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 graph of f has a vertical intercept at
f(0) = a(−r1)m1 ⋯ (−rk)mk;
that is, the vertical intercept 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.

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.
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.
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