Question

HELP!!! What is the asymptote for the above function y=2x^2/3x^2-16???

Answer 1

the asymptote is at 3x^2=16, therefore it is sqrt(5&1/3)

Answer 2

the horizontal asymptote is\[\frac{2}{3}\] as x goes to \[\pm \infty\]

the vertical asymptote is
\[\pm \infty\]
as x goes to
\[\pm \frac{4}{[\sqrt{3}]}\]