AP Precalculus Quiz: Polynomial Functions And End Behavior

What this quiz covers

This quiz focuses on Polynomial Functions And End Behavior, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Precalculus.

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.

Question 1

1. It decreases without bound because the leading coefficient is negative.
2. It increases without bound because the leading term is positive degree five. (correct answer)
3. It approaches a constant because factors cancel at large $x$.
4. It oscillates between values because it has multiple real zeros.

Explanation: This question tests AP Precalculus skills: understanding polynomial functions and their end behavior. Polynomial end behavior is determined by the degree and leading coefficient; higher degree means steeper growth or decline, while the sign of the leading coefficient indicates the direction. Given S(x)=(x-1)(x+1)(2x-4)(x²+9), the degree is 1+1+1+2=5 (odd) and the leading coefficient is 1·1·2·1=2 (positive). Choice B is correct because with odd degree and positive leading coefficient, as x→∞, the function increases without bound. Choice A incorrectly assumes a negative leading coefficient, Choice C wrongly suggests factors cancel (they don't in polynomials), and Choice D confuses zeros with end behavior. To help students: emphasize that end behavior depends only on the leading term, not on zeros or other features; practice identifying degrees and leading coefficients in factored form; and use graphing technology to verify predictions. Common misconceptions include thinking that zeros affect end behavior or that the constant terms in factors matter.

Question 2

1. $T(x)\to\infty$ because the degree is odd.
2. $T(x)\to-\infty$ because the leading term is negative quintic. (correct answer)
3. $T(x)$ approaches a constant because factors balance out.
4. $T(x)$ is symmetric about the $y$-axis, so both ends match.

Explanation: This question tests AP Precalculus skills: understanding polynomial functions and their end behavior. Polynomial end behavior is determined by the degree and leading coefficient; higher degree means steeper growth or decline, while the sign of the leading coefficient indicates the direction. Given T(x) = -(x-2)(x+2)(x²-1)(x+1), note that x²-1 = (x-1)(x+1), so we have -(x-2)(x+2)(x-1)(x+1)², giving degree 1+1+1+2 = 5 (odd) with negative leading coefficient. Choice B is correct because with odd degree and negative leading coefficient, as x→∞, T(x)→-∞ (the negative quintic term dominates). Choice A is incorrect because it only considers the odd degree without the negative sign's effect. To help students: Always factor completely to identify repeated roots and accurate degree, and remember that a negative sign in front reverses all end behavior. Practice recognizing difference of squares patterns like x²-1.

Question 3

1. Degree $5$, leading coefficient $-2$; left rises and right falls. (correct answer)
2. Degree $4$, leading coefficient $-2$; both ends fall.
3. Degree $5$, leading coefficient $2$; left falls and right rises.
4. Degree $3$, leading coefficient $-2$; both ends rise.

Explanation: This question tests AP Precalculus skills: understanding polynomial functions and their end behavior. Polynomial end behavior is determined by the degree and leading coefficient; higher degree means steeper growth or decline, while the sign of the leading coefficient indicates the direction. Given S(x)=(x²-9)(-2x+6)(x²+1), the degree is 2+1+2=5 (odd) and the leading coefficient is found by multiplying the leading coefficients: 1·(-2)·1=-2. Choice A is correct because it identifies degree 5 with leading coefficient -2, which means left end rises and right end falls for odd degree with negative leading coefficient. Choice B incorrectly counts degree as 4, Choice C has the wrong sign for the leading coefficient, and Choice D has both degree and behavior wrong. To help students: carefully extract the leading coefficient from each factor (especially -2 from -2x+6), add degrees systematically, and remember the pattern for odd degree polynomials. Common errors include missing the negative in the middle factor or miscounting total degree.

Question 4

1. Left end up and right end down, matching odd degree with negative leading coefficient. (correct answer)
2. Left end down and right end up, matching odd degree with positive leading coefficient.
3. Both ends up because the squared factor forces upward growth.
4. Both ends down because adding $4x$ dominates at extremes.

