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 exponent on the 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;

the 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. Then 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 parabola with vertex (0, 3) 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 additional points.

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