AP Precalculus Quiz
Practice Rational Functions And Zeros in AP Precalculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.
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A rational function is a quotient of polynomials, f(x)=q(x)p(x). Zeros are the x-values where f(x)=0, which happens when p(x)=0 and q(x)=0. Vertical asymptotes occur where q(x)=0 and the factor does not cancel. Consider f(x)=x−1x2−1. Factoring gives x2−1=(x−1)(x+1), so (x−1) cancels for x=1, leaving f(x)=x+1 with a hole at x=1. The remaining zero is where x+1=0, so x=−1 is a zero.
Using the function provided, what are the zeros of f(x)=x−1x2−1?
This quiz focuses on Rational Functions And Zeros, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Precalculus.
Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.
A rational function is a quotient of polynomials, f(x)=q(x)p(x). Zeros are the x-values where f(x)=0, which happens when p(x)=0 and q(x)=0. Vertical asymptotes occur where q(x)=0 and the factor does not cancel. Consider f(x)=x−1x2−1. Factoring gives x2−1=(x−1)(x+1), so (x−1) cancels for x=1, leaving f(x)=x+1 with a hole at x=1. The remaining zero is where x+1=0, so x=−1 is a zero.
Using the function provided, what are the zeros of f(x)=x−1x2−1?
Explanation: This question tests understanding of zeros in rational functions with common factor cancellation. Zeros occur where the numerator equals zero and the denominator does not. For f(x) = (x²-1)/(x-1), factoring the numerator gives (x-1)(x+1)/(x-1), which simplifies to f(x) = x+1 for x≠1 after canceling the common factor (x-1). The zero of this simplified function occurs where x+1=0, giving x=-1. Choice B is correct because x=-1 is the only zero of the function. Choice A is incorrect because x=1 is not a zero but rather a point where the function has a hole due to the cancellation. Students should carefully factor and simplify rational functions to identify zeros correctly, remembering that holes are not zeros.
A rational function is a quotient of polynomials. For f(x)=x−2x2−4, zeros occur where f(x)=0 and the function is defined; vertical asymptotes come from non-canceling denominator zeros. Since x2−4=(x−2)(x+2), the function behaves like x+2 except at x=2, where it is undefined. Based on the function described, which statement best describes the behavior at its zero?
Explanation: This question tests understanding of how rational functions behave at their zeros, particularly when the function has been simplified. For f(x) = (x²-4)/(x-2), after factoring and canceling, we get f(x) = x+2 for x≠2, which has a zero at x=-2. Choice A is correct because at x=-2, the function value is f(-2) = -2+2 = 0, meaning the graph crosses the x-axis at this point. Choice B is incorrect because there is no vertical asymptote at x=-2; the function is well-defined and continuous there. Students should evaluate the simplified function at its zeros to verify the behavior, and remember that zeros are where the graph crosses the x-axis, not where vertical asymptotes occur.
A rational function is a quotient of two polynomials. For f(x)=x−2x2−4, zeros satisfy f(x)=0, and vertical asymptotes typically occur where the denominator is zero. Since x2−4=(x−2)(x+2), the factor (x−2) cancels, leaving a removable discontinuity at x=2 (a hole), not a vertical asymptote. Based on the function described, determine the vertical asymptotes of f(x)=x−2x2−4.
Explanation: This question tests understanding of vertical asymptotes in rational functions and the distinction between vertical asymptotes and removable discontinuities. Vertical asymptotes occur where the denominator equals zero and the factor does not cancel with the numerator. In this function f(x) = (x²-4)/(x-2), we can factor the numerator as (x-2)(x+2), so the function becomes f(x) = (x-2)(x+2)/(x-2). Choice B is correct because the (x-2) factor cancels completely, leaving no vertical asymptotes - only a removable discontinuity (hole) at x=2. Choice A is incorrect because x=2 is not a vertical asymptote but a hole. Students should always factor both numerator and denominator completely to identify which factors cancel, as canceling factors create holes rather than vertical asymptotes.