Explanation: This question tests AP Precalculus skills: understanding polynomial functions and their end behavior. Polynomial end behavior is determined by the degree and leading coefficient; higher degree means steeper growth or decline, while the sign of the leading coefficient indicates the direction. Given M(x)=-(x-2)²(x+3)(x²+1)+4x, the leading term comes from multiplying the highest degree terms: -x²·x·x² = -x⁵, giving degree 5 with negative leading coefficient. Choice A is correct because with odd degree and negative leading coefficient, the left end goes up (as x→-∞) and the right end goes down (as x→∞). Choice B incorrectly identifies a positive leading coefficient, while C and D incorrectly assume even degree behavior. To help students: practice identifying leading terms in factored form, remember that added linear terms like +4x don't affect end behavior, and use the mnemonic 'odd degree = opposite ends, even degree = same ends.' Watch for students who focus on the +4x term rather than the leading term.

Question 5

1. As $x\to\pm\infty$, $S(x)\to\infty$ because the leading term is $x^5$. (correct answer)
2. As $x\to\infty$, $S(x)\to-\infty$; as $x\to-\infty$, $S(x)\to\infty$.
3. As $x\to\pm\infty$, $S(x)\to-\infty$ because the degree is odd.
4. As $x\to\pm\infty$, $S(x)$ approaches a horizontal line.

Explanation: This question tests AP Precalculus skills: understanding polynomial functions and their end behavior. Polynomial end behavior is determined by the degree and leading coefficient; higher degree means steeper growth or decline, while the sign of the leading coefficient indicates the direction. Given S(x) = (x-1)²(x+2)(x²+1), the degree is 2+1+2 = 5 (odd) and the leading coefficient is 1×1×1 = 1 (positive). Choice A is correct because with odd degree and positive leading coefficient, as x→-∞, S(x)→-∞ (left end falls) and as x→∞, S(x)→∞ (right end rises). Choice B is incorrect because it reverses the end behavior, confusing the pattern for negative leading coefficients with positive ones. To help students: Create a reference chart showing all four combinations of odd/even degree with positive/negative leading coefficient. Practice identifying these patterns quickly from factored form without full expansion.

Question 6

1. As $x\to\infty$, $M(x)\to-\infty$ because the degree is odd.
2. As $x\to\infty$, $M(x)\to\infty$ because the leading term is positive. (correct answer)
3. As $x\to\infty$, $M(x)$ approaches a constant because factors balance.
4. As $x\to\infty$, $M(x)$ oscillates since it has three real zeros.

Explanation: This question tests AP Precalculus skills: understanding polynomial functions and their end behavior. Polynomial end behavior is determined by the degree and leading coefficient; higher degree means steeper growth or decline, while the sign of the leading coefficient indicates the direction. Given M(x)=(x-1)(x+2)(x³-3x), we can factor out x from the last term to get (x-1)(x+2)·x(x²-3), giving total degree 1+1+1+2=5 (odd) with positive leading coefficient 1·1·1·1=1. Choice B is correct because with odd degree and positive leading coefficient, as x→∞, M(x)→∞. Choice A incorrectly suggests negative infinity, Choice C wrongly claims the function approaches a constant, and Choice D confuses zeros with oscillation. To help students: factor completely to see all degrees clearly, remember that x³-3x has degree 3 not 1, and use the fact that odd degree with positive leading coefficient means left down, right up. Watch for students who don't recognize that x³-3x contributes degree 3 to the product.

Question 7

1. Both ends rise because the degree is even and leading coefficient is negative.
2. Left end rises and right end falls because the degree is odd and leading coefficient is negative. (correct answer)
3. Both ends fall because the degree is odd and leading coefficient is negative.
4. Left end falls and right end rises because the degree is odd and leading coefficient is positive.

Explanation: This question tests AP Precalculus skills: understanding polynomial functions and their end behavior. Polynomial end behavior is determined by the degree and leading coefficient; higher degree means steeper growth or decline, while the sign of the leading coefficient indicates the direction. Given h(t)=-(2t-2)(t²+1)(t²-4), the degree is 1+2+2=5 (odd) and the leading coefficient is -2·1·1=-2 (negative). Choice B is correct because with odd degree and negative leading coefficient, the left end rises (as t→-∞) and the right end falls (as t→∞). Choice A incorrectly identifies even degree, Choice C wrongly claims both ends fall (only happens with even degree and negative coefficient), and Choice D has the wrong sign for the leading coefficient. To help students: systematically add degrees from each factor, carefully track the negative sign and the coefficient 2 from (2t-2), and memorize the four end behavior patterns. Common errors include losing track of the initial negative sign or the coefficient 2 in the linear factor.

