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

degThe graph off=D=m_{1}+ ⋯ +m_{k}.

This monomial is thef(x) ~ax^{D}.

which is the product of the roots (counting them with their multiplicities) if the degree is even, or the opposite of this product if the degree is odd. Then the vertical intercept of the graph is (0,f(0) = (−1)^{D}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 1 in Section 5.2 Objective 1 on pages 346&347
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) = (−1)^{3}(−½)^{1}3^{2}= 9.

**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

This is what the graph is like at the extreme corners. Sincef(x) ~ 2x^{3}.

(0, 9).Near the 1st root,

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 additional information, you could draw a graph about as nice as the one in the textbook without having to calculate any additional points.f(x) ≈ 2 (3 + ½) (x− 3)^{2}= 7 (x− 3)^{2},

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