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Curve Analysis and Sketching · Tutorial 849 of 1495

Matching Roots, Turning Points, and Multiplicity

Use root multiplicities and derivative degree to test which combinations of polynomial roots, crossings, and turning points are possible.

Intermediate 11 min read

What You'll Learn

  • Define the multiplicity of a polynomial root and connect it to graph crossing behavior.
  • Determine when a repeated root is a turning point and when a stationary root is not.
  • Bound the number of turning points using polynomial degree.
  • Use Rolle’s theorem to relate distinct real roots to critical points between them.
  • Test proposed polynomial graph features for compatibility with degree and multiplicity.

Why Roots and Turning Points Must Match

A polynomial’s graph cannot combine roots, crossings, and turning points arbitrarily. Its degree limits how many times the graph can meet the horizontal axis and how many distinct places its derivative can be zero. Root multiplicities impose additional restrictions: they determine whether the graph crosses the axis or touches it and turns.

These restrictions are useful when sketching a polynomial and when checking whether a proposed sketch could belong to a polynomial of a specified degree. The key is to count carefully: a repeated root uses up degree, and a turning point requires a zero of the derivative.

Roots, Multiplicity, and Crossing Behavior

Definition: A real number \(r\) is a root of a polynomial \(p\) with multiplicity \(m\) if \(m\) is the largest positive integer for which \(p(x)=(x-r)^m q(x)\), where \(q\) is a polynomial and \(q(r)\ne 0\).

Near \(r\), the factor \(q(x)\) keeps the same sign as \(q(r)\), since it is continuous and nonzero there. The sign of \(p(x)\) on either side of \(r\) is therefore controlled by \((x-r)^m\). If \(m\) is even, that factor has the same sign on both sides: the graph touches the axis but does not cross it. If \(m\) is odd, its sign changes: the graph crosses the axis.

Multiplicity also says something about the slope at the root. When \(m=1\), the root is simple and \(p'(r)\ne 0\). When \(m\ge 2\), \(p'(r)=0\), so the root is a critical point. But not every critical point is a turning point: at an odd-multiplicity root such as a triple root, the graph can cross the axis while continuing in the same direction.

Key distinction: An even-multiplicity root is a local maximum or minimum. An odd-multiplicity root crosses the axis; if its multiplicity is at least three, it is also a critical point, but it is not a local maximum or minimum.

Degree Limits the Number of Turning Points

A nonconstant polynomial of degree \(n\) has a derivative of degree \(n-1\). A nonzero polynomial of degree \(n-1\) has at most \(n-1\) distinct real roots. Every local maximum or minimum of a differentiable function at an interior point of its domain has derivative zero. Consequently, a degree-\(n\) polynomial can have at most \(n-1\) distinct turning points.

This is an upper bound, not a promise that all \(n-1\) points occur. A derivative zero can fail to be a turning point if the derivative has the same sign on both sides. Also, if a domain is restricted, a boundary extremum need not have derivative zero; here, “turning point” means a local maximum or minimum at an interior point of the polynomial’s domain, which is all of \(\mathbb R\).

Worked Example: Counting Possible Turns from the Degree

Suppose a polynomial has degree \(5\). Its derivative has degree \(4\), so there can be at most four distinct real zeros of the derivative. Any turning point must be among those zeros. Thus the polynomial can have at most four distinct turning points. A proposed sketch with five turning points cannot represent a degree-\(5\) polynomial.

The bound does not say that the five turning points must be replaced by four. It only rules out five or more; a particular degree-\(5\) polynomial may have fewer than four turns. Checking the derivative’s sign on either side of each critical point is necessary to decide which zeros actually give turns.

How Roots Force Critical Points Between Them

Distinct real roots impose another constraint. If \(a<b\) are consecutive distinct real roots of a polynomial, then the polynomial has no root in \((a,b)\). It has a constant, nonzero sign throughout that interval. Since its value is zero at both endpoints, it must attain a positive maximum inside the interval if it is positive there, or a negative minimum inside if it is negative there. That interior extremum is a turning point. Equivalently, Rolle’s theorem guarantees at least one derivative zero in \((a,b)\).

