AP Precalculus Polynomial and Rational Functions — Worked Answer Explanations

Unit 1 · 12 questions explained

Below is a complete answer key for our AP Precalculus Polynomial and Rational Functions practice questions. For each question you'll find the correct choice, a full written explanation of how to get there, and — for every wrong answer — a short note on exactly why it's tempting and where it goes wrong. Reading these straight through is one of the fastest ways to find the gaps in a unit before exam day.

Prefer to test yourself first? Take the timed Polynomial and Rational Functions practice test and come back here to review, or head back to the Polynomial and Rational Functions unit overview.

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Question 1 · Easy
What is the degree of the polynomial $f (x) = 3 x^{4} - 2 x^{3} + 5 x - 7$ ?
- A
  $3$
  Why not A: Counted only the second-highest exponent.
- B
  $4$ Correct
- C
  $5$
  Why not C: Confused a coefficient with the degree.
- D
  $7$
  Why not D: Used the constant term, not the highest exponent.
Explanation
The degree of a polynomial is the highest exponent of $x$ that appears with a nonzero coefficient. Here $3 x^{4}$ is the highest-degree term, so the degree is $4$ .

Key takeaway
Degree = highest exponent with a nonzero coefficient.
Question 2 · Easy
What is the zero of the linear function $f (x) = 2 x - 8$ ?
- A
  $x = - 4$
  Why not A: Solved $2 x = - 8$ instead of $2 x - 8 = 0$ .
- B
  $x = - 8$
  Why not B: Used the constant term directly without solving.
- C
  $x = 4$ Correct
- D
  $x = 8$
  Why not D: Forgot to divide by the coefficient of $x$ .
Explanation
A zero of $f$ is an input that makes $f (x) = 0$ . Solving $2 x - 8 = 0$ gives $2 x = 8$ , so $x = 4$ . Verify: $f (4) = 2 (4) - 8 = 0$ .

Key takeaway
To find a zero, set the function equal to 0 and solve for $x$.
Question 3 · Easy
What are the real zeros of $f (x) = x^{2} - 9$ ?
- A
  $x = 3$ only
  Why not A: Took only the positive square root and missed the negative one.
- B
  $x = \pm 3$ Correct
- C
  $x = \pm 9$
  Why not C: Forgot to take the square root after isolating $x^{2}$ .
- D
  $x = 0$
  Why not D: Confused the $y$ -intercept ( $f (0) = - 9$ ) with a zero.
Explanation
Factor: $x^{2} - 9 = (x - 3) (x + 3)$ . Setting each factor equal to $0$ gives $x = 3$ or $x = - 3$ . Both are real zeros, so the answer is $x = \pm 3$ .

Key takeaway
Difference of squares: $a^2 - b^2 = (a - b)(a + b)$ — two real zeros when $b^2 > 0$.
Question 4 · Easy
As $x \to \infty$ , what is the end behavior of $f (x) = - 2 x^{3} + 5 x^{2} - x + 4$ ?
- A
  $f (x) \to \infty$
  Why not A: Ignored the negative leading coefficient.
- B
  $f (x) \to - \infty$ Correct
- C
  $f (x) \to 4$
  Why not C: Used the constant term — only relevant for $x = 0$ , not the limit.
- D
  $f (x) \to 0$
  Why not D: A nonconstant polynomial never has a finite, nonzero limit at infinity.
Explanation
End behavior of a polynomial is controlled by the leading term. Here that term is $- 2 x^{3}$ . Since the degree is odd and the leading coefficient is negative, as $x \to \infty$ , $- 2 x^{3} \to - \infty$ , so $f (x) \to - \infty$ .

Key takeaway
End behavior is determined by sign and parity of the leading term: odd degree + negative coefficient → $f \to -\infty$ as $x \to \infty$.
Question 5 · Medium
The polynomial $f (x) = (x - 2)^{2} (x + 1)$ has a zero at $x = 2$ . Which best describes the graph of $f$ at $x = 2$ ?
- A
  Crosses the $x$ -axis at $x = 2$ .
  Why not A: True for odd multiplicity zeros, but here the zero has even multiplicity.
- B
  Touches the $x$ -axis and turns around at $x = 2$ .Correct
- C
  Has a vertical asymptote at $x = 2$ .
  Why not C: Polynomials have no vertical asymptotes — that behavior belongs to rational functions.
- D
  Is undefined at $x = 2$ .
  Why not D: Polynomials are defined for all real $x$ .
Explanation
The factor $(x - 2)^{2}$ contributes a zero of multiplicity $2$ at $x = 2$ . Zeros of even multiplicity make the graph touch the $x$ -axis but not cross it — the sign of $f$ does not change there.

