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.
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.
This compares favorably with the actual
graph of the polynomial:
Some of my details (like my max and min
points) were a little off, but my overall sketch was still pretty good.
