First I'll find any vertical
asymptotes, by setting the denominator equal to zero and solving:

x
– 2 = 0
x
= 2

So I have a vertical
asymptote at x
= 2.

I'll dash this
in:

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Looking at the degrees
of the numerator and denominator, I see that the numerator is a quadratic
while the denominator is linear. Since the degree is one greater in
the numerator, I know that I will have a slant asymptote. But I need
to do the long division to find out what the equation of the slant asymptote
is going to be.

Hmm... The slant asymptote
is clearly y
= x + 1, but
there is no remainder; the division came out evenly. That's because
the numerator happens to factor as (x
– 2)(x + 1),
so the x
– 2 factor divides
out evenly and there is no remainder.

In any case, the
slant (not horizontal) asymptote is at y
= x + 1:

y
= 0: 0 = (x^{2}
– x – 2) / (x – 2)
_{
}0
= x^{2} – x – 2
0 = (x
– 2)(x + 1)
x
= 2, x = –1

I know that I can't
have x
= 2, as an
x-intercept,
because this is actually the vertical asymptote. So I'll ignore
that "solution" of the above equation.

The intercepts
then are (0,
1) and (–1,
0):

Now I'll plot a few more
points. However, since this example is actually a special case of slant-asymptote
problem (and something of a trick question), I'll also show you the
points for the slant asymptote only.

x

y
= ^{(x2 – x – 2)}/_{(x
– 2)}

y
= x + 1

–6

y
= ^{(36 + 6 – 2)}/_{(–6 – 2)} = –5

y
= –5

–5

y
= ^{(25 + 5 – 2)}/_{(–5 – 2)} =
–4

y
= –4

–4

y
= ^{(16 + 4 – 2)}/_{(–4 – 2)} =
–3

y
= –3

–3

y
= ^{(9 + 3 – 2)}/_{(–3 – 2)} =
–2

y
= –2

–2

y
= ^{(4 + 2 – 2)}/_{(–2 – 2)} =
–1

y
= –1

–1

y
= ^{(1 + 1 – 2)}/_{(–1 – 2)} = 0

y
= 0

0

y
= ^{(0 – 0 – 2)}/_{(0 – 2)} = 1

y
= 1

1

y
= ^{(1 – 1 – 2)}/_{(1 – 2)} = 2

y
= 2

2

—
y is not defined —

y
= 3

3

y
= ^{(9 – 3 – 2)}/_{(3 – 2)} = 4

y
= 4

4

y
= ^{(16 – 4 – 2)}/_{(4 – 2)} = 5

y
= 5

Do you see how that, except
for the one point where the rational function isn't defined (at x
= 2), the two lines
are the same? In general, this is not true for rationals. But in this
special case — where the long division works out with no remainder — you
don't actually even need the asyptote lines dashed in. Instead, you find
the slant asymptote equation (in this case, y
= x + 1), and
you draw that in for the rational
graph.

The only difference
between the slant asymptote of the rational function and the rational
function itself is that the rational function isn't defined at
x
= 2. To account
for this, I leave a nice big open circle at the point where x
= 2, showing
that I know that this point is not actually included on the graph,
because of the zero in the denominator of the rational.

Warning: This graph is
not the norm for rationals. But you should expect to encounter one like
this, perhaps on the test. I've only ever seen this done with slant asymptotes,
but it could be done with horizontal asymptotes, too. Keep in mind that
you can't just "cancel off" duplicate factors when graphing
rationals; you still need to account for all the zeroes of the denominator.

Note that your calculator
will probably not show this hole in the graph, and "TRACE" will
probably miss it too, since it is unlikely that x
= 2 will be exactly
on the pixel that the calculator is processing. But even if the graph
does not show the gap in the line, the calculator will show no value for
y
when x
is two.

When I
asked my calculator to evaluate y
for x
= 2, I got this:

If it just so happens that
the vertical asymptote falls on a pixel (rather than between
two pixels), then the gap in the graph will display on the calculator's
pretty picture.

For instance,
when I changed the window settings on my graphing calculator so
the x-values
ran from –7
to 7
(rather than the default –10
to 10),
the gap was (just barely) visible:

When you do these graphs,
get in the habit of working methodically through all the steps (vertical
asymptote(s), horizontal or slant asymptote, intercept(s), plotted points,
and graph), so you can do the graphing easily when you get to the test.
And, as you can see above, you will need to know this material well enough
that your calculator doesn't mess you up.

Stapel, Elizabeth.
"Graphing Rational Functions: The Special Case with the 'Hole'." Purplemath. Available from http://www.purplemath.com/modules/grphrtnl4.htm.
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