Question

What is the horizontal asymptote for the above function? y=x/x^2-16y=​

Answer 1

$\bf y = \cfrac{\stackrel{\textit{degree of 1}}{x^1}}{\underset{\textit{degree of 2}}{x^2-16}}$   when the degree of the denominator is greater than that of the numerator, the only horizontal asymptote occurs at y = 0.