---
title: "AP Precalculus Unit 1 Practice Test"
description: "A timed AP Precalculus Unit 1 (Polynomial and Rational Functions) practice test in the AP format, drawn from 1,174 multiple-choice questions and 13 FRQs written for the unit. The standard version takes about 85 minutes."
canonical: "https://fiveable.me/ap-pre-calc/unit-test/1"
type: "unit-test"
subject: "AP Precalculus"
unit: "Unit 1"
---

# AP Precalculus Unit 1 Practice Test

## Overview

A timed AP Precalculus Unit 1 (Polynomial and Rational Functions) practice test in the AP format, drawn from 1,174 multiple-choice questions and 13 FRQs written for the unit. The standard version takes about 85 minutes.

## Test Format

- 20 Multiple Choice Questions (50 minutes)
- 1 FRQ – FRQ 1 – Function Concepts (Calculator) (18 minutes)
- 1 FRQ – FRQ 2 – Modeling a Non-Periodic Context (Calculator) (18 minutes)

## Unit FRQs by Type

- [FRQ 1 – Function Concepts (Calculator)](/ap-pre-calc/frq-practice?type=frq-1-function-concepts-calculator&id=698f8622ef48559d16cd86ad): 5 questions
- [FRQ 2 – Modeling a Non-Periodic Context (Calculator)](/ap-pre-calc/frq-practice?type=frq-2-modeling-a-non-periodic-context-calculator&id=6a72d0738dbb41238f18eb96): 8 questions

## Practice Questions

### Question 1

A polynomial function $$p(x)$$ has standard form $$p(x) = -2x^4 + 5x^3 + 8x - 3$$. Which statement correctly describes the end behavior and provides valid mathematical reasoning?

- (A) As $$x \to \infty$$, $$p(x) \to -\infty$$, and as $$x \to -\infty$$, $$p(x) \to -\infty$$, because the leading term is $$-2x^4$$ with even degree and negative coefficient.
- (B) As $$x \to \infty$$, $$p(x) \to \infty$$, and as $$x \to -\infty$$, $$p(x) \to -\infty$$, because the degree is even and the leading coefficient is negative.
- (C) As $$x \to \infty$$, $$p(x) \to -\infty$$, and as $$x \to -\infty$$, $$p(x) \to \infty$$, because the leading coefficient is negative.
- (D) As $$x \to \infty$$, $$p(x) \to \infty$$, and as $$x \to -\infty$$, $$p(x) \to \infty$$, because the polynomial has four terms.

**Answer: (A)** Correctly applies the rule that even-degree polynomials with negative leading coefficients approach $$-\infty$$ in both directions. The leading term $$-2x^4$$ dominates end behavior, making this reasoning sound.

### Question 2

A rational function $$f(x) = \frac{2x^2 + 3x - 2}{x^2 + 1}$$ has numerator and denominator degrees that are equal. Which statement correctly identifies the horizontal asymptote and provides sound mathematical reasoning?

- (A) The horizontal asymptote is $$y = 2$$ because the degrees of numerator and denominator are equal, so the asymptote is the ratio of leading coefficients: $$\frac{2}{1} = 2$$.
- (B) The horizontal asymptote is $$y = 0$$ because the numerator degree equals the denominator degree, which means the function approaches zero at infinity.
- (C) There is no horizontal asymptote because the numerator and denominator both have degree 2, which means the function grows without bound.
- (D) The horizontal asymptote is $$y = 3$$ because the numerator has a constant term of $$-2$$ and the denominator has a constant term of $$1$$, so the ratio is $$\frac{3}{1} = 3$$.

**Answer: (A)** Applies the rule that when numerator and denominator have equal degree, the horizontal asymptote equals the ratio of leading coefficients. The leading coefficient of the numerator is 2 and the denominator is 1, yielding $$y = 2$$.

### Question 3

The polynomial $$p(x) = (x-1)^2(x+3)$$ is given in factored form. Which statement correctly identifies the x-intercepts and explains what the repeated factor reveals?

- (A) The x-intercepts are $$x = 1$$ and $$x = -3$$. The factor $$(x-1)^2$$ indicates that the graph touches the x-axis at $$x = 1$$ without crossing, because the zero has even multiplicity.
- (B) The x-intercepts are $$x = 1$$ and $$x = -3$$. The factor $$(x-1)^2$$ indicates that the graph crosses the x-axis at $$x = 1$$ with a steep slope because the exponent is 2.
- (C) The x-intercepts are $$x = 1, 1, -3$$ because $$(x-1)$$ appears twice in the factorization, creating three distinct intercepts.
- (D) The x-intercepts are $$x = 1$$ and $$x = -3$$. The factor $$(x-1)^2$$ indicates a vertical asymptote at $$x = 1$$ because the factor is squared.

**Answer: (A)** Correctly identifies both x-intercepts from the factored form and interprets multiplicity: even multiplicity (2) means the graph touches but does not cross the axis. This demonstrates understanding of how factored form reveals zero behavior.

