Graphing polynomials (§5.1)

Suppose that you have a polynomial function f that you can write fully factored:
f(x) = a(x − r1)m1 ⋯ (x − rk)mk.
That is, we have a leading coefficient a (which might be 1 or negative or a fraction), and a list of k distinct roots (or zeroes) r1, …, rk, where the ith root ri has multiplicity mi. In this case, the degree of the polynomial is the sum of the multiplicities:
deg f = D = m1 + ⋯ + mk.
The graph of f appears on a large scale like a monomial:
f(x) ~ axD.
This monomial is the leading term of the polynomial, the monomial with the same leading coefficient and degree as the original polynomial. On a smaller scale, the initial value of f is
f(0) = (−1)Dr1 ⋯ rk,
which is the product of the roots if the degree is even, or the opposite of the product of roots if the degree is odd. Then the vertical intercept of the graph is (0, f(0)).

Besides that, the graph has k horizontal intercepts, one for each root; the ith horizontal intercept is (ri, 0). Furthermore, the graph near that intercept will be close to a linear coordinate transformation of the power function with exponent mi. More specifically, if you take the factored formula for f(x) and substitute ri for each appearance of x except for the one in the factor (x − ri), then you get

f(x) ≈ a (ri − r1)m1 ⋯ (ri − ri−1)mi−1 (ri − ri+1)mi+1 ⋯ (ri − rk)mk (x − ri)mi.
That is, the power function with exponent mi is transformed ri places to the right (producing (x − ri)mi), and then multiplied vertically by all of that stuff out front (in which you'll notice that the variable x doesn't appear, so this will be some specific number in any specific example).

For example, consider f(x) = (2x + 1) (x − 3)2. Rewrite (2x + 1) as 2(x + ½), so

f(x) = 2(x + ½) (x − 3)2 = 2 (x − (−½))1 (x − 3)2;
then the leading cofficient is a = 2, the number of distinct roots is k = 2, the 1st root (i = 1) is r1 = −½ with multiplicity m1 = 1, the 2nd root (i = 2) is r2 = 3 with multiplicity m2 = 2, and the degree is D = m1 + m2 = 1 + 2 = 3. At this point, I could draw a very rough graph:

The remainder of these notes are optional. I can draw a more precise graph as follows: Since a = 2 and D = 3, the large-scale behaviour is

f(x) ~ 2x3.
This is what the graph is like at the extreme corners. Since f(0) = 9 (which I calculated in the very last step in the previous paragraph), the vertical intercept is
(0, 9).
Near the 1st root,
f(x) ≈ 2(x + ½)(−½ − 3)2 = ⁴⁹⁄₂(x + ½),
so approximately the line through (0, −½) with slope ⁴⁹⁄₂. Finally, near the 2nd root,
f(x) ≈ 2(3 + ½)(x − 3)2 = 7(x − 3)2,
so approximately the upwards-opening parabola with vertex (3, 0) stretched vertically by a factor of 7. With this additional information, you could draw a graph about as nice as any in the textbook without having to calculate any additional points.
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