Algebra 2 Quiz: Zeros Of Polynomials To Construct Graphs
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Zeros Of Polynomials To Construct GraphsQuestion 1 of 20

How does the graph behave at each zero of g(x)=(x−3)2(x+1)3?g(x)=(x-3)^2(x+1)^3? Include multiplicities and the end behavior.

Zeros: x=3x=3 (mult. 2) and x=−1x=-1 (mult. 3). Touches at x=3x=3; crosses (flattened) at x=−1x=-1. End behavior: left down, right up.
Zeros: x=3x=3 (mult. 2) and x=−1x=-1 (mult. 3). Crosses at x=3x=3; touches at x=−1x=-1. End behavior: both ends up.
Zeros: x=3x=3 (mult. 3) and x=−1x=-1 (mult. 2). Crosses (flattened) at x=3x=3; touches at x=−1x=-1. End behavior: left up, right down.
Zeros: x=3x=3 (mult. 2) and x=−1x=-1 (mult. 3). Touches at both zeros. End behavior: left down, right up.
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Algebra 2 Quiz

Algebra 2 Quiz: Zeros Of Polynomials To Construct Graphs

Practice Zeros Of Polynomials To Construct Graphs in Algebra 2 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 Zeros Of Polynomials To Construct Graphs, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 2.

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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.

All questions

Question 1

How does the graph behave at each zero of g(x)=(x−3)2(x+1)3?g(x)=(x-3)^2(x+1)^3? Include multiplicities and the end behavior.

  1. Zeros: x=3x=3 (mult. 2) and x=−1x=-1 (mult. 3). Touches at x=3x=3; crosses (flattened) at x=−1x=-1. End behavior: left down, right up. (correct answer)
  2. Zeros: x=3x=3 (mult. 2) and x=−1x=-1 (mult. 3). Crosses at x=3x=3; touches at x=−1x=-1. End behavior: both ends up.
  3. Zeros: x=3x=3 (mult. 3) and x=−1x=-1 (mult. 2). Crosses (flattened) at x=3x=3; touches at x=−1x=-1. End behavior: left up, right down.
  4. Zeros: x=3x=3 (mult. 2) and x=−1x=-1 (mult. 3). Touches at both zeros. End behavior: left down, right up.
Explanation: This question tests your ability to identify zeros from a polynomial's factored form and use them, along with multiplicity information and end behavior, to construct a rough sketch of the polynomial's graph. Zeros from factored form p(x) = a(x - r₁)(x - r₂)... are found by setting each factor equal to zero: from (x - r), the zero is x = r. Multiplicity (how many times a factor appears) determines behavior at that zero: odd multiplicity means the graph crosses the x-axis, even multiplicity means it touches and bounces back. For example, (x - 2)² makes the graph touch at x = 2, while (x - 2)³ makes it cross but with a flattened shape. End behavior depends only on the leading term (highest degree): for even-degree polynomials, both ends go the same direction (both up if positive leading coefficient, both down if negative). For odd-degree, ends go opposite directions (positive leading coefficient: left down, right up). This plus the zeros gives you the skeleton of the graph! For g(x) = (x - 3)^2(x + 1)^3, zeros are x = 3 (multiplicity 2, touches), x = -1 (multiplicity 3, crosses flattened), degree 5 odd positive, left down right up. Choice A correctly identifies zeros with multiplicities, shows proper crossing and touching, and has correct end behavior. Choice B swaps the behaviors, but even multiplicity touches and odd crosses—use the 'odd crossers, even bouncers' aid to remember! The four-step polynomial sketching strategy: (1) Find all zeros by setting each factor (x - r) equal to zero (watch signs!), (2) Determine multiplicity of each zero (count how many times factor appears) and whether graph crosses (odd) or touches (even), (3) Find end behavior using degree (even = same both ends, odd = opposite ends) and leading coefficient sign (positive eventually goes up, negative eventually goes down), (4) Mark zeros on x-axis and connect with smooth curve showing proper behavior at each zero and correct end behavior. Rough shape is all you need! Multiplicity memory aid: Think 'odd crossers, even bouncers.' Odd multiplicity (1, 3, 5...) = graph crosses through the x-axis. Even multiplicity (2, 4, 6...) = graph bounces off the x-axis without crossing. Higher multiplicity = flatter at that zero. So (x - 3)² touches and turns around, (x - 3)³ crosses but flattens, (x - 3)⁴ touches with even more flattening. The pattern is consistent!

Question 2

Sketch f(x)=(x−3)(x+1)(x−2)2f(x)=(x-3)(x+1)(x-2)^2 showing zeros, multiplicities, and end behavior. Which option matches the correct sketch description?

  1. Zeros: x=3x=3 (mult. 1), x=−1x=-1 (mult. 1), x=2x=2 (mult. 2). Crosses at x=3x=3 and x=−1x=-1, touches at x=2x=2. End behavior: as x→−∞x\to -\infty, f(x)→+∞f(x)\to +\infty; as x→+∞x\to +\infty, f(x)→+∞f(x)\to +\infty. (correct answer)
  2. Zeros: x=3x=3 (mult. 1), x=−1x=-1 (mult. 1), x=2x=2 (mult. 2). Crosses at all three zeros. End behavior: as x→−∞x\to -\infty, f(x)→+∞f(x)\to +\infty; as x→+∞x\to +\infty, f(x)→+∞f(x)\to +\infty.
  3. Zeros: x=3x=3 (mult. 1), x=−1x=-1 (mult. 1), x=2x=2 (mult. 2). Crosses at x=3x=3 and x=−1x=-1, touches at x=2x=2. End behavior: as x→−∞x\to -\infty, f(x)→−∞f(x)\to -\infty; as x→+∞x\to +\infty, f(x)→+∞f(x)\to +\infty.
  4. Zeros: x=3x=3 (mult. 2), x=−1x=-1 (mult. 1), x=2x=2 (mult. 1). Touches at x=3x=3, crosses at x=−1x=-1 and x=2x=2. End behavior: as x→−∞x\to -\infty, f(x)→+∞f(x)\to +\infty; as x→+∞x\to +\infty, f(x)→+∞f(x)\to +\infty.
Explanation: This question tests your ability to identify zeros from a polynomial's factored form and use them, along with multiplicity information and end behavior, to construct a rough sketch of the polynomial's graph. Zeros from factored form p(x) = a(x - r₁)(x - r₂)... are found by setting each factor equal to zero: from (x - r), the zero is x = r. Multiplicity (how many times a factor appears) determines behavior at that zero: odd multiplicity means the graph crosses the x-axis, even multiplicity means it touches and bounces back. For example, (x - 2)² makes the graph touch at x = 2, while (x - 2)³ makes it cross but with a flattened shape. For f(x) = (x-3)(x+1)(x-2)², we find zeros: x-3=0 gives x=3 (multiplicity 1), x+1=0 gives x=-1 (multiplicity 1), and (x-2)²=0 gives x=2 (multiplicity 2). The graph crosses at x=3 and x=-1 (odd multiplicities) and touches at x=2 (even multiplicity). The polynomial has degree 1+1+2=4 (even) with positive leading coefficient (1), so both ends go up: as x→±∞, f(x)→+∞. Choice A correctly identifies zeros x=3 (mult. 1), x=-1 (mult. 1), x=2 (mult. 2), shows crossing at x=3 and x=-1, touching at x=2, and has both ends going to +∞. Choice B incorrectly claims the graph crosses at all three zeros, missing that x=2 with even multiplicity should touch, not cross. The four-step polynomial sketching strategy: (1) Find all zeros by setting each factor (x - r) equal to zero (watch signs!), (2) Determine multiplicity of each zero (count how many times factor appears) and whether graph crosses (odd) or touches (even), (3) Find end behavior using degree (even = same both ends, odd = opposite ends) and leading coefficient sign (positive eventually goes up, negative eventually goes down), (4) Mark zeros on x-axis and connect with smooth curve showing proper behavior at each zero and correct end behavior. Rough shape is all you need! Multiplicity memory aid: Think 'odd crossers, even bouncers.' Odd multiplicity (1, 3, 5...) = graph crosses through the x-axis. Even multiplicity (2, 4, 6...) = graph bounces off the x-axis without crossing. Higher multiplicity = flatter at that zero. So (x - 3)² touches and turns around, (x - 3)³ crosses but flattens, (x - 3)⁴ touches with even more flattening. The pattern is consistent!

