Posted by Paisley on Wednesday, March 18, 2009 at 3:43pm.
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.
by the way sqrt 48 = 4 sqrt 3
Related Questions
calculus - f(x)= x^3/x^2-16 defined on the interval [-19,16]. Enter points, such...
math help please - 2. A teacher asks her class of 22 students, “What is ...
calc - The function f is defined by f(x) = x^3 - x^2 - 4x + 4 The point (a,b) is...
ALGEBRA 1 - Reiny Thanks for your help. But you lost me. I still end up with...
Calculus - What is the point of inflection of the function f(x)=x^3-8x^2+5x+50? ...
Math: Calculus - Answer the following questions for the function f(x) = \frac{ x...
calc. - I'm having a lot of trouble with this problem: Sketch the graph and ...
calc - one more question!! what values of c deos the polynomial f(x) = x^4 + cx^...
calculus - The function f is defined by f(x) = x^3 - x^2 - 4x + 4 The point (a,b...
calc - how do i find the derivative of 16^(sinx)
For Further Reading
