Hello, euclid2!
I use a very primitive approach.
But it has always worked well for me.
Graph: .$\displaystyle f(x) \:=\:\frac{x}{x^24} $
i know it has V.A. at $\displaystyle x = \pm2$ and H.A. at $\displaystyle y = 0$
. . because (deg. of denom'r) > (deg of num'r) . . . . Right!
Sketch the asymptotes . . . Code:

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The only intercept is at (0,0).
The $\displaystyle x$axis is partitioned into four intervals.
Evaluate a point in each interval.
. . $\displaystyle \begin{array}{ccccc}f(\text{}3) &=& \frac{3}{5} & \Rightarrow & (3,\:\frac{3}{5}) \\ \\[3mm]
f(\text{}1) &=& +\frac{1}{5} & \Rightarrow & (1,\:\frac{1}{5})\\ \\[3mm]
f(1) &=& \frac{1}{3}& \Rightarrow & (1,\:\frac{1}{3}) \\ \\[3mm]
f(3) &=& +\frac{3}{5} & \Rightarrow & (3,\:\frac{3}{5})\end{array}$
Plot these points.
Using your knowledge of asymptotes, sketch the graph.
Code:
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