Question 3

Sketch f(x)=(x+4)(x+1)(x−2)2f(x)=(x+4)(x+1)(x-2)^2 showing all zeros, their multiplicities, and the correct end behavior. Which description matches?

  1. Zeros: x=4x=4 (mult. 1), x=1x=1 (mult. 1), x=2x=2 (mult. 2). Crosses at 44 and 11, touches at 22. End behavior: both ends up.
  2. Zeros: x=−4x=-4 (mult. 1), x=−1x=-1 (mult. 1), x=2x=2 (mult. 2). Touches at all zeros. End behavior: left down, right up.
  3. Zeros: x=−4x=-4 (mult. 1), x=−1x=-1 (mult. 1), x=2x=2 (mult. 2). Crosses at −4-4 and −1-1, touches at 22. End behavior: both ends down.
  4. Zeros: x=−4x=-4 (mult. 1), x=−1x=-1 (mult. 1), x=2x=2 (mult. 2). Crosses at −4-4 and −1-1, touches at 22. End behavior: both ends up. (correct answer)
Explanation: This question tests your ability to identify zeros from a polynomial's factored form and use them, along with multiplicity information and end behavior, to construct a rough sketch of the polynomial's graph. Zeros from factored form p(x) = a(x - r₁)(x - r₂)... are found by setting each factor equal to zero: from (x - r), the zero is x = r. Multiplicity (how many times a factor appears) determines behavior at that zero: odd multiplicity means the graph crosses the x-axis, even multiplicity means it touches and bounces back. For example, (x - 2)² makes the graph touch at x = 2, while (x - 2)³ makes it cross but with a flattened shape. End behavior depends only on the leading term (highest degree): for even-degree polynomials, both ends go the same direction (both up if positive leading coefficient, both down if negative). For odd-degree, ends go opposite directions (positive leading coefficient: left down, right up). This plus the zeros gives you the skeleton of the graph! For f(x) = (x + 4)(x + 1)(x - 2)^2, zeros are x = -4 (multiplicity 1, crosses), x = -1 (multiplicity 1, crosses), x = 2 (multiplicity 2, touches), degree 4 even positive, both ends up. Choice A correctly identifies zeros with multiplicities, shows proper crossing and touching, and has correct end behavior. Choice B misidentifies zeros like x = 4 instead of x = -4—always solve factors with attention to signs! The four-step polynomial sketching strategy: (1) Find all zeros by setting each factor (x - r) equal to zero (watch signs!), (2) Determine multiplicity of each zero (count how many times factor appears) and whether graph crosses (odd) or touches (even), (3) Find end behavior using degree (even = same both ends, odd = opposite ends) and leading coefficient sign (positive eventually goes up, negative eventually goes down), (4) Mark zeros on x-axis and connect with smooth curve showing proper behavior at each zero and correct end behavior. Rough shape is all you need! Multiplicity memory aid: Think 'odd crossers, even bouncers.' Odd multiplicity (1, 3, 5...) = graph crosses through the x-axis. Even multiplicity (2, 4, 6...) = graph bounces off the x-axis without crossing. Higher multiplicity = flatter at that zero. So (x - 3)² touches and turns around, (x - 3)³ crosses but flattens, (x - 3)⁴ touches with even more flattening. The pattern is consistent!

Question 4

Sketch f(x)=−(x+3)2(x−2)(x−1)f(x)=- (x+3)^2(x-2)(x-1) showing all zeros with multiplicities, whether the graph crosses or touches at each, and the end behavior.

  1. Zeros: x=3x=3 (mult. 2), x=1x=1 (mult. 1), x=2x=2 (mult. 1). Touches at 33, crosses at 11 and 22. End behavior: both ends down.
  2. Zeros: x=−3x=-3 (mult. 2), x=1x=1 (mult. 1), x=2x=2 (mult. 1). Crosses at −3-3, touches at 11 and 22. End behavior: left up, right down.
  3. Zeros: x=−3x=-3 (mult. 2), x=1x=1 (mult. 1), x=2x=2 (mult. 1). Touches at −3-3, crosses at 11 and 22. End behavior: both ends up.
  4. Zeros: x=−3x=-3 (mult. 2), x=1x=1 (mult. 1), x=2x=2 (mult. 1). Touches at −3-3, crosses at 11 and 22. End behavior: both ends down. (correct answer)
Explanation: This question tests your ability to identify zeros from a polynomial's factored form and use them, along with multiplicity information and end behavior, to construct a rough sketch of the polynomial's graph. Zeros from factored form p(x) = a(x - r₁)(x - r₂)... are found by setting each factor equal to zero: from (x - r), the zero is x = r. Multiplicity (how many times a factor appears) determines behavior at that zero: odd multiplicity means the graph crosses the x-axis, even multiplicity means it touches and bounces back. For example, (x - 2)² makes the graph touch at x = 2, while (x - 2)³ makes it cross but with a flattened shape. End behavior depends only on the leading term (highest degree): for even-degree polynomials, both ends go the same direction (both up if positive leading coefficient, both down if negative). For odd-degree, ends go opposite directions (positive leading coefficient: left down, right up). This plus the zeros gives you the skeleton of the graph! For f(x) = - (x + 3)^2 (x - 2)(x - 1), the zeros are x = -3 (multiplicity 2, from (x + 3)^2 = 0), x = 1 (multiplicity 1, from x - 1 = 0), and x = 2 (multiplicity 1, from x - 2 = 0); the graph touches at x = -3 due to even multiplicity and crosses at x = 1 and x = 2 due to odd multiplicity, with the degree 4 (even) and negative leading coefficient indicating both ends down. Choice A correctly identifies the zeros with multiplicities, shows proper touching and crossing, and has correct end behavior. Choice B fails because it incorrectly states both ends up, ignoring the negative leading coefficient that makes both ends down. The four-step polynomial sketching strategy: (1) Find all zeros by setting each factor (x - r) equal to zero (watch signs!), (2) Determine multiplicity of each zero (count how many times factor appears) and whether graph crosses (odd) or touches (even), (3) Find end behavior using degree (even = same both ends, odd = opposite ends) and leading coefficient sign (positive eventually goes up, negative eventually goes down), (4) Mark zeros on x-axis and connect with smooth curve showing proper behavior at each zero and correct end behavior. Rough shape is all you need! Multiplicity memory aid: Think 'odd crossers, even bouncers.' Odd multiplicity (1, 3, 5...) = graph crosses through the x-axis. Even multiplicity (2, 4, 6...) = graph bounces off the x-axis without crossing. Higher multiplicity = flatter at that zero. So (x - 3)² touches and turns around, (x - 3)³ crosses but flattens, (x - 3)⁴ touches with even more flattening. The pattern is consistent!

