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#.