Math 3 Quiz: Zero Multiplicity And Graph Behavior
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Zero Multiplicity And Graph BehaviorQuestion 1 of 18

A polynomial function has zeros at x=−2x = -2 (multiplicity 3), x=1x = 1 (multiplicity 2), and x=4x = 4 (multiplicity 1). If the leading coefficient is negative, which statement best describes the behavior of the graph near x=1x = 1?

The graph crosses the x-axis and changes from increasing to decreasing as x increases through 1
The graph touches the x-axis and changes from decreasing to increasing as x increases through 1
The graph touches the x-axis and changes from increasing to decreasing as x increases through 1
The graph crosses the x-axis and changes from decreasing to increasing as x increases through 1
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Math 3 Quiz

Math 3 Quiz: Zero Multiplicity And Graph Behavior

Practice Zero Multiplicity And Graph Behavior in Math 3 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Zero Multiplicity And Graph Behavior, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 3.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

A polynomial function has zeros at x=−2x = -2 (multiplicity 3), x=1x = 1 (multiplicity 2), and x=4x = 4 (multiplicity 1). If the leading coefficient is negative, which statement best describes the behavior of the graph near x=1x = 1?

  1. The graph crosses the x-axis and changes from increasing to decreasing as x increases through 1
  2. The graph touches the x-axis and changes from decreasing to increasing as x increases through 1
  3. The graph touches the x-axis and changes from increasing to decreasing as x increases through 1 (correct answer)
  4. The graph crosses the x-axis and changes from decreasing to increasing as x increases through 1
Explanation: At x = 1, the multiplicity is 2 (even), so the graph touches but doesn't cross the x-axis. With a negative leading coefficient and degree 6, the function approaches negative infinity as x approaches positive infinity. The behavior near x = 1 shows the graph touching the axis and changing from increasing to decreasing. A: Incorrect because even multiplicity means touching, not crossing. B: Incorrect direction of concavity change. D: Incorrect because even multiplicity means touching, not crossing.

Question 2

The graph of a polynomial function touches the x-axis at x=−3x = -3 and x=2x = 2, and crosses the x-axis at x=5x = 5. If this polynomial has the minimum possible degree, what is the multiplicity of the zero at x=−3x = -3?

  1. The multiplicity could be 2, 4, or any other even positive integer
  2. The multiplicity is exactly 2 based on the minimum degree constraint (correct answer)
  3. The multiplicity could be 1, 3, or any other odd positive integer
  4. The multiplicity is exactly 1 since the graph touches the axis
Explanation: Since the graph touches (doesn't cross) at x = -3, the multiplicity must be even. For minimum degree, we use the smallest even multiplicity, which is 2. The same reasoning applies to x = 2 (touches = even = minimum 2), and x = 5 (crosses = odd = minimum 1). Total minimum degree is 2 + 2 + 1 = 5. A: Incorrect because we need minimum degree. C: Incorrect because touching requires even multiplicity. D: Incorrect because multiplicity 1 would cause crossing, not touching.

Question 3

A polynomial function has the form h(x)=(x−a)m(x−b)n(x−c)ph(x) = (x-a)^m(x-b)^n(x-c)^p where a<b<ca < b < c and m,n,pm, n, p are positive integers. If the graph shows exactly two local extrema at x-intercepts, which combination of multiplicities is possible?

  1. m=2,n=1,p=2m = 2, n = 1, p = 2 because two even multiplicities create exactly two extrema (correct answer)
  2. m=4,n=3,p=2m = 4, n = 3, p = 2 because higher multiplicities create more pronounced extrema
  3. m=1,n=2,p=1m = 1, n = 2, p = 1 because the even multiplicity creates one extremum at each endpoint
  4. m=3,n=2,p=3m = 3, n = 2, p = 3 because odd multiplicities greater than 1 create local extrema
Explanation: Local extrema at x-intercepts occur only when the multiplicity is even and ≥ 2. With exactly two local extrema at x-intercepts, exactly two of the multiplicities must be even and ≥ 2. Option A has m = 2 and p = 2 (both even), with n = 1 (odd, no extremum). B: Would create extrema at x = a and x = c, but also an inflection at x = b. C: Only creates one extremum at x = b. D: Odd multiplicities ≥ 3 create inflection points, not extrema.

