Math 3 Quiz: Interpreting Polynomial Graphs
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Interpreting Polynomial GraphsQuestion 1 of 7

A polynomial function has zeros at x=−3,1,4x = -3, 1, 4 with multiplicities 2, 1, and 3 respectively. What can be concluded about the end behavior?

As x→−∞x \to -\infty, f(x)→+∞f(x) \to +\infty and as x→+∞x \to +\infty, f(x)→+∞f(x) \to +\infty
As x→−∞x \to -\infty, f(x)→−∞f(x) \to -\infty and as x→+∞x \to +\infty, f(x)→+∞f(x) \to +\infty
As x→−∞x \to -\infty, f(x)→+∞f(x) \to +\infty and as x→+∞x \to +\infty, f(x)→−∞f(x) \to -\infty
The end behavior cannot be determined without knowing the leading coefficient sign
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Math 3 Quiz

Math 3 Quiz: Interpreting Polynomial Graphs

Practice Interpreting Polynomial Graphs 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 Interpreting Polynomial Graphs, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 3.

How to use this quiz

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.

All questions

Question 1

A polynomial function has zeros at x=−3,1,4x = -3, 1, 4 with multiplicities 2, 1, and 3 respectively. What can be concluded about the end behavior?

  1. As x→−∞x \to -\infty, f(x)→+∞f(x) \to +\infty and as x→+∞x \to +\infty, f(x)→+∞f(x) \to +\infty
  2. As x→−∞x \to -\infty, f(x)→−∞f(x) \to -\infty and as x→+∞x \to +\infty, f(x)→+∞f(x) \to +\infty
  3. As x→−∞x \to -\infty, f(x)→+∞f(x) \to +\infty and as x→+∞x \to +\infty, f(x)→−∞f(x) \to -\infty
  4. The end behavior cannot be determined without knowing the leading coefficient sign (correct answer)
Explanation: The total degree is 2+1+3=6 (even), so both ends go in the same direction. However, whether they both go to +∞ or both go to -∞ depends on whether the leading coefficient is positive or negative. Choices A, B, and C incorrectly assume specific leading coefficient signs or odd degree behavior.

Question 2

A polynomial has zeros at x=−3,1,4x = -3, 1, 4 with the graph touching the x-axis at x=1x = 1 and crossing at the other two zeros. If the y-intercept is positive, what can be determined about the leading coefficient?

  1. The leading coefficient must be positive
  2. The leading coefficient must be negative (correct answer)
  3. The leading coefficient could be either positive or negative
  4. The leading coefficient cannot be determined without knowing the exact degree
Explanation: The zeros give us factors (x+3)(x-1)²(x-4) with minimum degree 4. The y-intercept is f(0) = (3)(-1)²(-4) = -12 times the leading coefficient. Since the y-intercept is positive and -12 is negative, the leading coefficient must be negative to make the product positive. Choice A gets the sign wrong. Choice C doesn't use the constraint properly. Choice D ignores that we can determine this from the given information.

Question 3

A cubic polynomial function intersects the x-axis at exactly two points. At one intersection point, the graph crosses the axis, and at the other, it appears to be tangent to the axis. What can be determined about the nature of all zeros of this polynomial?

  1. The polynomial has two real zeros and one complex zero with distinct values
  2. The polynomial has one real zero of multiplicity 1 and one real zero of multiplicity 2 (correct answer)
  3. The polynomial has three distinct real zeros, but one is not visible on the given domain
  4. The polynomial has one real zero of multiplicity 3 at the tangent point only
Explanation: A cubic polynomial has exactly 3 zeros (counting multiplicity). If the graph crosses the x-axis at one point, that's a zero of odd multiplicity (likely 1). If it's tangent at another point, that's a zero of even multiplicity (likely 2). This accounts for all 3 zeros: 1 + 2 = 3 total multiplicity.

Question 4

A degree-5 polynomial has exactly 3 x-intercepts. Two of the intercepts show the graph crossing the x-axis, while at the third intercept, the graph touches the x-axis but does not cross. What must be true about the multiplicities and the nature of the remaining zeros?