Even-multiplicity roots give turning points at the roots themselves, while consecutive distinct roots force a turn strictly between them. These locations do not overlap: the roots are endpoints of the intervals, and the forced turns are inside them. Thus, if a polynomial has \(k\) distinct real roots and \(e\) of those roots have even multiplicity, it has at least \(k-1+e\) distinct turning points. In particular, this count cannot exceed \(n-1\).

Root-and-turn bound: For a nonconstant degree-\(n\) polynomial with \(k\) distinct real roots, of which \(e\) have even multiplicity, \(k-1+e\le n-1\). Each interval between consecutive distinct roots forces a turn, and each even-multiplicity root is itself a turn.

Multiplicity also gives a more detailed derivative count. If the distinct real roots are \(r_1,\ldots,r_k\), with multiplicities \(m_1,\ldots,m_k\), then each root \(r_i\) contributes a derivative root of multiplicity \(m_i-1\) when \(m_i\ge2\). Rolle’s theorem supplies at least one additional derivative root in each of the \(k-1\) open intervals between consecutive roots. Together these account for at least \(\sum_{i=1}^{k}(m_i-1)+(k-1)=\sum_{i=1}^{k}m_i-1\) roots of the derivative, counted with multiplicity. Since the derivative has degree \(n-1\), this total cannot exceed \(n-1\).

Why Multiplicity Determines the Local Shape

Write a polynomial near a root \(r\) of multiplicity \(m\) as \(p(x)=(x-r)^m q(x)\), where \(q(r)\ne0\). Differentiating gives \(p'(x)=m(x-r)^{m-1}q(x)+(x-r)^m q'(x)\). Factoring out the common power yields:

$$ p'(x)=(x-r)^{m-1}\bigl(mq(x)+(x-r)q'(x)\bigr). $$

The bracketed factor has value \(mq(r)\ne0\) at \(r\). Thus \(p'\) has a root of multiplicity exactly \(m-1\) at \(r\). If \(m\) is even, then \(m-1\) is odd, so \(p'\) changes sign there and the polynomial turns. If \(m\) is odd and at least three, then \(m-1\) is even, so \(p'\) does not change sign there: the graph has a horizontal tangent but does not turn.

Worked Example: A Touch, a Crossing, and Three Turning Points

Consider \(p(x)=(x+2)^2(x-1)^3\). Its degree is \(2+3=5\). It has a root at \(-2\) of multiplicity \(2\), so the graph touches the axis and turns there. It has a root at \(1\) of multiplicity \(3\), so it crosses the axis there with a horizontal tangent, but does not turn at that root.

Differentiate using the product rule:

$$ \begin{aligned} p'(x) &=2(x+2)(x-1)^3+3(x+2)^2(x-1)^2\\ &=(x+2)(x-1)^2\bigl(2(x-1)+3(x+2)\bigr)\\ &=(x+2)(x-1)^2(5x+4). \end{aligned} $$

The derivative zeros are \(x=-2\), \(x=1\), and \(x=-\frac45\). The degree bound allows at most four distinct turning points, so these three critical points are within the limit. To classify them, inspect the signs of the derivative factors. At \(-2\), \(p'\) changes from positive to negative, so there is a local maximum. At \(-\frac45\), it changes from negative to positive, so there is a local minimum. At \(1\), the factor \((x-1)^2\) does not change sign and the other factors are positive nearby; \(p'\) stays positive, confirming that this is a stationary crossing, not a turn.

As a numerical check on the non-root critical point, \(p(-\frac45)=(\frac65)^2(-\frac95)^3=-\frac{26244}{3125}\). The factorization also checks the critical-point calculation: substituting \(-\frac45\) into \(5x+4\) gives \(5(-\frac45)+4=0\).

Worked Example: Four Roots Force at Least Three Turns

Consider \(q(x)=(x+2)^2x(x-3)\), a degree-\(4\) polynomial. Its distinct roots are \(-2\), \(0\), and \(3\); \(-2\) has multiplicity \(2\), and the other two roots are simple. The even root gives one turn at \(-2\), and the intervals \((-2,0)\) and \((0,3)\) each force another turn. Therefore at least three distinct turning points are required. A degree-\(4\) polynomial can have at most three, so this count reaches the upper bound.