A rational function is a quotient of two polynomials, such as f(x)=x−2x2−4. Zeros are x-values where f(x)=0, typically where the numerator is zero (and the denominator is not). Vertical asymptotes occur at zeros of the denominator that do not cancel. Here, x2−4=(x−2)(x+2), so the factor (x−2) cancels, giving f(x)=x+2 for x=2. The function has a zero at x=−2 and is undefined at x=2, creating a removable hole rather than a vertical asymptote. Using the function provided, what are the zeros of f(x)=x−2x2−4?
Explanation: This question tests understanding of rational functions and their zeros, specifically how to identify zeros when factors cancel. Rational functions have zeros where the numerator equals zero and the denominator does not. In this case, the function f(x) = (x²-4)/(x-2) can be factored as f(x) = (x-2)(x+2)/(x-2), which simplifies to f(x) = x+2 for x≠2. Choice C is correct because when we set the simplified function equal to zero: x+2 = 0, we get x = -2 as the only zero. Choice D is incorrect because while x²-4 = 0 gives x = ±2, the factor (x-2) cancels out, eliminating x = 2 as a zero and creating a hole instead. Students should practice factoring and simplifying rational functions before finding zeros, recognizing that canceled factors create holes rather than zeros.
A rational function is a quotient of two polynomials. For f(x)=x−2x2−4, zeros are found where the numerator equals zero while the denominator is nonzero. Vertical asymptotes occur at denominator zeros that do not cancel. Factoring x2−4=(x−2)(x+2) shows (x−2) cancels, so f(x)=x+2 for x=2. The function is undefined at x=2 (a hole) and has a zero at x=−2. Using the function provided, what are the zeros of f(x)=x−2x2−4?
Explanation: This question tests understanding of finding zeros of rational functions through factoring and simplification. Zeros occur where the numerator equals zero and the denominator does not. For f(x) = (x²-4)/(x-2), factoring gives (x-2)(x+2)/(x-2), and after canceling the common factor, f(x) = x+2 for x≠2. Choice C is correct because setting the simplified function equal to zero gives x+2 = 0, so x = -2 is the only zero. Choice A is incorrect because while x²-4 = 0 yields x = ±2, the x = 2 value is eliminated when the (x-2) factor cancels, creating a hole rather than a zero. Students should always simplify by canceling common factors first, then find zeros from the reduced form. Practice recognizing that original numerator zeros may not all be function zeros if they coincide with denominator zeros.
A rational function is a quotient of two polynomials, written f(x)=q(x)p(x). Zeros are the x-values where f(x)=0, which requires p(x)=0 and q(x)=0. Vertical asymptotes occur at real values where q(x)=0 (unless the factor cancels). For example, f(x)=x2−1x2+1 has denominator x2−1=(x−1)(x+1), so it is undefined at x=±1, giving vertical asymptotes there. The numerator x2+1=0 leads to x=±i, which are complex zeros, so the graph has no real x-intercepts.
Based on the function described, what are the zeros of f(x)=x2−1x2+1?
Explanation: This question tests understanding of zeros in rational functions, including complex zeros. Zeros of a rational function occur where the numerator equals zero and the denominator does not. For f(x) = (x²+1)/(x²-1), the numerator x²+1=0 gives x²=-1, which has solutions x=±i (complex numbers). The denominator x²-1≠0 at these values since (±i)²-1=-1-1=-2≠0. Choice B is correct because the zeros are x=±i, even though they are complex numbers. Choice D is incorrect because while there are no real zeros (the graph doesn't cross the x-axis), the function does have complex zeros. Students should remember that zeros can be complex numbers and practice solving equations like x²+1=0 to find complex solutions.
A rational function is a quotient of two polynomials. For f(x)=x−2x2−4, zeros occur where the function equals zero, typically where the numerator is zero and the denominator is not. Vertical asymptotes occur at x-values that make the denominator zero and do not cancel with the numerator. Since x2−4=(x−2)(x+2), the factor (x−2) cancels, so f(x)=x+2 for x=2. The point x=2 is still excluded from the domain, creating a hole instead of a vertical asymptote. Using the function provided, determine the vertical asymptotes of f(x)=x−2x2−4.
