## Graphing asymptotes: y = (-x + 1) / (x^2 - 2x)

Complex numbers, rational functions, logarithms, sequences and series, matrix operations, etc.
angryrabbit
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### Graphing asymptotes: y = (-x + 1) / (x^2 - 2x)

Hi. I am curious to know if there is a way to graph the function...

$y = \left(\frac{-x + 1}{x^2 - 2x}\right)$

... without manually plotting each point. I imagine using something like the transformation rules, but I find it difficult to mentally parse which transformations to apply. For example, given...

$y = x^2 + 2$

... I know to shift the parabola up by two units. And, say...

$y = (x + 2)^2$

... shifts leftward two units.

But when it comes to the rational function, I'm confused as to what constitutes a change to the x value versus the y value. Any ideas?

stapel_eliz
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angryrabbit wrote:Hi. I am curious to know if there is a way to graph the function...

$y = \left(\frac{-x + 1}{x^2 - 2x}\right)$

... without manually plotting each point.

Since there are infinitely-many points on the curve, plotting "each" point is obviously impossible.

Trying to figure out transformations from some base function would probably be much too much trouble. Instead, try first finding the vertical and horizontal asymptotes and the intercepts. Then find just enough points in between the "interesting" bits to give you the general idea of what the graph looks like.

angryrabbit
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Joined: Tue Mar 17, 2009 9:21 pm
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### Re:

Wow, I got a reply from THE Elizabeth.

stapel_eliz wrote:Since there are infinitely-many points on the curve, plotting "each" point is obviously impossible.

Ah, yes. I should have worded that better. Clearly, attempting to plot every point would keep me occupied until death came knocking.

stapel_eliz wrote:Trying to figure out transformations from some base function would probably be much too much trouble. Instead, try first finding the vertical and horizontal asymptotes and the intercepts. Then find just enough points in between the "interesting" bits to give you the general idea of what the graph looks like.

I ended up plotting a total of 4 points. One on the far left, and one on the far right of the graph ("outside" the two vertical asymptotes). The x-intercept acted as a point between the two vertical asymptotes. There was no y-intercept in this problem. Finally, I plotted two more points between the vertical asymptotes. I was curious if I could avoid plotting even that many points though.

I suppose in practice it wasn't that tough, just mildly tedious. Oh well. Thank you for all your help Elizabeth. Your math instruction is second-to-none.