Question 5

Use zeros to construct a rough graph of the polynomial p(x)=(x+2)(x−1)2(x−4).p(x)=(x+2)(x-1)^2(x-4). Identify the zeros (with multiplicities), state whether the graph crosses or touches the x-axis at each zero, and determine the end behavior.

  1. Zeros: x=2x=2 (mult. 1), x=1x=1 (mult. 2), x=4x=4 (mult. 1). Crosses at all zeros. End behavior: as x→−∞x \to -\infty, p(x)→+∞p(x) \to +\infty and as x→+∞x \to +\infty, p(x)→+∞p(x) \to +\infty.
  2. Zeros: x=−2x=-2 (mult. 1), x=1x=1 (mult. 2), x=4x=4 (mult. 1). Touches at −2-2 and 44, crosses at 11. End behavior: as x→−∞x \to -\infty, p(x)→−∞p(x) \to -\infty and as x→+∞x \to +\infty, p(x)→−∞p(x) \to -\infty.
  3. Zeros: x=−2x=-2 (mult. 1), x=1x=1 (mult. 2), x=4x=4 (mult. 1). Crosses at −2-2 and 44, touches at 11. End behavior: as x→−∞x \to -\infty, p(x)→+∞p(x) \to +\infty and as x→+∞x \to +\infty, p(x)→+∞p(x) \to +\infty. (correct answer)
  4. Zeros: x=−2x=-2 (mult. 1), x=1x=1 (mult. 1), x=4x=4 (mult. 1). Crosses at all zeros. End behavior: as x→−∞x \to -\infty, p(x)→+∞p(x) \to +\infty and as x→+∞x \to +\infty, p(x)→+∞p(x) \to +\infty.
Explanation: This question tests your ability to identify zeros from a polynomial's factored form and use them, along with multiplicity information and end behavior, to construct a rough sketch of the polynomial's graph. Zeros from factored form p(x)=a(x−r1)(x−r2)…p(x) = a(x - r_1)(x - r_2) \dots are found by setting each factor equal to zero: from (x−r)(x - r), the zero is x=rx = r. Multiplicity (how many times a factor appears) determines behavior at that zero: odd multiplicity means the graph crosses the x-axis, even multiplicity means it touches and bounces back. For example, (x−2)2(x - 2)^2 makes the graph touch at x=2x = 2, while (x−2)3(x - 2)^3 makes it cross but with a flattened shape. End behavior depends only on the leading term (highest degree): for even-degree polynomials, both ends go the same direction (both up if positive leading coefficient, both down if negative). For odd-degree, ends go opposite directions (positive leading coefficient: left down, right up). This plus the zeros gives you the skeleton of the graph! For p(x)=(x+2)(x−1)2(x−4)p(x) = (x + 2)(x - 1)^2(x - 4), the zeros are x=−2x = -2 (multiplicity 1, from x+2=0x + 2 = 0), x=1x = 1 (multiplicity 2), and x=4x = 4 (multiplicity 1); the graph crosses at x=−2x = -2 and x=4x = 4 due to odd multiplicity and touches at x=1x = 1 due to even multiplicity, with the degree 4 (even) and positive leading coefficient indicating both ends up. Choice A correctly identifies the zeros with multiplicities, shows proper crossing and touching, and has correct end behavior. Choice B fails because it incorrectly lists a zero at x=2x = 2 instead of x=−2x = -2 and states crossing at all zeros, ignoring the even multiplicity at x=1x = 1. The four-step polynomial sketching strategy: (1) Find all zeros by setting each factor (x−r)(x - r) equal to zero (watch signs!), (2) Determine multiplicity of each zero (count how many times factor appears) and whether graph crosses (odd) or touches (even), (3) Find end behavior using degree (even = same both ends, odd = opposite ends) and leading coefficient sign (positive eventually goes up, negative eventually goes down), (4) Mark zeros on x-axis and connect with smooth curve showing proper behavior at each zero and correct end behavior. Rough shape is all you need! Multiplicity memory aid: Think 'odd crossers, even bouncers.' Odd multiplicity (1, 3, 5...) = graph crosses through the x-axis. Even multiplicity (2, 4, 6...) = graph bounces off the x-axis without crossing. Higher multiplicity = flatter at that zero. So (x−3)2(x - 3)^2 touches and turns around, (x−3)3(x - 3)^3 crosses but flattens, (x−3)4(x - 3)^4 touches with even more flattening. The pattern is consistent!

Question 6

For the polynomial p(x)=2(x+3)(x−1)2(x−4)p(x)=2(x+3)(x-1)^2(x-4), identify all zeros with their multiplicities, determine whether the graph crosses or touches the xx-axis at each zero, and state the end behavior.

