# Graphing polynomials (§§4.1&4.2)

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 degree of the polynomial is the sum of the multiplicities:
deg f = D = m1 + ⋯ + mk.
The graph of f appears on a large scale like a monomial:
f(x) ~ axD.
This monomial is the leading term of the polynomial, the monomial with the same leading coefficient and degree as the original polynomial. On a smaller scale, the initial value of f is
f(0) = (−1)Dr1 ⋯ rk,
which is the product of the roots if the degree is even, or the opposite of the product of the roots if the degree is odd. Then the vertical intercept of the graph is (0, f(0)).

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 1 on pages 346&347 in Section 4.2 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, the 2nd root (i = 2) is r2 = 3 with multiplicity m2 = 2, and the degree is D = m1 + m2 = 1 + 2 = 3. 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 ignoring 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 only 1).
• Since there are no more roots, I head off leftwards 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) = (−1)3 (−½)1 32 = 9.
Compare the graph near the bottom of page 347 at the end of the example 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: Since a = 2 and D = 3, the large-scale behaviour is

f(x) ~ 2x3.
This is what the graph is like at the extreme corners. Since f(0) = 9 (which I calculated in the very last step in the previous paragraph), the vertical intercept is
(0, 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 additional information, you could draw a graph about as nice as the one in the textbook without having to calculate any additional points.
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