If I have f(x) = (2x^2+x-2)/(x^2-1), I know I have a horizontal asymptote at y=2. However, with x=0, y=2. How is y=2 an asymptote if this function has a point at (0,2)?
This is a very good question to ask, as it really susses out the distinction between vertical asymptotes and horizontal asymptotes. Vertical asymptotes are interested in the limiting values of the inputs. So, a vertical asymptote will be a line of the form x = (something), where as x approaches this value, the corresponding y-values "blow up" to either positive or negative infinity (often times the function will go to positive infinity when approaching from one side of the asymptote and to negative infinity when approaching from the other side).
By contrast, horizontal asymptotes are interested in the limiting values of the outputs. Just as the y-values "blew up" to infinity near vertical asymptote, so too do the x-values near a horizontal asymptote. What this basically means is that the function grows ever closer and closer to the horizontal asymptote but never actually reaches it. However, there's a huge thing that needs to be noticed here. This behavior of approaching-but-never-reaching the line only applies when the x-values are going off towards infinity (or negative infinity). For any finite x-value, the function is free to touch and/or cross the horizontal asymptote as many times as it wishes, so long as it, in the limit as x approaches infinity, settles down into an asymptotic behavior with the line. A result of this is that there can be infinitely many vertical asymptotes, there can only be at most two horizontal asymptotes (one as x approaches infinity, and the other as x approaches negative infinity).
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