# What are the points of inflection, if any, of f(x)=x^3 - 12x^2 ?

Nov 18, 2016

$T h e$ point of inflexion (POI) is ( 4, -`128 ). I could not drag the graph to $y < - 90$, tosee the POI that is far down below.

#### Explanation:

$y = {x}^{3} - 12 {x}^{2} = {x}^{2} \left(x - 12\right) = 0$, for x = 0 and x = 12.

$y ' = 3 {x}^{2} - 24 x = 3 x \left(x - 8\right) = 0$, for x = 0 and x = 8.

These are turning points. The first point is (0, 0). The second is far

down at $\left(8 , - 256\right)$, to rise for $y \to \infty , a s x \to \infty$, through (12, 0)..

y''=6x-24=0, for x = 4. This point ( 4, -128) is a candidate for point of

inflexion (POI).

y'''= 6. y''' not 0 when y'' = 0 is a sufficient condition,

for this point (4, -128) to be a POI.

Of course, as $x \to - \infty , y \to - \infty$.

Using contracting scale 1 for 20 , exclusively for y and 1 for 2 in x,

this POI and other features that are not revealed by this graph,

could be brought lo light, from hiding.

In this edition, I have managed contraction in another graph ( the

first below ) to reveal more features. Yet, for the the POI (point of

inllexion ) location, local zooming has to be made at $\left(4 , - 128\right)$.

graph{y=x^3-12x^2 [-324.5, 324.5, -166.8, 157.7]}
graph{y=x^3-12x^2 [-9.915, 10.085, -9.52, 0.48]}