Curve Sketching with $y''=0$ but NO inflexion.

Curve Sketching with $y''=0$ but NO inflexion.

Consider the curve, 
      $$y=x^4-8x^3+24x^2-288x+5$$

      $y'=4x^3-24x^2+48x-288$

      $y'=4x^2(x-6)+48(x-6)$
      $y'=(4x^2+48)(x-6)$

      $y'=4(x^2+12)(x-6)$

      $y''=12x^2-48x+48$

      $y''=12(x^2-4x+4)$

      $y''=12(x-2)^2$

Stationary Points: Solve $y'=0$.
$\therefore x=6$ and so the stationary point is $(6,-1291)$.

When $x=6$, $y''=12(6-2)^2>0$ and hence it's a minimum turning point. 

Possible Inflexions: Solve $y''=0$.
$\therefore x=2$ and the possible inflexion point is $(2, -523)$.

Check the sign of $y''$ either side of the possible inflexion. 

$x$	$1$	$2$	$3$
$y’’$	$+$	$0$	$+$

The sign of $y''$ stays positive, and so this is NOT an inflexion.
The curve just flattens out at the point where $x=2$. See graph.



Here are a list of curves with similar features, in that y''=0 but not inflexion!

• $y=x^4-8x^3+24x^2-288x+5$ (above)

• $y=3x^4-12x^3+18x^2-108x+1$

• $y=x^4-4x^3+6x^2-36x-10$

• $y=3x^4-8x^3+8x^2-32x+11$

• $y=3x^4-20x^3+50x^2-500x$

• $y=3x^4-4x^3+2x^2-56x+2$ ($y'$ has a $x-2$ as a factor, hint :-))