  1. Zeros: x=−3x=-3 (mult. 1), x=1x=1 (mult. 2), x=4x=4 (mult. 1). Crosses at −3-3 and 44, touches at 11. End behavior: as x→−∞x\to -\infty, p(x)→+∞p(x)\to +\infty; as x→+∞x\to +\infty, p(x)→+∞p(x)\to +\infty. (correct answer)
  2. Zeros: x=−3x=-3 (mult. 2), x=1x=1 (mult. 1), x=4x=4 (mult. 1). Touches at −3-3, crosses at 11 and 44. End behavior: as x→−∞x\to -\infty, p(x)→+∞p(x)\to +\infty; as x→+∞x\to +\infty, p(x)→+∞p(x)\to +\infty.
  3. Zeros: x=3x=3 (mult. 1), x=1x=1 (mult. 2), x=4x=4 (mult. 1). Crosses at 33 and 44, touches at 11. End behavior: as x→−∞x\to -\infty, p(x)→+∞p(x)\to +\infty; as x→+∞x\to +\infty, p(x)→+∞p(x)\to +\infty.
  4. Zeros: x=−3x=-3 (mult. 1), x=1x=1 (mult. 2), x=4x=4 (mult. 1). Crosses at all zeros. End behavior: as x→−∞x\to -\infty, p(x)→−∞p(x)\to -\infty; as x→+∞x\to +\infty, p(x)→+∞p(x)\to +\infty.
Explanation: This question tests your ability to identify zeros from a polynomial's factored form and use them, along with multiplicity information and end behavior, to construct a rough sketch of the polynomial's graph. Zeros from factored form p(x) = a(x - r₁)(x - r₂)... are found by setting each factor equal to zero: from (x - r), the zero is x = r. Multiplicity (how many times a factor appears) determines behavior at that zero: odd multiplicity means the graph crosses the x-axis, even multiplicity means it touches and bounces back. For example, (x - 2)² makes the graph touch at x = 2, while (x - 2)³ makes it cross but with a flattened shape. For p(x) = 2(x+3)(x-1)²(x-4), we find zeros by setting each factor to zero: x+3=0 gives x=-3 (multiplicity 1), (x-1)²=0 gives x=1 (multiplicity 2), and x-4=0 gives x=4 (multiplicity 1). Since multiplicities 1 are odd, the graph crosses at x=-3 and x=4; since multiplicity 2 is even, the graph touches at x=1. The polynomial has degree 1+2+1=4 (even) with positive leading coefficient (2), so both ends go up. Choice A correctly identifies zeros x=-3 (mult. 1), x=1 (mult. 2), x=4 (mult. 1), shows crossing at -3 and 4, touching at 1, and has both ends going to +∞. Choice C incorrectly identifies x=3 as a zero instead of x=-3, missing the sign in the factor (x+3). The four-step polynomial sketching strategy: (1) Find all zeros by setting each factor (x - r) equal to zero (watch signs!), (2) Determine multiplicity of each zero (count how many times factor appears) and whether graph crosses (odd) or touches (even), (3) Find end behavior using degree (even = same both ends, odd = opposite ends) and leading coefficient sign (positive eventually goes up, negative eventually goes down), (4) Mark zeros on x-axis and connect with smooth curve showing proper behavior at each zero and correct end behavior. Rough shape is all you need! Multiplicity memory aid: Think 'odd crossers, even bouncers.' Odd multiplicity (1, 3, 5...) = graph crosses through the x-axis. Even multiplicity (2, 4, 6...) = graph bounces off the x-axis without crossing. Higher multiplicity = flatter at that zero. So (x - 3)² touches and turns around, (x - 3)³ crosses but flattens, (x - 3)⁴ touches with even more flattening. The pattern is consistent!

Question 7

Consider the factored polynomial p(x)=−2(x+1)3(x−3)2(x−5)p(x) = -2(x+1)^3(x-3)^2(x-5). Which statement correctly describes both the end behavior and the behavior at x=3x = 3?

  1. As x→∞x \to \infty, p(x)→−∞p(x) \to -\infty, and the graph touches the x-axis at x=3x = 3 without crossing (correct answer)
  2. As x→∞x \to \infty, p(x)→+∞p(x) \to +\infty, and the graph touches the x-axis at x=3x = 3 without crossing
  3. As x→∞x \to \infty, p(x)→−∞p(x) \to -\infty, and the graph crosses the x-axis at x=3x = 3
  4. As x→∞x \to \infty, p(x)→+∞p(x) \to +\infty, and the graph crosses the x-axis at x=3x = 3
Explanation: The polynomial has degree 3+2+1=63+2+1=6 (even) with leading coefficient −2-2 (negative), so as x→∞x \to \infty, p(x)→−∞p(x) \to -\infty. At x=3x=3, the factor (x−3)2(x-3)^2 has even multiplicity, so the graph touches but doesn't cross the x-axis. Choice B has wrong end behavior (should be −∞-\infty, not +∞+\infty). Choices C and D wrongly claim the graph crosses at x=3x=3.

Question 8

A polynomial function has zeros at x=−2x = -2 (multiplicity 2), x=1x = 1 (multiplicity 1), and x=4x = 4 (multiplicity 3). If the leading coefficient is positive, which interval contains a point where the function value is negative?

  1. (−∞,−2)(-\infty, -2) because the function starts negative for large negative x-values
  2. (1,4)(1, 4) because the function changes sign at both endpoints of this interval (correct answer)
  3. (4,∞)(4, \infty) because the function has odd multiplicity at x=4x = 4
  4. (−2,1)(-2, 1) because the function bounces off the x-axis at x=−2x = -2 staying negative
Explanation: The polynomial has degree 2+1+3=62+1+3=6 (even) with positive leading coefficient, so it goes to +∞+\infty as x→±∞x \to \pm\infty. Sign changes occur only at zeros with odd multiplicity (x=1x=1 and x=4x=4). Starting from the right: positive for x>4x>4, negative for 1<x<41<x<4, positive for x<1x<1. Choice A is wrong because the function is positive for x<−2x<-2. Choice C is wrong because the function is positive for x>4x>4. Choice D is wrong because the function is positive in (−2,1)(-2,1).

Question 9

Describe the graph's behavior based on factored form: h(x)=(x+1)4(x−3).h(x)=(x+1)^4(x-3). Identify the zeros with multiplicities, state whether the graph crosses or touches at each zero, and determine the end behavior.

  1. Zeros: x=−1x=-1 (mult. 4), x=3x=3 (mult. 1). Touches at −1-1; crosses at 33. End behavior: left down, right up. (correct answer)
  2. Zeros: x=−1x=-1 (mult. 4), x=3x=3 (mult. 1). Crosses at −1-1; touches at 33. End behavior: both ends up.
  3. Zeros: x=1x=1 (mult. 4), x=3x=3 (mult. 1). Touches at 11; crosses at 33. End behavior: left down, right up.
  4. Zeros: x=−1x=-1 (mult. 4), x=3x=3 (mult. 1). Touches at −1-1; crosses at 33. End behavior: both ends down.
Explanation: This question tests your ability to identify zeros from a polynomial's factored form and use them, along with multiplicity information and end behavior, to construct a rough sketch of the polynomial's graph. Zeros from factored form p(x) = a(x - r₁)(x - r₂)... are found by setting each factor equal to zero: from (x - r), the zero is x = r. Multiplicity (how many times a factor appears) determines behavior at that zero: odd multiplicity means the graph crosses the x-axis, even multiplicity means it touches and bounces back. For example, (x - 2)² makes the graph touch at x = 2, while (x - 2)³ makes it cross but with a flattened shape. End behavior depends only on the leading term (highest degree): for even-degree polynomials, both ends go the same direction (both up if positive leading coefficient, both down if negative). For odd-degree, ends go opposite directions (positive leading coefficient: left down, right up). This plus the zeros gives you the skeleton of the graph! For h(x) = (x + 1)^4 (x - 3), the zeros are x = -1 (multiplicity 4, from (x + 1)^4 = 0) and x = 3 (multiplicity 1, from x - 3 = 0); the graph touches at x = -1 due to even multiplicity and crosses at x = 3 due to odd multiplicity, with degree 5 (odd) and positive leading coefficient indicating left down and right up. Choice A correctly identifies the zeros with multiplicities, shows proper touching and crossing, and has correct end behavior. Choice D fails because it incorrectly states both ends down, which would require a negative leading coefficient for this even-like behavior, but the leading is positive and degree odd. The four-step polynomial sketching strategy: (1) Find all zeros by setting each factor (x - r) equal to zero (watch signs!), (2) Determine multiplicity of each zero (count how many times factor appears) and whether graph crosses (odd) or touches (even), (3) Find end behavior using degree (even = same both ends, odd = opposite ends) and leading coefficient sign (positive eventually goes up, negative eventually goes down), (4) Mark zeros on x-axis and connect with smooth curve showing proper behavior at each zero and correct end behavior. Rough shape is all you need! Multiplicity memory aid: Think 'odd crossers, even bouncers.' Odd multiplicity (1, 3, 5...) = graph crosses through the x-axis. Even multiplicity (2, 4, 6...) = graph bounces off the x-axis without crossing. Higher multiplicity = flatter at that zero. So (x - 3)² touches and turns around, (x - 3)³ crosses but flattens, (x - 3)⁴ touches with even more flattening. The pattern is consistent!

