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Calculus/Vertical Asymptote

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Expert: - 12/16/2008


Question
I am tryng to find the vertical asymptote of the following equation but I am confused by a negative answer.  

5t /(.01t^2 +3.3)

Answer
To find the vertical assymptote, that would be where the denominator is 0.

Looking at the problem, we need to find where 0.01t² + 3.3 is 0.
Here, 0.01t² is always positive and 3.3 is always positive, so the sum of the two is always positive.  This would mean that there is no vertical assymptote.

However, graphing the function in Excel shows me that it is concave up when x≤0 and concave down when x≥0.

Taking the derivative (quotient rule) gives (lo d hi - hi d lo)/lo².
5(0.01t²+3.3) - 5t(d lo)/(0.01t²+3.3)².

Evaluating this at t = 0 gives 5(3.3)/3.3² = 5/3.3 = C.
I'll let you compute C.

The assymptote that does occur is at y=Cx.

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