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},

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