Posted by **Paisley** on Wednesday, March 18, 2009 at 3:43pm.

f(x)= x^3/x^2-16 defined on [-19, 16]

How would you find the vertical asymptopes? and the inflection point?

I think you take the derivative and set it to zero? but i cant figure out the correct derivative or something??

- calc -
**Damon**, Wednesday, March 18, 2009 at 3:51pm
assume you mean (x^2-16) on the bottom

vertical asymptotes are where the denominator goes to zero

that is when x = -4 and x = +4

derivative of fraction =

(bottom * derivative of top - top * derivative of bottom)

all over bottom squared

[(x^2-16)*3 x^2 - x^3(2x) ]/bottom^2

look for zeros of numerator

3 x^4 - 48 x^2 - 2 x^4

=x^4 - 48 x^2

= x^2 (x^2-48)

= x^2 (x-sqrt 48)(x+sqrt 48)

so zero at -sqrt 48, 0 and + sqrt 48

check my arithmetic, I did that fast.

- calc -
**Damon**, Wednesday, March 18, 2009 at 3:52pm
by the way sqrt 48 = 4 sqrt 3

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