Question 10

A cubic polynomial s(x)s(x) has exactly two x-intercepts visible on its graph, with the graph crossing the x-axis at x=−1x = -1 and appearing to just touch the x-axis at x=3x = 3. Which factored form best represents this polynomial?

  1. s(x)=a(x+1)3s(x) = a(x+1)^3 where aa is a nonzero constant and the zero at x=3x=3 is imaginary
  2. s(x)=a(x+1)2(x−3)s(x) = a(x+1)^2(x-3) where aa is a nonzero constant
  3. s(x)=a(x+1)(x−3)(x−b)s(x) = a(x+1)(x-3)(x-b) where aa and bb are nonzero constants
  4. s(x)=a(x+1)(x−3)2s(x) = a(x+1)(x-3)^2 where aa is a nonzero constant (correct answer)
Explanation: When analyzing polynomial graphs, the key is understanding how the behavior at each x-intercept reveals information about the multiplicity of that zero. If the graph crosses the x-axis, the zero has odd multiplicity. If it touches but doesn't cross (creating a "bounce"), the zero has even multiplicity. Since the graph crosses at x=−1x = -1, this zero has odd multiplicity. Since it touches but doesn't cross at x=3x = 3, this zero has even multiplicity. For a cubic polynomial with only two visible x-intercepts, you need the multiplicities to add up to 3. The only way to achieve this is with multiplicities of 1 and 2. Since x=−1x = -1 crosses (odd), it has multiplicity 1. Since x=3x = 3 touches (even), it has multiplicity 2. This gives us s(x)=a(x+1)(x−3)2s(x) = a(x+1)(x-3)^2, which is answer choice D. Choice A is incorrect because it only has one real zero at x=−1x = -1 with multiplicity 3, not two visible x-intercepts. Choice B reverses the multiplicities—it would show touching at x=−1x = -1 and crossing at x=3x = 3, opposite of what's described. Choice C suggests three distinct real zeros, which would produce three x-intercepts, not two. Remember this pattern: crossing the x-axis means odd multiplicity, while touching (bouncing off) means even multiplicity. For cubic polynomials, always check that the sum of multiplicities equals 3.

Question 11

A polynomial function f(x)f(x) has the factored form f(x)=a(x+4)(x−1)2(x−6)f(x) = a(x+4)(x-1)^2(x-6) where aa is a nonzero constant. If the y-intercept of the graph is 48, what is the value of aa, and how does this affect the graph's behavior at x=1x = 1?

  1. a=2a = 2, and the graph has a local minimum at x=1x = 1 touching the x-axis
  2. a=−2a = -2, and the graph has a local maximum at x=1x = 1 touching the x-axis (correct answer)
  3. a=2a = 2, and the graph has a local maximum at x=1x = 1 touching the x-axis
  4. a=−2a = -2, and the graph has a local minimum at x=1x = 1 touching the x-axis
Explanation: The y-intercept is f(0)=a(4)(−1)2(−6)=a(4)(1)(−6)=−24a=48f(0) = a(4)(-1)^2(-6) = a(4)(1)(-6) = -24a = 48, so a=−2a = -2. At x=1x=1, we have (x−1)2(x-1)^2 with even multiplicity, so the graph touches the x-axis. Since a<0a<0, the parabolic behavior near x=1x=1 opens downward, creating a local maximum. Choice A has wrong sign for aa and wrong type of extremum. Choice C has wrong type of extremum. Choice D has wrong type of extremum.

Question 12

For g(x)=x2(x−5)3,g(x)=x^2(x-5)^3, identify the zeros and their multiplicities, describe how the graph behaves at each zero (crosses or touches, and whether it flattens), and state the end behavior.

  1. Zeros: x=0x=0 (mult. 2) and x=5x=5 (mult. 3). Touches/bounces at 00; crosses with flattening at 55. End behavior: left down, right up. (correct answer)
  2. Zeros: x=0x=0 (mult. 3) and x=5x=5 (mult. 2). Crosses with flattening at 00; touches at 55. End behavior: left down, right up.
  3. Zeros: x=0x=0 (mult. 2) and x=5x=5 (mult. 3). Crosses at 00; touches at 55. End behavior: both ends up.
  4. Zeros: x=0x=0 (mult. 2) and x=−5x=-5 (mult. 3). Touches at 00; crosses with flattening at −5-5. End behavior: left up, right down.
Explanation: This question tests your ability to identify zeros from a polynomial's factored form and use them, along with multiplicity information and end behavior, to construct a rough sketch of the polynomial's graph. Zeros from factored form p(x) = a(x - r₁)(x - r₂)... are found by setting each factor equal to zero: from (x - r), the zero is x = r. Multiplicity (how many times a factor appears) determines behavior at that zero: odd multiplicity means the graph crosses the x-axis, even multiplicity means it touches and bounces back. For example, (x - 2)² makes the graph touch at x = 2, while (x - 2)³ makes it cross but with a flattened shape. End behavior depends only on the leading term (highest degree): for even-degree polynomials, both ends go the same direction (both up if positive leading coefficient, both down if negative). For odd-degree, ends go opposite directions (positive leading coefficient: left down, right up). This plus the zeros gives you the skeleton of the graph! For g(x) = x^2 (x - 5)^3, the zeros are x = 0 (multiplicity 2, from x2x^2 = 0) and x = 5 (multiplicity 3, from (x - 5)^3 = 0); the graph touches and bounces at x = 0 due to even multiplicity and crosses with flattening at x = 5 due to odd multiplicity greater than 1, with the degree 5 (odd) and positive leading coefficient indicating left down and right up. Choice A correctly identifies the zeros with multiplicities, shows proper touching and crossing with flattening, and has correct end behavior. Choice C fails because it swaps the multiplicities, stating x = 0 has multiplicity 3 and x = 5 has 2, which reverses the behaviors at those zeros. The four-step polynomial sketching strategy: (1) Find all zeros by setting each factor (x - r) equal to zero (watch signs!), (2) Determine multiplicity of each zero (count how many times factor appears) and whether graph crosses (odd) or touches (even), (3) Find end behavior using degree (even = same both ends, odd = opposite ends) and leading coefficient sign (positive eventually goes up, negative eventually goes down), (4) Mark zeros on x-axis and connect with smooth curve showing proper behavior at each zero and correct end behavior. Rough shape is all you need! Multiplicity memory aid: Think 'odd crossers, even bouncers.' Odd multiplicity (1, 3, 5...) = graph crosses through the x-axis. Even multiplicity (2, 4, 6...) = graph bounces off the x-axis without crossing. Higher multiplicity = flatter at that zero. So (x - 3)² touches and turns around, (x - 3)³ crosses but flattens, (x - 3)⁴ touches with even more flattening. The pattern is consistent!