Question 4

A polynomial p(x)p(x) has degree 6 and exactly four distinct real zeros. If the graph touches the x-axis at exactly two points and crosses at exactly two points, which statement about the multiplicities must be true?

  1. Three zeros have multiplicity 1, and one zero has multiplicity 3
  2. One zero has multiplicity 1, one has multiplicity 3, and two have multiplicity 1
  3. One zero has multiplicity 2, one has multiplicity 2, and two have multiplicity 1
  4. Two zeros have multiplicity 1, and two zeros have multiplicity 2 (correct answer)
Explanation: When you encounter polynomial questions about zeros and their behavior at the x-axis, focus on the relationship between multiplicity and graph behavior. A zero with odd multiplicity causes the graph to cross the x-axis, while even multiplicity causes it to touch (but not cross) the x-axis. Given information: degree 6 polynomial, four distinct real zeros, touches at two points, crosses at two points. Since the sum of all multiplicities must equal the degree, we need multiplicities totaling 6. The graph touches at two points, meaning two zeros have even multiplicity. The graph crosses at two points, meaning two zeros have odd multiplicity. Since we need the simplest multiplicities that satisfy these conditions while totaling 6: two zeros with multiplicity 2 (even, causing touches) and two zeros with multiplicity 1 (odd, causing crosses). This gives us 2+2+1+1=62 + 2 + 1 + 1 = 6, matching our degree. Choice A is incorrect because it lists four individual zeros but claims three have multiplicity 1 and one has multiplicity 3, which would only account for 1+1+1+3=61 + 1 + 1 + 3 = 6 but doesn't match the touching/crossing pattern described. Choice B is incorrect due to unclear phrasing and mathematical inconsistency—it appears to double-count zeros. Choice C is incorrect because having two zeros of multiplicity 2 and two of multiplicity 1 would give the right total (2+2+1+1=62 + 2 + 1 + 1 = 6), but this matches choice D's description, not C's wording. Remember: odd multiplicity = crosses x-axis, even multiplicity = touches x-axis. Always verify that multiplicities sum to the polynomial's degree.

Question 5

A polynomial function s(x)s(x) can be written as s(x)=xn+lower degree termss(x) = x^n + \text{lower degree terms} where nn is a positive integer. If s(x)s(x) has exactly three distinct real zeros, and the graph touches the x-axis at exactly one of these zeros, what is the smallest possible value of nn?

  1. 4 (correct answer)
  2. 5
  3. 6
  4. 7
Explanation: Three distinct zeros with one touching (even multiplicity) and two crossing (odd multiplicity). Minimum multiplicities: 2, 1, 1. This gives degree 2+1+1=42 + 1 + 1 = 4. Since the leading coefficient is positive (xnx^n term), we need to verify this configuration is possible. With degree 4 and three distinct zeros having multiplicities 2, 1, 1, the polynomial would be s(x)=(x−a)2(x−b)(x−c)s(x) = (x-a)^2(x-b)(x-c) which indeed has degree 4. This satisfies all conditions: exactly three distinct real zeros, touches at one zero (multiplicity 2), crosses at two zeros (multiplicity 1 each).

Question 6

The polynomial r(x)=(x−a)2(x−b)3(x−c)r(x) = (x-a)^2(x-b)^3(x-c) where a<b<ca < b < c has exactly one local maximum and one local minimum. Based on this information, what can be determined about the relationship between the zeros and their multiplicities?

