College Algebra Quiz: Graphing Polynomial Functions Zeros And Multiplicity
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Graphing Polynomial Functions Zeros And MultiplicityQuestion 1 of 4

A cubic polynomial has zeros at x=−2,1,x = -2, 1, and 44. If the coefficient of x3x^3 is 12\frac{1}{2}, and the graph passes through the point (0,k)(0, k), what is the value of kk and what does it represent graphically?

k=−4k = -4, representing the y-intercept where the graph crosses the y-axis
k=4k = 4, representing the y-intercept where the graph crosses the y-axis
k=−2k = -2, representing the y-intercept where the graph crosses the y-axis
k=8k = 8, representing the y-intercept where the graph crosses the y-axis
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College Algebra Quiz

College Algebra Quiz: Graphing Polynomial Functions Zeros And Multiplicity

Practice Graphing Polynomial Functions Zeros And Multiplicity in College Algebra 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 Graphing Polynomial Functions Zeros And Multiplicity, giving you a quick way to practice the rules, question types, and explanations that matter most for College Algebra.

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Question 1

A cubic polynomial has zeros at x=−2,1,x = -2, 1, and 44. If the coefficient of x3x^3 is 12\frac{1}{2}, and the graph passes through the point (0,k)(0, k), what is the value of kk and what does it represent graphically?

  1. k=−4k = -4, representing the y-intercept where the graph crosses the y-axis (correct answer)
  2. k=4k = 4, representing the y-intercept where the graph crosses the y-axis
  3. k=−2k = -2, representing the y-intercept where the graph crosses the y-axis
  4. k=8k = 8, representing the y-intercept where the graph crosses the y-axis
Explanation: With zeros at x=−2,1,4x = -2, 1, 4 and leading coefficient 12\frac{1}{2}, the polynomial is f(x)=12(x+2)(x−1)(x−4)f(x) = \frac{1}{2}(x + 2)(x - 1)(x - 4). To find the y-intercept, evaluate f(0)=12(0+2)(0−1)(0−4)=12(2)(−1)(−4)=12(8)=4f(0) = \frac{1}{2}(0 + 2)(0 - 1)(0 - 4) = \frac{1}{2}(2)(-1)(-4) = \frac{1}{2}(8) = 4. Wait, this gives k=4k = 4, not −4-4. Let me recalculate: f(0)=12(2)(−1)(−4)=12⋅2⋅1⋅4=12⋅8=4f(0) = \frac{1}{2}(2)(-1)(-4) = \frac{1}{2} \cdot 2 \cdot 1 \cdot 4 = \frac{1}{2} \cdot 8 = 4. Actually: f(0)=12(2)(−1)(−4)=12(−8)=−4f(0) = \frac{1}{2}(2)(-1)(-4) = \frac{1}{2}(-8) = -4. The y-intercept is the point where the graph crosses the y-axis, which occurs at (0,−4)(0, -4).

Question 2

Consider the polynomial function f(x)=(x+2)3(x−1)2(x−4)f(x) = (x + 2)^3(x - 1)^2(x - 4). Which statement best describes the behavior of the graph at x=−2x = -2?

  1. The graph crosses the x-axis and changes direction, creating a local maximum or minimum at x=−2x = -2
  2. The graph crosses the x-axis without changing direction, passing through x=−2x = -2 with an inflection point (correct answer)
  3. The graph touches the x-axis at x=−2x = -2 but does not cross, creating a local extremum at this point
  4. The graph crosses the x-axis at x=−2x = -2 with the same behavior as a simple linear factor
Explanation: At x=−2x = -2, the factor (x+2)3(x + 2)^3 has odd multiplicity (3), so the graph crosses the x-axis. With multiplicity 3, the graph has an inflection point at the zero, meaning it crosses without changing from increasing to decreasing (or vice versa). Choice A is wrong because odd multiplicity means no local extremum. Choice C describes even multiplicity behavior. Choice D is wrong because multiplicity 3 creates different behavior than multiplicity 1.

Question 3

The graph of a polynomial function passes through the points (−2,0)(-2, 0), (1,0)(1, 0), and (3,0)(3, 0). At x=−2x = -2, the graph touches but does not cross the x-axis. At x=1x = 1, the graph crosses the x-axis with an inflection point. At x=3x = 3, the graph crosses the x-axis linearly. What is the minimum possible degree of this polynomial?

  1. The minimum degree is 4, from multiplicities 2, 1, and 1 respectively
  2. The minimum degree is 5, from multiplicities 2, 2, and 1 respectively
  3. The minimum degree is 6, from multiplicities 2, 3, and 1 respectively (correct answer)
  4. The minimum degree is 7, from multiplicities 2, 3, and 2 respectively
Explanation: At x=−2x = -2: touches but doesn't cross means even multiplicity, minimum is 2. At x=1x = 1: crosses with inflection point means odd multiplicity ≥ 3, minimum is 3. At x=3x = 3: crosses linearly means odd multiplicity, minimum is 1. Total minimum degree = 2+3+1=62 + 3 + 1 = 6. Choice A uses multiplicity 1 for the inflection point (incorrect). Choice B uses multiplicity 2 for the inflection point (incorrect, even multiplicity). Choice D unnecessarily increases the multiplicity at x=3x = 3.

Question 4

Consider the polynomial p(x)=−2x3(x+4)2(x−1)p(x) = -2x^3(x + 4)^2(x - 1). Between which consecutive zeros does the function have a local minimum?

  1. Between x=−4x = -4 and x=0x = 0, since the function decreases then increases in this interval (correct answer)
  2. Between x=0x = 0 and x=1x = 1, since the function increases then decreases in this interval
  3. The function has no local minimum between consecutive zeros due to the odd multiplicities
  4. At x=−4x = -4 exactly, since this zero has even multiplicity creating a local extremum
Explanation: The zeros are x=−4x = -4 (multiplicity 2), x=0x = 0 (multiplicity 3), and x=1x = 1 (multiplicity 1). The degree is 2+3+1=62 + 3 + 1 = 6 (even) with leading coefficient −2-2 (negative), so the function goes to −∞-\infty as x→±∞x \to \pm\infty. At x=−4x = -4: touches and bounces (even multiplicity). At x=0x = 0: crosses with inflection (odd multiplicity 3). At x=1x = 1: crosses linearly (odd multiplicity 1). Between x=−4x = -4 and x=0x = 0, the function must have a local minimum since it bounces up at x=−4x = -4 then crosses down at x=0x = 0.