Question 13

A polynomial has zeros at x=−3,1,1,4x = -3, 1, 1, 4 (listed with repetition according to multiplicity). If this polynomial has a negative leading coefficient, in which interval must the function have a positive value?

  1. (−3,1)(-3, 1) because the function changes sign at x=−3x = -3 and approaches the x-axis at x=1x = 1 (correct answer)
  2. (1,4)(1, 4) because the function stays on the same side of the x-axis between these consecutive zeros
  3. (4,∞)(4, \infty) because the function must eventually become positive for large x-values
  4. (−∞,−3)(-\infty, -3) because the negative leading coefficient makes the left end behavior positive
Explanation: The polynomial is f(x)=a(x+3)(x−1)2(x−4)f(x) = a(x+3)(x-1)^2(x-4) where a<0a<0. Degree is 4 (even) with negative leading coefficient, so both ends go to −∞-\infty. Sign changes only occur at zeros with odd multiplicity: x=−3x=-3 and x=4x=4. Working from right to left: negative for x>4x>4, positive for −3<x<4-3<x<4, negative for x<−3x<-3. Since x=1x=1 has even multiplicity, no sign change occurs there. Choice B includes the interval (1,4)(1,4) but the entire interval (−3,4)(-3,4) is positive. Choices C and D are incorrect about end behavior.

Question 14

The graph of y=g(x)y = g(x) passes through the point (2,0)(2, 0) and has the property that g(x)=(x−2)2⋅h(x)g(x) = (x-2)^2 \cdot h(x) where h(x)h(x) is a polynomial with h(2)≠0h(2) \neq 0. Based on this information and the factor (x−2)2(x-2)^2, what can be concluded about the graph near x=2x = 2?

  1. The graph crosses the x-axis at x=2x = 2 because (x−2)2(x-2)^2 contains the factor (x−2)(x-2)
  2. The graph has a local minimum at x=2x = 2 and touches the x-axis without crossing
  3. The graph touches the x-axis at x=2x = 2 but whether it has a minimum or maximum depends on h(x)h(x) (correct answer)
  4. The graph has a local maximum at x=2x = 2 and touches the x-axis without crossing
Explanation: Since (x−2)2(x-2)^2 gives x=2x=2 multiplicity 2 (even), the graph touches but doesn't cross the x-axis. However, whether there's a local min or max depends on the sign of h(2)h(2): if h(2)>0h(2)>0, there's a local minimum; if h(2)<0h(2)<0, there's a local maximum. Choice A is wrong because even multiplicity means no crossing. Choices B and D are wrong because they definitively state min/max without knowing the sign of h(2)h(2).

Question 15

The polynomial q(x)=x4−5x3+6x2q(x) = x^4 - 5x^3 + 6x^2 can be factored by first factoring out the greatest common factor. After complete factorization, which statement about the x-intercepts and their effect on the graph is correct?

  1. The graph has x-intercepts at x=0,2,3x = 0, 2, 3 and crosses the x-axis at x=2x = 2 and x=3x = 3 only
  2. The graph has x-intercepts at x=0,2,3x = 0, 2, 3 and crosses the x-axis at all three points
  3. The graph has x-intercepts at x=0,2,3x = 0, 2, 3 and touches without crossing at x=0x = 0 only (correct answer)
  4. The graph has x-intercepts at x=0,2,3x = 0, 2, 3 and touches without crossing at x=2x = 2 and x=3x = 3 only
Explanation: When you encounter a polynomial and need to analyze its x-intercepts and graphing behavior, start by factoring completely, then examine the multiplicity of each root to determine whether the graph crosses or touches the x-axis at each intercept. Let's factor q(x)=x4−5x3+6x2q(x) = x^4 - 5x^3 + 6x^2 step by step. First, factor out the greatest common factor of x2x^2: q(x)=x2(x2−5x+6)q(x) = x^2(x^2 - 5x + 6). Next, factor the quadratic x2−5x+6x^2 - 5x + 6 by finding two numbers that multiply to 6 and add to -5: those are -2 and -3. So x2−5x+6=(x−2)(x−3)x^2 - 5x + 6 = (x - 2)(x - 3). The complete factorization is q(x)=x2(x−2)(x−3)q(x) = x^2(x - 2)(x - 3). The x-intercepts occur where q(x)=0q(x) = 0, giving us x=0,2,3x = 0, 2, 3. The key insight is examining the multiplicity of each root. The factor x2x^2 means x=0x = 0 has multiplicity 2 (even), while (x−2)(x - 2) and (x−3)(x - 3) each have multiplicity 1 (odd). When a root has even multiplicity, the graph touches the x-axis without crossing. When a root has odd multiplicity, the graph crosses the x-axis. Choice A incorrectly states the graph doesn't cross at x=0x = 0, but it actually touches there. Choice B incorrectly claims the graph crosses at all three points, ignoring the even multiplicity at x=0x = 0. Choice D incorrectly suggests x=2x = 2 and x=3x = 3 have even multiplicities when they're actually odd. Remember: even multiplicity means "touch without crossing," odd multiplicity means "cross through." Always check the exponent of each factor after complete factorization.

Question 16

What is the end behavior of p(x)=3(x−2)(x+1)(x−4)2?p(x)=3(x-2)(x+1)(x-4)^2? (You may also use the degree and leading coefficient to decide.)

  1. As x→−∞x\to-\infty, p(x)→∞p(x)\to\infty and as x→∞x\to\infty, p(x)→∞p(x)\to\infty. (correct answer)
  2. As x→−∞x\to-\infty, p(x)→∞p(x)\to\infty and as x→∞x\to\infty, p(x)→−∞p(x)\to-\infty.
  3. As x→−∞x\to-\infty, p(x)→−∞p(x)\to-\infty and as x→∞x\to\infty, p(x)→−∞p(x)\to-\infty.
  4. As x→−∞x\to-\infty, p(x)→−∞p(x)\to-\infty and as x→∞x\to\infty, p(x)→∞p(x)\to\infty.
Explanation: This question tests your ability to identify zeros from a polynomial's factored form and use them, along with multiplicity information and end behavior, to construct a rough sketch of the polynomial's graph. End behavior depends only on the leading term (highest degree): for even-degree polynomials, both ends go the same direction (both up if positive leading coefficient, both down if negative). For odd-degree, ends go opposite directions (positive leading coefficient: left down, right up). This plus the zeros gives you the skeleton of the graph! For p(x)=3(x−2)(x+1)(x−4)2p(x) = 3(x-2)(x+1)(x-4)^2, the degree is 1+1+2=41+1+2=4 (even). When expanded, the leading term will be 3x43x^4, which has positive coefficient. For even degree with positive leading coefficient, both ends go up: as x→−∞x \to -\infty, p(x)→∞p(x) \to \infty and as x→∞x \to \infty, p(x)→∞p(x) \to \infty. Choice C correctly states this end behavior. Choice A shows the end behavior for odd degree (opposite ends), Choice B shows odd degree with negative leading coefficient, and Choice D shows even degree with negative leading coefficient. The four-step polynomial sketching strategy: (1) Find all zeros by setting each factor (x−rx - r) equal to zero (watch signs!), (2) Determine multiplicity of each zero (count how many times factor appears) and whether graph crosses (odd) or touches (even), (3) Find end behavior using degree (even = same both ends, odd = opposite ends) and leading coefficient sign (positive eventually goes up, negative eventually goes down), (4) Mark zeros on x-axis and connect with smooth curve showing proper behavior at each zero and correct end behavior. Rough shape is all you need!

