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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 of the polynomial: 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 with all real-number roots, and the root 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 (the "flexing" being that bendy part of the graph) right in the area of x = 5. This flexing is what tells you that the multiplicity of x = 5 had to be more than just 1. In this particular case, the multiplicity couldn't have been 5 or 7 or more, because the degree of the whole polynomial was only 5,but the multiplicity certainly had to be more than just 1. Keep this in mind: Any odd-multiplicity zero that flexes at the crossing point, like this graph did at x = 5, is of multiplicity 3 or more.

Notw: If you get that odd flexing behavior at some location on the graph that is off the x-axis (above or below the axis), then you're probably looking at the effect of complex zeroes; namely, the zeroes that you'd find by using the Quadratic Formula, the zeroes that don't correspond to the graph crossing the x-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 real-number zeroes (namely, zeroes at x = b and at x = c), I know that Graph C can't be right: it only crosses the x-axis once. So Graph C may be of odd degree, but it doesn't have enough zeroes.

From the end behavior, I can see that Graph D is of odd degree. Also, I know that the negative zero has an even multiplicity because the graph just touches the axis; this zero could correspond to x = b. But there is no flexing where the graph crosses the positive x-axis, so the odd zero here must have a multiplicity of only 1, and I need the multiplicity of this zero to be more than just 1. So Graph D might have the right overall degree (if the zero x = b is of multiplicity 1), but the multiplicities of the two zeroes don't match up with what I need.

On the other hand, the ends of the graph tell me that Graph B is of odd degree, and the way the graph touches or crosses the x-axis at the two graphed x-intercepts tells me that the polynomial being graphed has one even-multiplicity zero and one odd-and-more-than-1-multiplicity zero. This matches what I need.

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, telling me that 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 multiplicity 4 or multiplicity 6 or something bigger.

Working backwards from the zeroes, I get the following expression for the polynomial:

y = a(x + 4)²(x + 1)³(x – 4)²

They marked that one point on the graph so that 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 exercise 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 a completely acceptable form of the answer.

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