AP Precalculus Flashcards: Rational Functions And Zeros

What this deck covers

This deck focuses on Rational Functions And Zeros, giving you a quick way to review the definitions, rules, and examples that matter most for AP Precalculus.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

Flashcard 1: Identify a removable discontinuity in $f(x) = \frac{x^2 - 1}{x - 1}$.

Answer: Removable discontinuity at $x = 1$. The factor $(x-1)$ cancels from numerator and denominator.

Flashcard 2: What is the slant asymptote of $f(x) = \frac{x^2 + 1}{x}$?

Answer: The slant asymptote is $y = x$. Divide $x^2 + 1$ by $x$ using polynomial long division.

Flashcard 3: What is the horizontal asymptote of $f(x) = \frac{7x}{2x + 3}$?

Answer: The horizontal asymptote is $y = \frac{7}{2}$. Equal degrees give horizontal asymptote $y = \frac{7}{2}$.

Flashcard 4: Identify the hole in $f(x) = \frac{x^2 - 1}{x - 1}$.

Answer: Hole at $x = 1$. Factor $x^2 - 1 = (x-1)(x+1)$ and cancel the common factor.

Flashcard 5: What is a zero of a rational function?

Answer: A value of $x$ for which $f(x) = 0$. A zero occurs when the numerator equals zero but the denominator doesn't.

Flashcard 6: What is the degree of the polynomial $p(x) = 3x^4 - 2x^2 + 1$?

Answer: The degree is 4. The highest power term determines the degree of a polynomial.

Flashcard 7: Find the vertical asymptote of $f(x) = \frac{2x + 3}{x - 4}$.

Answer: Vertical asymptote at $x = 4$. Set the denominator $x - 4 = 0$ to find the vertical asymptote.

Flashcard 8: Identify the vertical asymptote of $f(x) = \frac{2x}{x^2 - 4}$.

Answer: Vertical asymptotes are $x = 2$ and $x = -2$. Factor $x^2 - 4 = (x-2)(x+2)$ to find where the denominator is zero.

Flashcard 9: What is the behavior of $f(x) = \frac{1}{x}$ as $x \to 0$?

Answer: As $x \to 0$, $f(x) \to \pm \infty$. The function has a vertical asymptote at $x = 0$.

Flashcard 10: What is the horizontal asymptote for $f(x) = \frac{x + 1}{x + 2}$?

Answer: The horizontal asymptote is $y = 1$. When degrees are equal, the horizontal asymptote is $\frac{1}{1} = 1$.

Flashcard 11: What is the horizontal asymptote for $f(x) = \frac{2x^2}{x^2 + 1}$?

Answer: The horizontal asymptote is $y = 2$. Equal degrees give horizontal asymptote $y = \frac{2}{1} = 2$.

Flashcard 12: Which term denotes values not in the domain of a rational function?

Answer: These are the poles or vertical asymptotes. Points where the denominator equals zero make the function undefined.

Flashcard 13: State the horizontal asymptote for $f(x) = \frac{4x}{x + 2}$.

Answer: The horizontal asymptote is $y = 4$. As $x \to \infty$, the function approaches $\frac{4}{1} = 4$.

Flashcard 14: What is the domain of $f(x) = \frac{x}{x^2 - 4x}$?

Answer: Domain: $x \neq 0, x \neq 4$. Factor the denominator: $x^2 - 4x = x(x - 4)$.

Flashcard 15: What is the behavior of $f(x) = \frac{x}{x^2 - 1}$ as $x \to 1$?

Answer: As $x \to 1$, $f(x) \to \infty$. The denominator $(x-1)$ approaches 0 while numerator approaches 1.

Flashcard 16: What are the zeros of $f(x) = \frac{x^2 - 9}{x + 2}$?

Answer: The zeros are $x = 3$ and $x = -3$. Factor the numerator: $x^2 - 9 = (x-3)(x+3)$.

Flashcard 17: State the condition for a vertical asymptote in a rational function.

Answer: Occurs where $q(x) = 0$ and $p(x) \neq 0$. The denominator is zero while the numerator is non-zero.

Flashcard 18: Which term describes $x$ values making $f(x)$ undefined?

Answer: These are the excluded values or domain restrictions. Values that make the denominator zero are excluded from the domain.

Flashcard 19: Find the zero of $f(x) = \frac{2x - 4}{x + 3}$.

Answer: The zero is $x = 2$. Set $2x - 4 = 0$ to find where the numerator equals zero.

