Why is the inflection point of cubic functions on x=0 when the quadratic term is removed from the equation?

Question

How is this possible and what's its relationship with the second derivative of a cubic equation because the second derivative of a cubic equation, which is $-\frac{a}{3}$ for $x^3+ax^2+bx+c=0$, gives the inflection point on the graph and used as a substitute to solve cubic equations by removing the quadratic term.

Thanks in advance.

Answer 1

For the root of any polynomial to be at zero, it must have no constant term. The quadratic term in a cubic provides the constant term to the second derivative.

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The inflection point is the point at which the second derivative is zero. The second derivative of your cubic function is a line: $$f’’(x) = 6x + 2a$$

The root of this line is $-a/3$. To put the inflection point on zero you can do a horizontal shift of the whole function. As suggested by Mohammed, the substitution happens to cancel the quadratic term. This makes sense, as the root of the second derivative of a cubic equation with no quadratic term is a line passing through zero:

$$g(x) = x^3 + \beta x + \gamma$$ $$ g’’(x) = 6x $$

Answer 2

You shift your equation via the $$x\to (x-a/3 )$$ so the inflection point is moved from $x=a/3$ to $x=0$

That is the shift which eliminates the $x^2$ term.

Note that the $x^2$ term vanishes in $$(x-a/3)^3 + a(x-a/3)^2 +b(x-a/3)+c=0$$

$$3x^2(-a/3)+ax^2 = 0$$

Answer 3

It is indeed "easier" to solve $x^3+px+q=0$ than $x^3+ax^2+bx+c$. Comparing the second derivatives, you get $6x$ vs. $6x+2a$, and that corresponds to a shift of the variable by $-\dfrac a3$. (As everything shifts, so does the inflection point, which moves to $x=0$.)

Indeed,

$$\left(x-\frac a3\right)^3+a\left(x-\frac a3\right)^2+b\left(x-\frac a3\right)^3+c= x^3+\left(b-\frac{a^2}3\right)x+c-\frac{ab}3+\frac{2a^3}{27}$$

giving the values of $p$ and $q$.

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The whole story:

Let us try to decompose the roots as a sum $x=u+v$. Now the depressed equation yields,

$$(u+v)^3+p(u+v)+q=u^3+v^3+(3uv+p)(u+v)+q=0$$

and if we set

$$3uv+p=0$$

we obtain the system

$$\begin{cases}u^3+v^3+q=0,\\27u^3v^3=-p^3 \end{cases}.$$

Multiplying by $27v^3$,

$$27u^6+27qu^3-p^3=0$$ which is quadratic in $u^3$.