Explanation: This question tests understanding of vertical asymptotes in rational functions and how they differ from removable discontinuities. Vertical asymptotes occur at x-values where the denominator equals zero and this zero does not cancel with the numerator. In this case, f(x) = (x²-4)/(x-2) = (x-2)(x+2)/(x-2), where the (x-2) factor cancels completely. Choice A is correct because after cancellation, there are no remaining zeros in the denominator, meaning no vertical asymptotes exist. Choice B is incorrect because while x = 2 makes the original denominator zero, this factor cancels with the numerator, creating a removable discontinuity (hole) rather than a vertical asymptote. Students should distinguish between vertical asymptotes and holes by checking whether denominator factors cancel. Practice factoring both numerator and denominator completely before identifying asymptotes.
A rational function is a quotient of polynomials, f(x)=q(x)p(x). Zeros are the x-values where f(x)=0, so p(x)=0 and q(x)=0. Vertical asymptotes occur at real solutions of q(x)=0 that do not cancel with the numerator. Consider the rational function f(x)=x2−1x2+1. Here, x2−1=(x−1)(x+1), so the function is undefined at x=±1, producing vertical asymptotes at x=1 and x=−1 because nothing cancels. The numerator x2+1=0 has solutions x=±i, which are complex zeros; this means there are no real x-intercepts.
Using the function provided, determine the vertical asymptotes of f(x)=x2−1x2+1.
Explanation: This question tests understanding of vertical asymptotes in rational functions where no factor cancellation occurs. Vertical asymptotes happen where the denominator equals zero and the factor doesn't cancel with the numerator. For f(x) = (x²+1)/(x²-1), the denominator x²-1 factors as (x-1)(x+1), equaling zero at x=±1. Since the numerator x²+1 has no real factors (its zeros are x=±i), nothing cancels with the denominator factors. Choice C is correct because vertical asymptotes occur at both x=1 and x=-1 where the denominator equals zero. Choice D is incorrect because x=±i are the complex zeros of the function, not vertical asymptotes. Students should practice factoring both numerator and denominator to identify which factors cancel and which create vertical asymptotes.
A rational function is a quotient of polynomials, such as f(x)=x−2x2−4. Zeros occur where f(x)=0, while vertical asymptotes occur where the denominator is zero (unless a factor cancels). Here x2−4=(x−2)(x+2), so f(x)=x+2 for x=2, creating a hole at x=2 rather than a vertical asymptote. Using the function provided, what are the zeros of f(x)=x−2x2−4?
Explanation: This question tests understanding of rational functions and their zeros, specifically how to identify zeros when factors cancel. Rational functions have zeros where the numerator equals zero and the denominator does not. In this case, the function f(x) = (x²-4)/(x-2) can be factored as f(x) = (x-2)(x+2)/(x-2), which simplifies to f(x) = x+2 for x≠2. Choice A is correct because when we set the simplified function equal to zero, we get x+2=0, which gives x=-2 as the only zero. Choice C is incorrect because it includes x=2, which is not a zero but rather a hole in the function where it is undefined. Students should practice factoring and simplifying rational functions before finding zeros, and remember that zeros come from the simplified numerator, not the original expression.
A rational function is a quotient of two polynomials. For f(x)=x−2x2−4, zeros occur where the function equals zero, and vertical asymptotes occur at denominator zeros that do not cancel. Because x2−4=(x−2)(x+2), the common factor cancels and f(x)=x+2 for x=2. Thus x=2 is a removable discontinuity and not a vertical asymptote. Using the function provided, determine the vertical asymptotes of f(x)=x−2x2−4.
Explanation: This question tests understanding of vertical asymptotes and how they differ from removable discontinuities in rational functions. Vertical asymptotes occur where the denominator has zeros that do not cancel with the numerator. For f(x) = (x²-4)/(x-2) = (x-2)(x+2)/(x-2), the (x-2) factor appears in both numerator and denominator and cancels completely. Choice D is correct because after cancellation, no zeros remain in the denominator, meaning there are no vertical asymptotes. Choice A is incorrect because while x = 2 is excluded from the domain, the canceling factor creates a hole rather than a vertical asymptote. Students should factor completely and cancel before identifying vertical asymptotes, remembering that canceled factors create holes. Practice distinguishing between these two types of discontinuities through factoring and graphing.
