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This deck focuses on Rational Functions And Holes, giving you a quick way to review the definitions, rules, and examples that matter most for AP Precalculus.
Study Rational Functions And Holes in AP Precalculus with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.
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Perform long division on f(x)=x+1x2+3x+2.
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Quotient is x+2, remainder is 0. x2+3x+2=(x+1)(x+2) divides evenly.
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This deck focuses on Rational Functions And Holes, giving you a quick way to review the definitions, rules, and examples that matter most for AP Precalculus.
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.
Answer: Quotient is x+2, remainder is 0. x2+3x+2=(x+1)(x+2) divides evenly.
Answer: Removable discontinuity at x=3. x−3(x−3)(x+3) cancels at x=3.
Answer: Occurs as x→∞, based on the degrees of p(x) and q(x). Determined by comparing degrees of numerator and denominator.
Answer: Determine where the graph crosses axes. Show where function crosses or touches coordinate axes.
Answer: Function approaches a limit but is not defined at the hole. Limiting value exists despite the discontinuity.
Answer: Horizontal asymptote at y=0. Denominator degree exceeds numerator, so limit is zero.
Answer: Occurs at x where q(x)=0 and p(x) does not cancel. Denominator zero with no cancellation creates vertical line.
Answer: Occurs when degree of p(x) is one more than q(x). Creates diagonal asymptote from polynomial long division.
Answer: Describes f(x) as x→±∞, often related to asymptotes. Behavior of function values as x approaches infinity.
Answer: Determine where the graph crosses axes. Show where function crosses or touches coordinate axes.
Answer: Hole at x=5. x(x−5)(x−5)(x+5) has (x−5) common factor.
Answer: They determine the horizontal asymptote when degrees are equal. When degrees are equal, their ratio gives horizontal asymptote.
Answer: Approaches x+1, but undefined at x=1. x−1(x−1)(x+1) approaches x+1=2 as x→1.
Answer: A vertical asymptote in a rational function. Infinite discontinuity that cannot be removed by cancellation.
Answer: Factor p(x) and q(x), find common factors. Cancel common factors to reveal removable discontinuities.
Answer: f(x)=x−2 for x=2. x−2(x−2)2 simplifies to (x−2) with hole.
Answer: Dominates q(x), leading to polynomial-like behavior. Function grows without bound as x increases.
Answer: Function approaches a limit but is not defined at the hole. Limiting value exists despite the discontinuity.
Answer: Slant asymptote is y=x+1. Divide x2+1 by x−1 using long division.
Answer: A hole occurs if p(x) and q(x) have a common factor. Common factors cancel, creating removable discontinuities.
Answer: Horizontal asymptote at y=23. Equal degrees: ratio of leading coefficients is 23.
Answer: f(0)=q(0)p(0), if defined. Evaluate function at x=0 if denominator nonzero.
Answer: Hole at x=1, as (x−1) is a common factor. (x−1)2(x−1)(x+1) has (x−1) in common.
Answer: Quotient determines slant asymptote if degrees differ by 1. Provides slant asymptote when numerator degree exceeds by one.
Answer: Occurs at x where q(x)=0 and p(x) does not cancel. Denominator zero with no cancellation creates vertical line.
Answer: Hole at x=2, as x−2 is a common factor. Factor: x−2(x−2)(x+2), so (x−2) cancels.
Answer: No common factors between p(x) and q(x). All common factors have been canceled from the fraction.
Answer: When degree of p(x) is greater than q(x). Numerator degree exceeds denominator degree.
Answer: As x→±∞, f(x)→±∞. x2x3=x grows linearly to infinity.
Answer: Removable discontinuity at x=3. x−3(x−3)(x+3) cancels at x=3.
Answer: Creates a hole in the graph at the factor's root. Removable discontinuity where factors cancel out.
Answer: Describes f(x) as x→±∞, often related to asymptotes. Behavior of function values as x approaches infinity.
Answer: Hole at x=5. x(x−5)(x−5)(x+5) has (x−5) common factor.
