find all horizontal asymptote for f(x)= 6x/ (sqrt x^2 -3)
plese help me. i dont even know what to do.
2007-10-20 11:30:29 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
to find horizontal asymptotes, take the limit as the function approaches +- infinity. If the limits don't exist, no horiz. asymptotes.
lim(x->infinity)6x/sqrt(x^2-3) = lim(x->infinity)6x/x = 6
lim(x-> -infinity)6x/sqrt(x^2-3) = lim(x-> -infinity) 6x/-x = -6
2007-10-20 11:36:05 · answer #1 · answered by holdm 7 · 1⤊ 0⤋
f(x)= 6x/ (sqrt( x^2 -3))
First you find where the function is defined
it is defined on the interval ]-infinity ; -sqrt(3)[ and ]sqrt(3) ; +infinity[
horizontal asymptotes are found (if they exist) when x ---> plus, minus infinity
as x--->+infinity , sqrt(x^2 -3) behaves as x
as x--->-infinity, sqrt(x^2 -3) behaves like -x
hence
as x--->+infinity, we are left with 6x/x =6
as x--->-infinity, we are left with -6x/x =-6
those are the 2 horizontal asymptotes
the graph of your function is between y =-6 and y =+6
2007-10-20 18:44:05 · answer #2 · answered by Any day 6 · 0⤊ 0⤋
Horizontal asymptotes are y = 6 and y = -6.
Divide 6 by +/- 1
2007-10-20 18:40:15 · answer #3 · answered by ironduke8159 7 · 0⤊ 0⤋