Hello, euclid2!
I use a very primitive approach.
But it has always worked well for me.
Sketch the asymptotes . . . Code:
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The only intercept is at (0,0).
The
-axis is partitioned into four intervals.
Evaluate a point in each interval.
. . ![\begin{array}{ccccc}f(\text{-}3) &=& -\frac{3}{5} & \Rightarrow & (-3,\:-\frac{3}{5}) \\ \\[-3mm]<br />
f(\text{-}1) &=& +\frac{1}{5} & \Rightarrow & (-1,\:\frac{1}{5})\\ \\[-3mm]<br />
f(1) &=& -\frac{1}{3}& \Rightarrow & (1,\:-\frac{1}{3}) \\ \\[-3mm]<br />
f(3) &=& +\frac{3}{5} & \Rightarrow & (3,\:\frac{3}{5})\end{array}](http://latex.codecogs.com/png.latex?\begin{array}{ccccc}f(\text{-}3) &=& -\frac{3}{5} & \Rightarrow & (-3,\:-\frac{3}{5}) \\ \\[-3mm]<br />
f(\text{-}1) &=& +\frac{1}{5} & \Rightarrow & (-1,\:\frac{1}{5})\\ \\[-3mm]<br />
f(1) &=& -\frac{1}{3}& \Rightarrow & (1,\:-\frac{1}{3}) \\ \\[-3mm]<br />
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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