Question 8

1. As $t\to\infty$, $P(t)\to\infty$; as $t\to-\infty$, $P(t)\to-\infty$. (correct answer)
2. As $t\to\infty$, $P(t)\to-\infty$; as $t\to-\infty$, $P(t)\to\infty$.
3. As $t\to\pm\infty$, $P(t)\to\infty$ because the constant is positive.
4. End behavior is undefined because like terms cancel completely.

Explanation: This question tests AP Precalculus skills: understanding polynomial functions and their end behavior. Polynomial end behavior is determined by the degree and leading coefficient; higher degree means steeper growth or decline, while the sign of the leading coefficient indicates the direction. Given P(t) = 4(t-1)^5 - 2(t-1)^5 + 7, this simplifies to P(t) = 2(t-1)^5 + 7, which has degree 5 and positive leading coefficient 2. Choice A is correct because with a positive leading coefficient and odd degree, as t→∞, P(t)→∞ and as t→-∞, P(t)→-∞. Choice C is incorrect because it attributes end behavior to the constant term - constants affect vertical position but not end behavior, which is determined solely by the leading term. To help students: Focus on combining like terms before analyzing end behavior, use algebraic simplification to identify the true leading term, and practice recognizing that constants don't affect end behavior. Emphasize simplifying expressions first.

Question 9

1. Left end down and right end up, matching a positive leading coefficient and odd degree. (correct answer)
2. Left end up and right end up, matching a positive leading coefficient and even degree.
3. Left end up and right end down, matching a negative leading coefficient and odd degree.
4. Both ends approach $0$ because the factors create horizontal leveling.

Explanation: This question tests AP Precalculus skills: understanding polynomial functions and their end behavior. Polynomial end behavior is determined by the degree and leading coefficient; higher degree means steeper growth or decline, while the sign of the leading coefficient indicates the direction. Given f(x) = (x-2)(x+1)(x^3-4), when expanded, the highest degree term comes from multiplying x·x·x^3 = x^5 with a positive coefficient. Choice A is correct because with a positive leading coefficient and odd degree (5), the left end goes down (as x→-∞, f(x)→-∞) and the right end goes up (as x→∞, f(x)→∞). Choice D is incorrect because it confuses end behavior with zeros - polynomial end behavior is determined by the leading term, not by where the function crosses the x-axis. To help students: Focus on identifying leading terms through multiplication, use graphing tools to visualize end behavior, and practice distinguishing polynomial characteristics from other functions. Emphasize that factored form still reveals degree and leading coefficient.

Question 10

1. $f(x)\to\infty$ because the highest power is $x^5$.
2. $f(x)\to-\infty$ because the simplified leading coefficient is negative. (correct answer)
3. $f(x)\to 0$ because fractional coefficients reduce long-term growth.
4. $f(x)$ approaches a constant because lower-degree terms dominate eventually.

Explanation: This question tests AP Precalculus skills: understanding polynomial functions and their end behavior. Polynomial end behavior is determined by the degree and leading coefficient; higher degree means steeper growth or decline, while the sign of the leading coefficient indicates the direction. Given f(x) = 3x^5 - (9/2)x^5 + x^2, combining like terms gives f(x) = -3/2·x^5 + x^2, with leading term -3/2·x^5. Choice B is correct because the simplified leading coefficient is -3/2 (negative) with odd degree 5, so as x→∞, f(x)→-∞. Choice A is incorrect because it only looks at the highest power without combining like terms - the coefficients of x^5 must be combined to find the true leading coefficient. To help students: Focus on combining all terms of the same degree before analyzing end behavior, practice arithmetic with fractions when combining coefficients, and always simplify completely before determining end behavior. Emphasize that like terms must be combined first.

Question 11

1. Both ends rise because the leading coefficient is positive and degree is four. (correct answer)
2. Both ends fall because the leading coefficient is negative and degree is four.
3. Left falls and right rises because the degree is odd.
4. End behavior is determined by the $x$-intercepts at $x=\pm1$.

Explanation: This question tests AP Precalculus skills: understanding polynomial functions and their end behavior. Polynomial end behavior is determined by the degree and leading coefficient; higher degree means steeper growth or decline, while the sign of the leading coefficient indicates the direction. Given R(x)=½(4x-2)(x+5)(x²-1), we can factor x²-1=(x-1)(x+1), giving total degree 1+1+1+1=4 (even) with leading coefficient ½·4·1·1=2 (positive). Choice A is correct because with even degree and positive leading coefficient, both ends rise to positive infinity. Choice B has the wrong sign for the leading coefficient, Choice C incorrectly identifies odd degree, and Choice D wrongly claims intercepts determine end behavior. To help students: always factor completely before counting degree, track all coefficients including fractions, and remember that x²-1 contributes degree 2. Common errors include forgetting to factor difference of squares or losing track of the ½ coefficient.

