Question

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

Answer 1

See explanation.

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]}