Key takeaway
Even multiplicity: touch and turn. Odd multiplicity: cross. Higher multiplicity flattens the graph near the zero.
Question 6 · Medium
Identify the vertical asymptote(s) of $f (x) = \frac{x + 3}{x ^{2} - 4}$ .
- A
  $x = - 3$ only
  Why not A: $- 3$ is a zero of the numerator, not the denominator — that gives a zero of $f$ , not an asymptote.
- B
  $x = 2$ only
  Why not B: Found one denominator zero but missed the other.
- C
  $x = 2$ and $x = - 2$ Correct
- D
  $x = \pm 3$
  Why not D: Used the numerator's value to set up the denominator equation incorrectly.
Explanation
Vertical asymptotes occur where the denominator is zero and the numerator is nonzero. Factor: $x^{2} - 4 = (x - 2) (x + 2)$ , so the denominator is zero at $x = 2$ and $x = - 2$ . Numerator $x + 3$ is nonzero at both, so both are vertical asymptotes.

Key takeaway
Vertical asymptote = denominator zero where the numerator is nonzero (otherwise it's a hole).
Question 7 · Medium
What is the horizontal asymptote of $f (x) = \frac{3 x ^{2} + 7}{2 x ^{2} - x + 5}$ ?
- A
  $y = 0$
  Why not A: Applies only when the numerator has lower degree than the denominator.
- B
  $y = \frac{2}{3}$
  Why not B: Inverted the leading-coefficient ratio.
- C
  $y = \frac{3}{2}$ Correct
- D
  No horizontal asymptote.
  Why not D: Applies only when numerator degree exceeds denominator degree.
Explanation
When the numerator and denominator of a rational function have the same degree, the horizontal asymptote is the ratio of their leading coefficients: here $\frac{3}{2}$ .

Key takeaway
Same-degree top and bottom → horizontal asymptote = ratio of leading coefficients.
Question 8 · Medium
Find the quotient when $f (x) = x^{3} + 2 x^{2} - 5 x - 6$ is divided by $x - 2$ .
- A
  $x^{2} + 4 x + 3$ Correct
- B
  $x^{2} - 4 x + 3$
  Why not B: Flipped the sign on the linear term — likely divided by $x + 2$ by mistake.
- C
  $x^{2} + 2 x - 3$
  Why not C: Coefficient arithmetic error during the division step.
- D
  $x^{2} + 4 x + 5$
  Why not D: Carried a wrong remainder into the constant term.
Explanation
Using synthetic division with $c = 2$ : bring down $1$ ; $1 \cdot 2 = 2$ , add to $2$ : $4$ ; $4 \cdot 2 = 8$ , add to $- 5$ : $3$ ; $3 \cdot 2 = 6$ , add to $- 6$ : $0$ (remainder). The quotient coefficients $1, 4, 3$ give $x^{2} + 4 x + 3$ .

Key takeaway
Synthetic division: bring down, multiply by $c$, add — last entry is the remainder.
Question 9 · Medium
A linear function $g$ has constant rate of change $2$ , and a quadratic function $q (x) = x^{2}$ has rate of change that varies with $x$ . Over the interval $[1, 5]$ , what is the average rate of change of $q$ , and how does it compare to $g$ 's rate?
- A
  Average rate of $q$ is $4$ ; faster than $g$ 's rate.
  Why not A: Correct conclusion, but the rate value is wrong — likely used $q (2) - q (1) = 3$ over wrong interval.
- B
  Average rate of $q$ is $6$ ; faster than $g$ 's rate.Correct
- C
  Average rate of $q$ is $6$ ; slower than $g$ 's rate.
  Why not C: Got the average rate right but reversed the comparison.
- D
  Average rate of $q$ is $24$ ; faster than $g$ 's rate.
  Why not D: Computed $q (5) - q (1) = 24$ but forgot to divide by $5 - 1 = 4$ .
Explanation
Average rate of change of $q$ on $[1, 5]$ is $\frac{q ( 5 ) - q ( 1 )}{5 - 1} = \frac{25 - 1}{4} = 6$ . Since $6 > 2$ , $q$ is changing faster (on average) than $g$ over this interval.