  1. The relative positions of the zeros cannot be determined from the given information
  2. The zero with multiplicity 2 must be located between the other two zeros
  3. The zeros with odd multiplicities must be at the endpoints of the interval containing all zeros
  4. The zero with multiplicity 3 must be located between the other two zeros (correct answer)
Explanation: When analyzing polynomials with given local extrema, you need to connect the behavior of the derivative to the multiplicities and positions of the zeros. The derivative of r(x)=(x−a)2(x−b)3(x−c)r(x) = (x-a)^2(x-b)^3(x-c) will have zeros that determine where local maxima and minima occur. Using the product rule, r′(x)r'(x) will be a polynomial of degree 5 with zeros related to aa, bb, and cc. Since r(x)r(x) has exactly one local maximum and one local minimum, r′(x)r'(x) must have exactly two sign changes. The key insight is understanding how multiplicity affects the derivative. At x=ax = a (multiplicity 2), the derivative will have aa as a simple zero. At x=bx = b (multiplicity 3), the derivative will have bb as a zero of multiplicity 2. At x=cx = c (multiplicity 1), the derivative will have cc as a simple zero. For r′(x)r'(x) to have exactly two sign changes (giving one max and one min), the zero of multiplicity 2 in the derivative must be positioned to "absorb" one potential sign change. This occurs when bb (the zero with multiplicity 3) lies between aa and cc, making answer D correct. A is wrong because the constraint of exactly two extrema does determine the relative positions. B is incorrect because placing the multiplicity-2 zero between the others would create more than two extrema. C misidentifies which zeros have odd multiplicities and their required positions. Remember: when counting local extrema, always consider how the multiplicities of zeros in the original polynomial translate to the behavior of its derivative.

Question 7

The polynomial g(x)=(x+3)2(x−1)3(x−5)g(x) = (x + 3)^2(x - 1)^3(x - 5) has a zero at x=1x = 1 with multiplicity 3. Based on this information alone, which statement about the behavior of g(x)g(x) near x=1x = 1 is most accurate?

  1. The graph crosses the x-axis and has a horizontal tangent line at x=1x = 1
  2. The graph touches the x-axis but does not cross, creating a local minimum or maximum
  3. The graph crosses the x-axis with an inflection point at x=1x = 1 (correct answer)
  4. The graph crosses the x-axis and has the same slope as y=xy = x at x=1x = 1
Explanation: A zero with odd multiplicity (3) means the graph crosses the x-axis. For multiplicity 3, the graph has the characteristic 'S-curve' shape through the zero, which creates an inflection point. The graph doesn't just cross linearly but has a flattened S-shape. Choice A describes even multiplicity behavior. Choice B also describes even multiplicity. Choice D describes simple multiplicity 1 behavior.

Question 8

Consider a polynomial t(x)t(x) with leading coefficient −2-2 and zeros at x=1x = 1 (multiplicity 3) and x=−2x = -2 (multiplicity 2). Which statement correctly describes the end behavior and local behavior of t(x)t(x)?

  1. As x→±∞x \to \pm\infty, t(x)→+∞t(x) \to +\infty; the graph crosses at x=1x = 1 and touches at x=−2x = -2
  2. As x→−∞x \to -\infty, t(x)→+∞t(x) \to +\infty and as x→+∞x \to +\infty, t(x)→−∞t(x) \to -\infty; the graph crosses at x=1x = 1 and touches at x=−2x = -2 (correct answer)
  3. As x→−∞x \to -\infty, t(x)→−∞t(x) \to -\infty and as x→+∞x \to +\infty, t(x)→+∞t(x) \to +\infty; the graph touches at x=1x = 1 and crosses at x=−2x = -2
  4. As x→±∞x \to \pm\infty, t(x)→−∞t(x) \to -\infty; the graph crosses at x=1x = 1 and touches at x=−2x = -2
Explanation: t(x)=−2(x−1)3(x+2)2t(x) = -2(x-1)^3(x+2)^2 has degree 3+2=53 + 2 = 5 (odd) with negative leading coefficient −2-2. End behavior: as x→−∞x \to -\infty, t(x)→+∞t(x) \to +\infty and as x→+∞x \to +\infty, t(x)→−∞t(x) \to -\infty. Local behavior: multiplicity 3 (odd) at x=1x = 1 means the graph crosses the x-axis; multiplicity 2 (even) at x=−2x = -2 means the graph touches but doesn't cross. Choice A has wrong end behavior (even degree pattern). Choice C has wrong end behavior and wrong local behavior. Choice D has wrong end behavior (even degree pattern).

Question 9

Consider the polynomial f(x)=(x+1)4(x−3)2(x−7)f(x) = (x + 1)^4(x - 3)^2(x - 7). At which x-intercept does the graph exhibit a local maximum or minimum?

