AP Precalculus -1.6 Polynomial Functions and End Behavior- MCQ Exam Style Questions - Effective Fall 2023
AP Precalculus -1.6 Polynomial Functions and End Behavior- MCQ Exam Style Questions – Effective Fall 2023
AP Precalculus -1.6 Polynomial Functions and End Behavior- MCQ Exam Style Questions – AP Precalculus- per latest AP Precalculus Syllabus.
Question
(B) The degree of \( f \) is even, and the leading coefficient is positive.
(C) The degree of \( f \) is odd, and the leading coefficient is negative.
(D) The degree of \( f \) is odd, and the leading coefficient is positive.
▶️ Answer/Explanation
End behavior: as \( x \to -\infty \), \( f(x) \to \infty \) (up to the left); as \( x \to \infty \), \( f(x) \to -\infty \) (down to the right).
For odd degree polynomials: ends go in opposite directions.
Leading coefficient negative ⇒ up-left, down-right. Leading coefficient positive ⇒ down-left, up-right.
Given up-left, down-right ⇒ odd degree, negative leading coefficient.
✅ Answer: (C)
Question
(B) The sign of the leading term of \( p \) is negative, and the degree of the leading term of \( p \) is odd; therefore, \( \lim_{x \to -\infty} p(x) = \infty \) and \( \lim_{x \to \infty} p(x) = -\infty \).
(C) The sign of the leading term of \( p \) is positive, and the degree of the leading term of \( p \) is odd; therefore, \( \lim_{x \to -\infty} p(x) = -\infty \) and \( \lim_{x \to \infty} p(x) = \infty \).
(D) The sign of the leading term of \( p \) is negative, and the degree of the leading term of \( p \) is odd; therefore, \( \lim_{x \to -\infty} p(x) = -\infty \) and \( \lim_{x \to \infty} p(x) = \infty \).
▶️ Answer/Explanation
Leading term: \( -4x^5 \) → negative coefficient, degree 5 (odd).
For odd degree, negative leading coefficient: up-left, down-right.
So \( \lim_{x \to -\infty} p(x) = \infty \) and \( \lim_{x \to \infty} p(x) = -\infty \).
This matches option (B).
✅ Answer: (B)
Question
(B) \( f \) is a polynomial of degree 2 with a leading coefficient of 9.
(C) \( f \) is a polynomial of degree 3 with a leading coefficient of 3.
(D) \( f \) is a polynomial of degree 3 with a leading coefficient of 9.
▶️ Answer/Explanation
Factor \( (3x – 1)^2 \) has degree 2, factor \( (x – 4) \) has degree 1 ⇒ total degree = 3.
Leading term from expanding: \( (3x)^2 \cdot x = 9x^3 \) ⇒ leading coefficient 9.
✅ Answer: (D)
Question
(B) \( j = 2k \) and \( k \) is even.
(C) \( j + k \) is even.
(D) \( j \cdot k \) is even.
▶️ Answer/Explanation
Even function means \( p(-x) = p(x) \).
This requires \( j \) and \( k \) both even, because \( (-x)^n = x^n \) only if \( n \) is even.
Check options:
(A) \( k=0 \) (0 is even) but \( j \) could be odd ⇒ not guaranteed even.
(B) \( j = 2k \) and \( k \) even ⇒ \( j \) even (since even × integer = even). Both exponents even ⇒ \( p \) even.
(C) \( j+k \) even: e.g., \( j=1, k=3 \) ⇒ both odd ⇒ \( p \) not even.
(D) \( j \cdot k \) even: e.g., \( j=1, k=2 \) ⇒ \( j \) odd ⇒ not even.
Thus only (B) guarantees both exponents even.
✅ Answer: (B)
Question
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▶️ Answer/Explanation
Given \( P(x) = 3x(x + 1)^2(x – a) \).
For both \( x = -1 \) and \( x = 0 \) to have even multiplicity (graph tangent to x-axis at both), \( x = 0 \) must be multiplicity 2 ⇒ \( a = 0 \) makes \( P(x) = 3x^2(x + 1)^2 \), so both zeros multiplicity 2.
The correct graph must show tangency at \( x = 0 \) and \( x = -1 \), no crossing there, and degree 4, positive leading coefficient 3 ⇒ ends up.
Match to the option with that shape.
✅ Answer: (D)
Question
(B) \( f \) has either a global maximum or a global minimum, but not both.
(C) \( f \) has neither a global maximum nor a global minimum.
(D) The nature of a global maximum or a global minimum for \( f \) cannot be determined without more information about \( b, c, d, \) and \( k \).
▶️ Answer/Explanation
\( f \) is a quartic polynomial (degree 4, even).
For even-degree polynomials with leading coefficient \( a \neq 0 \):
– If \( a > 0 \), as \( x \to \pm\infty \), \( f(x) \to +\infty \) ⇒ global minimum exists, no global maximum.
– If \( a < 0 \), as \( x \to \pm\infty \), \( f(x) \to -\infty \) ⇒ global maximum exists, no global minimum.
Thus \( f \) has either a global max or a global min, but not both.
✅ Answer: (B)
