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 exponent on the monomial is thef(x) ~ax^{m1+⋯+mk};

that is, the vertical intercept 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

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

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

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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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