Answer: Vertical asymptotes at x=3 and x=−3. x2−9=(x−3)(x+3)=0 when x=±3.
Answer: f(x)=q(x)p(x) where p(x) and q(x) are polynomials. Standard notation where both numerator and denominator are polynomials.
Answer: A vertical asymptote in a rational function. Infinite discontinuity that cannot be removed by cancellation.
Answer: As x→±∞, f(x)→±∞. x2x3=x grows linearly to infinity.
Answer: Factor p(x) and q(x), find common factors. Cancel common factors to reveal removable discontinuities.
Answer: Hole at x=1, as (x−1) is a common factor. (x−1)2(x−1)(x+1) has (x−1) in common.
Answer: x-intercept at x=3. x+3(x−3)(x+3) has zero at x=3.
Answer: Creates a hole in the graph at the factor's root. Removable discontinuity where factors cancel out.
Answer: A point where the function is not defined due to a hole. Gap in graph that can be 'filled' by canceling factors.
Answer: Vertical asymptotes at x=1 and x=−1. x2−1=(x−1)(x+1)=0 when x=±1.
Answer: f(0)=q(0)p(0), if defined. Evaluate function at x=0 if denominator nonzero.
Answer: Slant asymptote is y=x+1. Divide x2+1 by x−1 using long division.
Answer: As x→±∞, f(x)→±∞. Numerator degree exceeds denominator, so no horizontal limit.
Answer: When degree of p(x) is greater than q(x). Numerator degree exceeds denominator degree.
Answer: Dominates q(x), leading to polynomial-like behavior. Function grows without bound as x increases.
Answer: Common factor is x. Factor x appears in both numerator and denominator.
Answer: Horizontal asymptote at y=23. Equal degrees: ratio of leading coefficients is 23.
Answer: f(x)=x−2 for x=2. x−2(x−2)2 simplifies to (x−2) with hole.
Answer: x-intercept at x=3. x+3(x−3)(x+3) has zero at x=3.
Answer: Common factor is x. Factor x appears in both numerator and denominator.
Answer: Hole at x=2, as x−2 is a common factor. Factor: x−2(x−2)(x+2), so (x−2) cancels.
Answer: All real numbers except where q(x)=0. Excludes values that make the denominator zero.
Answer: Vertical asymptotes at x=3 and x=−3. x2−9=(x−3)(x+3)=0 when x=±3.
Answer: As x→±∞, f(x)→±∞. Numerator degree exceeds denominator, so no horizontal limit.
Answer: A hole occurs if p(x) and q(x) have a common factor. Common factors cancel, creating removable discontinuities.
Answer: All real numbers except where q(x)=0. Excludes values that make the denominator zero.
Answer: Horizontal asymptote at y=0. Denominator degree exceeds numerator, so limit is zero.
Answer: Quotient is x+2, remainder is 0. x2+3x+2=(x+1)(x+2) divides evenly.
Answer: Occurs as x→∞, based on the degrees of p(x) and q(x). Determined by comparing degrees of numerator and denominator.
Answer: They determine the horizontal asymptote when degrees are equal. When degrees are equal, their ratio gives horizontal asymptote.
Answer: Horizontal asymptote at y=2. Equal degrees: 24=2 gives horizontal asymptote.
Answer: No common factors between p(x) and q(x). All common factors have been canceled from the fraction.
Answer: Occurs when degree of p(x) is one more than q(x). Creates diagonal asymptote from polynomial long division.
Answer: f(x)=q(x)p(x) where p(x) and q(x) are polynomials. Standard notation where both numerator and denominator are polynomials.
Answer: Vertical asymptotes at x=1 and x=−1. x2−1=(x−1)(x+1)=0 when x=±1.
Answer: Horizontal asymptote at y=2. Equal degrees: 24=2 gives horizontal asymptote.
Answer: A point where the function is not defined due to a hole. Gap in graph that can be 'filled' by canceling factors.