A rational function is a quotient of two polynomials, and its zeros are the x-values where the function equals zero. Consider f(x)=x−2x2−4. Factoring the numerator gives x2−4=(x−2)(x+2). After canceling the common factor (x−2), the simplified rule is f(x)=x+2 for x=2. Vertical asymptotes come from denominator zeros that do not cancel; here the denominator’s zero cancels, so there is no vertical asymptote, but the function is still undefined at x=2 (a hole). Based on the function described, which statement best describes the behavior of the function at its zero?
Explanation: This question tests understanding of how rational functions behave at their zeros, distinguishing between x-intercepts and other discontinuities. A zero of a rational function occurs where the function value equals zero, which happens when the numerator is zero and the denominator is not. For f(x) = (x²-4)/(x-2) = (x-2)(x+2)/(x-2) = x+2 (for x≠2), the zero occurs when x+2 = 0, giving x = -2. Choice A is correct because at x = -2, the function equals zero, meaning the graph crosses the x-axis at this point. Choice B is incorrect because x = -2 is not a vertical asymptote; it's where the simplified numerator equals zero, creating an x-intercept. Students should evaluate the simplified form to find zeros while remembering that vertical asymptotes come from non-canceling denominator factors. Practice graphing rational functions to visualize the difference between zeros and asymptotes.
A rational function is a quotient of polynomials. For f(x)=x−2x2−4, zeros require the numerator to be zero while the function remains defined. Vertical asymptotes correspond to denominator zeros that do not cancel. Since (x−2) cancels from numerator and denominator, x=2 is a removable discontinuity. Using the function provided, which x-value is a root of the numerator but not a zero of f(x)?
Explanation: This question tests understanding of the distinction between roots of the numerator and zeros of the rational function. The numerator x²-4 has roots at x=2 and x=-2, but not all roots of the numerator are zeros of the function. Choice B is correct because x=2 is a root of the numerator (since 2²-4=0) but is not a zero of f(x) because the function is undefined at x=2 due to the canceling factor. Choice A is incorrect because x=-2 is both a root of the numerator and a zero of the function. Students should understand that when a factor cancels from both numerator and denominator, the corresponding x-value creates a hole rather than a zero, even though it satisfies the numerator equation.
A rational function is a quotient of polynomials. For f(x)=x−2x2−4, zeros come from numerator zeros that are not excluded by the denominator. Vertical asymptotes come from non-canceling denominator zeros. Factoring gives f(x)=x+2 for x=2, so x=2 is excluded. Based on the function described, what is the set of zeros of f(x)?
Explanation: This question tests understanding of expressing the complete set of zeros for a rational function using set notation. After simplifying f(x) = (x²-4)/(x-2) to f(x) = x+2 for x≠2, we find zeros by solving x+2=0. Choice A is correct because this gives only x=-2 as the zero, which we express as the set {-2}. Choice C is incorrect because it includes x=2, which is not a zero but rather where the function is undefined. Students should practice using proper set notation and remember that the set of zeros includes only those x-values where the function equals zero and is defined, excluding any holes or vertical asymptotes.
A rational function is the quotient of two polynomials, f(x)=q(x)p(x). Zeros are the x-values where f(x)=0, so the numerator must be zero while the denominator is nonzero. Vertical asymptotes occur at real x where q(x)=0 and does not cancel. Consider f(x)=x+3x2−9. Since x2−9=(x−3)(x+3), the factor (x+3) cancels, producing a hole at x=−3 rather than a vertical asymptote. The function is still undefined at x=−3 in the original expression.
Using the function provided, identify the x-value for which f(x)=x+3x2−9 is undefined.