  1. Two zeros have multiplicity 1, one zero has multiplicity 3, and there are no complex zeros (correct answer)
  2. Two zeros have multiplicity 1, one zero has multiplicity 2, and there is one complex zero
  3. Three zeros have multiplicity 1, one zero has multiplicity 2, and there are no complex zeros
  4. Two zeros have multiplicity 1, one zero has multiplicity 2, and there are two complex conjugate zeros
Explanation: The graph crosses at two points (each multiplicity 1) and touches without crossing at one point (even multiplicity). For a degree-5 polynomial, we need total multiplicity of 5. If the touching point has multiplicity 2, we get 1+1+2=4, leaving one more zero. But a degree-5 polynomial with real coefficients cannot have exactly one complex zero (complex zeros come in conjugate pairs). Therefore, the touching point must have multiplicity 3, giving us 1+1+3=5 total, accounting for all zeros as real.

Question 5

A polynomial function has the following characteristics: it has a y-intercept of -12, zeros at x=−2x = -2 and x=3x = 3, and as x→∞x \to \infty, f(x)→−∞f(x) \to -\infty. If the function has the minimum possible degree, what is the multiplicity of the zero at x=3x = 3?

  1. The multiplicity is 1 because the y-intercept and end behavior determine a unique odd-degree polynomial
  2. The multiplicity is 2 because the negative y-intercept requires an even number of negative factors (correct answer)
  3. The multiplicity is 3 because the end behavior requires the rightmost zero to dominate the sign
  4. The multiplicity cannot be determined without additional information about the graph's behavior at x=3x = 3
Explanation: Let f(x) = a(x+2)^m(x-3)^n. The y-intercept is f(0) = a(2)^m(-3)^n = -12. For minimum degree, try m=n=1: f(0) = a(2)(-3) = -6a = -12, so a = 2. But then f(x) = 2(x+2)(x-3) has positive leading coefficient, contradicting the end behavior. We need negative leading coefficient, so a < 0. If m=1, n=2: f(0) = a(2)(9) = 18a = -12, so a = -2/3 < 0. This gives the correct end behavior for a cubic with negative leading coefficient.

Question 6

If a polynomial function has the factored form f(x)=−2(x+1)3(x−2)2(x−4)f(x) = -2(x + 1)^3(x - 2)^2(x - 4), what happens to the graph at x=2x = 2?

  1. The graph crosses the x-axis with an inflection point at x=2x = 2
  2. The graph touches the x-axis and bounces off without crossing at x=2x = 2 (correct answer)
  3. The graph has a vertical asymptote at x=2x = 2
  4. The graph crosses the x-axis with a local maximum or minimum at x=2x = 2
Explanation: The factor (x-2)² has even multiplicity 2, which means the graph touches the x-axis at x = 2 but doesn't cross it (bounces off). Choice A describes odd multiplicity behavior. Choice C describes rational function behavior, not polynomial. Choice D incorrectly suggests the graph crosses the axis.

Question 7

A quartic polynomial function has zeros at x=−2,0,3x = -2, 0, 3 and has a y-intercept of 0. What must be true about the multiplicity of the zero at x=0x = 0?

  1. The zero at x=0x = 0 has multiplicity 1, and there is another zero not given
  2. The zero at x=0x = 0 has multiplicity 2, accounting for the total degree of 4 (correct answer)
  3. The zero at x=0x = 0 has multiplicity 3, making this actually a quintic polynomial
  4. The zero at x=0x = 0 could have multiplicity 1 or 2, depending on the leading coefficient
Explanation: A y-intercept of 0 confirms that x = 0 is indeed a zero. For a quartic (degree 4) with zeros at x = -2, 0, 3, we need the multiplicities to sum to 4. If x = -2 and x = 3 each have multiplicity 1 (minimum), then x = 0 must have multiplicity 2 to reach degree 4. Choice A would give degree 3 or require a fourth distinct zero. Choice C makes it degree 5. Choice D misunderstands that the y-intercept being 0 constrains the factorization.