Question 12

1. Degree $5$ with leading coefficient $3$, so right end rises and left end falls. (correct answer)
2. Degree $6$ with leading coefficient $3$, so both ends rise.
3. Degree $5$ with leading coefficient $-3$, so right end falls and left end rises.
4. Degree $3$ with leading coefficient $3$, so end behavior depends on zeros.

Explanation: This question tests AP Precalculus skills: understanding polynomial functions and their end behavior. Polynomial end behavior is determined by the degree and leading coefficient; higher degree means steeper growth or decline, while the sign of the leading coefficient indicates the direction. Given h(t) = (t^2-4)(3t^3-6), when expanded, the highest degree term comes from t^2·3t^3 = 3t^5, giving degree 5 and leading coefficient 3. Choice A is correct because with degree 5 (odd) and positive leading coefficient (3), the right end rises (as t→∞, h(t)→∞) and the left end falls (as t→-∞, h(t)→-∞). Choice B is incorrect because it miscalculates the degree as 6 instead of 5 - when multiplying polynomials, degrees add (2+3=5), not multiply. To help students: Focus on correctly adding degrees when multiplying polynomials, use the distributive property to find leading terms, and practice identifying degree and leading coefficient from factored forms. Emphasize the degree addition rule for polynomial multiplication.

Question 13

1. As $t\to\infty$, $h(t)\to-\infty$; as $t\to-\infty$, $h(t)\to\infty$.
2. As $t\to\infty$, $h(t)\to\infty$; as $t\to-\infty$, $h(t)\to-\infty$. (correct answer)
3. As $t\to\pm\infty$, $h(t)\to\infty$ because the degree is odd.
4. End behavior depends on the zeros, so $h(t)$ approaches $0$ for large $|t|$.

Explanation: This question tests AP Precalculus skills: understanding polynomial functions and their end behavior. Polynomial end behavior is determined by the degree and leading coefficient; higher degree means steeper growth or decline, while the sign of the leading coefficient indicates the direction. Given h(t) = 2t^5 - 10t^3 + 8t, the leading term 2t^5 dictates that as t approaches infinity, the function behaves like a positive odd-degree polynomial. Choice B is correct because with a positive leading coefficient (2) and odd degree (5), as t→∞, h(t)→∞ and as t→-∞, h(t)→-∞. Choice C is incorrect because it misinterprets odd degree behavior - odd degree polynomials have opposite end behaviors, not both ends going to infinity. To help students: Focus on identifying leading terms, use graphing tools to visualize end behavior, and practice distinguishing polynomial characteristics from other functions. Watch for misconceptions about degree and leading coefficient impacts.

Question 14

1. Both ends rise because the leading term is $x^4$ with positive coefficient.
2. Both ends fall because the simplified leading term is $-4x^4$. (correct answer)
3. Left end falls and right end rises because the degree is odd.
4. End behavior depends on the $2x$ term, not the $x^4$ terms.

Explanation: This question tests AP Precalculus skills: understanding polynomial functions and their end behavior. Polynomial end behavior is determined by the degree and leading coefficient; higher degree means steeper growth or decline, while the sign of the leading coefficient indicates the direction. Given S(x) = (x^2+2x)^2 - 5x^4, expanding (x^2+2x)^2 gives x^4+4x^3+4x^2, so S(x) = x^4+4x^3+4x^2-5x^4 = -4x^4+4x^3+4x^2, with leading term -4x^4. Choice B is correct because the simplified leading term is -4x^4, giving negative leading coefficient and even degree, so both ends fall: as x→±∞, S(x)→-∞. Choice A is incorrect because it fails to combine like terms - the x^4 terms must be combined before determining the leading coefficient. To help students: Focus on expanding expressions completely before identifying leading terms, practice combining like terms systematically, and verify end behavior by checking the final simplified form. Always simplify polynomials before analyzing end behavior.

Question 15

1. $P(t)\to\infty$ because the degree is $5$.
2. $P(t)\to-\infty$ because the leading term is negative and degree is odd. (correct answer)
3. $P(t)\to 0$ because squared factors dominate growth.
4. $P(t)$ oscillates between positive and negative values indefinitely.