Explanation: This question tests understanding of undefined points in rational functions, specifically identifying holes. Rational functions are undefined where the original denominator equals zero, even if simplification occurs. For f(x) = (x²-9)/(x+3), the denominator x+3 equals zero when x=-3, making the function undefined at this point. Although the function simplifies to f(x) = x-3 after factoring and canceling, the original function remains undefined at x=-3, creating a removable discontinuity. Choice B is correct because x=-3 is where the original denominator equals zero. Choice A is incorrect because x=3 is actually a zero of the function, not an undefined point. Students should remember to check the original denominator for undefined points, not just the simplified form.
A rational function is a quotient of polynomials, f(x)=q(x)p(x). Zeros occur where f(x)=0, so p(x)=0 while q(x)=0. Vertical asymptotes occur at real x-values where q(x)=0 (and the factor does not cancel). Consider the simple rational function f(x)=x−2x2−4. Here, x2−4=(x−2)(x+2), so the factor (x−2) cancels for x=2, giving f(x)=x+2 with a hole at x=2 rather than a vertical asymptote. The remaining zero comes from x+2=0, so x=−2 is a zero. The function is undefined at x=2 because the original denominator is zero there.
Using the function provided, what are the zeros of f(x)=x−2x2−4?
Explanation: This question tests understanding of rational functions and their zeros, specifically when simplification occurs due to common factors. Rational functions have zeros where the numerator equals zero and the denominator does not. In this case, f(x) = (x²-4)/(x-2) can be factored as f(x) = (x-2)(x+2)/(x-2), which simplifies to f(x) = x+2 for x≠2, creating a hole at x=2 rather than a vertical asymptote. The zero occurs where the simplified function equals zero: x+2=0, giving x=-2. Choice B is correct because it identifies the only zero at x=-2. Choice C is incorrect because x=2 is not a zero but rather a point where the function is undefined due to the hole. Students should practice factoring and simplifying rational functions to identify zeros correctly, remembering that zeros come from the numerator after any cancellation has occurred.
A rational function is a quotient of polynomials. For f(x)=x−2x2−4, zeros are x-values where f(x)=0, and vertical asymptotes occur where the denominator is zero after simplification. Since x2−4=(x−2)(x+2), the function simplifies to f(x)=x+2 for x=2. The canceled denominator factor means x=2 does not create a vertical asymptote; instead, the function has a removable hole there. Using the function provided, what are the zeros of f(x)=x−2x2−4?
Explanation: This question tests understanding of finding zeros of rational functions after simplification. Zeros occur where the function equals zero, which happens when the numerator is zero and the denominator is not. For f(x) = (x²-4)/(x-2), factoring gives (x-2)(x+2)/(x-2), which simplifies to f(x) = x+2 for x≠2. Choice B is correct because setting the simplified function equal to zero gives x+2 = 0, so x = -2 is the only zero. Choice D is incorrect because while the original numerator x²-4 = 0 gives x = ±2, the x = 2 solution is eliminated when the (x-2) factor cancels, leaving only x = -2 as a zero. Students should always simplify rational functions first by canceling common factors, then find zeros from the simplified form. Practice recognizing that canceled factors create holes, not zeros.
A rational function is a quotient of two polynomials, f(x)=q(x)p(x). Zeros are x-values where f(x)=0, meaning p(x)=0 and q(x)=0. Vertical asymptotes occur where q(x)=0 and the factor does not cancel. Consider f(x)=x+3x2−9. Factoring gives x2−9=(x−3)(x+3), so the (x+3) cancels for x=−3, leaving f(x)=x−3 with a hole at x=−3 (not a vertical asymptote). The zero comes from x−3=0, so x=3 is a zero.
Based on the function described, what are the zeros of f(x)=x+3x2−9?
Explanation: This question tests understanding of zeros in rational functions with factor cancellation. Zeros occur where the numerator equals zero and the denominator does not. For f(x) = (x²-9)/(x+3), factoring gives (x-3)(x+3)/(x+3), which simplifies to f(x) = x-3 for x≠-3 after canceling the common factor (x+3). The zero of the simplified function occurs where x-3=0, giving x=3. Choice B is correct because x=3 is the only zero of the function. Choice A is incorrect because x=-3 is not a zero but rather a point where the function has a hole due to the cancellation. Students should practice factoring and simplifying rational functions, remembering that zeros come from the simplified numerator after cancellation.
