How do you find all points of inflection given $y=x^3+9x^2+24x+22$ ?

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Answer 1 · 2016-11-22T16:14:17

$( -3, 4 )$

graph{y=x^3+9x^2+24x+22 [-20, 20, -10, 10]}

$y'' = 6x+18=0$ , when $x = -3 and f'''=6$ ( not 0 ) reveal that

$x=-3$ is the point of inflexion (POI).

See the graph and locate this POI $( -3, 4 )$ , at which the

tangent crosses the curve.

Clue: It near in the middle-part middle. In the other two parts,

$y to +-oo$ , as $x to +-oo$ .

How do you find all points of inflection given y=x^3+9x^2+24x+22y=x^3+9x^2+24x+22?