3 years ago

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

1 answer:

8 0

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

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

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