How do you sketch the general shape of #f(x)=x^3-10x^2+33x-38# using end behavior?

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Answer 1 · 2017-02-10T11:58:19

See explanation.

graph{x^3-10x^2+33x-38 [0, 5, -5, 5]}

Polynomials #x^n+a_1x^(n-1)+..+a_n# are differentiable,

for #x in (-oo, oo)#.

Here, as #x to +-oo, f=x^3(1-10.x=33/x^2-38/x^3) to +-oo#

The not-to-scale graph above reveals turning points and point of

inflexion..

#f' =3x^2-20x+33=0, at x= 3 and 11/3#

#f''=6x-20=0, at x = 10/3#. Here, and everywhere, #f'''=6 ne 0#.

So, the POI is at x =10/3.

Overall rise-and-fall graph is inserted below.

graph{x^3-10x^2+33x-38 [-30, 20, -15, 10]}