(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
Now determine their nature using a y' box for each point.
For (-1, 32).
[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)
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