simple one variable polynomial function, point of inflexion: finding coefficients $$f(x) = a x^3 + b  x^2 + c x + d$$
The graph of $f$ definitely has a point of inflexion. Which of the coefficients  $a, b, c, d$ determines that the point of inflexion of the function $f$ lies on the vertical axis? Write down this coefficient and the corresponding condition."
I tried to calculate it, the answer doesn't seem to me that simple.
the second derivative is:
$$f''(x)=6ax+2b$$
so the coordinates of the point of inflexion are $(-b/(3a),f(-b/(3a))$
And I know that $f(b/(3a))=0$, because it lies on the $x$ axis.
But this is one equation with four variables.
Where did I overcomplicate it?
Thank you
 A: With
$f(x) = ax^3 + bx^2 + cx + d, \tag 1$
we have
$f'(x) = 3ax^2 + 2bx + c, \tag 2$
and
$f''(x) = 6ax + 2b; \tag 3$
points of inflection occur at those $x$ for which
$f'(x) = f''(x) = 0, \tag 4$
that is, where
$6ax + 2b = 0, \tag 5$
and
$3ax^2 + 2bx + c = 0; \tag 6$
(5) is possessed of the unique solution
$x = -\dfrac{b}{3a}, \tag 7$
if there is an inflection point, this $x$ must also satisfy (6), whence
$\dfrac{b^2}{3a} - \dfrac{2b^2}{3a} + c = c - \dfrac{b^2}{3a} = 0. \tag 8$
Now I assume
$a \ne 0, \tag 9$
necessary for $f(x)$ to be a bona fide cubic; thus the expressions (7)-(8) make sense; for the inflection point to lie on the vertical axis $(0, y)$ we must have
$x = -\dfrac{b}{3a} = 0 \Longrightarrow b = 0,\tag{10}$
and then via (8)
$c = 0 \tag{11}$
as well.
It thus appears that $b$ is the coefficient of (1) determinative of whether an inflection point lies on the vertical axis; we observe that $b$ also specifies $c$ via (8), so that (1) becomes
$f(x) = ax^3 + d, \tag{13}$
a simple cubic translated up or down in $y$ according to $d$.