  1. At x=−1x = -1 only, because the multiplicity 4 creates a turning point
  2. At x=3x = 3 only, because even multiplicity with degree 2 creates extrema
  3. At both x=−1x = -1 and x=3x = 3, since both have even multiplicities (correct answer)
  4. At x=7x = 7 only, because odd multiplicity creates the steepest local change
Explanation: Local extrema at x-intercepts occur when the multiplicity is even and greater than 1. At x = -1 (multiplicity 4) and x = 3 (multiplicity 2), the graph touches the x-axis and creates local extrema. At x = 7 (multiplicity 1), the graph crosses the axis without creating a local extremum. A: Incorrect because x = 3 also has even multiplicity > 1. B: Incorrect because x = -1 also has even multiplicity > 1. D: Incorrect because odd multiplicity causes crossing without local extrema.

Question 10

For the polynomial f(x)=(x+1)5(x−2)2f(x) = (x+1)^5(x-2)^2, which statement best describes the relative flatness at the two x-intercepts?

  1. The graph is flatter at x=2x = 2 because lower multiplicities require less steep approaches to reach zero
  2. The graph is flatter at x=2x = 2 because even multiplicities create more gradual axis approaches than odd ones
  3. Both intercepts show equal flatness because multiplicity differences only affect crossing versus touching behavior
  4. The graph is flatter at x=−1x = -1 because multiplicity 5 creates more gradual axis contact than multiplicity 2 (correct answer)
Explanation: When you encounter a polynomial in factored form, the multiplicity of each factor tells you how the graph behaves at that x-intercept. Higher multiplicities create "flatter" contact with the x-axis because the function and its derivatives approach zero more gradually. For f(x)=(x+1)5(x−2)2f(x) = (x+1)^5(x-2)^2, you have x-intercepts at x=−1x = -1 (multiplicity 5) and x=2x = 2 (multiplicity 2). At x=−1x = -1, the factor (x+1)5(x+1)^5 means the function approaches zero very gradually - imagine the graph gently kissing the x-axis before crossing. At x=2x = 2, the factor (x−2)2(x-2)^2 creates a sharper approach to zero, like a parabola touching and bouncing off the axis. Choice A incorrectly suggests lower multiplicities create flatter approaches, which is backwards - higher multiplicities create more gradual contact. Choice B focuses on the even/odd distinction, which affects crossing versus touching behavior but not relative flatness between different multiplicities. Choice C claims equal flatness, missing that multiplicity directly controls how gradually the function approaches zero. Choice D correctly identifies that multiplicity 5 creates more gradual axis contact than multiplicity 2. The higher the multiplicity, the more "flat" the graph appears at that intercept. Study tip: Remember that multiplicity controls flatness - think of it as how many times the function "tries" to reach zero at that point. More attempts (higher multiplicity) means a gentler, flatter approach to the x-axis.

Question 11

A student claims that if a polynomial has a zero of multiplicity 6 at x=cx = c, then the graph must have a local minimum at (c,0)(c, 0). Which response best evaluates this claim?

  1. The claim is correct because even multiplicities always create local minima at x-intercepts
  2. The claim is incorrect because the sign of the leading coefficient determines whether it's a minimum or maximum (correct answer)
  3. The claim is incorrect because multiplicity 6 is too high to guarantee a local extremum occurs
  4. The claim is correct because multiplicity 6 creates the strongest possible attraction to the x-axis
Explanation: Even multiplicity ≥ 2 guarantees a local extremum at the x-intercept, but whether it's a minimum or maximum depends on the sign of the leading coefficient and the relative position of this zero. If the leading coefficient is negative, or if there's an odd number of zeros to the right, the local extremum could be a maximum instead of a minimum. A: Incorrect because even multiplicities create extrema, but not necessarily minima. C: Incorrect because multiplicity 6 definitely creates a local extremum. D: Incorrect reasoning about 'attraction strength'.

Question 12

The polynomial f(x)=−2x(x+3)2(x−1)4f(x) = -2x(x+3)^2(x-1)^4 has a y-intercept at the origin. Near x=0x = 0, which description best characterizes the local behavior?

