Once you know the basic
behavior of polynomial graphs, you can use this knowledge to quickly sketch
rough graphs, if required. This can save you the trouble of trying to
plot a zillion points for a degree-seven polynomial, for instance. Once
the graph starts heading off to infinity, you know that the graph is going
to keep going, so you can just draw the line heading off the top or bottom
of the graph; you don't need to plot a bunch of actual points.

Without plotting any
points other than intercepts, draw a graph of the following polynomial:

This polynomial has already
been put into factored form, which saves me the trouble of doing the
solving for the zeroes. I'll just solve the factors, noting the multiplicities
as I go. The zeroes will be:

x
= – 5, with multiplicity 2 (so
the graph will be just touching the x-axis
here) x = –1,
with multiplicity 1 (so
the graph will be crossing the axis here) x
= 4, with multiplicity 3 (so
the graph will be crossing the axis here, but also flexing) x
= 7, with multiplicity 1 (so
the graph will be just crossing the axis here)

Also,
adding the degrees of the factors, I see that this is a polynomial
of degree seven (that is, an odd degree), so the ends will head
off in opposite directions. Because the leading coefficient
is negative, the left-hand end will be "up" (coming
down from the top of the graph) and the right-hand end will
go "down" (heading off the bottom of the graph). So
I can start my graph by pencilling in the zeroes, the behavior
near the zeroes, and the ends, like this:

If
I multiplied this polynomial out (and I'm not going to, so don't
hold your breath), the constant terms of the factors would give
me 5
× 5 × 1 × (–4) × (–4) × (–4) × (–7) = 11 200,
which is rather large. This would explain the large denominator
of the leading coefficient: by dividing the polynomial by a
sufficiently-large number, they made this polynomial graphable.
Otherwise, the graph would likely go off the picture between
the zeroes. (Not all texts notice this, so don't worry about
this consideration if it doesn't come up in class.) When x
= 0, I get
a y-value
of (–1/5600)(11
200) = –2,
so I can pencil this in, too:

I'm not supposed to find
other plot points, so I'll just sketch in a rough guess as to what the
graph looks like. I'll go further from the axis where there is more
space between the zeroes, and I won't be so primitive as to assume that
the y-intercept
point is the minimum point, what with the midpoint between the two nearest
zeroes, x
= –1 and x
= 4, being at x
= 1.5. Granted, the
flexy zero at x
= 4 will probably push
the graph a little to the left, but the bump is still probably to the
right of the y-axis.

So I'll "rough in"
an approximate drawing, and then draw my final answer as a heavier
line, erasing my preliminary sketch-marks before I hand in my
solution.