That is, we have af(x) =a(x−r_{1})^{m1}⋯ (x−r_{k})^{mk}.

the coefficient on that monomial is the leading coefficient of the polynomial, and the exponent on that monomial is thef(x) ~ax^{m1+⋯+mk};

thus, the vertical intercept of the graph is (0,f(0) =a(−r_{1})^{m1}⋯ (−r_{k})^{mk};

Besides that, the graph has *k* horizontal intercepts, one for each root;
the *i*th 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

That is, the power function with exponentf(x) ≈a(r_{i}−r_{1})^{m1}⋯ (r_{i}−r_{i−1})^{mi−1}(r_{i}−r_{i+1})^{mi+1}⋯ (r_{i}−r_{k})^{mk}(x−r_{i})^{mi}.

For example, consider *f*(*x*) =
(2*x* + 1)(*x* − 3)^{2}
(which is Example 5.1.10 on page 334 of the textbook).
Rewrite 2*x* + 1 as 2(*x* + ½),
so

then the leading cofficient isf(x) = 2(x+ ½)(x− 3)^{2}= 2(x− −½)^{1}(x− 3)^{2};

- 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
*r*_{2}= 3 is the largest root). - I bounce off of the horizontal axis
(since this root's multiplicity,
*m*_{2}= 2, is even). - Always moving leftwards,
I curve back down to (−½, 0)
(since
*r*_{1}= −½ is the only other root). - I go through the horizontal axis
(since this root's multiplicity,
*m*_{1}= 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) = 9.

*The remainder of these notes are optional.*
I can draw a more precise graph as follows:
The degree is
*m*_{1} + *m*_{2} = 1 + 2 = 3,
so the large-scale behaviour is

This is what the graph is like at the extreme corners. The vertical intercept is atf(x) ~ 2x^{3}.

Near the 1st root,f(0) = 2(½)(−3)^{2}= 9.

so approximately the line through (0, −½) with slope ⁴⁹⁄₂. Finally, near the 2nd root,f(x) ≈ 2(x+ ½)(−½ − 3)^{2}= ⁴⁹⁄₂(x+ ½),

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.f(x) ≈ 2(3 + ½)(x− 3)^{2}= 7(x− 3)^{2},

Go back to the course homepage.

This web page was written in 2015 and 2016 by Toby Bartels, last edited on 2016 September 1. Toby reserves no legal rights to it.

The permanent URI of this web page
is
`http://tobybartels.name/MATH-1150/2017SP/polynomials/`

.