Question 17

Sketch f(x)=−2(x+3)2(x−1)(x−4)f(x)=-2(x+3)^2(x-1)(x-4) showing zeros, their multiplicities, and end behavior. Which description matches the correct rough graph?

  1. Zeros: x=−3x=-3 (mult. 2), x=1x=1 (mult. 1), x=4x=4 (mult. 1). Touches at −3-3, crosses at 11 and 44. End behavior: left down, right up.
  2. Zeros: x=−3x=-3 (mult. 2), x=1x=1 (mult. 2), x=4x=4 (mult. 1). Touches at −3-3 and 11, crosses at 44. End behavior: both ends down.
  3. Zeros: x=3x=3 (mult. 2), x=1x=1 (mult. 1), x=4x=4 (mult. 1). Touches at 33, crosses at 11 and 44. End behavior: both ends down.
  4. Zeros: x=−3x=-3 (mult. 2), x=1x=1 (mult. 1), x=4x=4 (mult. 1). Touches at −3-3, crosses at 11 and 44. 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. (correct answer)
Explanation: This question tests your ability to identify zeros from a polynomial's factored form and use them, along with multiplicity information and end behavior, to construct a rough sketch of the polynomial's graph. Zeros from factored form p(x) = a(x - r₁)(x - r₂)... are found by setting each factor equal to zero: from (x - r), the zero is x = r. Multiplicity (how many times a factor appears) determines behavior at that zero: odd multiplicity means the graph crosses the x-axis, even multiplicity means it touches and bounces back. For example, (x - 2)² makes the graph touch at x = 2, while (x - 2)³ makes it cross but with a flattened shape. End behavior depends only on the leading term (highest degree): for even-degree polynomials, both ends go the same direction (both up if positive leading coefficient, both down if negative). For odd-degree, ends go opposite directions (positive leading coefficient: left down, right up). This plus the zeros gives you the skeleton of the graph! For f(x) = -2(x + 3)^2(x - 1)(x - 4), zeros are x = -3 (multiplicity 2, touches), x = 1 (multiplicity 1, crosses), x = 4 (multiplicity 1, crosses), with degree 4 (even) and negative leading coefficient, so both ends down. Choice A correctly identifies zeros with multiplicities, shows proper crossing and touching, and has correct end behavior. A distractor like choice B might get the end behavior wrong by assuming odd degree, but count the total degree as 4 (even) and note the negative coefficient flips it downward—keep practicing to spot this! The four-step polynomial sketching strategy: (1) Find all zeros by setting each factor (x - r) equal to zero (watch signs!), (2) Determine multiplicity of each zero (count how many times factor appears) and whether graph crosses (odd) or touches (even), (3) Find end behavior using degree (even = same both ends, odd = opposite ends) and leading coefficient sign (positive eventually goes up, negative eventually goes down), (4) Mark zeros on x-axis and connect with smooth curve showing proper behavior at each zero and correct end behavior. Rough shape is all you need! Multiplicity memory aid: Think 'odd crossers, even bouncers.' Odd multiplicity (1, 3, 5...) = graph crosses through the x-axis. Even multiplicity (2, 4, 6...) = graph bounces off the x-axis without crossing. Higher multiplicity = flatter at that zero. So (x - 3)² touches and turns around, (x - 3)³ crosses but flattens, (x - 3)⁴ touches with even more flattening. The pattern is consistent!

Question 18

For t(x)=(x−1)3(x+4)2,t(x)=(x-1)^3(x+4)^2, identify the zeros and determine how the graph behaves at each zero (crosses or touches, and whether it flattens). Also determine the end behavior.

  1. Zeros: x=1x=1 (mult. 3), x=−4x=-4 (mult. 2). Crosses with flattening at 11; touches at −4-4. End behavior: left down, right up. (correct answer)
  2. Zeros: x=1x=1 (mult. 3), x=−4x=-4 (mult. 2). Touches at 11; crosses at −4-4. End behavior: both ends up.
  3. Zeros: x=1x=1 (mult. 2), x=−4x=-4 (mult. 3). Touches at 11; crosses with flattening at −4-4. End behavior: left down, right up.
  4. Zeros: x=1x=1 (mult. 3), x=−4x=-4 (mult. 2). Crosses with flattening at 11; touches at −4-4. End behavior: both ends down.
Explanation: This question tests your ability to identify zeros from a polynomial's factored form and use them, along with multiplicity information and end behavior, to construct a rough sketch of the polynomial's graph. Zeros from factored form p(x) = a(x - r₁)(x - r₂)... are found by setting each factor equal to zero: from (x - r), the zero is x = r. Multiplicity (how many times a factor appears) determines behavior at that zero: odd multiplicity means the graph crosses the x-axis, even multiplicity means it touches and bounces back. For example, (x - 2)² makes the graph touch at x = 2, while (x - 2)³ makes it cross but with a flattened shape. End behavior depends only on the leading term (highest degree): for even-degree polynomials, both ends go the same direction (both up if positive leading coefficient, both down if negative). For odd-degree, ends go opposite directions (positive leading coefficient: left down, right up). This plus the zeros gives you the skeleton of the graph! For t(x) = (x - 1)^3 (x + 4)^2, the zeros are x = 1 (multiplicity 3, from (x - 1)^3 = 0) and x = -4 (multiplicity 2, from (x + 4)^2 = 0); the graph crosses with flattening at x = 1 due to odd multiplicity greater than 1 and touches at x = -4 due to even multiplicity, with degree 5 (odd) and positive leading coefficient indicating left down and right up. Choice A correctly identifies the zeros with multiplicities, shows proper crossing with flattening and touching, and has correct end behavior. Choice B fails because it incorrectly states touching at x=1 (which is odd multiplicity, so crosses) and both ends up (which is for even degree). The four-step polynomial sketching strategy: (1) Find all zeros by setting each factor (x - r) equal to zero (watch signs!), (2) Determine multiplicity of each zero (count how many times factor appears) and whether graph crosses (odd) or touches (even), (3) Find end behavior using degree (even = same both ends, odd = opposite ends) and leading coefficient sign (positive eventually goes up, negative eventually goes down), (4) Mark zeros on x-axis and connect with smooth curve showing proper behavior at each zero and correct end behavior. Rough shape is all you need! Multiplicity memory aid: Think 'odd crossers, even bouncers.' Odd multiplicity (1, 3, 5...) = graph crosses through the x-axis. Even multiplicity (2, 4, 6...) = graph bounces off the x-axis without crossing. Higher multiplicity = flatter at that zero. So (x - 3)² touches and turns around, (x - 3)³ crosses but flattens, (x - 3)⁴ touches with even more flattening. The pattern is consistent!

