How do you find the Vertical, Horizontal, and Oblique Asymptote given $g(t) = (t − 6) / (t^(2) + 36)$ ?

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Answer 1 · 2018-08-07T22:53:31

This function only has a horizontal asymptote $y=0$

Given:

$g(t) = (t-6)/(t^2+36)$

Note that the denominator is positive for any real value of $t$ .

Hence this rational function has no vertical asymptotes and no holes.

The degree of the denominator is also greater than the numerator. So it has no blique asymptotes - only the horizontal asymptote $y=0$ .

$lim_(t->+-oo) (t-6)/(t^2+36) = lim_(t->+-oo) (1/t-6/t^2)/(1+36/t^2) = (0-0)/(1+0) = 0$

graph{(x-6)/(x^2+36) [-20, 30, -0.3, 0.3]}

How do you find the Vertical, Horizontal, and Oblique Asymptote given g(t) = (t − 6) / (t^(2) + 36)g(t) = (t − 6) / (t^(2) + 36)?