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

## Answer This Question

## Related Questions

- calc. - I'm having a lot of trouble with this problem: Sketch the graph and show...
- calc - one more question!! what values of c deos the polynomial f(x) = x^4 + cx^...
- Calc - Profit=p(-p^2+33p+9)-9(-p^2+33p+9)+100 How do I simplify this so I can ...
- calc - The function f is defined by f(x) = x^3 - x^2 - 4x + 4 The point (a,b) is...
- Calc - 1. Find values of a,b,c, and d such that g(x) = a(x^3)+b(x^2)+cx+d has a ...
- Math - Find the intervals on which the function is increasing, decreasing, ...
- Math - find the the 2nd derivative and set equal to zero in order to find ...
- Calculus - For y=(1/4)x^4-(2/3)x^3+(1/2)x^2-3, find the exact intervals on which...
- Calculus - What is the point of inflection of the function f(x)=x^3-8x^2+5x+50? ...
- Calc Help - Find two functions that this could be the derivative of: y'=4x+7 ...

More Related Questions