Flashcard 20: Find the domain of $f(x) = \frac{4x + 1}{x^2 - 9}$.

Answer: Domain: $x \neq 3, x \neq -3$. Set $x^2 - 9 = 0$ to find where the function is undefined.

Flashcard 21: Identify the horizontal asymptote of $f(x) = \frac{5x}{x^2 + 4}$.

Answer: The horizontal asymptote is $y = 0$. Numerator degree is less than denominator degree, so $y = 0$.

Flashcard 22: What is the degree of $q(x) = x^3 - 7x$?

Answer: The degree is 3. The highest power term $x^3$ has degree 3.

Flashcard 23: Find the zeros of $f(x) = \frac{3x^2 - 12}{x + 2}$.

Answer: The zeros are $x = 2$ and $x = -2$. Factor the numerator: $3x^2 - 12 = 3(x^2 - 4) = 3(x-2)(x+2)$.

Flashcard 24: State the horizontal asymptote of $f(x) = \frac{2x^2 + 3}{x^2 + 1}$.

Answer: The horizontal asymptote is $y = 2$. Equal degrees give horizontal asymptote $y = \frac{2}{1} = 2$.

Flashcard 25: Find the zero of $f(x) = \frac{x - 4}{x + 1}$.

Answer: The zero is $x = 4$. Set the numerator $x - 4 = 0$ to find where $f(x) = 0$.

Flashcard 26: State the vertical asymptote of $f(x) = \frac{2x + 1}{x^2 - 16}$.

Answer: Vertical asymptotes at $x = 4$ and $x = -4$. Factor the denominator: $x^2 - 16 = (x-4)(x+4)$.

Flashcard 27: Identify the horizontal asymptote of $f(x) = \frac{x^2 + 3}{x^2 - 1}$.

Answer: The horizontal asymptote is $y = 1$. When degrees are equal, the horizontal asymptote equals the ratio of leading coefficients.

Flashcard 28: What is the end behavior of $f(x) = \frac{3x^3 + 4}{5x^3 + 2}$?

Answer: As $x \to \pm \infty$, $f(x) \to \frac{3}{5}$. When degrees are equal, divide leading coefficients: $\frac{3}{5}$.

Flashcard 29: What is the behavior of $f(x) = \frac{2x}{x + 5}$ as $x \to -5$?

Answer: As $x \to -5$, $f(x) \to \pm \infty$. The function has a vertical asymptote at $x = -5$.

Flashcard 30: Identify the vertical asymptote of $f(x) = \frac{5}{x^2 - 25}$.

Answer: Vertical asymptotes at $x = 5$ and $x = -5$. Factor the denominator: $x^2 - 25 = (x-5)(x+5)$.

Flashcard 31: For $f(x) = \frac{x^2 - 4}{x - 2}$, what is the discontinuity?

Answer: Discontinuity (hole) at $x = 2$. Factor and cancel: $(x-2)$ appears in both numerator and denominator.

Flashcard 32: What is the intercept of $f(x) = \frac{3}{x-2}$?

Answer: Vertical intercept at $(0, -\frac{3}{2})$. Substitute $x = 0$ into the function to find the y-intercept.

Flashcard 33: How do you determine a hole in a rational function?

Answer: If $p(x)$ and $q(x)$ share a common factor. Common factors can be cancelled, creating a removable discontinuity.

Flashcard 34: Describe when a rational function has no horizontal asymptote.

Answer: When the degree of $p(x)$ is greater than $q(x)$. When numerator degree exceeds denominator degree, no horizontal asymptote exists.

Flashcard 35: What happens to $f(x) = \frac{1}{x-3}$ as $x \to 3$?

Answer: $f(x) \to \pm \infty$, vertical asymptote at $x = 3$. As $x$ approaches 3, the denominator approaches 0 while numerator stays 1.

Flashcard 36: What is the general form of a rational function?

Answer: $f(x) = \frac{p(x)}{q(x)}$ where $p(x)$ and $q(x)$ are polynomials. This defines a rational function as a ratio of two polynomial functions.

Flashcard 37: What is the y-intercept of $f(x) = \frac{2x - 5}{x + 3}$?

Answer: The y-intercept is $(0, -\frac{5}{3})$. Substitute $x = 0$ to get $f(0) = \frac{-5}{3}$.

Flashcard 38: What is the domain of $f(x) = \frac{1}{x(x + 3)}$?

Answer: Domain: $x \neq 0, x \neq -3$. Set $x(x + 3) = 0$ to find where the function is undefined.