Differentiate to locate them:

$$ \begin{aligned} q(x)&=x^4+x^3-8x^2-12x,\\ q'(x)&=4x^3+3x^2-16x-12\\ &=(x+2)(4x^2-5x-6)\\ &=(x+2)(4x+3)(x-2). \end{aligned} $$

The derivative zeros are \(-2\), \(-\frac34\), and \(2\). They lie at the even root and inside the two required intervals, respectively. Each is simple as a derivative root, so the derivative changes sign at each: the three points are turns. Direct substitution confirms the two between-root values: \(q(-\frac34)=(\frac54)^2(-\frac34)(-\frac{15}{4})=\frac{1125}{256}\), and \(q(2)=4^2\cdot2\cdot(-1)=-32\). At the root \(-2\), \(q(-2)=0\), as required.

Testing Whether a Proposed Sketch Is Possible

A proposed feature list should be checked against both the degree bound and the root pattern. For example, four distinct roots already force at least three turns between consecutive roots. If one of those roots has even multiplicity, it contributes an additional turn at the root. The maximum-turn bound may then leave no room for any extra turns elsewhere.

Worked Example: Rejecting an Impossible Feature Combination

Suppose a sketch is claimed to represent a degree-\(4\) polynomial with four distinct real roots and four distinct turning points. The degree bound gives at most \(4-1=3\) distinct turning points. The claim of four turns contradicts that bound, so no degree-\(4\) polynomial can have all the proposed features.

The roots alone are compatible with degree \(4\) if they are all simple: their multiplicities already sum to \(4\). For instance, \(s(x)=(x^2-9)(x^2-1)\) has roots \(-3,-1,1,3\), each simple. But \(s'(x)=4x^3-20x=4x(x^2-5)\), which has only three distinct zeros, \(-\sqrt5,0,\sqrt5\). In fact these give three turns, not four. Substitution gives \(s(0)=9\) and \(s(\pm\sqrt5)=(5-9)(5-1)=-16\), consistent with the three intervening extrema. This example confirms that four simple roots can occur while the extra proposed turn cannot.

A Common Mistake: Treating Every Critical Point as a Turn

A frequent error is to see \(p'(r)=0\) and conclude that the graph must have a local maximum or minimum at \(r\). A zero derivative is necessary at an interior turn, but it is not sufficient. For a concrete counterexample, let \(p(x)=(x-1)^3\). Then \(p'(x)=3(x-1)^2\), so \(p'(1)=0\). But \(p'(x)>0\) on both sides of \(1\), and \(p(x)\) changes from negative to positive there. The graph crosses with a horizontal tangent and does not turn.

The correction is to check how the derivative behaves on either side of the critical point. A change from positive to negative indicates a local maximum; a change from negative to positive indicates a local minimum. If there is no sign change, the critical point is not a turning point. For polynomial roots, multiplicity predicts this behavior: even multiplicity gives a turn, while odd multiplicity gives a crossing.

Using the Constraints in Practice

When checking a polynomial sketch, keep three counts separate: distinct roots, total root multiplicity, and distinct turning points. A degree-\(n\) polynomial has at most \(n\) roots counted with multiplicity and at most \(n-1\) distinct turning points. Its consecutive distinct real roots force at least one turn in each gap, and each even-multiplicity root adds a turn at that root. Odd roots of multiplicity at least three are stationary crossings, not turns.

These are necessary constraints, not a complete construction method. Passing the counts does not by itself prove that a polynomial with every proposed feature exists. But failing any one of them proves that the proposed combination is impossible. That makes degree and multiplicity a useful first consistency check before attempting a detailed sketch.

Check Your Understanding

Use degree, multiplicity, and derivative behavior to answer the following.

  1. What is the maximum number of distinct turning points a degree-\(6\) polynomial can have?
  2. At a root of multiplicity \(4\), does a polynomial cross the axis or touch it? Is that root a turning point?
  3. A degree-\(5\) polynomial has four distinct real roots, one of which has even multiplicity. What is the minimum number of distinct turning points forced by these features?
  4. Why does a triple root have zero derivative without being a local maximum or minimum?
  5. Can a degree-\(3\) polynomial have four distinct real roots? Explain using multiplicity and degree.

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