I seem to be having a lot of difficulty finding whether for a horizontal asymptote, whether the curve approaches the asymptote from above or below.(adsbygoogle = window.adsbygoogle || []).push({});

For example, for the problem [tex]y = \frac{6x + 1}{1 - 2x}[/tex], I know that:

For the vertical asymptote, x = 1/2, and that [tex]\lim_{x \to \frac{1}{2}^{+}} = - \infty[/tex], while [tex]\lim_{x \to \frac{1}{2}^{-}} = \infty[/tex].

Meanwhile, for the horizontal asymptote, there is one at y = -3. However, I do not know understand whether it approaches from above or below.

This is how I tried to find it:

As x --> -3 from the left, eg. x = -10, [tex]\frac{6(-3)+1}{1-2(-10)}[/tex] = [tex]\frac{-17}{21}[/tex] = -0.81

therefore approaches from above, because -0.81 is greater than -3.

As x --> -3 from the right, eg. x = 10, [tex]\frac{6(-3)+1}{1-2(10)}[/tex] = [tex]\frac{-17}{-19}[/tex] = 0.95

therefore also approaches from above, because -0.81 is greater than -3.

However, my graphing calculator says otherwise. The graph shows that, left of the vertical asymptote, the curve approaches from above; to the right, it approaches from the bottom. What am I doing wrong?

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# Homework Help: Horizontal asymptotes - approaches from above or below?

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