A rational function is defined as a quotient of polynomials, f(x)=q(x)p(x). Zeros are the real (or complex) values of x that make f(x)=0, which requires p(x)=0 while q(x)=0. Vertical asymptotes are real x-values where q(x)=0 and the factor does not cancel with the numerator. Consider f(x)=x−2x2−4. Factoring gives x2−4=(x−2)(x+2), so the denominator factor cancels, leaving f(x)=x+2 for x=2. Because the problematic factor cancels, x=2 produces a hole, not a vertical asymptote. The only zero occurs where x+2=0, so x=−2 is the x-intercept.
Based on the function described, determine the vertical asymptotes of f(x)=x−2x2−4.
Explanation: This question tests understanding of vertical asymptotes in rational functions and how they differ from removable discontinuities. Vertical asymptotes occur where the denominator equals zero and the factor does not cancel with the numerator. In f(x) = (x²-4)/(x-2), factoring gives (x-2)(x+2)/(x-2), where the (x-2) factor cancels, leaving f(x) = x+2 for x≠2. Because the problematic factor cancels completely, x=2 creates a hole (removable discontinuity) rather than a vertical asymptote. Choice D is correct because there are no vertical asymptotes in this function. Choice A is incorrect because the cancellation prevents a vertical asymptote at x=2, creating only a hole instead. Students should practice identifying when factors cancel to distinguish between vertical asymptotes and holes, remembering that vertical asymptotes only occur when factors in the denominator do not cancel.
A rational function is a quotient of polynomials, f(x)=q(x)p(x). Zeros are values of x that make f(x)=0, so the numerator must be zero while the denominator is not. Vertical asymptotes occur where the denominator equals zero and does not cancel with the numerator. Consider f(x)=x2−1x2+1. Since x2−1=0 at x=±1, the function is undefined there and has vertical asymptotes at those x-values. The numerator x2+1 has no real roots, so there are no real zeros.
Using the function provided, identify the x-values for which f(x)=x2−1x2+1 is undefined.
Explanation: This question tests understanding of where rational functions are undefined, focusing on the denominator. Rational functions are undefined wherever the denominator equals zero, regardless of what happens with the numerator. For f(x) = (x²+1)/(x²-1), the denominator x²-1=0 when x²=1, giving x=±1. At these points, division by zero occurs, making the function undefined. Choice A is correct because the function is undefined at both x=1 and x=-1. Choice B is incorrect because x=±i are zeros of the numerator, not points where the function is undefined. Students should focus on the denominator to find undefined points and remember that these create either vertical asymptotes or holes depending on whether factors cancel.
A rational function is a quotient of two polynomials: f(x)=q(x)p(x). Zeros are x-values where f(x)=0, which means the numerator is zero while the denominator is not. Vertical asymptotes occur where q(x)=0 and the zero does not cancel. For the example f(x)=x−2x2−4, we factor x2−4=(x−2)(x+2), so f(x)=x+2 for x=2. The cancellation means x=2 is a removable discontinuity (a hole), not a vertical asymptote. The function crosses the x-axis where x+2=0.
Using the function provided, what is the zero of the simplified form of f(x) (with the original domain restriction kept)?
Explanation: This question tests understanding of zeros in simplified rational functions while maintaining domain restrictions. When a rational function simplifies due to factor cancellation, the zeros come from the simplified form, but domain restrictions from the original function remain. For f(x) = (x²-4)/(x-2), the simplified form is f(x) = x+2 with x≠2. The zero of this simplified function occurs where x+2=0, giving x=-2. Choice A is correct because x=-2 is where the simplified function equals zero. Choice B is incorrect because x=2 is not a zero but rather a point excluded from the domain due to the hole. Students should remember that simplification doesn't change the domain restrictions, and zeros are found from the simplified numerator while respecting the original domain.