Explanation: This question tests AP Precalculus skills: understanding polynomial functions and their end behavior. Polynomial end behavior is determined by the degree and leading coefficient; higher degree means steeper growth or decline, while the sign of the leading coefficient indicates the direction. Given P(t) = -(t-3)^2(t+2)^3, the degree is 2+3=5 (odd) and the leading coefficient is negative (-1). Choice B is correct because with a negative leading coefficient and odd degree, as t→∞, P(t)→-∞. Choice A is incorrect because it only considers the degree but ignores the negative sign in front, which reverses the typical odd-degree behavior. To help students: Focus on identifying both degree (sum of exponents) and sign of leading coefficient in factored form, use graphing tools to visualize end behavior, and practice with various factored polynomials. Emphasize that the negative sign in front affects the entire polynomial's end behavior.

Question 16

1. $S(x)\to-\infty$ on both ends because the constant term is positive.
2. $S(x)\to\infty$ on both ends because the leading term is positive and degree is even. (correct answer)
3. $S(x)\to\infty$ as $x\to\infty$ and $S(x)\to-\infty$ as $x\to-\infty$.
4. End behavior cannot be determined without finding all zeros first.

Explanation: This question tests AP Precalculus skills: understanding polynomial functions and their end behavior. Polynomial end behavior is determined by the degree and leading coefficient; higher degree means steeper growth or decline, while the sign of the leading coefficient indicates the direction. Given S(x) = ½x^4 - 3x^2 + 5/2, the leading term ½x^4 dictates end behavior with positive coefficient (½) and even degree (4). Choice B is correct because with a positive leading coefficient and even degree, both ends of the graph go up: as x→±∞, S(x)→∞. Choice A is incorrect because it confuses the constant term's sign with end behavior - the constant term affects vertical shift but not end behavior, which is determined by the leading term. To help students: Focus on identifying the highest degree term regardless of other terms, use graphing tools to visualize end behavior, and practice distinguishing between terms that affect end behavior versus those that don't. Emphasize that only the leading term matters for end behavior.

Question 17

1. Left end down and right end up, consistent with an odd degree and positive leading coefficient.
2. Left end up and right end down, consistent with an odd degree and negative leading coefficient. (correct answer)
3. Both ends up, consistent with an even degree and negative leading coefficient.
4. Both ends down, consistent with an even degree and positive leading coefficient.

Explanation: This question tests AP Precalculus skills: understanding polynomial functions and their end behavior. Polynomial end behavior is determined by the degree and leading coefficient; higher degree means steeper growth or decline, while the sign of the leading coefficient indicates the direction. Given h(t) = -t^5 + 4t^3 - t, the leading term -t^5 has negative coefficient (-1) and odd degree (5). Choice B is correct because with a negative leading coefficient and odd degree, the left end goes up (as t→-∞, h(t)→∞) and the right end goes down (as t→∞, h(t)→-∞). Choice A is incorrect because it describes the behavior of a positive leading coefficient with odd degree - the negative sign reverses the typical odd-degree pattern. To help students: Focus on the sign of the leading coefficient, use mnemonic devices like 'negative odd flips the pattern', and practice graphing to visualize how negative leading coefficients affect end behavior. Emphasize that negative leading coefficients reverse expected patterns.

Question 18

1. Left end up and right end down, consistent with odd degree and negative leading coefficient. (correct answer)
2. Left end down and right end up, consistent with odd degree and positive leading coefficient.
3. Both ends up because the squared factor forces positive outputs.
4. Both ends down because the constant $+4$ in $(t^2+4)$ dominates.

Explanation: This question tests AP Precalculus skills: understanding polynomial functions and their end behavior. Polynomial end behavior is determined by the degree and leading coefficient; higher degree means steeper growth or decline, while the sign of the leading coefficient indicates the direction. Given P(t)=-(t-2)²(t+3)(t²+4), the degree is 2+1+2=5 (odd) and the leading coefficient is negative due to the minus sign in front. Choice A is correct because with odd degree and negative leading coefficient, the left end goes up (as t→-∞, P(t)→∞) and the right end goes down (as t→∞, P(t)→-∞). Choice B has the ends reversed, while C and D incorrectly assume even degree behavior. To help students: use the power rule for (t-2)² to count degree 2, remember that t²+4 cannot be factored over reals but still contributes degree 2, and practice the 'opposite ends for odd degree' rule. Watch for confusion about which end is 'left' versus 'right' on the graph.