cal
posted by cal .
Determine all values of x, (if any), at which the graph of the function has a horizontal tangent.
y(x) = 6x/(x9)^2
when i workout this problem I get this:
by quotient rule:
dy/dx = ( (x9)^2 (6)  6x(2)(x9))/(x9)^4
= 0 at a horizontal tangent
6(x9)^2  12x(x9) = 0
6(x9)[x9  2] = 0
6(x9)(x11) = 0
x=9 or x=11 , but xâ‰ 9 , there is a vertical asymptote at x=9
x = 11
but I have these answer choices to choose from:
A. x=9 and x=6
B. x=9
C. x=9 and x=6
D. x= 6
E. The graph has no horizontal tangents.
is it E then

you lost an x there when factoring out the 6(x9)
y' = 6(x+9)/(x9)^3
y'=0 at x = 9