can read from the graph are on that line. While this was given as a homework assignment, it wasn't to me, and I have been struggling with it for days now, since I was asked to help. The deadline has long passed. I have, in the meantime, studied several of my old text books, to no avail.
The question, as stated, was: given the graph below, determine the cubic function expression on the form $f(x) = ax^3 + bx^2 + cx + d$.

Note that the graph crosses $(0,1)$, has a local maximum at $x=1$ and minimum at $x=3$, and satisfies $f(3)=1$.
This corresponds to $f'(x) = 3ax^2 + 2bx + c$ and $f''(x) = 6ax + 2b$.  I know that the $a$ value must be positive, given the graph.
I have determined visually that $f(0) = 1$, $f(3) = 1$, $f'(1) = 0$, $f'(3) = 0$, $f''(2) = 0$.
The exact values of the root or the value at $f(1)$ are impossible to tell.
This gives me this set of equations:
$$d = f(0) = 1 \tag{1}$$
$$f(3) = 1 \Rightarrow 27a + 9b + 3c + 1 = 1 \Rightarrow 27a + 9b + 3c = 0 \tag{2}$$
$$f'(1) = 0 \Rightarrow 3a + 2b + c = 0 \Rightarrow c=-3a-2b\tag{3}$$
$$f'(3) = 0 \Rightarrow 27a + 6b + c = 0 \tag{4}$$
$$f''(2) = 0 \Rightarrow 12a + 2b = 0 \Rightarrow b = -6a \tag{5}$$
Substituting $(3)$ and $(5)$ into $(2)$ gives me:
$$27a + 9(-6a) + 3(-3a-2(-6a)) = 0 \Rightarrow 27a - 54a + 27a = 0\Rightarrow 0=0\tag{6}$$
I get a similar result if I substitute into $(4)$:
$$27a + 6(-6a) + (-3a -2(-6a)) = 0 \Rightarrow 27a-36a-3a+12a = 0\Rightarrow0=0\tag{7}$$
But $(6)$ and $(7)$ are useless. I realize that all of the points I'm using are on $y = 1$ and the derivatives hold true for that line, too. Given that the graph isn't precise enough to actually read the values from either the root or the value of the local maximum - how do I proceed here to find the value of $a$? It seems to me I've pretty much found the rest once I lock down that one.
 [/INST] The derivative of the cubic function has two zeroes at $x = 1 $ and $x = 3$.  Therefore,
$f'(x) = A (x - 1)(x - 3) = A (x^2 - 4 x + 3) $
Integrate that from $0$ to $x$,
$f(x) = f(0) + A (\dfrac{x^3}{3} - 2 x^2 + 3 x ) $
Now from the graph, $f(0) = 1$ and  $f(3) = 1$ , hence,
$f(x) = 1 + A (\dfrac{x^3}{3} - 2 x^2 + 3 x ) $
Substitute $f(3) = 1$
$ 1 = 1 + A ( 9 - 18 + 9 ) $
which is $0 = 0$ , so $A$ cannot be determined from the information used so far.
So let's consider $x = 2$ and let's assume that the value of $y$ at that $x$ is 1.75
Then $1.75 = 1 + A (\dfrac{8}{3} - 8 + 6 ) $
From which, $ 0.75 = A (\dfrac{2}{3} )$, which means that $A = \dfrac{9}{8} $
Hence the cubic function equation is
$ y = f(x) = 1 + \dfrac{9}{8} (\dfrac{x^3}{3} - 2 x^2 + 3 x ) $
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