### Question 4

A rational function $$g(x) = \frac{x^3 - 8}{x - 2}$$ can be rewritten by factoring the numerator as $$g(x) = \frac{(x-2)(x^2+2x+4)}{x-2}$$. Which statement correctly describes the simplified form and the function's behavior?

- (A) For $$x \neq 2$$, the function simplifies to $$g(x) = x^2 + 2x + 4$$, and there is a hole at $$(2, 12)$$ because the factor $$(x-2)$$ cancels and $$2^2 + 2(2) + 4 = 12$$.
- (B) For $$x \neq 2$$, the function simplifies to $$g(x) = x^2 + 2x + 4$$, and there is a vertical asymptote at $$x = 2$$ because the original denominator is zero there.
- (C) The function cannot be simplified because the numerator and denominator have different degrees, so there is a vertical asymptote at $$x = 2$$.
- (D) For $$x \neq 2$$, the function simplifies to $$g(x) = x^2 + 2x + 4$$, and the hole is at $$(2, 8)$$ because $$x - 2 = 0$$ when $$x = 2$$.

**Answer: (A)** Correctly simplifies by canceling the common factor and identifies the hole location by evaluating the simplified form at $$x = 2$$. This demonstrates understanding of how factored form reveals removable discontinuities and their coordinates.

### Question 5

Given $$f(x) = x^3 + 5x^2 + 8x + 4$$ and $$g(x) = x + 1$$, which of the following represents the quotient $$\frac{f(x)}{g(x)}$$ using polynomial long division?

- (A) $$x^2 + 4x + 4$$, with no remainder
- (B) $$x^2 + 4x + 4 + \frac{1}{x+1}$$
- (C) $$x^2 + 5x + 8$$, with remainder 4
- (D) $$x^2 + 6x + 2$$, with no remainder

**Answer: (A)** Polynomial long division divides the leading term $$x^3$$ by $$x$$ to get $$x^2$$, then systematically reduces. The process yields quotient $$x^2 + 4x + 4$$ with remainder 0, confirming that $$f(x) = g(x)(x^2 + 4x + 4)$$.

### Question 6

A rational function is defined as $$h(x) = \frac{2x^3 - 3x^2 - 8x + 12}{x - 2}$$. Using polynomial long division, which expression represents $$h(x)$$ and identifies any slant asymptote?

- (A) $$h(x) = 2x^2 + x - 6$$, with no slant asymptote
- (B) $$h(x) = 2x^2 + x - 6 + \frac{0}{x-2}$$, with slant asymptote $$y = 2x^2 + x - 6$$
- (C) $$h(x) = 2x^2 - x + 6$$, with slant asymptote $$y = 2x^2 - x$$
- (D) $$h(x) = 2x + 1 + \frac{10}{x-2}$$, with slant asymptote $$y = 2x + 1$$

**Answer: (A)** Polynomial long division yields quotient $$2x^2 + x - 6$$ with remainder 0. Since the remainder is zero, the numerator factors completely and no slant asymptote exists; the function simplifies to a polynomial.

### Question 7

Given $$p(x) = x^4 - 1$$ and $$q(x) = x^2 - 1$$, which expression represents the quotient $$\frac{p(x)}{q(x)}$$ after polynomial long division and simplification?

- (A) $$x^2 + 1$$, with domain $$x \neq \pm 1$$
- (B) $$x^2 + 1$$, with no domain restrictions
- (C) $$x^2 - 1$$, with domain $$x \neq \pm 1$$
- (D) $$\frac{x^4 - 1}{x^2 - 1}$$ cannot be simplified by polynomial long division

**Answer: (A)** Polynomial long division of $$x^4 - 1$$ by $$x^2 - 1$$ yields $$x^2 + 1$$ with remainder 0. Although the original rational function has domain restrictions at $$x = \pm 1$$ (zeros of the denominator), the quotient itself is defined everywhere.

### Question 8

A rational function $$r(x) = \frac{x^3 + 2x^2 - 5x - 6}{x + 3}$$ is divided using polynomial long division. Which of the following correctly represents the result and identifies the type of discontinuity at $$x = -3$$?

- (A) $$r(x) = x^2 - x - 2$$, with a removable discontinuity (hole) at $$x = -3$$
- (B) $$r(x) = x^2 - x - 2 + \frac{0}{x+3}$$, with a vertical asymptote at $$x = -3$$
- (C) $$r(x) = x^2 + x - 2$$, with a removable discontinuity at $$x = -3$$
- (D) $$r(x) = x^2 - x - 2 + \frac{5}{x+3}$$, with a vertical asymptote at $$x = -3$$

**Answer: (A)** Polynomial long division yields quotient $$x^2 - x - 2$$ with remainder 0, indicating the numerator factors as $$(x+3)(x^2 - x - 2)$$. The common factor $$(x+3)$$ cancels, creating a hole at $$x = -3$$, not a vertical asymptote.

