# Ask Uncle Colin: The Turning Points of a Cubic

Dear Uncle Colin,

How can I find the turning points of a cubic (for example, $y = x^3 - x$) without calculus?

- Calculus Unnecessary (But It’s Clever)

Hi, CUBIC, and thanks for your message. This one gave me a proper ‘oo’ when I realised how it worked!

The key insight here is to translate the cubic vertically until it just grazes the $x$-axis, use the fact that this gives you a repeated root to solve for $x$, and translate it back to solve for $y$.

In this example, translating the curve vertically upwards by $k$ gives a graph $y = x^3 - x + k$. This has a double root, so we can write it as $y= (x-a)^2(x-b)$ for some values of $a$ and $b$.

Expanding, it’s $y = x^3 - (2a+b)x^2 + (a^2 + 2ab)x - a^2b$.

Matching coefficients, $2a+b = 0$, $a^2 + 2ab = - 1$ and $-a^2 b = k.

The first of those gives $b = -2a$; substituting into the other two gives $-3a^2 = -1$ and $-2a^3 = k$.

The two possibilities for $a$ (which is the $x$-value we’re after) are $\pm \frac{1}{\sqrt{3}}$, and the corresponding values of $k$ are $\mp \frac{2}{3\sqrt{3}}$.

The turning points are therefore at $\left( \frac{1}{\sqrt{3}}, -\frac{2}{3\sqrt{3}}\right)$ and $\left( -\frac{1}{\sqrt{3}}, \frac{2}{3\sqrt{3}}\right)$.

Take that, Newton!

Hope that helps,

- Uncle Colin