  1. The graph passes through the origin with the same steepness as y=−2xy = -2x
  2. The graph passes through the origin with modified steepness due to the nearby zeros (correct answer)
  3. The graph touches the origin and creates a local extremum due to nearby even multiplicities
  4. The graph passes through the origin with an inflection point due to the overall degree
Explanation: At x = 0, the multiplicity is 1, so the graph crosses the axis. However, the local behavior is not simply like y = -2x because the other factors (x+3)² and (x-1)⁴ evaluate to non-zero values at x = 0, affecting the steepness. Specifically, f(0) = 0 and f'(0) = -2(3)²(1)⁴ = -18, not -2. A: Incorrect because other factors modify the slope. C: Incorrect because multiplicity 1 means crossing, not touching. D: Incorrect because simple crossing occurs, not an inflection point.

Question 13

Consider the polynomial g(x)=a(x−r)4(x−s)2(x−t)g(x) = a(x-r)^4(x-s)^2(x-t) where a>0a > 0 and r<s<tr < s < t. If the graph has exactly one local maximum between consecutive zeros, where must this maximum occur?

  1. Between x=rx = r and x=sx = s, since both even multiplicities create local extrema that frame this region
  2. Between x=sx = s and x=tx = t, since the transition from even to odd multiplicity creates the maximum
  3. Between x=rx = r and x=sx = s, since the multiplicities 4 and 2 create compatible upward behavior
  4. Between x=sx = s and x=tx = t, since the local minimum at x=sx = s must rise to cross at x=tx = t (correct answer)
Explanation: With a > 0: at x = r (multiplicity 4, even) there's a local minimum; at x = s (multiplicity 2, even) there's a local minimum; at x = t (multiplicity 1, odd) the graph crosses. Between r and s, both endpoints are local minima so no local maximum exists there. Between s and t, the function must rise from the local minimum at s to cross the axis at t, creating exactly one local maximum in this interval.

Question 14

If a polynomial function has a zero at x=kx = k with multiplicity mm, and the graph appears to have a horizontal tangent at (k,0)(k, 0), what is the minimum value of mm?

  1. The minimum value is m=2m = 2 because horizontal tangents require even multiplicity
  2. The minimum value is m=2m = 2 because the derivative equals zero at the intercept (correct answer)
  3. The minimum value is m=3m = 3 because odd multiplicities can also create horizontal tangents
  4. The minimum value is m=1m = 1 because any zero can theoretically have a horizontal tangent
Explanation: A horizontal tangent at (k, 0) means f(k) = 0 and f'(k) = 0. If f(x) = (x-k)^m · g(x) where g(k) ≠ 0, then f'(x) = m(x-k)^(m-1)g(x) + (x-k)^m g'(x). For f'(k) = 0, we need m(k-k)^(m-1)g(k) + (k-k)^m g'(k) = 0, which requires m ≥ 2. A: Correct value but wrong reasoning - odd multiplicities ≥ 3 can also create horizontal tangents. C: m = 3 works but is not the minimum. D: m = 1 gives a non-zero slope at the intercept.

Question 15

If g(x)=x2(x−4)3(x+2)2g(x) = x^2(x-4)^3(x+2)^2, which statement correctly describes the behavior at x=4x = 4?

  1. The graph crosses the x-axis with an inflection point, creating an S-shaped curve through the axis (correct answer)
  2. The graph touches the x-axis and immediately bounces back without changing concavity
  3. The graph crosses the x-axis with a flattened appearance but no inflection point occurs
  4. The graph touches the x-axis and creates a local extremum with changed concavity
Explanation: At x = 4, the multiplicity is 3 (odd and > 1), so the graph crosses the x-axis. Odd multiplicity ≥ 3 creates an inflection point at the x-intercept, resulting in an S-shaped curve as the graph passes through the axis. B: Incorrect because odd multiplicity means crossing, not touching. C: Incorrect because multiplicity 3 does create an inflection point. D: Incorrect because odd multiplicity means crossing, and the inflection point is the key feature, not a local extremum.

Question 16

A degree 5 polynomial has exactly three distinct zeros. If the graph crosses the x-axis at exactly one point, what can be concluded about the multiplicities?

