Calculus
posted by Anonymous .
let f(x)=(x^2+2x15)/(3x^2+12x15)
Find the horizontal and vertical asymptotes of f(x).

To have a vertical asymptote , the denominator must be zero
so, 3x^2 + 12x  15 = 0
x^2 + 4x  5 = 0
(x+5)(x1) = 0
x = 5 or x = 1
so we have two vertical asymptotes,
x = 5 and x = 1
for the horizontal asymptote, let x become large
then we are left with x^2/3x^2 or 1/3
V.A. : y = 1/3 
I did the exact same thing as you did for the verticle asymptotes, but my online homework says that 5, or 1 are wrong. I even tried putting both of them in list form.

sorry, but I am 100% confident that what I did was correct. Since I can't see what type of input your online website expects, we are at an impasse.