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 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 ⋯ (−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;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.
This web page was written in 2015 by Toby Bartels, last edited on 2015 November 24. Toby reserves no legal rights to it.
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