- A polynomial function has the form h(x)=(x−a)m(x−b)n(x−c)p where a<b<c and m,n,p are positive integers. If the graph shows exactly two local extrema at x-intercepts, which combination of multiplicities is possible?
Answer: m=2,n=1,p=2 because two even multiplicities create exactly two extrema
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. - Consider the polynomial g(x)=a(x−r)4(x−s)2(x−t) where a>0 and r<s<t. If the graph has exactly one local maximum between consecutive zeros, where must this maximum occur?
Answer: Between x=s and x=t, since the local minimum at x=s must rise to cross at x=t
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. - If a polynomial function has a zero at x=k with multiplicity m, and the graph appears to have a horizontal tangent at (k,0), what is the minimum value of m?
Answer: The minimum value is m=2 because the derivative equals zero at the intercept
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−1g(x) + (x−k)m g'(x). For f'(k) = 0, we need m(k−k)m−1g(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. - If g(x)=x2(x−4)3(x+2)2, which statement correctly describes the behavior at x=4?
Answer: The graph crosses the x-axis with an inflection point, creating an S-shaped curve through the axis
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.
- 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?
Answer: One zero has multiplicity 1, and the other two zeros each have multiplicity 2
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.
- Consider the polynomial h(x)=x4−6x3+9x2. After factoring completely, the multiplicities of all zeros can be determined. What is the sum of all multiplicities?
Answer: 4
First factor: h(x)=x2(x2−6x+9)=x2(x−3)2. The zeros are x=0 with multiplicity 2 and x=3 with multiplicity 2. The sum of all multiplicities is 2+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. - A polynomial 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?
Answer: Two zeros have multiplicity 1, and two zeros have multiplicity 2
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=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=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=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. - The graph of a polynomial function touches the x-axis at x=−3 and x=2, and crosses the x-axis at x=5. If this polynomial has the minimum possible degree, what is the multiplicity of the zero at x=−3?
Answer: The multiplicity is exactly 2 based on the minimum degree constraint
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.
- If a polynomial function p(x) has a zero of multiplicity 4 at x=c, which statement about p′(x) (the derivative of p(x)) is necessarily true?
Answer: p′(c)=0 and p′(x) has a zero of multiplicity 3 at x=c
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) has a zero of multiplicity 4 at x=c, then p(x)=(x−c)4⋅q(x) where q(x) is some polynomial with q(c)=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)].
Since q(c)=0, the bracketed expression 4q(c)+(c−c)q′(c)=4q(c)=0 when evaluated at x=c. This means p′(x)=(x−c)3⋅(nonzero factor), so p′(x) has exactly a zero of multiplicity 3 at x=c, and p′(c)=0.
Choice A suggests uncertainty about whether p′(x) has a zero at x=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)=0, contradicting our calculation that shows p′(c)=0. Choice D correctly identifies both that p′(c)=0 and that the multiplicity decreases from 4 to 3.
Remember this key pattern: when a polynomial has a zero of multiplicity n at some point, its derivative has a zero of multiplicity n−1 at that same point. Differentiation always reduces multiplicity by exactly one. - The polynomial r(x)=(x−a)2(x−b)3(x−c) where a<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?
Answer: The zero with multiplicity 3 must be located between the other two zeros
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) will have zeros that determine where local maxima and minima occur. Using the product rule, r′(x) will be a polynomial of degree 5 with zeros related to a, b, and c. Since r(x) has exactly one local maximum and one local minimum, r′(x) must have exactly two sign changes.
The key insight is understanding how multiplicity affects the derivative. At x=a (multiplicity 2), the derivative will have a as a simple zero. At x=b (multiplicity 3), the derivative will have b as a zero of multiplicity 2. At x=c (multiplicity 1), the derivative will have c as a simple zero.
For 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 b (the zero with multiplicity 3) lies between a and c, 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. - A polynomial function s(x) can be written as s(x)=xn+lower degree terms where n is a positive integer. If 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 n?
Answer: 4
Three distinct zeros with one touching (even multiplicity) and two crossing (odd multiplicity). Minimum multiplicities: 2, 1, 1. This gives degree 2+1+1=4. Since the leading coefficient is positive (xn 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) 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). - Consider a polynomial t(x) with leading coefficient −2 and zeros at x=1 (multiplicity 3) and x=−2 (multiplicity 2). Which statement correctly describes the end behavior and local behavior of t(x)?
Answer: As x→−∞, t(x)→+∞ and as x→+∞, t(x)→−∞; the graph crosses at x=1 and touches at x=−2
t(x)=−2(x−1)3(x+2)2 has degree 3+2=5 (odd) with negative leading coefficient −2. End behavior: as x→−∞, t(x)→+∞ and as x→+∞, t(x)→−∞. Local behavior: multiplicity 3 (odd) at x=1 means the graph crosses the x-axis; multiplicity 2 (even) at x=−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). - A polynomial function has zeros at x=−2 (multiplicity 3), x=1 (multiplicity 2), and x=4 (multiplicity 1). If the leading coefficient is negative, which statement best describes the behavior of the graph near x=1?
Answer: The graph touches the x-axis and changes from increasing to decreasing as x increases through 1
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.
- Consider the polynomial f(x)=(x+1)4(x−3)2(x−7). At which x-intercept does the graph exhibit a local maximum or minimum?
Answer: At both x=−1 and x=3, since both have even multiplicities
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. - For the polynomial f(x)=(x+1)5(x−2)2, which statement best describes the relative flatness at the two x-intercepts?
Answer: The graph is flatter at x=−1 because multiplicity 5 creates more gradual axis contact than multiplicity 2
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)2, you have x-intercepts at x=−1 (multiplicity 5) and x=2 (multiplicity 2). At x=−1, the factor (x+1)5 means the function approaches zero very gradually - imagine the graph gently kissing the x-axis before crossing. At x=2, the factor (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. - The polynomial f(x)=−2x(x+3)2(x−1)4 has a y-intercept at the origin. Near x=0, which description best characterizes the local behavior?
Answer: The graph passes through the origin with modified steepness due to the nearby zeros
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.
- The polynomial g(x)=(x+3)2(x−1)3(x−5) has a zero at x=1 with multiplicity 3. Based on this information alone, which statement about the behavior of g(x) near x=1 is most accurate?
Answer: The graph crosses the x-axis with an inflection point at x=1
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. - A student claims that if a polynomial has a zero of multiplicity 6 at x=c, then the graph must have a local minimum at (c,0). Which response best evaluates this claim?
Answer: The claim is incorrect because the sign of the leading coefficient determines whether it's a minimum or maximum
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'.