### Question 9

Given $$f(x) = x^4 + 3x^3 - x - 3$$ and $$g(x) = x + 3$$, polynomial long division produces quotient $$q(x)$$ and remainder $$r$$. Which expression correctly represents $$f(x)$$ in the form $$f(x) = g(x) \cdot q(x) + r$$?

- (A) $$f(x) = (x+3)(x^3 - 1) + 0$$
- (B) $$f(x) = (x+3)(x^3 + 1) + 0$$
- (C) $$f(x) = (x+3)(x^3 - 1) - 6$$
- (D) $$f(x) = (x+3)(x^2 + 1) + (x-3)$$

**Answer: (A)** Polynomial long division of $$x^4 + 3x^3 - x - 3$$ by $$x + 3$$ yields quotient $$x^3 - 1$$ and remainder 0. This can be verified: $$(x+3)(x^3-1) = x^4 + 3x^3 - x - 3$$, confirming the decomposition.

### Question 10

The rational function $$h(x) = \frac{x^2 - 2x - 3}{x^2 - 9}$$ can be simplified by canceling a common factor. Which of the following represents the simplified form and identifies the hole?

- (A) Simplified: $$h(x) = \frac{x + 1}{x + 3}$$; hole at $$(3, \frac{2}{3})$$
- (B) Simplified: $$h(x) = \frac{x - 1}{x - 3}$$; hole at $$(3, 1)$$
- (C) Simplified: $$h(x) = \frac{x + 1}{x + 3}$$; hole at $$(-3, 0)$$
- (D) Simplified: $$h(x) = \frac{x + 1}{x + 3}$$; no hole exists

**Answer: (A)** Factoring: $$x^2 - 2x - 3 = (x - 3)(x + 1)$$ and $$x^2 - 9 = (x - 3)(x + 3)$$. The common factor $$(x - 3)$$ cancels, giving $$h(x) = \frac{x + 1}{x + 3}$$ for $$x \neq 3$$. The hole is at $$x = 3$$. Evaluating the simplified form: $$h(3) = \frac{3 + 1}{3 + 3} = \frac{4}{6} = \frac{2}{3}$$.

## Review the Unit

- [1.1 Change in Tandem](/ap-pre-calc/unit-1/change-tandem/study-guide/eQFiTo22fpkDFsnj)
- [1.2 Rates of Change](/ap-pre-calc/unit-1/rates-change/study-guide/P6aTsM1tBCZtaEPy)
- [1.3 Rates of Change in Linear and Quadratic Functions](/ap-pre-calc/unit-1/rates-change-linear-quadratic-functions/study-guide/8cCFDC3VHLyBZGbA)
- [1.4 Polynomial Functions and Rates of Change](/ap-pre-calc/unit-1/polynomial-functions-rates-change/study-guide/tQN39nNwYGsKoKj1)
- [1.5 Polynomial Functions and Complex Zeros](/ap-pre-calc/unit-1/polynomial-functions-complex-zeros/study-guide/Ex6Y5wBlobCpxdVr)
- [1.6 Polynomial Functions and End Behavior](/ap-pre-calc/unit-1/polynomial-functions-end-behavior/study-guide/d9SQc9MbLi6ocGAY)
- [1.7 Rational Functions and End Behavior](/ap-pre-calc/unit-1/rational-functions-end-behavior/study-guide/JzDbW7NX1lJKUE96)
- [1.8 Rational Functions and Zeros](/ap-pre-calc/unit-1/rational-functions-zeros/study-guide/tLF1Ul5EL1rbz8Ye)
- [1.9 Rational Functions and Vertical Asymptotes](/ap-pre-calc/unit-1/rational-functions-vertical-asymptotes/study-guide/lNBy0zlDvb8r6tn1)
- [1.10 Rational Functions and Holes](/ap-pre-calc/unit-1/rational-functions-holes/study-guide/XgQqsfMcOkHxszGG)
- [1.11 Equivalent Representations of Polynomial and Rational Expressions](/ap-pre-calc/unit-1/equivalent-representations-polynomial-rational-expressions/study-guide/NRzwc7vjmULoqIyP)
- [1.12 Transformations of Functions](/ap-pre-calc/unit-1/transformations-functions/study-guide/6S5lhzaXAYrpVwQz)
- [1.13 Function Model Selection and Assumption Articulation](/ap-pre-calc/unit-1/function-model-selection-assumption-articulation/study-guide/tuHPqpA5XkfN1iRD)
- [1.14 Function Model Construction and Application](/ap-pre-calc/unit-1/function-model-construction-application/study-guide/n3ZaYWJqkvxnoJEt)

## Other Unit Tests

- [Unit 2: Exponential and Logarithmic Functions](/ap-pre-calc/unit-test/2)
- [Unit 3: Trigonometric and Polar Functions](/ap-pre-calc/unit-test/3)

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