  1. One zero has multiplicity 1, and the other two zeros each have multiplicity 2 (correct answer)
  2. One zero has multiplicity 3, and the other two zeros each have multiplicity 1
  3. Two zeros have multiplicity 1 each, and one zero has multiplicity 3
  4. All three zeros must have odd multiplicities to ensure crossing behavior
Explanation: Since the graph crosses the x-axis at exactly one point, only one zero has odd multiplicity. With degree 5 and three distinct zeros, the multiplicities must sum to 5. The only way to have exactly one odd multiplicity is 1 + 2 + 2 = 5, where multiplicity 1 corresponds to the crossing point. B and C: Both give two odd multiplicities (causing two crossing points). D: Incorrect because having all odd multiplicities would cause three crossing points, not one.

Question 17

If a polynomial function p(x)p(x) has a zero of multiplicity 4 at x=cx = c, which statement about p′(x)p'(x) (the derivative of p(x)p(x)) is necessarily true?

  1. p′(c)=0p'(c) = 0 but p′(x)p'(x) may or may not have a zero at x=cx = c
  2. p′(c)=0p'(c) = 0 and p′(x)p'(x) has a zero of multiplicity 4 at x=cx = c
  3. p′(c)≠0p'(c) \neq 0 but p′(x)p'(x) changes sign at x=cx = c
  4. p′(c)=0p'(c) = 0 and p′(x)p'(x) has a zero of multiplicity 3 at x=cx = c (correct answer)
Explanation: When you encounter questions about zeros and their multiplicities in polynomial derivatives, think about how differentiation affects the structure of repeated factors. If p(x)p(x) has a zero of multiplicity 4 at x=cx = c, then p(x)=(x−c)4⋅q(x)p(x) = (x-c)^4 \cdot q(x) where q(x)q(x) is some polynomial with q(c)≠0q(c) \neq 0. Using the product rule to find the derivative: p′(x)=4(x−c)3⋅q(x)+(x−c)4⋅q′(x)=(x−c)3[4q(x)+(x−c)q′(x)]p'(x) = 4(x-c)^3 \cdot q(x) + (x-c)^4 \cdot q'(x) = (x-c)^3[4q(x) + (x-c)q'(x)]. Since q(c)≠0q(c) \neq 0, the bracketed expression 4q(c)+(c−c)q′(c)=4q(c)≠04q(c) + (c-c)q'(c) = 4q(c) \neq 0 when evaluated at x=cx = c. This means p′(x)=(x−c)3⋅(nonzero factor)p'(x) = (x-c)^3 \cdot (\text{nonzero factor}), so p′(x)p'(x) has exactly a zero of multiplicity 3 at x=cx = c, and p′(c)=0p'(c) = 0. Choice A suggests uncertainty about whether p′(x)p'(x) has a zero at x=cx = c, but we've shown it definitively does. Choice B incorrectly claims the multiplicity stays at 4, but differentiation always reduces multiplicity by exactly 1. Choice C wrongly states p′(c)≠0p'(c) \neq 0, contradicting our calculation that shows p′(c)=0p'(c) = 0. Choice D correctly identifies both that p′(c)=0p'(c) = 0 and that the multiplicity decreases from 4 to 3. Remember this key pattern: when a polynomial has a zero of multiplicity nn at some point, its derivative has a zero of multiplicity n−1n-1 at that same point. Differentiation always reduces multiplicity by exactly one.

Question 18

Consider the polynomial h(x)=x4−6x3+9x2h(x) = x^4 - 6x^3 + 9x^2. After factoring completely, the multiplicities of all zeros can be determined. What is the sum of all multiplicities?

  1. 3
  2. 4 (correct answer)
  3. 5
  4. 6
Explanation: First factor: h(x)=x2(x2−6x+9)=x2(x−3)2h(x) = x^2(x^2 - 6x + 9) = x^2(x - 3)^2. The zeros are x=0x = 0 with multiplicity 2 and x=3x = 3 with multiplicity 2. The sum of all multiplicities is 2+2=42 + 2 = 4. This also equals the degree of the polynomial, which provides a check. Students might incorrectly count the number of distinct zeros (choice A) or make arithmetic errors in factoring.