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 graph of f appears on a large scale like a monomial:
f(x) ~ axm1+⋯+mk;
the coefficient on that monomial is the leading coefficient of the polynomial, and the exponent on that monomial is the degree of the polynomial. On a smaller scale, the initial value of f is
f(0) = a(−r1)m1 ⋯ (−rk)mk;
thus, the vertical intercept of the graph is (0, f(0)), as given by that formula.

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 (which is Example 5.1.10 on page 334 of the textbook). 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, and the 2nd root (i = 2) is r2 = 3 with multiplicity m2 = 2. At this point, I could draw a very rough graph: Compare the graph on the bottom of page 334 in the textbook; if you travel along it from right to left, then this is what it does.

The remainder of these notes are optional. I can draw a more precise graph as follows: The degree is m1 + m2 = 1 + 2 = 3, so the large-scale behaviour is

f(x) ~ 2x3.
This is what the graph is like at the extreme corners. The vertical intercept is at
f(0) = 2(½)(−3)2 = 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 information, you could draw a graph about as nice as the one in the book without having to calculate any additional points.
Go back to the course homepage.
This web page was written in 2015 and 2016 by Toby Bartels, last edited on 2016 September 1. Toby reserves no legal rights to it.

The permanent URI of this web page is http://tobybartels.name/MATH-1150/2017SP/polynomials/.

HTML 5