Can someone help me with concavity? f(x) = (x^3)/(x^2-25)

[baphomet58]

f(x) = (x^3)/(x^2-25)
defined on the interval [-18,17].

Smallest x values must be entered first.

The function f(x) has vertical asympototes at _-5_ and _5_. <----- I think those are correct.

f(x) is concave up on the region __ to __and __to __.

The inflection point for this function is __.

Thanks, I don't know the other blanks.

 

[BigGlenntheHeavy]

Re: Can someone help me with concavity?

Vertical asymptotes, x = ±5 Horizontal asymptote, none, Oblique asymptote, y =x

Point of inflection (0,0)

concave up: (-5,0]µ(5,17], concave down: [-18,-5)µ[0,5)

note: if you graph the function, all this can be discover by inspection.

 

[baphomet58]

thanks. yes, i realize you can graph it to easily find the answers. However, for some strange reason the function wasn't graphing properly.