Key takeaway
Average rate of change = $\dfrac{f(b) - f(a)}{b - a}$. Quadratics speed up; linear functions have constant rate.
Question 10 · Hard
Which best describes the behavior of $f (x) = \frac{x ^{2} - 1}{x - 1}$ at $x = 1$ ?
- A
  Vertical asymptote at $x = 1$ .
  Why not A: True only if the numerator does not also vanish at $x = 1$ .
- B
  Hole (removable discontinuity) at $x = 1$ .Correct
- C
  $x$ -intercept at $x = 1$ .
  Why not C: $f$ is undefined at $x = 1$ , so it has no value — let alone $0$ — there.
- D
  Horizontal asymptote at $x = 1$ .
  Why not D: Horizontal asymptotes describe end behavior at $\pm \infty$ , not a specific $x$ .
Explanation
Factor: $\frac{x ^{2} - 1}{x - 1} = \frac{( x - 1 ) ( x + 1 )}{x - 1} = x + 1$ for $x \neq = 1$ . The $(x - 1)$ factor cancels, so $x = 1$ is a hole, not a vertical asymptote. The simplified line $y = x + 1$ would pass through $(1, 2)$ , but the original $f$ is undefined there.

Key takeaway
A common factor in numerator and denominator → hole. Denominator zero alone → vertical asymptote.
Question 11 · Hard
A polynomial $f (x)$ has degree $3$ , real coefficients, and zeros at $x = - 2$ , $x = 1$ , and $x = 4$ . If $f (0) = 16$ , what is $f (x)$ ?
- A
  $f (x) = (x + 2) (x - 1) (x - 4)$
  Why not A: Has the right zeros but $f (0) = (2) (- 1) (- 4) = 8$ , not $16$ — missed the leading coefficient.
- B
  $f (x) = 2 (x + 2) (x - 1) (x - 4)$ Correct
- C
  $f (x) = (x - 2) (x + 1) (x + 4)$
  Why not C: Flipped the sign on every zero — those are now zeros at $2, - 1, - 4$ .
- D
  $f (x) = 4 (x + 2) (x - 1) (x - 4)$
  Why not D: Used $f (0) = 16$ directly as the leading coefficient without dividing by the product of zeros.
Explanation
Zeros at $- 2, 1, 4$ mean $f (x) = a (x + 2) (x - 1) (x - 4)$ for some $a$ . Use $f (0) = 16$ : $a (2) (- 1) (- 4) = 8 a = 16$ , so $a = 2$ . Therefore $f (x) = 2 (x + 2) (x - 1) (x - 4)$ .

Key takeaway
Zeros fix the factors up to a leading coefficient — use one extra point to pin down $a$.
Question 12 · Hard
Let $f (x) = \frac{2 x ^{2} + 3 x + 1}{x ^{2} - 1}$ . Identify all asymptotes and any holes.
- A
  Vertical asymptotes at $x = \pm 1$ ; horizontal asymptote $y = 2$ .
  Why not A: Missed that $x = - 1$ is actually a hole, not a vertical asymptote, because the factor $(x + 1)$ cancels.
- B
  Hole at $x = - 1$ ; vertical asymptote $x = 1$ ; horizontal asymptote $y = 2$ .Correct
- C
  Hole at $x = 1$ ; vertical asymptote $x = - 1$ ; horizontal asymptote $y = 2$ .
  Why not C: Swapped which factor cancels — $(x + 1)$ is the common one, so the hole is at $- 1$ .
- D
  Vertical asymptotes at $x = \pm 1$ ; no horizontal asymptote.
  Why not D: Equal-degree rational functions always have a horizontal asymptote at the ratio of leading coefficients.
Explanation
Factor: $2 x^{2} + 3 x + 1 = (2 x + 1) (x + 1)$ and $x^{2} - 1 = (x - 1) (x + 1)$ . The $(x + 1)$ factor cancels, so $x = - 1$ is a hole (both numerator and denominator vanish there). The remaining denominator factor $(x - 1)$ gives a vertical asymptote at $x = 1$ . Same degree top and bottom with leading-coefficient ratio $2/1$ gives a horizontal asymptote $y = 2$ .

Key takeaway
Always factor numerator and denominator before classifying — a cancelling factor is a hole, a non-cancelling factor is a vertical asymptote.