# cal

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Determine all values of x, (if any), at which the graph of the function has a horizontal tangent.

y(x) = 6x/(x-9)^2

when i workout this problem I get this:

by quotient rule:

dy/dx = ( (x-9)^2 (6) - 6x(2)(x-9))/(x-9)^4
= 0 at a horizontal tangent

6(x-9)^2 - 12x(x-9) = 0
6(x-9)[x-9 - 2] = 0
6(x-9)(x-11) = 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

• cal -

you lost an x there when factoring out the 6(x-9)

y' = -6(x+9)/(x-9)^3
y'=0 at x = -9

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