STATIONARY POINTS EXAMPLE WITH AN INFLEXION AS WELL

QUESTION: Consider the curve $$y=x^3-3x^2-9x+27$$
(a) Find the coordinates of any stationary points and determine their nature.
(b) Find the coordinates of any points of inflexion (and verify using a y'' box that concavity changes)
(c) Get the $x$ and $y$ intercepts. (its possible here)
(d) Sketch it!
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Before we begin just find y' and y'' right away, factorise both of them ready for use later!

$y'=3x^2-6x-9$
  $y'=3(x^2-2x-3)$
  $y'=3(x-3)(x+1)$                                                (1)

$y''=6x-6$
$y''=6(x-1)$                                                          (2)


(a) For Stat Pts: solve $y'=0$.

 $3(x-3)(x+1)=0$                                      from (1)

$\therefore x=-1, 3 $

For $x=-1, y=(-1)^3-3(-1)^2-9(-1)+27=32$

For $x=3, y=(3)^3-3(3)^2-9(3)+27=0$

Stationary Points are  $(-1, 32)$  and  $(3,0)$.
Now determine their nature using a $y'$ box for each point.

For $(-1, 32)$.

 For $(3, 0)$.

[REMARK: Instead of a $y'$ box we could use $y''$ to test concavity at each $x$ value, to determine the nature. This works only if $y''\neq 0$.
Since $y''(-1)<0$ it's concave down at $x=-1$, so it's a MAX.
Since $y''(3)>0$ it's concave up at $x=3$, so it's a MIN. ]

(b)  For possible inflexion, solve $y''=0$

$6(x-1)=0$, 
$\therefore x=1$.

When $x=1$, $y=(1)^3-3(1)^2-9(1)+27=16$
Possible Inflexion is $(1,16)$.

Verify that is an inflexion using $y''$ box.

(c)  For $y$ intercept put $x=0$.  $\therefore y=27$ For $x$ intercept put $y=0$.  $\therefore 0=x^3-3x^2-9x+27$. By inspection we see that $x=-3,3$.

[REMARK: its often the case that you can't easily find the $x$ intercepts. Only do this as necessary or unless asked to do so in the question!]

(d)