#AP Pre-Calculus Study Guide: Rational Functions and Zeros

Hey there, future AP superstar! Let's break down rational functions and their zeros. This guide is designed to be your go-to resource the night before the exam. Let's get started! 💪

#1.8 Rational Functions and Zeros

# 😎 Real Zeros of Rational Functions

Think of it this way: a fraction is zero only when its top part (numerator) is zero. The bottom part (denominator) just tells us where the function is defined.

Steps to Find Real Zeros:

1. Factor: Factor both the numerator and the denominator.
2. Domain Check: Identify and exclude values where the denominator equals zero. These are not in the domain of the function.
3. Simplify: Cancel out common factors if possible.
4. Numerator Zeros: Set the simplified numerator equal to zero and solve for x. These are your real zeros! 0️⃣

Example:

Let's look at the function:

$$r(x) = \frac{x^2 - 4}{x - 2}$$

1. Factor: $$r(x) = \frac{(x - 2)(x + 2)}{x - 2}$$
2. Domain Check: x cannot be 2, because the denominator would be zero.
3. Simplify: $$r(x) = x + 2, \quad x \ne 2$$
4. Numerator Zeros: Set x + 2 = 0. This gives us x = -2. So, the real zero is -2. 🎉

# Zeros as Endpoints and Asymptotes

The zeros of the numerator and denominator are critical for understanding the behavior of a rational function. They act as either endpoints of intervals or asymptotes, telling us where the function is positive or negative. 🐎

Key Idea:

• Numerator Zeros: These are the x-values where the function equals zero (if they're in the domain). They are potential endpoints of intervals where the function changes sign.
• Denominator Zeros: These are the x-values where the function is undefined (vertical asymptotes). They also divide the x-axis into intervals where the function's sign might change. 🔨

Analyzing Inequalities:

To solve inequalities like $r(x) ≥ 0$ or $r(x) ≤ 0$, we look at the sign of the function on each side of these critical points (zeros and asymptotes). 👀

Example Walkthrough:

Let’s go back to our example:

$$r(x) = \frac{x^2 - 4}{x - 2}$$

1. Critical Points:

   • Numerator zeros: x = -2 and x = 2
   • Denominator zero: x = 2

2. Interval Analysis:

   • We know x cannot be 2. So, we test values in the intervals $(-\infty, -2)$, $(-2, 2)$, and $(2, \infty)$.
   • Test x = -3: $r(-3) = \frac{(-3)^2 - 4}{-3 - 2} = \frac{5}{-5} = -1$ (negative)
   • Test x = 0: $r(0) = \frac{0^2 - 4}{0 - 2} = \frac{-4}{-2} = 2$ (positive)
   • Test x = 3: $r(3) = \frac{3^2 - 4}{3 - 2} = \frac{5}{1} = 5$ (positive)

3. Conclusions:

   • The function is negative for $x < -2$ and positive for $x > -2$ (except at x=2 where it's undefined).
   • x = -2 is a zero of the function, and it's an endpoint where the function changes sign.
   • x = 2 is an asymptote, and the function changes sign around it. 🙅🏽


Graph displaying the zeros of the function ($x^2-4$) divided by ($x-2$)

Image Courtesy of CK-12

#Final Exam Focus

Alright, let's focus on what's most important for the exam:

• Finding Zeros: You must be able to find the real zeros of rational functions. This involves factoring, domain considerations, and simplifying. 💡
• Interval Analysis: Be comfortable using zeros and asymptotes to determine where a function is positive or negative. This is crucial for solving inequalities.
• Connections: Remember that rational functions combine polynomial concepts. Understanding polynomial zeros and end behavior is essential.

Last-Minute Tips:

• Time Management: Don't spend too long on one problem. If you're stuck, move on and come back later. ⏱️
• Common Pitfalls: Watch out for domain restrictions. Always exclude values that make the denominator zero.

* **Strategies:** Use test values to check your work. Graphing can also help you visualize the behavior of the function. 📈

#Practice Questions

Practice Question

Multiple Choice Questions:

1. What are the real zeros of the function $f(x) = \frac{x^2 - 9}{x + 2}$? (A) x = 3 only (B) x = -3 only (C) x = 3 and x = -3 (D) x = 3, x = -3, and x = -2

2. Which of the following intervals represents where the function $g(x) = \frac{x - 1}{x + 3}$ is negative? (A) $(-\infty, -3)$ (B) $(-3, 1)$ (C) $(1, \infty)$ (D) $(-\infty, -3) \cup (1, \infty)$

Free Response Question:

Consider the rational function:

$$h(x) = \frac{x^2 - 4x + 3}{x - 2}$$

(a) Find the real zeros of h(x). (b) Determine the vertical asymptote(s) of h(x). (c) Determine the intervals where h(x) > 0. (d) Sketch a graph of h(x), clearly labeling all zeros and asymptotes.

Scoring Breakdown:

(a) (2 points) Factoring the numerator correctly (1 point), identifying the zeros (1 point). (b) (1 point) Identifying the vertical asymptote. (c) (3 points) Identifying critical points (1 point), testing intervals (1 point), stating intervals correctly (1 point). (d) (3 points) Correctly plotting zeros and asymptotes (1 point), overall shape (1 point), correctly showing the function's sign in each interval (1 point).

You've got this! Go ace that AP Pre-Calculus exam! 🌟