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More About Zeroes and Their
     Multiplicities: "Flexing" at the Axis (page 3 of 5)

Sections: End behavior, Zeroes and their multiplicities, "Flexing", Turnings & "bumps", Graphing

There's an extra detail I'd like to mention regarding the multiplicity of a zero and the graph; in particular, you can tell from the graph whether an odd-multiplicity zero occurs only once or if it occurs more than once.

• What is the multiplicity of x = 5, given that the graph shows a fifth-degree polynomial and x = –5 has a multiplicity of 2?

The intercept at x = –5 is of multiplicity 2. The polynomial is of degree 5, so the zero at x = 5, the only other zero, must use up the rest of the multiplicities. Since 5 – 2 = 3, then

x = 5 must be of multiplicity 3.

The zero at x = 5 had to be of odd multiplicity, since the graph went through the x-axis. But the graph flexed a bit (that bendy part of the graph) right along x = 5. This tells you that the multiplicity of x = 5 had to be more than just 1. But it couldn't have been 5 or 7 or more, because the degree of the whole polynomial was only 5. Keep this in mind: Any odd-multiplicity zero that flexes like this graph did at
x = 5 is of multiplicity 3 or more. And if you get that odd flexing behavior off the x-axis, you're probably looking at the effect of complex zeroes (the ones you'd find using the Quadratic Formula; the ones that don't cross the axis).   Copyright © Elizabeth Stapel 2006-2008 All Rights Reserved

• Which of the following graphs could represent the polynomial
       f(x) = a(x – b)²(x – c)³?
     

Whatever this polynomial is, it is of degree 5. (I know this by adding the degrees on on the two repeated factors. If I multiplied everything out, the degree on the leading term would be 5.) Since the polynomial is of odd degree, Graph A can't be correct, because its ends both go the same direction, meaning it is an even-degree polynomial. Since my polynomial has two zeroes, Graph C can't be right, because it only crosses the x-axis once; Graph C may be of odd degree, but it doesn't have enough zeroes. Graph D is of odd degree and one zero has even multiplicity (the one where the graph just touches the axis), but the odd zero has a multiplicity of only 1 (because there's no flexing as the graph crosses the axis), and I need a polynomial of degree 5, not degree 3. On the other hand, Graph B is of odd degree, with one even-multiplicity zero and one odd-and-more-than-1-multiplicity zero.

The correct graph is Graph B.

• Find the degree-7 polynomial corresponding to the following graph, given that one of the zeroes has multiplicity 3.

From the graph, I can see that there are zeroes of even multiplicity at x = –4 and x = 4. The zero at x = –1 must be the zero of multiplicity 3. (This matches the graph, since the line goes through the axis, but flexes as it does so. The multiplicity must be odd, and must be more than 1.) Since the total degree of the polynomial is 7, and I already have multiplicities of 2, 2, and 3 (which adds up to 7), then the zeroes at –4 and 4 must be of multiplicity 2, rather than 4 or 6 or something bigger.

Working backwards from the zeroes, I get the polynomial y = a(x + 4)²(x + 1)³(x – 4)². The one marked point on the graph is there so I can figure out the exact polynomial; that is, so I can figure out the value of the leading coefficient "a". Plugging in these x- and y-values from the point (1, –2), I get:

a(1 + 4)²(1 + 1)³(1 – 4)² = a(25)(8)(9) = 1800a = –2

Then a = –1/900, and the polynomial, in factored form, is:

y = ( ^–1/₉₀₀ )(x + 4)²(x + 1)³(x – 4)²

The problem didn't say that I had to multiply this out, so I'm not going to. The factored form (especially for something as huge as this) should be fine.

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