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 − r_k)^m_k.

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, …, r_k, where the ith root r_i has multiplicity m_i. In this case, the graph of f appears on a large scale like a monomial:

f(x) ~ ax^m₁+⋯+m_k;

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 ⋯ (−r_k)^m_k;

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 (r_i, 0). Furthermore, the graph near that intercept will be close to a linear coordinate transformation of the power function with exponent m_i. More specifically, if you take the factored formula for f(x) and substitute r_i for each appearance of x except for the one in the factor (x − r_i), then you get

f(x) ≈ a(r_i − r₁)^m₁ ⋯ (r_i − r_i−1)^m_i−1(r_i − r_i+1)^m_i+1 ⋯ (r_i − r_k)^m_k(x − r_i)^m_i.

That is, the power function with exponent m_i is transformed r_i places to the right (producing (x − r_i)^m_i), 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:

I set up horizontal and vertical axes, marking the scale on the horizontal axis so that the roots, −½ and 3, appear, but ignorning the scale on the vertical axis.
The leading coefficient, a = 2, is positive (rather than negative), so I start in the upper right corner (rather than the lower right corner).
As I draw leftwards, I move to (3, 0) (since r₂ = 3 is the largest root).
I bounce off of the horizontal axis (since this root's multiplicity, m₂ = 2, is even).
Always moving leftwards, I curve back down to (−½, 0) (since r₁ = −½ is the only other root).
I go through the horizontal axis (since this root's multiplicity, m₁ = 1, is odd), in fact straight through since the multiplicity is 1.
Since there are no more roots, I head off into the corner (in this case the lower left corner).
Finally, I mark the scale on the vertical axis so that my graph crosses it at f(0) = 9.

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.

Go back to the course homepage.

This web page was written in 2015 and 2016 by Toby Bartels, last edited on 2016 May 18. Toby reserves no legal rights to it.

The permanent URI of this web page is http://tobybartels.name/MATH-1150/2016SP/polynomials/.