Question 19

Use zeros to construct a rough graph of the polynomial p(x)=(x+2)(x−1)2(x−4).p(x)=(x+2)(x-1)^2(x-4). Identify all zeros with multiplicities, state whether the graph crosses or touches the x-axis at each zero, and determine the end behavior.

  1. Zeros: x=2x=2 (mult. 1), x=1x=1 (mult. 2), x=4x=4 (mult. 1). Crosses at all zeros. End behavior: left down, right up.
  2. Zeros: x=−2x=-2 (mult. 2), x=1x=1 (mult. 1), x=4x=4 (mult. 1). Touches at −2-2, crosses at 11 and 44. End behavior: left up, right down.
  3. Zeros: x=−2x=-2 (mult. 1), x=1x=1 (mult. 2). Crosses at −2-2, touches at 11. End behavior: both ends up.
  4. Zeros: x=−2x=-2 (mult. 1), x=1x=1 (mult. 2), x=4x=4 (mult. 1). Crosses at −2-2 and 44, touches at 11. End behavior: as x→−∞x\to -\infty, p(x)→+∞p(x)\to +\infty and as x→+∞x\to +\infty, p(x)→+∞p(x)\to +\infty. (correct answer)
Explanation: This question tests your ability to identify zeros from a polynomial's factored form and use them, along with multiplicity information and end behavior, to construct a rough sketch of the polynomial's graph. Zeros from factored form p(x) = a(x - r₁)(x - r₂)... are found by setting each factor equal to zero: from (x - r), the zero is x = r. Multiplicity (how many times a factor appears) determines behavior at that zero: odd multiplicity means the graph crosses the x-axis, even multiplicity means it touches and bounces back. For example, (x - 2)² makes the graph touch at x = 2, while (x - 2)³ makes it cross but with a flattened shape. End behavior depends only on the leading term (highest degree): for even-degree polynomials, both ends go the same direction (both up if positive leading coefficient, both down if negative). For odd-degree, ends go opposite directions (positive leading coefficient: left down, right up). This plus the zeros gives you the skeleton of the graph! For p(x) = (x + 2)(x - 1)^2(x - 4), the zeros are x = -2 (multiplicity 1, crosses), x = 1 (multiplicity 2, touches), and x = 4 (multiplicity 1, crosses), with degree 4 (even) and positive leading coefficient, so both ends up. Choice A correctly identifies zeros with multiplicities, shows proper crossing and touching, and has correct end behavior. A common distractor like choice B might misidentify zeros (e.g., x = 2 instead of x = -2) due to sign errors, but remember to solve (x + 2) = 0 carefully for x = -2. The four-step polynomial sketching strategy: (1) Find all zeros by setting each factor (x - r) equal to zero (watch signs!), (2) Determine multiplicity of each zero (count how many times factor appears) and whether graph crosses (odd) or touches (even), (3) Find end behavior using degree (even = same both ends, odd = opposite ends) and leading coefficient sign (positive eventually goes up, negative eventually goes down), (4) Mark zeros on x-axis and connect with smooth curve showing proper behavior at each zero and correct end behavior. Rough shape is all you need! Multiplicity memory aid: Think 'odd crossers, even bouncers.' Odd multiplicity (1, 3, 5...) = graph crosses through the x-axis. Even multiplicity (2, 4, 6...) = graph bounces off the x-axis without crossing. Higher multiplicity = flatter at that zero. So (x - 3)² touches and turns around, (x - 3)³ crosses but flattens, (x - 3)⁴ touches with even more flattening. The pattern is consistent!

Question 20

Factor the polynomial p(x)=x4−5x2+4p(x)=x^4-5x^2+4 and use the zeros to sketch a rough graph (x-intercepts, crossing vs touching, and end behavior). Which option is correct?

  1. Factored form: (x−2)(x+2)(x−1)(x+1)(x-2)(x+2)(x-1)(x+1). Zeros: −2,−1,1,2-2,-1,1,2 (all mult. 1). Crosses at each intercept. End behavior: both ends up. (correct answer)
  2. Factored form: (x2−4)(x2+1)(x^2-4)(x^2+1). Zeros: −2,2-2,2 only (both mult. 1). Crosses at −2-2 and 22. End behavior: both ends up.
  3. Factored form: (x−2)2(x+1)2(x-2)^2(x+1)^2. Zeros: 22 (mult. 2), −1-1 (mult. 2). Touches at both zeros. End behavior: both ends up.
  4. Factored form: (x−2)(x+2)(x−1)(x+1)(x-2)(x+2)(x-1)(x+1). Zeros: −2,−1,1,2-2,-1,1,2 (all mult. 1). Touches at each intercept. End behavior: left down, right up.
Explanation: This question tests your ability to identify zeros from a polynomial's factored form and use them, along with multiplicity information and end behavior, to construct a rough sketch of the polynomial's graph. Zeros from factored form p(x) = a(x - r₁)(x - r₂)... are found by setting each factor equal to zero: from (x - r), the zero is x = r. Multiplicity (how many times a factor appears) determines behavior at that zero: odd multiplicity means the graph crosses the x-axis, even multiplicity means it touches and bounces back. For example, (x - 2)² makes the graph touch at x = 2, while (x - 2)³ makes it cross but with a flattened shape. End behavior depends only on the leading term (highest degree): for even-degree polynomials, both ends go the same direction (both up if positive leading coefficient, both down if negative). For odd-degree, ends go opposite directions (positive leading coefficient: left down, right up). This plus the zeros gives you the skeleton of the graph! First factor x^4 - 5x^2 + 4 as (x2x^2 - 4)(x2x^2 - 1) = (x - 2)(x + 2)(x - 1)(x + 1), so zeros x = -2, -1, 1, 2 (all multiplicity 1, crosses each), degree 4 even positive, both ends up. Choice A correctly identifies zeros with multiplicities, shows proper crossing and touching, and has correct end behavior. Choice B uses an incorrect factoring like (x2x^2 - 4)(x2x^2 + 1), which gives imaginary zeros and wrong polynomial—double-check factoring by expanding to verify! The four-step polynomial sketching strategy: (1) Find all zeros by setting each factor (x - r) equal to zero (watch signs!), (2) Determine multiplicity of each zero (count how many times factor appears) and whether graph crosses (odd) or touches (even), (3) Find end behavior using degree (even = same both ends, odd = opposite ends) and leading coefficient sign (positive eventually goes up, negative eventually goes down), (4) Mark zeros on x-axis and connect with smooth curve showing proper behavior at each zero and correct end behavior. Rough shape is all you need! Multiplicity memory aid: Think 'odd crossers, even bouncers.' Odd multiplicity (1, 3, 5...) = graph crosses through the x-axis. Even multiplicity (2, 4, 6...) = graph bounces off the x-axis without crossing. Higher multiplicity = flatter at that zero. So (x - 3)² touches and turns around, (x - 3)³ crosses but flattens, (x - 3)⁴ touches with even more flattening. The pattern is consistent!